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\textbf{Math 325: Differential Equations}
\hfill\makebox[2.5in][r]{Name:\hrulefill}

\textbf{Quiz 4}\quad \textsl{Nov. 8, 1996}

\begin{enumerate}

\item
The solution to the initial value problem, $\b{X}'(t) =
\b{A}\cdot\b{X}(t)$, $\b{X}(0) = \b{X}^{(0)}$, is given by
$\b{X}(t) = \exp(\b{A} t)\cdot\b{X}^{(0)}$,
where
\[
\exp(\b{A} t) = \b{I} + \b{A} t + \b{A}^2\frac{t^2}{2!} +
                  \b{A}^3\frac{t^3}{3!} + \dots
\]

Let
\[
    \b{A} = \begin{mtrx} 3 & -4\\ 5 & -6 \end{mtrx}
\]
Calculate the matrix $\exp(\b{A} t)$ another way using the
eigenvalues and eigenvectors of $\b{A}$.

\answerspace

\problemspace
\item
Suppose a fundamental solution for a homogeneous linear system
$\b{X}'(t) = \b{P}(t).\b{X}(t)$ is
\[
    \b{X}(t) =   c_1 \begin{vctr}  2 \\ -1 \end{vctr} t^2
               + c_2 \begin{vctr} -1 \\  1 \end{vctr} t^3
\]
Use the method of variation of parameters to find a particular solution to
the non-homogeneous equation
\[
    \b{X}'(t) = \b{P}(t)\cdot\b{X}(t) + \begin{vctr} t \\ -t \end{vctr}
\]

\answerspace

\end{enumerate}

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