Math292 Textbook Exercises

Section 1.4 Exercises

Exercises Exercises

1.

Solve $y^{″} - 6 y^{'} + 8 y = 1 .$

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*}

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2.

Solve $y^{″} + 5 y^{'} + 6 y = e^{- 3 t} .$

Hint

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

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3.

Solve $y^{″} + 5 y^{'} + 6 y = e^{- 2 t} .$

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4.

Solve $4 y^{″} + 4 y^{'} + y = g (t),$ where

\begin{array}{r} g (t) = {\begin{cases} 1 & 0 < t < 1, \\ 0 & t \geq 1. \end{cases} \end{array}

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)

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5.

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

Is it possible for $y (t)$ to solve this ODE while having $y (t) \to 0$ as $t \to - \infty ?$ 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?

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6.

Solve $y^{″} + 2 y^{'} + 1 = e^{t} .$

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

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7.

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

First, find the general solution to $y^{″} - y^{'} + 2 y = 0 .$
Note that $\cos t = Re {e^{i t}} .$ Prove that if $z (t)$ solves the ODE $z^{″} - z^{'} + 2 z = e^{i t},$ then $y (t) = Re {z (t)}$ solves the ODE $y^{″} - y^{'} + 2 y = Re {e^{i t}} = \cos t .$

This fact can actually be generalized for an ODE of the form $p (t) z^{″} + q (t) z^{'} + r (t) z = g (t) .$ Here, we'll take $t \in R$ 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?
Find a particular solution to the ODE $z^{″} - z^{'} + 2 z = e^{i t}$ using variation of parameters. (When performing derivatives or integrals with $e^{i t},$ just treat $i$ as a constant.)

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

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8.

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

Hint

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

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