MA 2930, Feb 9, 2011Worksheet 3

1.

A tank with a capacity of 500 gal originally contains 200 gal of water with100 lb of salt in solution. Water containing 1 lb of salt per gallon enters at arate of 3 gal/min, and the mixture flows out at a rate of 2 gal/min. In orderto find the amount of salt in the tank at any time prior to its overflow

(a) set up the differential equation for the quantity of interest,(b) identify the initial condition, and(c) solve the equation.(d) What is the concentration of salt at the moment of overflow?(a) It’s clear that the quantity of interest is the amount of salt in the

tank at any time, so let that amount be q(t) lb at time t min. The problemtells us how q(t) is changing with time.

The rate at which salt is coming into the tank at time t is (1 lb/gal)(3gal/min) = 3 lb/min.

The rate at which salt is leaving the tank is, however, not constant withtime. It depends on what the concentration of salt in the tank is at time t.The concentration of salt, c(t), can be calculated as follows: the volume of thesalt solution in the tank at time t, V (t), is equal to the initial volume + addedsolution in t minutes = 200+(3−2)t = 200+t gal. The amount of salt at timet is q(t), by definition. So, the concentration, c(t) = q(t)/V (t) = q(t)/(200+t)lb/gal. So, rate at which salt leaves the tank at time t is c(t) lb/gal × 2gal/min = 2q(t)/(200 + t) lb/min.

Therefore, since in rate - out rate = rate of change of amount of salt inthe tank,

dq

dt= 3− 2q

200 + t

(b) The initial condition is q(0) = 100, which is the initial amount of saltin the tank.

(c) The differential equation for q is first-order linear, written in thestandard form as

dq

dt+

2

200 + tq = 3

So we can solve it using the integrating factor µ(t) = exp(∫

2200+t

dt) =exp(2 ln(200 + t)) = exp(ln(200 + t)2) = (200 + t)2.

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Upon multiplication by the integrating factor our equation becomes

d

dt[(200 + t)2q] = 3(200 + t)2

So,

(200 + t)2q = 3

∫(200 + t)2dt = (200 + t)3 + c

Therefore,q(t) = 200 + t+ c(200 + t)−2

Now, we apply the initial condition to find the integration constant c:

100 = q(0) = 200 + c(200)−2 → c = (100− 200)(200)2 = −4000000

Hence,

q(t) = 200 + t− 4000000

(200 + t)2

(d) First let’s find out the moment of overflow, T . This is clearly the timewhen the volume of the solution in the tank reaches its capacity: V (T ) = 500.Since V (T ) = 200 + T from part (b), T = 300 min. What is the amount ofsalt in the tank at time T = 300? q(300) = 200 + 300 − 4000000/(500)2 =500 − 16 = 484 lb. Therefore, the concentration of salt at that moment is484/500 = 0.97 lb/gal.

2.

Find without solving the differential equation the largest t-interval on whichthe solution of the following initial value problem is guaranteed to exist:

(4− t2)y′ + 2ty = 3t2(4− t2)2, y(0) = 4

Now solve the equation and determine directly from the solution the largestinterval of its existence.

This is a first-order linear equation. First let’s write it in the standardform:

y′ +2t

4− t2y = 3t2(4− t2), y(0) = 4

The existence and uniqueness theorem for first-order linear equations saysthat there is guaranteed to be a unique solution to the equation in the largest

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interval containing the initial point t = 0 in which the functions p(t) = 2t4−t2

and q(t) = 3t2(4− t2) are continuous. q(t) is continuous everywhere, whereasthe points of discontinuity of p(t) on the entire t-line are also the pointsof its non-existence, namely t = −2 and t = 2. (Recall that a rationalfunction is continuous wherever it exists.) Thus, largest contiguous intervalsof continuity for both these functions together are (−∞,−2), (−2, 2) and(2,∞). The initial point t = 0 is contained in (−2, 2), so there is a uniquesolution through y(1) = −3 that extends throughout (−2, 2).

Whether the solution extends beyond it, the theorem can not say. To seeif it does we need to find the solution.

An integrating factor for the equation is exp(∫

2t4−t2dt) = exp(− ln(4 −

t2)) = 1/(4− t2).Upon multiplying by it the equation becomes

y

(4− t2)=

∫3t2 dt = t3 + c

So, y = (3t2 + c)(4− t2). Applying the initial condition gives 4 = y(0) = 4c.Therefore,

y = (3t2 + 1)(4− t2)

which exists for all t, not just on (−2, 2), showing that the largest interval ofexistence the theorem guarantees may be smaller than the actual one.

3.

At which points in the (t, y)-plane may y′ = ln(ty) not have a solution? Atwhich points does the solution exist, but may not be unique?

By the existence and uniqueness theorem for non-linear first-order equa-tions of the form y′ = f(t, y) we know that we should look at the points atwhich either f(t, y) = ln(ty) or its partial derivative fy = 1/y is discontinous.

ln(ty) exists and is continuous wherever ty > 0, i.e., when t > 0, y > 0 orwhen t < 0, y < 0. These are the first and the third open quadrants of the(t, y)-plane. So, by the theorem for any initial condition lying in these twoopen quadrants there is guaranteed to be a solution of the equation. Whetherit is unique depends on the continuity of fy = 1/y in these quadrants. But1/y is continuous at any of these points, so the solution is also bound to beunique.

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4.

Show that for any (not necessarily first order) homogeneous linear equationif y(t) is a solution, so is cy(t) for any constant c. Take a simple non-linearequation and verify that it need not have this property.

A homogeneous linear differential equation for y(t) looks like this:

fn(t)dny

dtn+ fn−1(t)

dn−1y

dtn−1+ · · ·+ f1(t)

dy

dt+ f0(t)y = 0

where the fi(t)’s can be any functions of t.Let’s see what the left hand side of the equation becomes for cy(t):

LHS = fn(t)dn(cy)

dtn+ fn−1(t)

dn−1(cy)

dtn−1+ · · ·+ f1(t)

d(cy)

dt+ f0(t)(cy)

But constants ”come out of the derivatives”, so:

LHS = c(fn(t)dny

dtn+ fn−1(t)

dn−1y

dtn−1+ · · ·+ f1(t)

dy

dt+ f0(t)y) = c(0) = 0

So cy(t) is also a solution for any constant c.To show that this is not necessarily true for a non-homogeneous linear or

a non-linear equation, consider the simple equation y′ = y2. The solutions areeasily found using separation of variables: −y−1 = t+c, so y(t) = −1/(t+c).It’s clear that, e.g., −1/t is a solution, but −2/t is not.

(NB: One can also easily show that if y1(t) and y2(t) are two solutions of alinear homogeneous differential equation, so is y1(t) + y2(t). Thus, arbitrarylinear combinations of solutions of linear homogeneous equations are alsosolutions. This is not true for non-homogeneous linear or non-linear equationseither.)

5.

Find y(2) if y′ = −ty+y3, y(1) = 2, first approximately using Euler’s methodwith step size h = 0.5, and then exactly by solving the equation.

Euler method works as follows: start with the initial condition and suc-cessively calculate the approximate value of y after each time step of size hby pretending that the y′ at the beginning of the step is the same throughout

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the step. In our case

y0 = y(1) = 2

y1 = y(1.5) ≈ y(1) + hy′(1) = 2 + 0.5[−1(2) + 23] = 5

y2 = y(2) ≈ y(1.5) + hy′(1.5) = 5 + 0.5[1.5(−4) + 43)] = 34

To solve the equation exactly note that it’s first order non-linear as wellas non-separable, but the non-linear is of the form yn with n = 3, so it is aBernoulli equation:

y′ + ty = y3

We can convert this equation to a first-order linear equation by means ofthe substitution u = y1−n = y−2. Then, u′ = −2y−3y′. (Note ’ meansdifferentiation with respect to t.) To put the equation in terms of u firstmultiply throughout by −2y−3:

−2y−3y′ − 2ty−2 = −2

This gives usu′ − 2tu = −2

which is indeed linear. An integrating factor is exp(∫−2tdt) = e−t

2. Using

it we get

e−t2

u = −2

∫e−t

2

dt

It is well-known that this Gaussian integral can not be expressed in termsof elementary functions, so our solutions are (after choosing t = 1 as thestarting point and defining c accordingly)

u(t) = −2et2

(

∫ t

1

e−τ2

dτ + c)

Therefore,

y(t) =1√u

=1√

−2et2(∫ t1e−τ2dτ + c)

Now, apply the initial condition:

2 = y(1) =1√−2ec

→ c = −1/8e

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Therefore, the solution to the initial-value problem is

y(t) =2√

−et2−1(8e∫ t1e−τ2dτ − 1)

The exact value of y(2) is:

y(2) =2√

−e3(8e∫ 2

1e−τ2dτ − 1)

I leave it to you to find this value using your fancy calculators or computerprograms (or by looking up a table of the values of the Gaussian integral ifyou can find one these days) and compare it with the approximate value of34 Euler’s method gave above.

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