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\newcommand{\ihat}{\hat\imath}
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\begin{document}

\centerline{
  \begin{tabular}{l}
    {\bf EML 4930/5930}\\{\it Closed book}
  \end{tabular}
  \hfill
  \begin{tabular}{c}
    {\bf Analysis in M.E. II}\\{\it Van Dommelen}
  \end{tabular}
  \hfill
  \begin{tabular}{r}
    \hspace{.75truein}{\bf 04/24/07}\\{\it 3:00-5:00 pm}
  \end{tabular}
}

Show all reasoning and intermediate results leading to your answer, or
credit will be lost. One book of mathematical tables, such as Schaum's
Mathematical Handbook, may be used, as well as a calculator and one
handwritten letter-size single formula sheet.

\begin{enumerate}
  
\item Solve the first order partial differential equation and initial
  condition:
  \begin{displaymath}
    y u_x - x u_y = y u \qquad u(x,0) = \sin(x)
  \end{displaymath}

\begin{rawhtml}
<A HREF="figures/1.jpg">Solution.</A>
\end{rawhtml}
  
\item Find the numerical value of u(0.3,3) for acoustics in a pipe of
  length one with open ends.  The partial differential equation is
  \begin{displaymath}
    u_{tt} = u_{xx}
  \end{displaymath}
  with boundary conditions and initial conditions
  \begin{displaymath}
    u(0,t)=u(1,t)=0 \qquad u(x,0)=x \quad u_t(x,0)=0
  \end{displaymath}

\begin{rawhtml}
<A HREF="figures/2.jpg">Solution.</A>
\end{rawhtml}
  
\item Solve the following problem using the method of separation of
  variable.  The partial differential equation is
  \begin{displaymath}
    u_{tt} = u_{xx}
  \end{displaymath}
  with boundary conditions and initial conditions
  \begin{displaymath}
    u(0,t)=2 \quad u(1,t)=1 \qquad u(x,0)=2 \quad u_t(x,0)=0
  \end{displaymath}

\begin{rawhtml}
<A HREF="figures/3.jpg">Solution.</A>
\end{rawhtml}

\item Solve the following problem of a viscous flow next to a plate.
  The partial differential equation is:
  \begin{displaymath}
    u_t = u_{xx}
  \end{displaymath}
  and the initial and boundary conditions are:
  \begin{displaymath}
    u(x,0)= 0 \qquad u(0,t)=\frac{1}{\sqrt{t}}
  \end{displaymath}

\begin{rawhtml}
<A HREF="figures/4.jpg">Solution.</A>
\end{rawhtml}

\end{enumerate}
\newpage
\begin{center}
  \renewcommand{\arraystretch}{2}
  {\bf Table 6.3: Properties of the Laplace Transform}\\
  \begin{tabular}{|l|l|}
    \hline\hline
    \multicolumn{1}{|c|}{$\displaystyle f(t)_{\strut}^{\strut}$} &
    \multicolumn{1}{c|}{$\displaystyle \hat f(s)
      =\int_{\strut0}^{\strut\infty}f(t)e^{-st}\D t$} \\
    \hline\hline
    {\bf 1.} $\displaystyle C_1 f_1(t) + C_2 f_2(t) $ &
    $\displaystyle C_1 \hat f_1(s) + C_2 \hat f_2(s)$ \\
    \hline
    {\bf 2.} $\displaystyle f(at)$ &
    $\displaystyle a^{-1} \hat f(s/a)\qquad (a>0)$ \\
    \hline
    {\bf 3.} $\displaystyle f^{(n)}(t)$ &
    $\displaystyle s^n \hat f(s) - s^{n-1} f(0) - \ldots -
     f^{(n-1)}(0) \qquad (n=1,2,\ldots)$ \\
    \hline
    {\bf 4.} $\displaystyle t^n f(t)$ &
    $\displaystyle (-1)^n
    \hat f^{(n)}(s)\qquad (n=1,2,\ldots)$ \\
    \hline
    {\bf 5.} $\displaystyle e^{ct} f(t)$ &
    $\displaystyle \hat f(s-c)\qquad (s = \mbox{const.})$ \\
    \hline
    {\bf 6.}
    \begin{tabular}[t]{l}
      $\displaystyle H(t-b) f(t-b)$, where \\
      {\renewcommand{\arraystretch}{1.5}
        $\displaystyle \quad H(t) =
        \bigg\{_{\strut}^{\strut}
        \begin{array}{lr} 0\ & t<0 \\ 1\ & t>0 \end{array}$}
    \end{tabular} &
    $\displaystyle e^{-bs} \hat f(s)\qquad (b>0)$ \\
    \hline
    {\bf 7.} $\displaystyle f * g(t)\equiv
    \int_{{\strut}0}^{{\strut}t} f(t-\tau)g(\tau)\D\tau$ &
    $\displaystyle \hat f(s)\hat g(s)$ \\
    \hline\hline
  \end{tabular}
  \\
  \vspace{5pt}
  {\bf Table 6.4: Laplace Transform Pairs}\\
  \begin{tabular}{|l|l|}
    \hline\hline
    \multicolumn{1}{|c|}{$\displaystyle f(t)$} &
    \multicolumn{1}{c|}{$\displaystyle \hat f(s)
      =\int_{\strut0}^{\strut\infty}f(t)e^{-st}\D t$} \\
    \hline\hline
    {\bf 1.} $\displaystyle 1 $ &
    $\displaystyle \frac{1}{s}_{\strut}$ \\
    \hline
    {\bf 2.} $\displaystyle t^n$ &
    $\displaystyle \frac{n!}{s^{n+1}}_{\strut}\qquad(n=1,2,\ldots)$ \\
    \hline
    {\bf 3.} $\displaystyle e^{kt}$ &
    $\displaystyle \frac1{s-k}_{\strut}$ \\
    \hline
    {\bf 4.} $\displaystyle \sin(at)$ &
    $\displaystyle \frac{a}{s^2+a^2}_{\strut}$ \\
    \hline
    {\bf 5.} $\displaystyle \cos(at)$ &
    $\displaystyle \frac{s}{s^2+a^2}_{\strut}$ \\
    \hline
    {\bf 6.} $\displaystyle \frac{1}{\sqrt{\pi t}}_{\strut}$ &
    $\displaystyle \frac{1}{\sqrt{s}}$ \\
    \hline
    {\bf 7.} $\displaystyle \frac{1}{\sqrt{\pi t}}_{\strut} e^{-k^2/4t}$ &
    $\displaystyle \frac{1}{\sqrt{s}} e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline
    {\bf 8.} $\displaystyle \frac{k}{\sqrt{4\pi t^3}}_{\strut} e^{-k^2/4t}$ &
    $\displaystyle e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline
    {\bf 9.} 
    \begin{tabular}[t]{l}
      $\displaystyle \hbox{erfc}\left(k/2\sqrt{t}\right)_{\strut}$,
      where\\
      $\displaystyle \hbox{erfc}(z) \equiv
      \frac2{\sqrt{\pi}}\int_{\strut z}^\infty e^{-u^2} \D u$
    \end{tabular} &
    $\displaystyle \frac1s e^{-k\sqrt{s}}\qquad(k>0)$ \\
    \hline\hline
  \end{tabular}
\end{center}



\end{document}
