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\begin{document}
\vspace*{-4em}
\begin{center}{\bf{\Large
MATHEMATICS 315, FINAL EXAM, FALL 1989
}}\end{center}

\vspace{1em}
\noindent NAME\rule[-.5em]{2in}{1.5pt} \hfill 
          ID\rule[-.5em]{1.5in}{1.5pt} \hfill
          ROW \rule[-.5em]{.5in}{1.5pt}

\vspace{1ex}
{\bf Two Hours are allowed for this exam. Books, notes and extra sheets of 
paper are not allowed. Calculators are not allowed. The problems are worth 15
points each.}

\Hline

\[
\begin{array}{llll}
\num &\Frac{dy}{dt}= ty^2 + 3ty  & \num &(t^2+2ty)\Frac{dy}{dt} + 2ty+y^2 =0 \\[1em]
\num &y'' + y =  \cos t          & \num &y'' + y= \sec t      \\[1em]
\num &y'''' + 16y = 0            & \num &y''' +3y' - 4y = 3+2t \\[2em]
\num & \mb{Y}'=
\left[\begin{array}{cc}
3 & -4 \\
1 & -1
\end{array}\right]\mb{Y} \ \ \ \ \ \ \ &\num &\mbox{Let   } A =
                                        \left[\begin{array}{cc}
                                        3 & 1 \\
                                        1 & 3
                                        \end{array}\right]
                                        \txt{Compute} e^{At} .
\end{array}
\]

\Hline 

\num Consider the equation 
\[
\frac{dy}{dt}=\sqrt[3]{2y+3t}
\]
Initial conditions of the form $y(t_0)=y_0$\ are given. For what values of
$t_0$\ and $y_0$\ might it possible for two different solutions to pass through
the same point $(t_0, y_0)$? Explain why you think so.

\Hline

\vspace{1em}
\num We will use the Improved Euler method to approximate the solution of an 
equation of the form $y' = f(t,y)$\ with boundary conditions $y(t_0)=y_0$.\\

a) First, simply write down the general formulas used to calculate with the 
Improved Euler method.\\

b) Next, use those formulas to approximate the solution at $t=.4$\ using the 
example
\[
y' = t + y \txt{with} y(0)=1 \txt{and} h=.2.
\]
Notice that, starting at $t=0$, you must to take two steps to approximate the 
solution at $t=.4$.

\newpage

\num I give you the fact that the fundamental matrix for the problem
\[
\mb{Y}'=
\left[\begin{array}{cc}
1 & 0 \\
1 & 2
\end{array}\right]\mb{Y}
\txt{is given by}
e^{At} =
\left[\begin{array}{cc}
e^t & 0 \\
e^{2t} - e^t & e^{2t}
\end{array}\right].
\]
Use this information and the \underline{Variation of Parameters Method} to find 
a particular solution of the equation
\[
\mb{Y}'=
\left[\begin{array}{cc}
1 & 0 \\
1 & 2
\end{array}\right]\mb{Y} + 
\left[\begin{array}{c}
3 \\
1
\end{array}
\right]
\]

\Hline

\vspace{1em}
\num Use power series methods to solve for \underline{all} of the coefficients 
in the series solution $y(x)$\ of the equation
\[
y'' -2x y' + 6 y=0, \txt{with boundary conditions} y(0)=0, \quad y'(0)=3.
\]

\Hline

\vspace{2em}
\num One solution of the equation $t^2y'' - 12y = 0$\ is given by 
$y_1(t) = t^4$. \\
\underline{Use the $y_2(t)=u(t)y_1(t)$\ method} to solve for $y_2(t)$.

\end{document}