Assignment 2

Due Thursday, September 10

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

1Finding solutions

Solve each initial-value problem by separating variables or undoing the product rule. You choose which approach to use!

For each solution, state its domain: the largest interval containing the initial time on which it solves the differential equation. (It may be all of $ℝ$ ! Or the solution may blow up at a finite time.)

(a)

$𝑦' + 2 𝑦 = 𝑒 𝑡$ , $𝑦 (0) = 1$ .

Solution

To make $𝑚 𝑦' + 2 𝑚 𝑦$ equal $(𝑚 𝑦)'$ , we need $𝑚' = 2 𝑚$ . Choose $𝑚 = 𝑒 2 𝑡$ and multiply:

𝑒 2 𝑡 𝑦' + 2 𝑒 2 𝑡 𝑦 = 𝑒 3 𝑡 ⟹ (𝑒 2 𝑡 𝑦)' = 𝑒 3 𝑡 .

Integrating gives $𝑒 2 𝑡 𝑦 = 13 𝑒 3 𝑡 + 𝐶$ , so $𝑦 = 13 𝑒 𝑡 + 𝐶 𝑒 - 2 𝑡$ . The initial condition says $1 = 13 + 𝐶$ , and therefore

𝑦 = 13 𝑒 𝑡 + 23 𝑒 - 2 𝑡 .

This solves the equation on $(- \infty, \infty)$ .

(b)

$𝑦' = 2 𝑡 (1 + 𝑦 2)$ , $𝑦 (0) = 0$ .

Solution

Since $1 + 𝑦 2$ is always positive, we may separate:

𝑑 𝑦 1 + 𝑦 2 = 2 𝑡 𝑑 𝑡 ⟹ a r c t a n 𝑦 = 𝑡 2 + 𝐶 .

The initial condition gives $𝐶 = 0$ , so $𝑦 = t a n (𝑡 2)$ . Starting from $𝑡 = 0$ , the first poles occur where $𝑡 2 = 𝜋 / 2$ . The domain is

(- \sqrt 𝜋 2, \sqrt 𝜋 2) .

Although the formula is defined again beyond these poles, those other pieces cannot extend the solution through its initial point.

(c)

$𝑡 𝑦' + 2 𝑦 = 𝑡 2$ , $𝑦 (1) = 0$ .

Solution

On an interval containing $1$ with $𝑡 > 0$ , divide by $𝑡$ to obtain $𝑦' + (2 / 𝑡) 𝑦 = 𝑡$ . Since $(𝑡 2)' / 𝑡 2 = 2 / 𝑡$ , multiplying by $𝑡 2$ creates a product derivative:

𝑡 2 𝑦' + 2 𝑡 𝑦 = 𝑡 3 ⟹ (𝑡 2 𝑦)' = 𝑡 3 .

Thus $𝑡 2 𝑦 = 𝑡 4 / 4 + 𝐶$ . Since $𝑦 (1) = 0$ , we have $𝐶 = - 1 / 4$ , giving

𝑦 = 𝑡 2 4 - 1 4 𝑡 2 .

The solution diverges as $𝑡 \to 0 +$ , so its domain is $(0, \infty)$ . The original equation is written without division by $𝑡$ , but this particular solution still cannot be continued through $0$ .

(d)

$𝑦' = 𝑒 𝑡 (1 - 𝑦)$ , $𝑦 (0) = 2$ .

Solution

The constant solution $𝑦 = 1$ does not meet our initial condition. For the solution we want, separating variables gives

𝑑 𝑦 1 - 𝑦 = 𝑒 𝑡 𝑑 𝑡 ⟹ - l n | 1 - 𝑦 | = 𝑒 𝑡 + 𝐶 .

Exponentiating and allowing a signed constant gives $1 - 𝑦 = 𝐴 𝑒 - 𝑒 𝑡$ . At $𝑡 = 0$ , $- 1 = 𝐴 𝑒 - 1$ , so $𝐴 = - 𝑒$ and

𝑦 = 1 + 𝑒 1 - 𝑒 𝑡 .

Its domain is $(- \infty, \infty)$ .

(e)

$𝑦' + 2 𝑡 𝑦 = 2 𝑡 3$ , $𝑦 (0) = 1$ .

Solution

We want $𝑚' = 2 𝑡 𝑚$ , so choose $𝑚 = 𝑒 𝑡 2$ . Multiplying gives

(𝑒 𝑡 2 𝑦)' = 2 𝑡 3 𝑒 𝑡 2 .

To integrate the right side, set $𝑣 = 𝑡 2$ , so $𝑑 𝑣 = 2 𝑡 𝑑 𝑡$ and $2 𝑡 3 𝑑 𝑡 = 𝑣 𝑑 𝑣$ . Integration by parts then gives

\int 2 𝑡 3 𝑒 𝑡 2 𝑑 𝑡 = \int 𝑣 𝑒 𝑣 𝑑 𝑣 = 𝑣 𝑒 𝑣 - 𝑒 𝑣 + 𝐶 = (𝑡 2 - 1) 𝑒 𝑡 2 + 𝐶 .

Therefore $𝑦 = 𝑡 2 - 1 + 𝐶 𝑒 - 𝑡 2$ . The initial condition gives $𝐶 = 2$ , so

𝑦 = 𝑡 2 - 1 + 2 𝑒 - 𝑡 2,

with domain $(- \infty, \infty)$ .

(f)

$𝑦' = 𝑡 𝑒 - 𝑦$ , $𝑦 (0) = 0$ .

Solution

Multiplying by $𝑒 𝑦$ separates the variables:

𝑒 𝑦 𝑑 𝑦 = 𝑡 𝑑 𝑡 ⟹ 𝑒 𝑦 = 𝑡 2 2 + 𝐶 .

The initial condition gives $𝐶 = 1$ , hence

𝑦 = l n (1 + 𝑡 2 2) .

The logarithm's argument is positive for every real $𝑡$ , so the domain is $(- \infty, \infty)$ .

2Find the combination

Consider

𝑦' = 𝑡 2 + 2 𝑡 𝑦 + 𝑦 2 - 3 𝑡 - 3 𝑦 + 1 .

Neither separating variables nor undoing the product rule seems to apply as written. But a combination of algebra and calculus will let us succeed.

(a)

Group the terms on the right-hand side to reveal the repeated combination $𝑡 + 𝑦$ . Set $𝑢 = 𝑡 + 𝑦$ , and derive a differential equation for $𝑢$ involving only $𝑢$ and its derivative. What can we now do that we could not do before?

Solution

The first three terms are $(𝑡 + 𝑦) 2$ , and the next two are $- 3 (𝑡 + 𝑦)$ . Thus the original equation says $𝑦' = 𝑢 2 - 3 𝑢 + 1$ . Since $𝑢 = 𝑡 + 𝑦$ , we have $𝑢' = 1 + 𝑦'$ , so

𝑢' = 1 + (𝑢 2 - 3 𝑢 + 1) = 𝑢 2 - 3 𝑢 + 2 = (𝑢 - 1) (𝑢 - 2) .

The equation for $𝑢$ has no explicit dependence on $𝑡$ , and its right side is a product of factors in $𝑢$ . We can now separate variables.

(b)

Solve the original equation with initial condition $𝑦 (0) = 0$ by first solving for $𝑢$ and then returning to $𝑦$ . State the domain of your solution.

Solution

The initial condition becomes $𝑢 (0) = 0$ . The constant solutions $𝑢 = 1$ and $𝑢 = 2$ do not meet this condition, so we separate the other solutions:

𝑑 𝑢 (𝑢 - 1) (𝑢 - 2) = 𝑑 𝑡 .

Partial fractions give

1 (𝑢 - 1) (𝑢 - 2) = 1 𝑢 - 2 - 1 𝑢 - 1,

and integrating yields

l n ∣ 𝑢 - 2 𝑢 - 1 ∣ = 𝑡 + 𝐶 .

Exponentiating and absorbing the sign into a constant,

𝑢 - 2 𝑢 - 1 = 𝐴 𝑒 𝑡 .

At $𝑡 = 0$ , $𝑢 = 0$ , so $𝐴 = 2$ . Solving for $𝑢$ gives

𝑢 (1 - 2 𝑒 𝑡) = 2 - 2 𝑒 𝑡 ⟹ 𝑢 = 2 - 2 𝑒 𝑡 1 - 2 𝑒 𝑡 .

Returning to $𝑦 = 𝑢 - 𝑡$ , we obtain

𝑦 = 2 - 2 𝑒 𝑡 1 - 2 𝑒 𝑡 - 𝑡 .

The denominator vanishes at $𝑡 = - l n 2$ , where the numerator is $1$ . The solution tends to $- \infty$ as $𝑡 \to - l n 2$ from the right, so it cannot extend through that time. Its domain is $(- l n 2, \infty)$ .

(c)

Find the constant solutions of the $𝑢$ -equation. Translate each into a solution of the original equation, and verify both by substitution.

Solution

Division by $(𝑢 - 1) (𝑢 - 2)$ excludes $𝑢 = 1$ and $𝑢 = 2$ . Since $𝑦 = 𝑢 - 𝑡$ , these give

𝑦 = 1 - 𝑡 a n d 𝑦 = 2 - 𝑡 .

Both have derivative $- 1$ . Substituting $𝑦 = 1 - 𝑡$ into the original right-hand side gives

(𝑡 + 𝑦) 2 - 3 (𝑡 + 𝑦) + 1 = 1 - 3 + 1 = - 1,

and substituting $𝑦 = 2 - 𝑡$ gives $4 - 6 + 1 = - 1$ . Both solve the original equation for every real $𝑡$ , although neither satisfies $𝑦 (0) = 0$ .