Math3d Winter Term 2002

Mid Term Solutions

1. (a) The homogeneous equation has constant coefficients and its general solution v can be computed by means of the characteristic equation

r^{2} + 2r + 1 = (r + 1)^{2} = 0 .

It has the double root r = −1 and therefore

v(t) = c_{1} e^{−t} + c_{2} te^{−t} .

A particular solution z to the inhomogeneous equation can now be computed either by the variation-of-parameters formula or by the Ansatz z(t) = A t^{p} te^{−t} for p = 1, 2. The case p = 0 can be excluded right away since it leads to a solution of the homogeneous equation. It turns out that p = 2 is the right choice. In fact

z′(t) = 3A t^{2}e^{−t} − A t^{3}e^{−t} , z′′(t) = 6A te^{−t} − 6A t^{2}e^{−t} + A t^{3}e^{−t}

and consequently

z′′(t) + 2z′(t) + z(t) = A e^{−t}[6t − 6t^{2} + t^{3} + 6t^{2} − 2t^{3} + t^{3}] = 6A te^{−t} .

Now, in order for the equation to be satisfied, we have to choose A = 1/6. Summarizing we obtain the general solution

y(t) = c_{1} e^{−t} + c_{2} te^{−t} + \frac{1}{6}t^{3}e^{−t} .

1. (b) This equation has non constant coefficients but doesn’t involve the function y at all. If we rename y′ to u the equation reads

u′ − tu = t .

Separating the variables we obtain

∫ \frac{(1 + u(t))′}{1 + u(t)} dt = c + ∫ t dt

which leads to u(t) = e^{c}e^{0.5t^{2}} = c_{1} e^{0.5t^{2}} − 1. Finally

y(t) = ∫ u dt = c_{2} + c_{1} ∫ e^{0.5t^{2}} dt − t .

2. Let us discuss the equation first. The initial value y_{0} is assumed to be positive. By inspecting the right-hand-side we can see that

\begin{cases}y′ < 0 , & if y > β/α \\ y′ = 0 , & if y = β/α \\ y′ > 0 , & if 0 < y < β/α \end{cases}

This already tells us that, if the solution starts out above β/α, it is going to decrease, that it is going to remain constant if y_{0} = β/α and that it is going to grow if y_{0} < β/α. This only leaves β/α as a candidate for lim_{t→∞} y(t). To analyze the solution’s convexity we need to compute y′′. We can do so using the equation to obtain

y′′ = ((α − βy) y)′ = y(α − 2βy)(α − βy) .

We therefore see that

y′′ < 0 , if β/2α < y < β/α
y′′ = 0 , if y = β/α , β/2α
y′′ > 0 , if 0 < y < β/2α or y > α/β

To compute the solution we observe that the equation is separable and that

\frac{1}{y(α − βy)} = \frac{1}{α} \frac{1}{y} + \frac{β}{α} \frac{1}{α − βy} .

This leads to

α t = α ∫_{0}^{t} \frac{y′}{y(α − βy)} dτ = ∫_{0}^{t} \frac{y′}{y} dτ + ∫_{0}^{t} \frac{βy′}{α − βy} dτ

= ln \frac{|y(t)|}{|y_{0}|} − ln \frac{|α − βy(t)|}{|α − βy_{0}|} = ln \frac{|α − βy_{0}|}{|y_{0}|} \frac{|y(t)|}{|α − βy(t)|}

Noticing that 0 < y(t) < α/β whenever 0 < y_{0} < α/β and that y(t) > α/β whenever y_{0} > α/β we can eliminate the absolute values and solve for y(t) obtaining

y(t) = \frac{αy_{0}e^{αt}}{α + βy_{0}(e^{αt} − 1)}

3. Assuming continuous compounding, the evolution of an initial amount of money K_{0} invested in an instrument with yearly return rate r and annual costs of C is described by

K′(t) = rK(t) − C , t > 0 , K(0) = K_{0} .

This equation was derived in class and its solution is given by

K(t) = \frac{C}{r} + (K_{0} − \frac{C}{r})e^{rt} .

(a) Answering the first question amounts to finding t > 0 such that

2K_{0} = \frac{C}{r} + (K_{0} − \frac{C}{r})e^{rt} = 750 + (K_{0} − 750)e^{0.08t} .

We therefore obtain

t = 12.5 ln(\frac{2K_{0} − 750}{K_{0} − 750}) .

(b) Here we need to compare

K_{1}(10) = 750 + (K_{0} − 750)e^{0.8}

to

K_{2}(10) = 1200 + (K_{0} − 1200)e^{1}

for K_{0} = 2000. This gives

K_{1}(10) ≈ 3500 and K_{2}(10) ≈ 3360

We would therefore pick the first fund for our investment.

4. By mere inspection we see that

y′ < 0 , if y > 1 or y < −1
y′ = 0 , if y = ±1
y′ > 0 , if − 1 < y < 1

We therefore conclude that

lim_{t→∞} y(t, y_{0}) = 1, provided y_{0} > −1

as well as lim_{t→∞} y(t, −1) = −1. After computing the solution we shall see that it blows up in finite time in the remaining case (y_{0} < −1). The equation is separable, in fact

\frac{y′}{1 − y^{2}} = 1 (y ≠ ±1) .

In order to integrate the left-hand-side of the above expression first notice that

\frac{1}{1 − y^{2}} = \frac{1}{(1 + y)(1 − y)} = \frac{1}{2} \frac{1}{1 + y} + \frac{1}{2} \frac{1}{1 − y} .

Then

2t = 2 ∫_{0}^{t} \frac{y′(τ)}{1 − y(τ)^{2}} dτ = ∫_{0}^{t}(\frac{y′(τ)}{1 + y(τ)} + \frac{y′(τ)}{1 − y(τ)}) dτ = ln(\frac{|1 + y(t)|}{|1 + y_{0}|} \frac{|1 − y_{0}|}{|1 − y(t)|})

Noticing that 1 ± y(t) always has the same sign as 1 ± y_{0} we can suppress the absolute values to obtain

\frac{1 + y(t)}{1 − y(t)} \frac{1 − y_{0}}{1 + y_{0}} = e^{2t}

from which follows that

y(t) = \frac{y_{0} − 1 + (1 + y_{0})e^{2t}}{1 − y_{0} + (1 + y_{0})e^{2t}} .

When y_{0} < −1 the denominator becomes zero if e^{2t} = \frac{y_{0}−1}{1+y_{0}}, that is, when t = \frac{1}{2} ln \frac{y_{0}−1}{1+y_{0}} .

5. The find the evolution of the fish population y we need to solve the given inhomogeneous first order differential equation. Since a solution of the homogeneous equation is given by e^{t} and the initial population amounts to a 1000 fish, we obtain

y(t) = 1000 e^{t} − F e^{t} ∫_{0}^{t} e^{−τ} (1 + cos(τ)) dτ .

using the variation-of-parameters formula. The integral can be computed by double integration by parts, for instance, to give

y(t) = (1000 − \frac{5}{2}F)e^{t} + F [1 − \frac{1}{2} sin(t) + \frac{1}{2} cos(t)] .

Now, the exponential term dominates for large times. To prevent extinction we therefore certainly need to at least impose that 1000 − \frac{5}{2}F ≥ 0 or, equivalently, that F ≤ 400. Otherwise, in fact, if F > 400 extinction is unavoidable. What if F = 400? Then

y(t) = 400[1 − \frac{1}{2} sin(t) + \frac{1}{2} cos(t)]

but 1 − \frac{1}{2} sin(t) + \frac{1}{2} cos(t) > 0 for all times and therefore the maximal allowed fishing quota is precisely F = 400 .

6. Since the equation is inhomogeneous and g(t) is not specified we can only compute the solution by the variation-of-parameters formula. First we compute the general solution of the homogeneous equation via the characteristic equation

r^{2} − α = 0

which leads to the two linearly independent solutions

y_{1}(t) = e^{\sqrt{α}t} and y_{2}(t) = e^{−\sqrt{α}t} .

Then a particular solution is given by

y_{p}(t) = −y_{1}(t) ∫_{0}^{t} \frac{y_{2}(τ)g(τ)}{W (y_{1}, y_{2})(τ)} dτ + y_{2}(t) ∫_{0}^{t} \frac{y_{1}(τ)g(τ)}{W (y_{1}, y_{2})(τ)} dτ

Since W (y_{1}, y_{2})(τ) = −2\sqrt{α} this amounts to

y_{p}(t) = \frac{1}{2\sqrt{α}}[e^{\sqrt{α}t} ∫_{0}^{t} e^{−\sqrt{α}τ} g(τ) dτ − e^{−\sqrt{α}t} ∫_{0}^{t} e^{\sqrt{α}τ} g(τ) dτ]

= \frac{1}{\sqrt{α}} ∫_{0}^{t}[\frac{e^{\sqrt{α}(t−τ)} − e^{−\sqrt{α}(t−τ)}}{2} ]g(τ) dτ = \frac{1}{\sqrt{α}} ∫_{0}^{t} sinh(\sqrt{α}(t − τ))g(τ) dτ

The general solution is therefore given by

y(t) = c_{1}y_{1}(t) + c_{2}y_{2}(t) + y_{p}(t) .

Finally we use the boundary conditions to obtain the following system

\begin{cases}0 = y(0) = c_{1} + c_{2} \\ 0 = y(1) = c_{1}e^{\sqrt{α}} + c_{2}e^{−\sqrt{α}} + \frac{1}{\sqrt{α}} ∫_{0}^{1} sinh(\sqrt{α}(1 − τ))g(τ) dτ \end{cases}

for c_{1} and c_{2}. Solving it gives

c_{1} = −c_{2} = −\frac{1}{2\sqrt{α} sinh(\sqrt{α})} ∫_{0}^{1} sinh(\sqrt{α}(1 − τ))g(τ) dτ

and thus

y(t) = −\frac{1}{\sqrt{α}} \frac{sinh(\sqrt{α} t)}{sinh(\sqrt{α})} ∫_{0}^{1} sinh(\sqrt{α}(1 − τ))g(τ) dτ + \frac{1}{\sqrt{α}} ∫_{0}^{t} sinh(\sqrt{α}(t − τ))g(τ) dτ