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{\bf Question}

Solve the initial value problem: $\displaystyle \frac{d^2x}{dt^2}
- \frac{dx}{dt} - 2x = 1$

subject to $\displaystyle x(0) = \frac{dx(0)}{dt} = 0$

\vspace{.25in} {\bf Answer}

$$ \frac{d^2x}{dt^2} - \frac{dx}{dt} - 2x = 1$$ with
$\displaystyle x(0) = \frac{dx(0)}{dt} = 0$

CF: \begin{eqnarray*} m^2 - m - 2 & = & 0 \\ (m - 2)(m + 1) & = &
0 \\ m & = & 2, \, -1 \\ x_c  & =  & Ae^{-t} + Be^{2t}
\end{eqnarray*}

Let $x^* = c$ substitute in $\displaystyle \frac{dx^*}{dt} =
\frac{d^2x^*}{dt^2} = 0 \Rightarrow x^* = -\frac{1}{2}$

Thus $x(t) = Ae^{-t} + Be^{2t} - \frac{1}{2}$

Find A and B
\begin{eqnarray*} x(0) & = & A + B - \frac{1}{2} = 0 \\
\frac{dx(t)}{dt} & = & -Ae^{-t} + 2Be^{2t} \\ \Rightarrow
\frac{dx(0)}{dt} & = & -A + 2B = 0 \\ {\rm Hence \ \ \ } A + B & =
& \frac{1}{2} \\ -A + 2B & = & 0 \Rightarrow A = 2B, \, B =
\frac{1}{6}, \, A = \frac{1}{3} \end{eqnarray*}

$$x(t) = \frac{1}{6} (2e^{-t} + e^{2t} - 3)$$



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