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Linear and Nonlinear EquationsModeling
Ordinary Differential Equations. Session 2
Dr. Marco A Roque Sol
17/01/2017
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t)
are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β)
that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0.
Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0,
there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t)
on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β)
to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g(t) are continuous real-valued functions on aninterval (α, β) that contains the point t0. Then, for any choice of(initial value) y0, there exists a unique solution y(t) on the wholeinterval (α, β) to the linear differential equation
dy(t)
dt+ p(t)y(t) = g(t)
for all t ∈ (α, β) and y(t0) = yo.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y)
and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y)
are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0).
Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ)
so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution
for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y) and∂f
∂y(t, y) are continuous functions
defined on a region R as
R = {(t, y) : t0 − δ < t < t0 + δ; y0 − ε < y < y0 + ε}
containing the point (t0, y0). Then there exists a number δ1(possibly smaller than δ) so that a solution y = f (t) to
y ′ = f (t, y) y(t0) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case,
both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1
and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1
are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y).
The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees
that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists
in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1,
and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique
in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
Example 21
Consider the ODE
y ′ = t − y + 1; y(1) = 2
In this case, both the function f (t, y) = ty + 1 and its partial
derivative∂f
∂y(t, y) = −1 are defined and continuous at all points
(x , y). The Theorem 2 guarantees that a solution to the ODEexists in some open interval centered at 1, and that this solution isunique in some (possibly smaller) interval centered at 1.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t.
In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words,
in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1
as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In fact, an explicit solution to this equation is
y(t) = t + e1t
This solution exists (and is the unique solution to the equation) forall real numbers t. In other words, in this example we may choosethe numbers δ and δ1 as large as we please.
Example 22
Consider the ODE
y ′ = 1 + y2; y(0) = 0
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2
and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y
are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous
at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees
that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists
insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0,
and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is unique
in some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating,
we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
In this case, both the function f (t, y) = 1 + y2 and its partial
derivative∂f
∂y(t, y) = 2y are defined and continuous at all points
(t, y).
The Theorem 2 guarantees that a solution to the ODE exists insome open interval centered at 0, and that this solution is uniquein some (possibly smaller) interval centered at 1.
By separating variables and integrating, we derive a solution to thisequation of the form
y(t) = tan(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t,
this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ...
However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain.
Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE.
In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2,
although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ,
may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Linear and Nonlinear Equations
As an abstract function of t, this is defined for allt 6= ...,−3π/2,−π/2, π/2, 3π/2, ... However, in order for thisfunction to be considered as a solution to this ODE, we mustrestrict the domain. Specifically, the function
y(t) = tan(t); −π/2 < x < π/2
is a solution to the above ODE. In this example we must chooseδ1 = π/2, although the initial value δ, may be chosen as large aswe please.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation.
Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations
that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job
are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time,
modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation
that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling
and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
We now move into one of the main applications of differentialequations.
Modeling is the process of writing a differential equation todescribe a physical situation. Almost all of the differentialequations that you will use in your job are there becausesomebody, at some time, modeled a situation to come up with thedifferential equation that you are using.
This section is designed to introduce you to the process ofmodeling and show you what is involved in modeling.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start
with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid.
Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank
may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not
contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it.
Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank
will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t)
gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance
dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t
we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved,
will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Mixing Problems
In these problems we will start with a substance that is dissolved ina liquid. Liquid will be entering and leaving a holding tank.
The liquid entering the tank may or may not contain more of thesubstance dissolved in it. Liquid leaving the tank will of coursecontain the substance dissolved in it.
If Q(t) gives the amount of the substance dissolved in the liquid inthe tank at any time t we want to develop a differential equationthat, when solved, will give us an expression for Q(t) .
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations
we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions
and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid,
but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here
is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid
is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank.
Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case,
but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank,
theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Note as well that in many situations we can think of air as a liquidfor the purposes of these kinds of discussions and so we dontactually need to have an actual liquid, but could instead use air asthe liquid.
The main assumption that well be using here is that theconcentration of the substance in the liquid is uniform throughoutthe tank. Clearly this will not be the case, but if we allow theconcentration to vary depending on the location in the tank, theproblem becomes very difficult.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The main equation that well be using to model this situation is :
Rate of change of Q(t) =
Rate at which Q(t) enters the tank -
Rate at which Q(t) exits the tank
where,
Rate of change of Q(t) =dQ(t)
dt= Q(t)′
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Rate at which Q(t) enters the tank = (flow rate of liquid entering)x (concentration of substance in liquid entering)
Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x(concentration of substance in liquid exiting)
Let’s take a look at the first problem
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Rate at which Q(t) enters the tank = (flow rate of liquid entering)x (concentration of substance in liquid entering)
Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x(concentration of substance in liquid exiting)
Let’s take a look at the first problem
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Rate at which Q(t) enters the tank = (flow rate of liquid entering)x (concentration of substance in liquid entering)
Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x(concentration of substance in liquid exiting)
Let’s take a look at the first problem
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Rate at which Q(t) enters the tank = (flow rate of liquid entering)x (concentration of substance in liquid entering)
Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x(concentration of substance in liquid exiting)
Let’s take a look at the first problem
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it.
Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank
has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal.
If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr,
how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water
enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank
to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point.
Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality,
but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 14
A 1500 gallon tank initially contains 600 gallons of water with 5lbs of salt dissolved in it. Water enters the tank at a rate of 9gal/hr and the water entering the tank has a salt concentration of15 (1 + cos(t)) lbs/gal. If a well mixed solution leaves the tank at arate of 6 gal/hr, how much salt is in the tank when it overflows?
We are going to assume that the instant the water enters the tankit somehow instantly disperses evenly throughout the tank to givea uniform concentration of salt in the tank at every point. Again,this will clearly not be the case in reality, but it will allow us to dothe problem.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now,
to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t)
wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering,
the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering,
the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and
the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tank
the concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank
and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
Now, to set up the IVP that we will need to solve to get Q(t) wewill need the flow rate of the water entering, the concentration ofthe salt in the water entering, the flow rate of the water leaving(weve got that) and the concentration of the salt in the waterexiting (we dont have this )
Since we are assuming a uniform concentration of salt in the tankthe concentration at any point in the tank and hence in the waterexiting is given by,
Concentration = (Amount of salt in the tank at any time t) / (Volume of water in the tank at any time t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t).
The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy.
We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons
and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters
and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave.
So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours,
everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank,
or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t
there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The amount at any time t is easy it’s just Q(t). The volume isalso pretty easy. We start with 600 gallons and every hour 9gallons enters and 6 gallons leave. So, if we use t in hours, everyhour 3 gallons enters the tank, or at any time t there is 600 + 3tgallons of water in the tank.
So, the IVP for this situation is,
Q ′(t) = (9)
(1
5(1 + cos(t))
)− 6
(Q(t)
600 + 3t
)Q(0) = 5
Q ′(t) =9
5(1 + cos(t))− 2
(Q(t)
200 + t
)Q(0) = 5
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation.
We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now,
multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equation
by the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
This is a linear differential equation. We will show most of thedetails, but leave the description of the solution process out.
Q ′(t) +
(2
Q(t)
200 + t
)=
9
5(1 + cos(t))
Now find the integrating factor:
µ(t) = e∫
2200+t
dt = e2ln(200+t) = e ln(200+t)2 = (200 + t)2
Now, multiply the rewritten differential equationby the integratingfactor.
(200 + t)2Q ′(t) + (200 + t)2(
2Q(t)
200 + t
)= (200 + t)2
9
5(1 + cos(t))
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.
∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
((200 + t)2Q(t)
)′= (200 + t)2
9
5(1 + cos(t))
Integrate both sides and solve for the solution.∫ ((200 + t)2Q(t)
)′dt =
∫9
5(200 + t)2 (1 + cos(t)) dt
(200 + t)2Q(t) =9
5[1
3(200 + t)3 + (200 + t)2sin(t) + ...
...+ 2(200 + t)cos(t)− 2sin(t)] + c
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)+
c
(200 + t)2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
And applying initial conditions
5 = Q(0) =9
5
(1
3(200) +
2
200
)+
c
(200)2
c = −4600720
The solution is then,
Q(t) =9
5
(1
3(200 + t)2 + sin(t) +
2cos(t)
200 + t− 2sin(t)
(200 + t)2
)− 4600720
(200 + t)2
Now, the tank will overflow at t = 300hrs. The amount of salt inthe tank at that time is.
Q(300) = 279.80lbs
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank
that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water
with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it.
Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr
and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it.
A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution
leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well.
When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off
and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr
while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.
Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 15
A 1000 gallon holding tank that catches runoff from some chemicalprocess initially has 800 gallons of water with 2 ounces of pollutiondissolved in it. Polluted water flows into the tank at a rate of 3gal/hr and contains 5 ounces/gal of pollution in it. A well mixedsolution leaves the tank at 3 gal/hr as well. When the amount ofpollution in the holding tank reaches 500 ounces the inflow ofpolluted water is cut off and fresh water will enter the tank at adecreased rate of 2 gal/hr while the outflow is increased to 4gal/hr.Determine the amount of pollution in the tank at any time t.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes.
If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed
there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well.
In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem.
One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation
whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank
and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached
and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Solution
The pollution in the tank will increase as time passes. If theamount of pollution ever reaches the maximum allowed there willbe a change in the situation.
This will necessitate a change in the differential equationdescribing the process as well. In other words, we’ll need two IVP’sfor this problem. One will describe the initial situation whenpolluted runoff is entering the tank and one for after the maximumallowed pollution is reached and fresh water is entering the tank.
Here are the two IVPs for this problem.
Q ′1(t) = (3)(5)− 3
(Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward
and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached.
We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.
Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time
sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy
with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term
as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one.
First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0.
We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm,
thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts.
Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank
and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero.
This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term,
and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = (2)(0)− 4
(Q2(t)
800− 2(t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first one is fairly straight forward and will be valid until themaximum amount of pollution is reached. We’ll call that time tm.Also, the volume in the tank remains constant during this time sowe dont need to do anything fancy with that this time in thesecond term as we did in the previous example.
We will need a little explanation for the second one. First noticethat we do not start over at t = 0. We start this one at tm, thetime at which the new process starts. Next, fresh water is flowinginto the tank and so the concentration of pollution in the incomingwater is zero. This will drop out the first term, and thats okay sodon’t worry about that.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now,
notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.
During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process
so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat.
However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample.
When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts up
there needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank
and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons
that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation.
So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume
we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times.
In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1)
we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons
in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Now, notice that the volume at any time looks a little funny.During this time frame we are losing two gallons of water everyhour of the process so we need the − 2 in there to account forthat. However, we cant just use t as we did in the previousexample. When this new process starts upthere needs to be 800gallons of water in the tank and if we just use t there we won’thave the required 800 gallons that we need in the equation. So, tomake sure that we have the proper volume we need to put in thedifference in times. In this way once we are one hour into the newprocess (i.e t - tm = 1) we will have 798 gallons in the tank asrequired.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty.
This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te .
Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400
since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty
400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well,
it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along
and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay,
now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of
here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Finally, the second process can’t continue forever as eventually thetank will empty. This is denoted in the time restrictions as te . Wecan also note that te = tm + 400 since the tank will empty 400hours after this new process starts up.
Well, it will end provided something doesn’t come along and startchanging the situation again.
Okay, now that we’ve got all the explanations taken care of here’sthe simplified version of the IVP’s that we’ll be solving.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′1(t) = 15−(
3Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Q ′2(t) = −(
2Q2(t)
400− (t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first IVP is a fairly simple linear differential equation so we willleave the details of the solution to you to check.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′1(t) = 15−(
3Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Q ′2(t) = −(
2Q2(t)
400− (t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first IVP is a fairly simple linear differential equation so we willleave the details of the solution to you to check.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′1(t) = 15−(
3Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Q ′2(t) = −(
2Q2(t)
400− (t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first IVP is a fairly simple linear differential equation
so we willleave the details of the solution to you to check.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′1(t) = 15−(
3Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Q ′2(t) = −(
2Q2(t)
400− (t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first IVP is a fairly simple linear differential equation so we willleave the details of the solution to you to check.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′1(t) = 15−(
3Q1(t)
800
)Q(0) = 2 0 ≤ t ≤ tm
Q ′2(t) = −(
2Q2(t)
400− (t − tm)
)Q(tm) = 500 tm ≤ t ≤ te
The first IVP is a fairly simple linear differential equation so we willleave the details of the solution to you to check.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now,
we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm.
We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500.
So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours.
Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake
here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q1(t) = 4000− 3998e−3t800
Now, we need to find tm. We need to do is determine when theamount of pollution reaches 500. So we need to solve.
Q1(t) = 4000− 3998e−3t800 = 500
tm = 35.475
So, the second process will pick up at 35.475 hours. Forcompleteness sake here is the IVP with this information inserted
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both
linear and separable and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable
and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable and I willleave the details to you again
to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Q ′2(t) = −(
2Q2(t)
435.475− t
)Q(35.475) = 50035.475 ≤ t ≤ 435.475
This differential equation is both linear and separable and I willleave the details to you again to check that we should get.
Q2(t) =(435.476− t)2
320
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
So, a solution that encompasses the complete running time of theprocess is
Q(t) =
{4000− 3998e−
3t800 0 ≤ t ≤ 35.475
(435.476−t)2320 35.475 ≤ t ≤ 435.4758
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
So, a solution that encompasses the complete running time of theprocess is
Q(t) =
{4000− 3998e−
3t800 0 ≤ t ≤ 35.475
(435.476−t)2320 35.475 ≤ t ≤ 435.4758
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
So, a solution that encompasses the complete running time of theprocess is
Q(t) =
{4000− 3998e−
3t800 0 ≤ t ≤ 35.475
(435.476−t)2320 35.475 ≤ t ≤ 435.4758
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Here is graph of the solution.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Here is graph of the solution.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Here is graph of the solution.
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16
Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation
only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth,
and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative
to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth.
If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances,
then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity
is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.
Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation
tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r
from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 16Escape Velocity
The model of constant gravitation only works when were close tothe surface of the earth, and the distances we’re dealing with aresmall relative to the radius of the earth. If we start to deal withlarger distances, then we must take into account that accelerationfrom gravity is weaker the farther we are away from the earth.Newton’s law of universal gravitation tells us that the force fromgravity experienced a distance r from the center of the earth willbe:
F = −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object,
M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth,
andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2.
Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth.
This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed
at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface
if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth
andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well,
we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth
along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center,
then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where m is the mass of the object, M is the mass of the earth, andG is Newtons gravitational constant G = 6.6710−11Nm2/kg2. Wecan use this relation to calculate an objects escape velocity onthe surface of the earth. This is the speed at which an object mustbe moving away from the earth at the earths surface if it is tobreak free from the gravitational attraction of the earth andcontinue to move away forever.
Well, we note that if we move away from the earth along a line thatgoes through the earths center, then Newtons second law tells us:
md2r
dt2= −GmM
r2
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we have
dvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
By the chain rule we havedvdt = dv
drdrdt , and if we note v = dr
dt then we can transform thisrelation into
mvdv
dr= −GmM
r2
If we integrate both sides with respect to r we get:
1
2mv2 =
GmM
r+ c
And applying initial conditions, v(R) = v0, we obtain:
v2 = v20 + 2GM
(1
r− 1
R
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,
then its velocity must always be positive as r →∞. This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,then its velocity must always be positive as r →∞.
This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,then its velocity must always be positive as r →∞. This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,then its velocity must always be positive as r →∞. This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,then its velocity must always be positive as r →∞. This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
If the object is to escape from the graviational action of the earth,then its velocity must always be positive as r →∞. This will bethe case if
v0 ≥√
2GM
R
For the earth the escape velocity is v0 = 11, 180m/s
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17
Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles,
with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles.
If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth
while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit,
can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid
using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 17Escape Velocity
Suppose that you are stranded -your rocket engine has failed- onan asteroid of diameter 3 miles, with density equal to that of theearth with radius 3960 miles. If you have enough spring in yourlegs to jump 4 feet straight up on earth while wearing your spacesuit, can you blast off from this asteroid using leg power alone ?
Solution
The escape velocity for the Earth is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vE =
√2GME
RE
Solving for ME in this equation we get:
ME =v2ERE
2G
The density of the earth is its mass divided by its volume
ME43πR
3E
=3v2E
8πGR2E
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid,
and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
A similar calculation can be done for the asteroid, and given boththe asteroid and the Earth have the same density we get:
3v2E8πGR2
E
=3v2A
8πGR2A
With a little algebra from this we can deduce the ratio:
v2Av2E
=R2A
R2E
So, the escape velocity from the asteroid is
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump,
all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2,
and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy
is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh.
So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
vA = vE
(RA
RE
)= 11, 180
(1.5
3960
)= 4.24m/s
At the begining of the jump, all the energy is initially kinetic,12mv2, and at the top of the jump all that energy is converted intopotential energy, mgh. So, the final height is given by the equation:
v =√
2gh
Plugging 4 feet in for h we get:
v =√
2(9.8)(4ft)(1m/3.28ft) = 4.89m/s > 4.24m/s
So, yes, you can get off the asteroid !!!!
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18
Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money.
Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.
We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations.
Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously.
If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,
then the rate of change of the initial deposit is dSdt , this quantity is
equal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt ,
this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues,
which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times
the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t).
Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 18Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.We can model the growth of an initial deposit with respect to theinterest rate r with differential equations. Let’s assume that theinitial deposit is compounded continuously. If t represents time,then the rate of change of the initial deposit is dS
dt , this quantity isequal to the rate at which interest accrues, which is the interestrate r times the current value of the investment S(t). Thus, themodel is given by
dS
dt= rS(t)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose
that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment
at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt
∫1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Suppose that we also know the value of the investment at someparticular time, say,
S(0) = S0
Integrating this separable IVP
dS
dt= rS(t)
1
SdS = rdt∫
1
SdS =
∫rdt
ln|S | = rt + K
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0. Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0,
we have that theconstant of integration is c = S0. Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0.
Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0. Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0. Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
S(t) = cert
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0. Therefore the solution to thisinitial value problem is:
S(t) = S0ert
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19
Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money.
Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.
Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously.
Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest,
dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k.
So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 19Compound Interest
Let S be an initial sum of money. Let r represent an interest rate.Assume that the initial deposit is compounded continuously. Let’suppose that there may be deposits or withdrawals in addition tothe accrual of interest, dividends, or capital gains and the depositsor withdrawals take place at a constant rate k. So the IVP is
dS
dt= rS(t) + k S(0) = S0
( k > 0 for deposits and k < 0 for withdrawals)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one,
with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt ,
so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0,
we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r .
Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
The above differential equation is a linear one, with integratingfactor e−rt , so its general solution is:
S(t) = cert − k
r
and using the initial condition that S(0) = S0, we have that theconstant of integration is c = S0 + k
r . Therefore the solution tothis initial value problem is:
S(t) = S0ert +
k
r
(ert − 1
)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20
Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states
that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a body
changes at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperature
between its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature
and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
Example 20Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a bodychanges at a rate proportional to the difference in temperaturebetween its own temperature and the temperature of itssurroundings.
We can therefore write
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time t
Ts= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)
T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the body
k = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
where,
T = Temperature of the body at any time tTs= Temperature of the surroundings (also called ambienttemperature)T0 = Initial temperature of the bodyk = constant of proportionality
Integrating this separable differential equation
dT
dt= −k (T − Ts)
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
1
T − TsdT = −kdt
∫1
T − TsdT = −
∫kdt
ln|T − Ts | = −kt + C
|T − Ts | = e−kt+C
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
1
T − TsdT = −kdt∫
1
T − TsdT = −
∫kdt
ln|T − Ts | = −kt + C
|T − Ts | = e−kt+C
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
1
T − TsdT = −kdt∫
1
T − TsdT = −
∫kdt
ln|T − Ts | = −kt + C
|T − Ts | = e−kt+C
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
1
T − TsdT = −kdt∫
1
T − TsdT = −
∫kdt
ln|T − Ts | = −kt + C
|T − Ts | = e−kt+C
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
1
T − TsdT = −kdt∫
1
T − TsdT = −
∫kdt
ln|T − Ts | = −kt + C
|T − Ts | = e−kt+C
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Modeling
T = Ts + ce−kt ; T > Ts (why?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts)e−kt
Dr. Marco A Roque Sol Ordinary Differential Equations. Session 2
Recommended