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%% MATH 54, Lenstra, Spring 96
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%% Solutions to Problem set 14
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%% by William Stein
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\author{For Lenstra's Math 54\\by William Stein}
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\title{Solutions to Problem Set 14 and Practice Final}
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\newcommand{\problem}[1]{\vspace{.3in}\par\noindent{\Large \bfseries #1. }}
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\newcommand{\soln}{\par {\em Solution.} }
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\begin{document}
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\maketitle
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This document contains the solutions to the 14th homework set.
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It also contains solutions to the practice final exam.
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The final examination is on Friday, May 17, 5--8 p.m. (room to
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be announced). The final examination covers differential
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equations.
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\section{Homework Solutions to 10.5.1, 10.5.7, 10.6.4}
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\problem{10.5.1} {\em Consider the conduction of
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heat in a copper rod $100\cm$ in length whose
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ends are maintained at $0^{\circ}\C$ for all $t>0$. Find an expression for
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the temperature $u(x,t)$ if the initial temperature distribution in the
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rod is given by
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\begin{itemize}
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\item[(a)] $u(x,0) = 50, \quad 0\leq x \leq 100$
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\item[(b)] $u(x,0) = \begin{cases} x, & 0\leq x < 50\\
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100 - x, & 50 \leq x \leq 100\end{cases}$
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\item[(c)] $u(x,0) = \begin{cases} 0, & 0\leq x < 25 \\
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50, & 25\leq x \leq 75 \\
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0, & 75 < x \leq 100\end{cases}$
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\end{itemize}
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}
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\soln
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{\em Summary.} (See page 543.) First find the
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sine series for $f(x)=u(x,0)$. Then the
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solution is $$u(x,t)=\sum_{n=1}^{\infty}b_ne^{-(n\pi\alpha/\ell)^2 t}
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\sin\frac{n\pi x}{\ell},$$
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where the coefficients $b_n$ are the same as in the series
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$$f(x)=\sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{\ell}.$$
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The $b_n$ are given by
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$$b_n=\frac{2}{\ell}\int_{0}^{\ell}f(x)\sin\frac{n\pi x}{\ell}\dx.$$
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(a) We compute the Fourier coefficients $b_n$.
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\begin{eqnarray*}
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b_n &=& \frac{2}{100}\int_{0}^{100} u(x,0)\sin\frac{n\pi x}{100}\dx\\
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&=& \frac{1}{50}\int_{0}^{100} 50 \sin\frac{n\pi x}{100}\dx\\
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&=& \Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_0^{100}
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= \begin{cases}\frac{200}{n\pi}&\text{$n$ odd}\\
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0&\text{$n$ even}\end{cases}
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\end{eqnarray*}
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Thus the solution is
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$$u(x,t) = \sum_{\text{$n$ odd}} \frac{200}{n\pi}e^{-(.0107n\pi)^2 t}
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\sin\frac{n\pi x}{100}.$$
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Here we have used the value $\alpha^2=1.14$ from page 513 to find
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$$\frac{\alpha}{\ell}=\frac{\sqrt{1.14}}{100}\approx .0107.$$
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(b) We compute
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\begin{eqnarray*}
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b_n &=& \frac{2}{100}\int_{0}^{100}u(x,0)\sin\frac{n\pi x}{100}\dx\\
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&=& \frac{1}{50}\int_{0}^{50}x\sin\frac{n\pi x}{100}\dx
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+\frac{1}{50}\int_{50}^{100}100\sin\frac{n\pi x}{100}\dx
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-\frac{1}{50}\int_{50}^{100}x\sin\frac{n\pi x}{100}\dx\\
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&=& \frac{1}{50}\Bigl\{ \Bigl[-\frac{100}{n\pi}x\cos\frac{n\pi x}{100}
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\Bigr]_0^{50}
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+\frac{100}{n\pi}\int_{0}^{50}\cos\frac{n\pi x}{100}\dx\Bigr\}
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+ 2\Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_{50}^{100} \\
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&& -\frac{1}{50}\Bigl\{\Bigl[-\frac{100}{n\pi}x\cos\frac{n\pi x}{100}
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\Bigr]_{50}^{100}
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+\frac{100}{n\pi}\int_{50}^{100}\cos\frac{n\pi x}{100}\dx\Bigr\}\\
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&=&-\frac{2}{n\pi}\Bigl(50\cos\frac{n\pi}{2}\Bigr)+\frac{2}{n\pi}
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\Bigl[\frac{100}{n\pi}\sin\frac{n\pi x}{100}\Bigr]_{0}^{50}
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-\frac{200}{n\pi}\Bigl(\cos n\pi - \cos \frac{n\pi}{2}\Bigr)\\
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& & +\frac{2}{n\pi}\Bigl(100\cos n\pi - 50 \cos\frac{n\pi}{2}\Bigr)
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-\frac{2}{n\pi}\Bigl[\frac{100}{n\pi}\sin\frac{n\pi x}{100}\Bigr]^{100}_{50}\\
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&=& -\frac{100}{n\pi}\cos\frac{n\pi}{2} +
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\frac{200}{n^2\pi^2}\sin\frac{n\pi}{2}
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- \frac{200}{n\pi}\cos n\pi + \frac{200}{n\pi}\cos \frac{n\pi}{2} \\
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&& + \frac{200}{n\pi}\cos n\pi - \frac{100}{n\pi}\cos \frac{n\pi}{2}
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+ \frac{200}{n^2\pi^2}\sin \frac{n\pi}{2}\\
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&=& \frac{400}{n^2\pi^2}\sin\frac{n\pi}{2}
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\end{eqnarray*}
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Thus the solution is
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$$u(x,t) = \sum_{n=1}^{\infty} \frac{400}{n^2\pi^2}\sin\frac{n\pi}{2}
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e^{-(.0107n\pi)^2 t} \sin\frac{n\pi x}{100}.$$
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(c) We compute
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\begin{eqnarray*}
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b_n&=&\frac{2}{100}\int_{0}^{100}u(x,0)\sin\frac{n\pi x}{100}\dx\\
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&=&\frac{2}{100}\int_{25}^{75}50\sin\frac{n\pi x}{100}\dx\\
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&=&\Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_{25}^{75}
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=\frac{100}{n\pi}\Bigl(\cos\frac{n\pi}{4}-\cos\frac{3n\pi}{4}\Bigr).
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\end{eqnarray*}
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The solution is thus
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$$u(x,t) = \sum_{n=1}^{\infty} \frac{100}{n\pi}
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\Bigl(\cos\frac{n\pi}{4}-\cos\frac{3\pi{}n}{4}\Bigr)
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e^{-(.0107n\pi)^2 t} \sin\frac{n\pi x}{100}.$$
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\problem{10.5.7} {\em Consider a uniform rod of length $\ell$ with an
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initial temperature given by $sin(\pi x/\ell)$, $0\leq x \leq \ell$. Assume
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that both ends of the bar are insulated. Find a formal series expansion
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for the temperature $u(x,t)$. What is the steady state temperature as
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$t\rightarrow\infty$?}
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\soln
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{\em Summary.} (See page 547.) Using separation of variables as in section 1
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Boyce-DiPrima derive the solution to the heat conduction problem
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with insulated ends. [Study this derivation. Understanding this type of
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derivation is essential if you want to do well on the final exam.]
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The solution is
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$$u(x,t)=\frac{c_0}{2}+\sum_{n=1}^{\infty} c_n e^{-(n\pi\alpha/\ell)^2t}
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\cos\frac{n\pi x}{\ell}.$$
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The coefficients $c_n$ are determined by the requirement that
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$$u(x,0)=\frac{c_0}{2} + \sum_{n=1}^{\infty} c_n\cos\frac{n\pi x}{\ell}=f(x).$$
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Thus to solve this problem we just need to find the cosine series
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expansion of $f(x)=\sin(\pi x/\ell)$.
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This is computed using equation (7) page 536. As it turns out,
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to integrate we must do the cases $n=1$ and $n\neq 1$ separately.
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First suppose $n\neq 1$. Then substituting
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$\sin\frac{\pi x}{\ell}$ for $f(x)$ we obtain
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\begin{eqnarray*}
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c_n&=&\frac{2}{\ell}\int_0^{\ell}
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\sin\frac{\pi x}{\ell}\cos\frac{n\pi x}{\ell}\dx\\
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&=&\frac{1}{\ell}\int_{0}^{\ell}\sin\frac{(n+1)\pi x}{\ell}
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+\sin\frac{(1-n)\pi x}{\ell}\dx\\
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&=&\frac{1}{\ell}\Bigl[-\frac{\ell}{(n+1)\pi}\cos\frac{(n+1)\pi x}{\ell}
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-\frac{\ell}{(1-n)\pi}
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\cos\frac{(1-n)\pi x}{\ell}\Bigr]^{\ell}_{0}\\
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&=&-\frac{1}{(n+1)\pi}\cos(n+1)\pi - \frac{1}{(1-n)\pi}\cos(1-n)\pi
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-\Bigl(-\frac{1}{(n+1)\pi}+\frac{1}{\pi(n-1)}\Bigr)\\
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&=& \begin{cases} 0&\text{if $n$ odd}\\
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-\frac{4}{(n^2-1)\pi}&\text{if $n$ even}
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\end{cases}
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\end{eqnarray*}
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Next suppose $n=1$. Then
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\begin{eqnarray*}
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c_1&=&\frac{2}{\ell}\int_0^{\ell}
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\sin\frac{\pi x}{\ell}\cos\frac{\pi x}{\ell}\dx\\
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&=&\frac{1}{\ell}\int_{0}^{\ell}\sin\frac{2\pi x}{\ell}\dx\\
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&=&-\frac{1}{2\pi}\Bigl[\cos\frac{2\pi x}{\ell}\Bigr]^{\ell}_{0}=0.
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\end{eqnarray*}
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To integrate $\sin\frac{\pi x}{\ell}\cos\frac{n\pi x}{\ell}$
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we have used the trigonometric identity
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$$\sin A \cos B = \frac{1}{2}(\sin (A+B) + \sin(A-B)).$$
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This identity follows by adding the two identities
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$$\sin(A+B)=\sin A \cos B + \sin B \cos A$$
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$$\sin(A-B)=\sin A \cos B - \sin B \cos A.$$
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The solution to this insulated heat conduction problem is thus
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$$u(x,t)=\frac{2}{\pi}-\frac{4}{\pi}\sum_{\text{$n$ even}}
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\frac{1}{n^2-1}
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e^{-(n\pi\alpha/\ell)^2t}
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\cos\frac{n\pi x}{\ell}.$$
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Finally $\lim_{t\rightarrow\infty}u(x,t)=\frac{2}{\pi}$ so
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the steady state temperature is $\frac{2}{\pi}$.
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\problem{10.6.4} {\em Find the displacement $u(x,t)$ in an elastic string of
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length $\ell$, fixed at both ends, that is set in motion from its straight
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equilibrium position with the initial velocity $g$ defined by
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$$g(x)=\begin{cases} Ax,&0\leq x \leq \ell/2\\
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A(\ell-x),& \ell/2 < x \leq \ell.\end{cases}$$}
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\soln
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{\em Summary.} (See pages 554-559.) Using separation of variables
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the authors find solutions to the elastic string problem
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$a^2u_{xx}=u_{tt}$.
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Their solution is
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$$u(x,t)=\sum_{n=1}^{\infty}\sin\frac{n\pi x}{\ell}
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\Bigl(c_n\sin\frac{n\pi a t}{\ell} +
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k_n\cos\frac{n\pi a t}{\ell}\Bigr).$$
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Let $f(x)$ be the initial displacement of the string and $g(x)$
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be the initial velocity of the string. Then
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$$k_n=\frac{2}{\ell}\int_{0}^{\ell} f(x)\sin\frac{n\pi x}{\ell}\dx$$
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and
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$$\frac{n\pi a}{\ell} c_n = \frac{2}{\ell}\int_{0}^{\ell}g(x)\sin\frac{n\pi x}{\ell}\dx.$$
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Note that $c_n$ is the $n$th coefficient of the sine series for $g(x)$
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{\em divided} by $\frac{n\pi a}{\ell}$.
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In our situation $f(x)=0$ so $k_n=0$ for all $n$. Thus to find the solution
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$u(x,t)$ we just need to find the sine series expansion of $g(x)$.
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Fortunately problem (10.5.1 b) involves almost the same integral.
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Looking at our work there we see that
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$b\frac{4\ell}{n^2\pi^2}\sin\frac{n\pi}{2}$
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is the $n$th Fourier coefficient
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of the sine series expansion of
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$$h(x)=\begin{cases}x,&0\leq x < \ell/2 \\
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\ell-x, & \ell/2\leq x \leq \ell\end{cases}.$$
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Multiplying by $A$ we see that the $n$th Fourier coefficient of $g(x)$
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is $$b_n=\frac{4A\ell}{n^2\pi^2}\sin\frac{n\pi}{2}.$$
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Thus $$c_n=\frac{\ell}{n\pi a}b_n
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=\frac{4\ell^2A}{n^3\pi^3a}\sin\frac{n\pi}{2}.$$
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So the solution to the elastic string problem is
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$$u(x,t)=\frac{4A\ell^2}{a\pi^3}\sum_{n=1}^{\infty}
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\frac{1}{n^3}\sin\frac{n\pi}{2}
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\sin\frac{n\pi x}{\ell}\sin\frac{n\pi a t}{\ell}.$$
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\section{Solutions to the Final Exam}
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\problem{Problem 1} Solve the system of differential equations
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$$x_1'(t)=3x_1(t)+3x_2(t),$$
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$$x_2'(t)=-2x_1(t)-4x_2(t)$$
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with the initial conditions
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$$x_1(0)=1, \quad x_2(0)=3.$$
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\soln
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Let $\vx(t)=(x_1(t),x_2(t))$. Then the system of equation becomes
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$\vx'=A\vx$ where
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$A=\Bigl(\begin{smallmatrix}3&3\\-2&-4\end{smallmatrix} \Bigr).$
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First compute the characteristic polynomial of $A$,
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$$|A-\lambda I|=\Bigl|
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\begin{matrix}3-\lambda&3\\-2&-4-\lambda\end{matrix}\Bigr|
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=\lambda^2+\lambda-6=(\lambda+3)(\lambda-2).$$
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Thus the eigenvalues are $-3$ and $2$. The eigenvalue $-3$ has associated
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eigenvector $v_1=(1,-2)$ and the eigenvalue $2$ has associated eigenvector
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$v_2=(3,-1)$. It follows that the general solution is
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$$\vx(t)=c_1e^{-3t}\Bigl[\begin{matrix}1\\-2\end{matrix}\Bigr]
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+c_2e^{2t}\Bigl[\begin{matrix}3\\-1\end{matrix}\Bigr].$$
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To find $c_1$ and $c_2$ use the initial condition $\vx(0)=(1,3)$.
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This gives $c_1=-2$ and $c_2=1$. Thus
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$$\vx(t)=\Bigl[\begin{matrix}-2e^{-3t}+3e^{2t}\\
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4e^{-3t}-e^{2t}\end{matrix}\Bigr]$$
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so $x_1(t)=-2e^{-3t}+3e^{2t}$ and
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$x_2(t)=4e^{-3t}-e^{2t}$.
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\problem{Problem 2}
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(a) Find three functions $y_1(x), y_2(x), y_3(x)$ defined on $(-\infty,\infty)$
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whose Wronskian is given by
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$$W(y_1,y_2,y_3)(x)=e^{4x}.$$
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(b) Are the functions $y_1,y_2,y_3$ that you found linearly independent
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on $(-\infty,\infty)$? Justify your answer.
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\soln
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(a) There are many ways to solve this problem. One way is to
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write down a matrix whose diagonal is $1,1,e^{4t}$
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$$\left|\begin{matrix}1&*&*\\
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0&1&*\\
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0&0&e^{4t}\end{matrix}\right|.$$
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Then integrate to fill in what the $*$'s must be to obtain
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$$\left|\begin{matrix}1&x&\frac{1}{16}e^{4t}\\
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0&1&\frac{1}{4}e^{4t}\\
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0&0&e^{4t}\end{matrix}\right|.$$
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Taking $y_1=1, y_2=x, y_3=\frac{1}{16}e^{4t}$ we see
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at once that $W(y_1,y_2,y_3)=e^{4t}$.
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(b) Yes, since their Wronskian is nonzero at some point.
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\problem{Problem 3}
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Find a homogeneous third-order linear differential equation with constant
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coefficients that has
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$$y(x)=3\cdot e^{-x}-cos(2x)$$
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as a solution. Explain how you found it.
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What is the general solution of that differential equation?
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\soln
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One way to solve this problem is to compute $y, y', y''$, and $y'''$ then
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try to find a linear relation between them. In general this amounts
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to solving a system of four equations in four unknowns.
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Another way to solve this is to note that $y(x)$ is a solution to a
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third order homogeneous linear differential equation which has fundamental
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solution set $e^{-x}$, $\cos(2x)$ and $\sin(2x)$. The roots of the
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characteristic equation associated to this differential equation must
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be $1$, $2i$ and $-2i$. Thus the characteristic equation is
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$$0=(r-1)(r-2i)(r+2i)=(r-1)(r^2+4)=r^3-r^2+4r-4$$
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It follows that the differential equation is
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$$y'''-y''+4y'-4y=0.$$
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The general solution is
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$$y=c_1e^{-x}+c_2\cos(2x)+c_3\sin(2x).$$
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\problem{Problem 4}
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For the function $f(x)=e^{x/\pi}$, draw a careful sketch of the graphs of the
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functions to which the following Fourier series converge on the interval
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$[0,4\pi]$:
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(a) the Fourier sine series of $f$ on $[0,\pi]$;
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(b) the Fourier cosine series of $f$ on $[0,\pi]$;
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(c) the ordinary Fourier series of $f$ on $[-\pi,\pi]$.
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\vspace{3in}
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\problem{Problem 5}
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Find a function $u=u(x,t)$, defined for $0\leq x\leq \pi$ and
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$t\geq 0$, satisfying the partial differential equation
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$$u_{xx}=-u_{tt}\quad (0<x<\pi, \quad t>0),$$
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with boundary conditions
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$$u(0,t)=u(\pi,t)=0\quad (t>0)$$
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and initial conditions
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$$u(x,0)=\sin(3x), \quad u_t(x,0)=0\quad (0<x<\pi).$$
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\soln
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Apply separation of variables. Let $u(x,t)=X(x)T(t)$.
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The initial conditions
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$$u(0,t)=u(\pi,t)=0\quad (t>0)$$
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imply $0=X(0)T(t)$ and $0=X(\pi)T(t)$ for all $t$ so, to avoid
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triviality, $X(0)=X(\pi)=0$.
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Substituting $u=XT$ into the differential equation yields
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$X''T=-XT''$ which becomes
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$$\frac{X''}{-X}=\frac{T''}{T}.$$
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Because the left hand side does not depend on $t$ and the right
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side does not depend on $x$ we conclude that there is a constant
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$a$ such that
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$$\frac{X''}{-X}=a \text{ and } \frac{T''}{T}=a.$$
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Thus
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$$X''+aX=0 \text{ and } T'' - aT=0.$$
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Since we are looking for a solution $u(x,t)$ for which
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$u(x,0)=\sin(3x)$
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we know that $a>0$ is probably going to be the right choice
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since it gives sine and cosine type solutions.
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Making this choice and solving yields
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$$X(x)=c_1\cos(\sqrt{a} x)+c_2\sin(\sqrt{a} x).$$
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Since $X$ must satisfy the initial conditions $X(0)=X(\pi)=0$
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it follows that
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$c_1=0$ and $c_2\sin(\sqrt{a}\pi)=0$. Thus
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$\sqrt{a}\pi=n\pi$ for some integer $n$. Thus $\sqrt{a}=n$
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so $a=n^2$. We see that $X(x)=\sin(nx)$ after setting the
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arbitrary factor $c_2$ equal to $1$.
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Next consider $T''-n^2T=0$. This has general solution
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$$T(t)=c_1e^{nt}+c_2e^{-nt}.$$
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Multiplying $X$ and $T$ together again we see that
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$$u_n=\sin (nx)e^{nt}\text{ and } v_n=\sin(nx)e^{-nt}$$
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for various integers $n$ give a bunch of solutions to the
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differential equation $u_{xx}=-u_{tt}$ satisfying the
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initial conditions $u(0,t)=u(\pi,t)=0$. [You should verify
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this assertion by substituting $u_n$ and then $v_n$ into
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$u_{xx}=-u_{tt}$.]
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Can we find some linear combination of the $u_n$ and $v_n$
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which satisfies the additional conditions $u(x,0)=\sin(3x)$
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and $u_t(x,0)=0$? Yes. Setting $t=0$ shows that it will have
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to be of the form
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$$c_1\sin(3x)e^{3t}+c_2\sin(3x)e^{-3t}$$
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for some constants $c_1$ and $c_2$.
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A little algebra shows that $c_1=c_2=\frac{1}{2}$ does the trick.
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Thus the solution is
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$$u(x,t)=\frac{1}{2}\sin(3x)(e^{3t}+e^{-3t}).$$
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\end{document}
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