Assignment 1

Due Thursday, September 3

Work these on your own paper. Show enough of the reasoning that a reader could follow it without you in the room.

1A loan, idealized

You take out a loan. You make regular payments on the loan, but at the same time it is accumulating interest. Let's model this by a differential equation for the rate of change of your loan balance $𝐿 (𝑡)$ .

To write a differential equation we think of the interest as compounding continuously at some rate, and of your payments as flowing out continuously at the steady rate, rather than in monthly chunks.

(a)

Write a differential equation for the amount $𝐿 (𝑡)$ you owe, in dollars, with $𝑡$ in years (define your own constants for interest and payment rates; and think about which one of these depends on the current size of the loan and which does not.)

Solution

Let $𝑟 > 0$ be the annual interest rate (in inverse years) and $𝑝 > 0$ the payment rate (in dollars per year). Interest adds $𝑟 𝐿$ dollars per year, while payments remove $𝑝$ :

𝐿' = 𝑟 𝐿 - 𝑝 .

(b)

Without solving anything, read the sign of $𝐿'$ from your equation: for which loan amounts is the debt shrinking, and for which is it growing? What happens at the border in-between?

Solution

Since $𝑟 > 0$ , the debt shrinks when $0 < 𝐿 < 𝑝 / 𝑟$ and grows when $𝐿 > 𝑝 / 𝑟$ . At $𝐿 = 𝑝 / 𝑟$ , payments exactly cover the interest, so the balance stays constant. The loan model stops when the debt reaches zero.

2The law of exponents, by uniqueness

Fix a number $𝑠$ , and consider the two functions

𝑓 (𝑡) = 𝑒 𝑠 + 𝑡 a n d 𝑔 (𝑡) = 𝑒 𝑠 𝑒 𝑡 .

(a)

Verify that both functions solve $𝑦' = 𝑦$ , and compute $𝑓 (0)$ and $𝑔 (0)$ .

Solution

By the chain rule $𝑓' (𝑡) = 𝑒 𝑠 + 𝑡 = 𝑓 (𝑡)$ , and since $𝑒 𝑠$ is a constant, $𝑔' (𝑡) = 𝑒 𝑠 𝑒 𝑡 = 𝑔 (𝑡)$ . Both functions solve $𝑦' = 𝑦$ , and

𝑓 (0) = 𝑒 𝑠 = 𝑔 (0) .

(b)

Check that the hypotheses of the existence–uniqueness theorem hold at every point.

Solution

Here $𝐹 (𝑡, 𝑦) = 𝑦$ and $𝜕 𝐹 / 𝜕 𝑦 = 1$ are continuous everywhere, so through each point there passes exactly one local solution.

(c)

What does the theorem now force? State the algebraic identity you have just proved.

Solution

The two functions solve the same equation from the same initial value. Uniqueness makes them agree throughout their common domain, which is all of $ℝ$ . Thus, for every $𝑠$ and $𝑡$ ,

𝑒 𝑠 + 𝑡 = 𝑒 𝑠 𝑒 𝑡 .

(d)

Run the same argument again to prove the law of logarithms: for a fixed $𝑠 > 0$ , show that $l n (𝑠 𝑡)$ and $l n 𝑠 + l n 𝑡$ solve the same initial-value problem, and conclude that $l n (𝑠 𝑡) = l n 𝑠 + l n 𝑡$ for every $𝑡 > 0$ .

Solution

Both functions have derivative $1 / 𝑡$ — the first by the chain rule, $𝑠 / (𝑠 𝑡) = 1 / 𝑡$ — so both solve $𝑦' = 1 / 𝑡$ , whose right side and its $𝑦$ -derivative are continuous for $𝑡 > 0$ . At $𝑡 = 1$ they agree: $l n (𝑠 \cdot 1) = l n 𝑠 = l n 𝑠 + l n 1$ . Two solutions of the same equation with the same value at $𝑡 = 1$ must coincide, so $l n (𝑠 𝑡) = l n 𝑠 + l n 𝑡$ for every $𝑡 > 0$ .

3Drawing a slope field

Consider the equation

𝑦' = 𝑦 2 - 𝑡 .

(a)

Find every point where the equation prescribes slope zero, and describe this set of points as a curve in the $(𝑡, 𝑦)$ -plane.

Solution

The slope is zero exactly when $𝑡 = 𝑦 2$ : a parabola through the origin, opening in the positive $𝑡$ direction.

(b)

On the region $0 \leq 𝑡 \leq 4$ , $- 2 \leq 𝑦 \leq 2$ , compute the prescribed slope at each integer grid point and sketch the slope field. Your curve from the previous part organizes the picture: which side of it has rising ticks, and which falling?

Solution

The slope at each grid point is $𝑦 2 - 𝑡$ :

$𝑦\𝑡$	$0$	$1$	$2$	$3$	$4$
$2$	$4$	$3$	$2$	$1$	$0$
$1$	$1$	$0$	$- 1$	$- 2$	$- 3$
$0$	$0$	$- 1$	$- 2$	$- 3$	$- 4$
$- 1$	$1$	$0$	$- 1$	$- 2$	$- 3$
$- 2$	$4$	$3$	$2$	$1$	$0$

Figure 1 The slope field at the integer grid points. The dashed parabola $𝑡 = 𝑦 2$ marks slope zero.

Ticks rise to the left of the parabola ( $𝑡 < 𝑦 2$ ), fall to its right ( $𝑡 > 𝑦 2$ ), and are horizontal on it.

(c)

Into your field, sketch the solutions starting from $𝑦 (0) = 2$ and from $𝑦 (0) = 0$ . Describe in a sentence or two how their fates differ.

Solution

The solution from $𝑦 (0) = 2$ begins with slope $4$ and climbs ever more steeply. The solution from $𝑦 (0) = 0$ begins flat, then decreases above the lower branch $𝑦 = - \sqrt 𝑡$ ; numerically, $𝑦 (4) \approx - 1.931$ . It follows that branch without lying on it: zero-slope points form a guide, not a solution curve.

4Solutions and impostors

In each part, decide whether the proposed function solves the differential equation. Explain your reasoning.

(a)

The equation $𝑦' = 𝑦 + 𝑒 𝑡$ , with the two candidates $𝑦 = 𝑡 𝑒 𝑡$ and $𝑦 = 2 𝑒 𝑡$ .

Solution

For $𝑦 = 𝑡 𝑒 𝑡$ , the product rule gives

𝑦' = 𝑒 𝑡 + 𝑡 𝑒 𝑡 = 𝑦 + 𝑒 𝑡,

so it is a solution. For $𝑦 = 2 𝑒 𝑡$ , the left side is $𝑦' = 2 𝑒 𝑡$ while the equation demands $𝑦 + 𝑒 𝑡 = 3 𝑒 𝑡$ . These are never equal, so $2 𝑒 𝑡$ is not a solution.

(b)

The system $𝑓' = 𝑓 + 𝑔$ , $𝑔' = 𝑔 - 𝑓$ , with the two candidate pairs $𝑓 = 𝑒 𝑡 c o s 𝑡$ , $𝑔 = - 𝑒 𝑡 s i n 𝑡$ and $𝑓 = 𝑒 𝑡$ , $𝑔 = 0$ .

Solution

For the first pair, the product rule gives

𝑓' = 𝑒 𝑡 c o s 𝑡 - 𝑒 𝑡 s i n 𝑡 = 𝑓 + 𝑔, 𝑔' = - 𝑒 𝑡 s i n 𝑡 - 𝑒 𝑡 c o s 𝑡 = 𝑔 - 𝑓,

so the pair solves both equations. The pair $𝑓 = 𝑒 𝑡$ , $𝑔 = 0$ satisfies the first equation, since $𝑓' = 𝑒 𝑡 = 𝑓 + 𝑔$ , but fails the second: $𝑔' = 0$ , while the equation demands $𝑔 - 𝑓 = - 𝑒 𝑡$ . A pair must satisfy both equations, so this pair is not a solution.

5Autonomous or not

Consider the three rules

(i) 𝑦' = 𝑦 2 - 4, (i i) 𝑦' = 𝑦 + 𝑡, (i i i) 𝑓' = 𝑓 + 𝑔, 𝑔' = 𝑔 - 𝑓 .

(a)

Which of these are autonomous? Answer with a sentence each.

Solution

Rules (i) and (iii) are autonomous: their instructions depend only on the current state, not on the time. Rule (ii) is not — the slope it prescribes changes with $𝑡$ even when $𝑦$ stays the same.

(b)

For rule (ii): show that the state $𝑦 = 0$ receives different instructions at two different times. Explain why this makes a phase line for (ii) impossible.

Solution

At $𝑦 = 0$ the prescribed slope is $0 + 𝑡 = 𝑡$ : zero at $𝑡 = 0$ , and $2$ at $𝑡 = 2$ . A phase line has one point for the state $𝑦 = 0$ , and that single point cannot carry both instructions at once. Only when the rule ignores the clock can the time direction be projected away.

6A phase line

Consider the autonomous equation

𝑦' = 𝑦 2 - 4 .

(a)

Find every value of $𝑦$ where the equation prescribes no motion, and draw the phase line, with arrows determined by the sign of $𝑦 2 - 4$ .

Solution

The rule vanishes at $𝑦 = \pm 2$ : two resting states. For $𝑦 < - 2$ the sign of $𝑦 2 - 4$ is positive, so the arrows point upward toward $- 2$ ; for $- 2 < 𝑦 < 2$ it is negative, so the arrows point downward toward $- 2$ ; for $𝑦 > 2$ it is positive, and the arrows point upward, away from $2$ .

↑ \circ 2 ↓ ∙ - 2 ↑

(b)

Using only your phase line, describe the entire future of the solution starting at $𝑦 (0) = 0$ , and of the solution starting at $𝑦 (0) = 3$ .

Solution

From $𝑦 (0) = 0$ the arrows point downward: the solution decreases, approaching the resting value $- 2$ but never arriving — reaching it would put two solutions through the same point, which uniqueness forbids. It stays between $- 2$ and $0$ and exists for every future time. From $𝑦 (0) = 3$ , the solution increases without bound: it cannot settle at a finite value where $𝑦 2 - 4$ is still positive. The phase line alone does not tell us whether this happens in finite or infinite time.

7Choosing the right guess

(a)

Consider $𝑡 2 𝑦 ″ = 2 𝑦$ . First try the guess $𝑦 = 𝑒 𝑟 𝑡$ and explain why no value of $𝑟$ can ever work. Then try a power, $𝑦 = 𝑡 𝑛$ : find every value of $𝑛$ that works, and verify the resulting solutions by substitution.

Solution

Substituting $𝑦 = 𝑒 𝑟 𝑡$ gives

𝑡 2 𝑟 2 𝑒 𝑟 𝑡 = 2 𝑒 𝑟 𝑡,

which would require $𝑡 2 𝑟 2 = 2$ at every time $𝑡$ . No fixed $𝑟$ can manage that: for $𝑟 \neq 0$ the left side varies with $𝑡$ , and for $𝑟 = 0$ it is zero.

Substituting $𝑦 = 𝑡 𝑛$ instead gives

𝑡 2 𝑛 (𝑛 - 1) 𝑡 𝑛 - 2 = 𝑛 (𝑛 - 1) 𝑡 𝑛 = 2 𝑡 𝑛,

so $𝑛 (𝑛 - 1) = 2$ , i.e. $𝑛 2 - 𝑛 - 2 = (𝑛 - 2) (𝑛 + 1) = 0$ : the powers $𝑛 = 2$ and $𝑛 = - 1$ . Directly, $𝑡 2 (𝑡 2) ″ = 2 𝑡 2$ and $𝑡 2 (1 / 𝑡) ″ = 𝑡 2 \cdot 2 / 𝑡 3 = 2 / 𝑡$ , so $𝑦 = 𝑡 2$ solves everywhere and $𝑦 = 1 / 𝑡$ solves away from $𝑡 = 0$ .

(b)

Consider $𝑦' + 𝑦 = 𝑒 2 𝑡$ . Here an exponential is clearly the right shape — but with a knob: try $𝑦 = 𝐴 𝑒 2 𝑡$ , and find the value of $𝐴$ that makes it a solution.

Solution

Substituting gives

2 𝐴 𝑒 2 𝑡 + 𝐴 𝑒 2 𝑡 = 3 𝐴 𝑒 2 𝑡 = 𝑒 2 𝑡,

so the guess works exactly when $𝐴 = 13$ : the solution $𝑦 = 13 𝑒 2 𝑡$ .

8Separating variables

Solve each initial-value problem. State its domain: the largest interval containing the initial time on which it solves the differential equation (be careful about division by zero, logarithms of negative numbers, etc.).

(a)

$𝑦' = - 𝑡 𝑦$ , $𝑦 (0) = 2$ .

Solution

No constant function solves this equation, so there are no equilibria to find. Separating,

𝑦 𝑑 𝑦 = - 𝑡 𝑑 𝑡 ⟹ 𝑦 2 2 = - 𝑡 2 2 + 𝐶,

so $𝑡 2 + 𝑦 2$ is constant: the solution curves are arcs of circles. The initial condition gives $𝑡 2 + 𝑦 2 = 4$ , and since $𝑦 (0) = 2 > 0$ we take the upper branch:

𝑦 = \sqrt 4 - 𝑡 2 .

The largest interval containing $0$ is $(- 2, 2)$ ; at its endpoints the solution reaches $𝑦 = 0$ , where the equation itself breaks down.

(b)

$𝑦' = 𝑡 𝑦 + 𝑡$ , $𝑦 (0) = 0$ .

Solution

Factoring the right side as $𝑡 (𝑦 + 1)$ shows the constant function $𝑦 \equiv - 1$ is an equilibrium solution. Our initial condition is not $- 1$ , so we may divide:

𝑑 𝑦 𝑦 + 1 = 𝑡 𝑑 𝑡 ⟹ l n | 𝑦 + 1 | = 𝑡 2 2 + 𝐶,

so $𝑦 + 1 = 𝐴 𝑒 𝑡 2 / 2$ . The initial condition gives $𝐴 = 1$ , and

𝑦 = - 1 + 𝑒 𝑡 2 / 2,

defined for all time.

(c)

$𝑦' = 𝑒 𝑡 + 𝑦$ , $𝑦 (0) = 0$ .

Solution

The right side splits: $𝑒 𝑡 + 𝑦 = 𝑒 𝑡 𝑒 𝑦$ , which is never zero, so there are no equilibria. Separating,

𝑒 - 𝑦 𝑑 𝑦 = 𝑒 𝑡 𝑑 𝑡 ⟹ - 𝑒 - 𝑦 = 𝑒 𝑡 + 𝐶,

so $𝑒 - 𝑦 = 𝐴 - 𝑒 𝑡$ . The initial condition gives $𝐴 = 2$ , and

𝑦 = - l n (2 - 𝑒 𝑡) .

This is defined only while $𝑒 𝑡 < 2$ : the largest interval containing $0$ is $(- \infty, l n 2)$ , and $𝑦 \to \infty$ as $𝑡 \to l n 2$ from below.