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{\bf Question}

Find the general solution to the first order system of equations
$$ \dot{\bf x}=A{\bf x} $$ where $A$ is the matrix given by
\begin{description}
\item[(a)] $\ds \pmatrix{3 &-2\cr 2 &-2\cr}$
\item[(b)]$\ds \pmatrix{4 &-3\cr 8 &-6\cr}$
\item[(c)] $\ds \pmatrix{2 &-1\cr 3 &-2\cr}$
\item[(d)] $\ds \pmatrix{1 &1\cr 4 &-2\cr}$
\item[(a)] $\ds \pmatrix{2 &-5\cr 1 &-2\cr}$
\item[(e)] $\ds \pmatrix{-1 &-4\cr 1 &-1\cr}$
\item[(f)] $\ds \pmatrix{5 &-1\cr 3 &1\cr}$
\item[(g)] $\ds \pmatrix{3 &-4\cr 1 &-1\cr}$
\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)] $\ds A=\pmatrix{3 &-2\cr 2 &-2\cr}$ has eigenvalues
$\lambda=-1$ and $\lambda=2$.

When $\lambda=-1$ the eigenvector is $\ds \pmatrix{1 \cr 2\cr}$.

When $\lambda=2$ the eigenvector is $\ds \pmatrix{2 \cr 1\cr}$.

Hence the solution is
$\ds {\bf x}(t)=c_1\pmatrix{1 \cr 2\cr}e^{-t}+c_2\pmatrix{2 \cr 1\cr}e^{2t}$.

\item[(b)]
$\ds A=\pmatrix{4 &-3\cr 8 &-6\cr}$ has eigenvalues $\lambda=0$ and
$\lambda=-2$.

When $\lambda=0$ the eigenvector is $\ds \pmatrix{3 \cr 4\cr}$.

When $\lambda=-2$ the eigenvector is $\ds \pmatrix{1 \cr 2\cr}$.

Hence the solution is
$\ds {\bf x}(t)=c_1\pmatrix{3 \cr 4\cr}+c_2\pmatrix{1 \cr 2\cr}e^{-2t}$.

\item[(c)]
$\ds A=\pmatrix{2 &-1\cr 3 &-2\cr}$ has eigenvalues $\lambda=-1$ and
$\lambda=1$.

When $\lambda=-1$ the eigenvector is $\ds \pmatrix{1 \cr 3\cr}$.

When $\lambda=1$ the eigenvector is $\ds \pmatrix{1 \cr 1\cr}$.

Hence the solution is
$\ds {\bf x}(t)=c_1\pmatrix{1 \cr 3\cr}e^{-t}+c_2\pmatrix{1 \cr 1\cr}e^{t}$.

\item[(d)]
$\ds A=\pmatrix{1 &1\cr 4 &-2\cr}$ has eigenvalues $\lambda=-3$ and
$\lambda=2$.

When $\lambda=-3$ the eigenvector is $\ds \pmatrix{1 \cr -4\cr}$.

When $\lambda=2$ the eigenvector is $\ds \pmatrix{1 \cr 1\cr}$.

Hence the solution is
$\ds {\bf x}(t)=c_1\pmatrix{1 \cr -4\cr}e^{-3t}+c_2\pmatrix{1 \cr 1\cr}e^{2t}$.

\item[(e)]
$\ds A=\pmatrix{2 &-5\cr 1 &-2\cr}$ has eigenvalues $\lambda=\pm i$.

When $\lambda=i$ the eigenvector is $\ds \pmatrix{2+i \cr 1\cr}$.

Hence the corresponding solution is given by \begin{eqnarray*}
{\bf x}(t)& = & \left( \begin{array}{c} 2+i\\ 1\end{array}
 \right)e^{it} \\ &= & \left( \begin{array}{c} 2+i\\ 1 \end{array}
 \right) (\cos t +i\sin t)\\ & = & \left( \begin{array}{c} 2\cos
t-\sin t \\ \cos t \end{array}  \right) +i\left(
\begin{array}{c}\cos t + 2\sin t\\ \sin t \end{array}
 \right) \end{eqnarray*}
Since the real and imaginary parts are each solutions to the
equation we find the general solution is given by:

$\ds {\bf x}(t)=c_1\pmatrix{2\cos t-\sin t \cr \cos t \cr}+
c_2\pmatrix{\cos t +2\sin t\cr \sin t \cr}$.

\item[(f)]
$\ds A=\pmatrix{-1 &-4\cr 1 &-1\cr}$ has eigenvalues $\lambda=-1\pm 2i$

When $\lambda=-1+2i$ the eigenvector is $\ds \pmatrix{2i \cr 1\cr}$.

Hence the corresponding solution is given by \begin{eqnarray*}
{\bf x}(t)&= & \pmatrix{2i\cr 1\cr}e^{-t+2it} \\ &=&
\pmatrix{2i\cr 1\cr}e^{-t}(\cos 2t +i\sin 2t)\\ &= &
\pmatrix{-2e^{-t}\sin 2t \cr e^{-t}\cos 2t} +i\pmatrix{2e^{-t}\cos
2t \cr e^{-t}\sin 2t\cr}\end{eqnarray*} Since the real and
imaginary parts are each solutions to the equation we find the
general solution is given by:

$\ds {\bf x}(t)=c_1\pmatrix{-2e^{-t}\sin 2t \cr e^{-t}\cos 2t}
+c_2\pmatrix{2e^{-t}\cos 2t \cr e^{-t}\sin 2t\cr}$.

\item[(g)]
$\ds A=\pmatrix{5 &-1\cr 3 &1\cr}$ has eigenvalues $\lambda=2$ and
$\lambda=4$.

When $\lambda=2$ the eigenvector is $\ds \pmatrix{1 \cr 3\cr}$.

When $\lambda=4$ the eigenvector is $\ds \pmatrix{1 \cr 1\cr}$.

Hence the solution is
$\ds {\bf x}(t)=c_1\pmatrix{1 \cr 3\cr}e^{2t}+c_2\pmatrix{1 \cr 1\cr}e^{4t}$.

\item[(h)]
$\ds A=\pmatrix{3 &-4\cr 1 &-1\cr}$ has eigenvalues $\lambda=1$ twice.

When $\lambda=1$ the eigenvector is $\ds {\bf a}=\pmatrix{2 \cr 1\cr}$.

To find the second solution we need to find a vector ${\bf b}$ such
that $(A-\lambda I){\bf b}={\bf a}$. Writing $\ds {\bf b}=\pmatrix{c
\cr d \cr}$ we want to solve
$$
\pmatrix{2&-4\cr 1&-2\cr}\pmatrix{c \cr d \cr}=\pmatrix{2 \cr 1 \cr}
$$
One solution to this is $\ds {\bf b}=\pmatrix{1 \cr 0\cr}$. Note this
solution is not unique and we can add on to it any multiple of ${\bf a}$.

Using ${\bf a}$ and ${\bf b}$ found above the solution is

$\ds {\bf x}(t)=c_1\pmatrix{2 \cr 1\cr}e^{t}+
c_2\left\{\pmatrix{2 \cr 1\cr}te^{t}+\pmatrix{1 \cr 0 \cr}e^{t}\right\}$.
\end{description}


\end{document}