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{\bf Question}

Solve the initial value problem for the forced damped system
$$\frac{d^2x}{dt^2}+ 2\frac{dx}{dt} + 4x = \cos 2t,$$ subject to
$\displaystyle x(0) = 1; \, \frac{dx(0)}{dt} = 0$

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


$$\frac{d^2x}{dt^2}+ 2\frac{dx}{dt} + 4x = \cos 2t$$

Auxiliary equation: \begin{eqnarray*} m^2 + 2m + 4  &amp; = &amp; 0 \\
\Rightarrow (m + 1)^2 + 3 &amp; = &amp; 0 \\ \Rightarrow m &amp; = &amp; -1 \pm
\sqrt3 j \\ \\ x_c(t) &amp; = &amp; e^{-t}[A \cos \sqrt 3 t + B \sin \sqrt
3 t] \end{eqnarray*}

PI: \begin{eqnarray*} x^*(t) &amp; = &amp; P \cos 2t + Q \sin 2t \\
\frac{dx^*}{dt} &amp; = &amp; 2Q \cos 2t - 2P \sin2t \\
\frac{d^2x^*}{dt^2} &amp; = &amp; -4P \cos 2t - 4Q \sin 2t \end{eqnarray*}

\begin{eqnarray*}  \frac{d^2x^*}{dt^2} + 2\frac{dx^*}{dt} +4x^* &amp; =
&amp; \cos 2t \{-4P + 4Q + 4P\} + \sin 2t \{-4Q -4P +4Q\} \\ &amp; = &amp;
4Q\cos 2t - 4P\sin 2t \\ &amp; \equiv &amp; \cos 2t \end{eqnarray*}

Hence $\displaystyle P = 0 \, Q = \frac{1}{4}$

General solution:

$$x(t)  =  e^{-t}[A \cos \sqrt 3 t + B \sin \sqrt 3 t] +
\frac{1}{4} \sin 2t\\$$ $$ \frac{dx}{dt}  =  e^{-t}[(\sqrt 3 B -
A) \cos \sqrt 3 t - (B + \sqrt 3 A) \sin \sqrt 3 t] + \frac{1}{2}
\cos 2t $$ $$ x(0) =  A = 1 \\ \frac{dx(0)}{dt} = \sqrt 3B - A +
\frac{1}{2} = 0 $$ $$  \Rightarrow  \sqrt 3 B - \frac{1}{2} = 0 \\
\Rightarrow  B = \frac{1}{2 \sqrt3}$$

Hence $$ x(t) = e^{-t}\left[ \cos \sqrt 3 t + \frac{1}{2 \sqrt3}
\sin \sqrt 3 t\right] + \frac{1}{4}  \sin 2t$$


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