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Day 4 Differential Equations(option chapter)
The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.)
So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.
If the rate of change is proportional to the amount present, the change can be modeled by:
dyky
dt
Recall from AP Calculus
dyky
dt
1 dy k dt
y
Rate of change is proportional to the amount present.
Divide both sides by y.
Integrate both sides.
1 dy k dt
y Integrate both sides.
Exponentiate both sides.
When multiplying like bases, add exponents. So added exponents can be written as multiplication.
ln y kt Ce e
C kty e e
Exponentiate both sides.
When multiplying like bases, add exponents. So added exponents can be written as multiplication.
C kty e ekty Ae Since is a constant, let .
Ce Ce A
C kty e e
kty Ae Since is a constant, let .Ce Ce A
At , .0t 0y y00
ky Ae
0y A
1
0kty y e This is the solution to our original initial
value problem.
0kty y eWe end with:
dyky
dtSo if we start with:
What if we have a series of differential equations?
We could solve these individually
y1 =c1ekt
y2 =c2ekt
y3 =c3ekt
Provided that we have initial conditions for each of these to solve for the constants.
If we define x
x1’(t) x’(t) = x2’(t) xn ’(t)
[ ]…
This yields the equation x’(t)= AxWhich is easy to solve in the case of a diagonal matrix.
x’1x’2x’3
[ ] = 3 0 00 -2 00 0 4[ ] x1
x2
x3[ ]
We can solve each of these as a separate differential equationx1’ = 3x1, x2’ = -2x2, x3’ = 4x3
x1 (t) = b1e3t, x2 (t) = b2e-2t, x3 (t) = b3e4t, This is the general solution. We can solve for the constants if given an initial condition.
First order homogeneous linear system of differential equations
x1’(t) = a11x1 (t) + a12 x2 (t) + … a1nxn (t)
x2’(t) = a21x1 (t) + a22 x2 (t) + … a2nxn (t)
xn ’(t) = an1x1 (t) + an2 x2 (t) + … annxn (t)
…We could write this in matrix form as: x1’ (t) a11 a12 …. a12
x(t) = x2’ (t) A = a21 a 22 … a2n
xn’ (t) an1 a n2 …anm
[ ]… [ ]…
What if our system is not diagonal?
du1 = -u1 + 2u2
dt
du2 = u1 – 2u2
dt
A = -1 2 1 -2 [ ]
The system at the left can be written as du/dt = Au with a as
How can we solve this system?Initial condition u(0) =
10[ ]
du/dt = Au
y = eAt
u(t) = c1eλ t x1 +c2e λ t x2+…+ cneλ t
xn
Check that each piece solves the given system
du/dt = Au
d (eλ t x1) = A eλ t
x1 λeλ t x1 = A eλ t
x1
dt λx1 = Ax1
1 2 n
Key Formulas
Difference Equations
Differential Equationsdu/dt = Au y = eAt
Solve the differential equations
A = -1 2 1 -2 [ ]The system at the left can be written as du/dt = Au with a as
What are the eigenvalues from inspection?Hint: A is singularThe trace is -3
Start by computing the eigenvalues and eigenvectors
Solve the differential equationsStep 1 find the eigenvalues and
eigenvectors
det -1-λ 2 1 -2-λ [ ]
We can a solve via finding the determinant of A - λI
Calculate the eigenvector associated with λ = 0, -3
By inspection: the matrix is singular therefore 0 is an eigenvalue the trace is -3 therefore the other eigenvalue is -3.
A = -1 2 1 -2[ ] For λ = 0 find a basis for the kernel of A
21[ ]
For λ= -3 find a basis for the kernel of A+3IA+ 3I = 2 2 1 1[ ] 1
-1[ ]
Solve the differential equations
A = -1 2 1 -2 [ ]
The system at the left can be written as du/dt = Au with a as
The form that we are expecting for the answer is y = c1 e λ t x1 + c2 e λ t x2
Note: the solutions of the equations are going to be e raised to
a power.
1 2
The eigenvalues are already telling us about the form of the solutions.A negative eigenvalue will mean that that portion goes to zero as x goes to infinity. An eigenvalue of zero will mean that we will have an e0 which will be a constant. We will call this type of system a steady state.
Solve the differential equations
A = -1 2 1 -2 [ ]
Solve by plugging in eigenvalues into expected equation and for λ1 and λ2. and the corresponding eigenvectors in x1 and x2
We find c1 and c2 by using the initial condition
y = c1 e0t 2 + c2 e -3t 1 1 -1[ ] [ ]
Initial condition u(0) = []10
Plugging in zero for t and the initial conditions yields:
1 = c1 2 + c2 10 1 -1[ ][ ][]
Recall:
c1 = 1/3c2 = 1/3
Solve the differential equations
The general solution isy = 1/3 2 + 1/3 e -3t 1 1 -1[ ] [ ]
We are interested in hat happens as time goes to infinityRecall our initial condition was 1 all of our quantity was in u1
0
Then as time progressed there was flow from u1 to u2. As time approaches infinity we end with the steady state 2/3 1/3[ ]
[]
The solution to y’ = ky is y = y0ekt
The solution to x’ = Au
is u = c0eAt
Homework: wkst 8.4 1-9 odd, 2 and 8
What if the matrix is not diagonal?
• White book p. 520 ex 3, 4, 5