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{\bf Question}

Write the second order differential equation

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

as a linear system of two first order equations, and solve it
using matrix methods.

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

Put $\displaystyle y = \frac{dx}{dt}$ so that $\displaystyle
\frac{dy}{dt} = \frac{d^2x}{dx^2} = 2x + \frac{dx}{dt} = 2x+ y$ so
we have a set of simultaneous first order differential equations

$\displaystyle \left. \begin{array} {ccl} \frac{dx}{dt} & = & y\\
\frac{dy}{dt} & = & 2x+y \end{array} \right\}$ or in matrix form
$\left(
\begin{array}{cc} \frac{dx}{dt}\\\frac{dy}{dt} \end{array}
\right)=\left(\begin{array}{cc}0 & 1\\2 & 1 \end{array} \right)
\left(\begin{array}{c}x\\y \end{array} \right)$

$\left|\begin{array}{rc} -\lambda & 1\\2 & 1-\lambda
\end{array} \right|=\lambda^2-\lambda-2=(\lambda-2)(\lambda+1)=0$
so $\lambda=-1,2$

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

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

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

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

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

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

General solution is given by $\left( \begin{array}{c} x \\ y
\end{array} \right) = \alpha \left( \begin{array}{c} 1 \\ -1
\end{array} \right)e^{-t} + \beta \left( \begin{array}{c} 1 \\ 2
\end{array} \right)e^{2t}$

Hence the general solution to the second order differential
equation is $ x = \alpha e^{-t} + \beta e^{2t}$
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