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Section 1.4 Exercises

Exercises Exercises

1.

Solve \(y'' - 6y' + 8y = 1\text{.}\)

Hint

What are the solutions to \(y'' - 6y' + 8y = 0\text{?}\) One might remember variation of parameters gives us a particular solution \(u_1y_1 + u_2y_2\) with

\begin{align*} \begin{bmatrix} y_1 \amp y_2 \\ y_1' \amp y_2' \end{bmatrix} \begin{bmatrix} u_1' \\ u_2' \end{bmatrix} = \begin{bmatrix} 0 \\ g(t)/p(t) \end{bmatrix}. \end{align*}
2.

Solve \(y'' + 5y' + 6y = e^{-3t}\text{.}\)

Hint

Solve the homogeneous version first. Then, use variation of parameters to find a particular solution of the inhomogeneous version.

3.

Solve \(y'' + 5y' + 6y = e^{-2t}\text{.}\)

4.

Solve \(4y'' + 4y' + y = g(t)\text{,}\) where

\begin{align*} g(t) = \begin{cases} 1 \amp 0 \lt t \lt 1, \\ 0 \amp t \geq 1. \end{cases} \end{align*}
Hint

Break up into cases:

When \(t \geq 1\text{,}\) the ODE to solve becomes \(4y'' + 4y' + y = 0\) (this is homogeneous!)

When \(0 \lt t \lt 1 \text{,}\) the ODE to solve becomes \(4y'' + 4y' + y = 1\) (this is inhomogeneous)

5.

Suppose \(y'' - 9y = 6\text{.}\) Find a condition on \(y(0)\) and \(y'(0)\) such that \(y(t) \to 0\) as \(t \to \infty\text{.}\)

Is it possible for \(y(t)\) to solve this ODE while having \(y(t) \to 0\) as \(t \to -\infty\text{?}\) If so, find a condition on \(y(0)\) and \(y'(0)\) that allows this to happen. What happens if both the condition found here and previously are put together?

6.

Solve \(y'' + 2y' + 1 = e^t\text{.}\)

Once you've done that, can you solve \(y'' + 2y' + 1 = e^{at}\) for any real constant \(a \neq 1\text{?}\) Which values of \(a\) allow all solutions \(y(t) \to 0\) as \(t \to \infty\text{?}\)

7.

It can be extremely useful to translate real ODEs into complex ones. Here is a guided exercise showing how one might solve \(y'' - y' + 2y = \cos t\text{.}\)

  1. First, find the general solution to \(y'' - y' + 2y = 0\text{.}\)

  2. Note that \(\cos t = \Re\{e^{it}\}\text{.}\) Prove that if \(z(t)\) solves the ODE \(z'' - z' + 2z = e^{it}\text{,}\) then \(y(t) = \Re\{z(t)\}\) solves the ODE \(y'' - y' + 2y = \Re\{e^{it}\} = \cos t\text{.}\)

    This fact can actually be generalized for an ODE of the form \(p(t) z'' + q(t) z' + r(t) z = g(t)\text{.}\) Here, we'll take \(t \in \mathbb{R}\) and \(p(t), q(t), r(t)\) to all be real-valued functions and \(g(t)\) to be a complex-valued function.

    Hint

    Set \(z(t) = y(t) + ix(t)\text{,}\) and substitute this into \(z'' - z' + 2z = e^{it}\text{.}\) What happens when we take the real part of both sides?

    For the generalized version, again set \(z(t) = y(t) + ix(t)\) and substitute this into the more general ODE. What happens when we take either the real or imaginary part of both sides?

  3. Find a particular solution to the ODE \(z'' - z' + 2z = e^{it}\) using variation of parameters. (When performing derivatives or integrals with \(e^{it}\text{,}\) just treat \(i\) as a constant.)

    Once you've found a particular solution \(Z(t)\) to \(z'' - z' + 2z = e^{it}\text{,}\) give an appropriate particular solution to \(z'' - z' + 2z = \cos t\text{,}\) and then state the general solution to \(z'' - z' + 2z = \cos t\text{.}\)

8.

Using the method described in the previous exercise, solve \(y'' + 7y' + 12y = e^{-t}\cos t\text{.}\)

Hint

Note that \(e^{-t}\cos t = \Re\{e^{(-1 + i)t}\}\text{.}\)

If you want even more exercises, check out this link. There are way more exercises there than the number a normal person should complete.