\documentclass[11pt]{amsart}
\setlength{\textheight}{10in}
\setlength{\topmargin}{-1in}
\usepackage{graphicx}
\pagestyle{empty}
\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}}
\newcommand{\dis}{\displaystyle}
\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Math 305 \hfill Homework Set 8  \hfill Spring 2017

\begin{center}Due Monday March 13 \end{center}

{\bf I.} For each differential equation below, find the general 
solution. 


\begin{enumerate}
\item $xy' + 2y + 3x = 0$

\item $4y'' + 17 y' + 4y = \sin t$

\item $y'' + 9y = 9 \sec^2(3t)$, $-\pi/6 < t < \pi/6$

\item $(e^x + 1)y' = y(1-e^x)$

\item  $\frac{d^2H}{d\theta^2} + \frac{dH}{d\theta} - 2H = \theta e^\theta$

\item $w'' + 9w = 9t - \cos 3t$

\item $3y''' - 2y'' +15 y' - 10 y = 0$

\item $y'''' - y = 0$

\item $y''' - y'' - 2y' = t^2 + t$
\end{enumerate}

{\bf II.}
\begin{enumerate}
\item Show that $y_1(t) = \tan t$ is a solution to $y'' - (2 \sec^2 t)y = 0$. Find a second solution $y_2(t)$ that is linearly independent from $y_1(t)$.


\item Let $g(t) = \left\{ \begin{array}{ccc}
0 & \mbox{for} & t<0 \\
1  & \mbox{for} & 0 \leq t < 5 \\
0 & \mbox{for} & t\geq 5
\end{array}\right.$

\vspace{.25in}

\noindent Let $h(t) = \left\{ \begin{array}{ccc}
0 & \mbox{for} & t<0 \\
1  & \mbox{for} & 0 \leq t < 4 \\
0 & \mbox{for} & t\geq 4
\end{array}\right.$

\vspace{.25in}

\noindent (a) Find a differentiable solution to $y'' + \pi^2y = g(t)$, $y(0)=y'(0)=0$, then graph it over $t \in [0,10]$

\vspace{.25in}

\noindent(b) Find a differentiable solution to $y'' + \pi^2y = h(t)$, $y(0)=y'(0)=0$, then graph it over $t \in [0,10]$

\vspace{.25in}



\item  Find the general solution to $y'' + 3y' + 2y = \cos^2 t$. Although this is not covered
by the method of undetermined coefficients given in your textbook, try
$y_p = A \cos^2 t + B \sin t \cos t + C \sin^2 t$, and see what 
happens!


\item Let $\epsilon(n) = \left\{ \begin{array}{rcl}
1 & \mbox{for} & n = 0,1,4,5,8,9,...\\
-1 & \mbox{for} & n=2,3,6,7,10,11,... \end{array} \right.$

\vspace{.25in}

\noindent Let $f(x) = \sum_{n=0}^\infty \epsilon(n) \frac{x^n}{n!} $.

\vspace{.25in}

\noindent Show by direct substitution that $y=f(x)$ solves the initial value problem $y'' + y = 0$, $y(0)=y'(0)=1$.

\end{enumerate}




\end{document}