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Math 305 \hfill Homework Set 9  \hfill Spring 2017

\begin{center}Due Monday March 27 \end{center}

{\bf O.} You should know how to do problems 1-27 in Section 5.1. Do not turn these in. You may be quizzed on them.


{\bf I.} For each initial value problem below, find the first 5 terms of the series solution.

\begin{enumerate} 

\item $y'' - (\sin x )y' - 3x^2 y = 0$, $y(0)=1$, $y'(0)=2$.

\item $e^x y'' - 2y' +xy = 0$, $y(0)=2$, $y'(0)=0$.

\item $y'' - x^2 y' - 3y = 0$, $y(3)=1$, $y'(3)=1$.

\end{enumerate}


{\bf II.} Find a lower bound for the radius of convergence of the series solution 
to each of the following. 
\begin{enumerate}

\item $x y'' + (x+3) y' + \frac{y}{1+x^2} = 0$, centered at $x_0 = 4$.

\item $y'' + \frac{1}{x-2} y' + \frac{x}{x^2+3} y = 0$, centered at $x_0 = 1$

\item $(1+x^3) y'' + 4x y' + y = 0$, centered at $x_0 = 3$.

\end{enumerate}


{\bf III.} Find the general series solution to each of the following, including a 
recursive formula for the coefficients. If you can find a non-recurvise formula do so. 
\begin{enumerate}

\item $y'' + xy' - y = 0$, centered at $x_0=0$.

\item $xy'' + xy' +3y = 0$, centered at $x_0=2$.

\item $(x^2+1)y'' + x^3y' - y = 0$, centered at $x_0=0$.

\end{enumerate}

{\bf IV.} Prove the following identities.
\begin{enumerate}

\item $[\sin t \ln (\sec t + \tan t) ]' -\cos t \ln (\sec t + \tan t) = \tan t$. Assume $-\pi/2 < t < \pi/2$.

\item $\sin^3 t = \frac{3}{4}\sin t - \frac{1}{4} \sin 3t$. Thus, if you had a $\sin^3 t$ forcing 
function you could convert it into a form were the method of undetermined coefficients would apply. 

\item For $m$ and $n$ both integers $\dis \int_{-L}^{L} \cos \left( \frac{m\pi x}{L} \right) \cos \left( \frac{n\pi x}{L} \right) \, dx = \left\{\begin{array}{ccc}
0 & \mbox{for} & m \neq n,\\
L & \mbox{for} & m=n. \end{array} \right.$

\item Show that for a $3 \times 3$ matrix switching the first and second rows changes only 
the sign of the determinant.

\end{enumerate}


{\bf V.} Do the following integrals.
\begin{enumerate}

\item $\int \frac{1}{v^2+1} \, dv$

\item $\int \frac{1}{v^2-1} \, dv$

\item $\int \frac{1}{v(v+1)(v-1)} \, dv$

\item $\int \frac{v^2}{v^2+1} \, dv$

\item $\int \frac{v^3+v^2+v+6}{v^2-v+3} \, dv$

\end{enumerate}

\end{document}
