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{\bf Question}

Express the following system of first order differential equations
in matrix form, and find the eigenvalues and eigenvectors if the
associated coefficient matrix:

$$\frac{dx}{dt}=2x + 3y$$

$$\frac{dy}{dt} = 2x + y.$$

Hence find the general solution to the equations.

{\bf Answer} \vspace{.2in}

In matrix form: $\left( \begin{array}{c} \frac{dx}{dt} \\
\frac{dy}{dt} \end{array} \right) = \left( \begin{array}{cc} 2 & 3
\\ 2 & 1 \end{array} \right)\left( \begin{array}{c} x \\
y \end{array} \right)$
\begin{eqnarray*}
 \left| \begin{array}{cc} {2 - \lambda} & {3} \\ {2} & {1 - \lambda} \\
\end{array} \right| & = & (2 - \lambda)(1 - \lambda) - 6 \\ & = &
\lambda^2 - 3\lambda - 4 \\ & = & (\lambda -  4)(\lambda + 1) = 0
\\ {\rm so \ \ \ } \lambda & = & -1, 4
\end{eqnarray*}

$ \underline {\lambda = -1}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} 3 & 3\\ 2 & 2\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $\left. \begin{array}{rcl} {3x + 3y} & = & 0 \\ {2x + 2y} & = &
0 \\ \end{array} \right\} \begin{array}{lrcl} {\rm let\ \ } & x &
= & \alpha \\ {\rm so\ \ } & y & = & -\alpha \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \alpha \\ -\alpha
\\ \end{array} \right)$

$ \underline {\lambda = 4}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} -2 & {3}\\ {2} & -3\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $\left. \begin{array}{rcl} {-2x + 3y} & = & 0 \\ {2x - 3y} & =
& 0 \\ \end{array} \right\} \begin{array}{lrcl} {\rm let\ \ } & x
& = & 3\beta \\ {\rm so\ \ } & y & = & 2\beta \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} 3\beta \\ 2\beta
\\ \end{array} \right)$

General solution to equations:
\begin{eqnarray*}
\left( \begin{array}{c} x \\ y \end{array} \right) & = & \left(
\begin{array}{c} \alpha \\ -\alpha \end{array} \right)e^{-t} +
\left( \begin{array}{c} 3\beta \\ 2\beta \end{array} \right)e^{4t}
\\ & = & \alpha \left( \begin{array}{c} 1 \\ -1 \end{array} \right)e^{-t}
+\beta \left( \begin{array}{c} 3 \\ 2 \end{array} \right)e^{4t}
\end{eqnarray*}


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