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Find the general solution to
(a)

$y^{\prime \prime}+3 y^{\prime}+2 y=260 \cos (3 t)$

(b) As $t$ increases, the solution settles into a periodic steady state oscillation (which does not depend on the initial conditions). Find it's period and amplitude.
Solution:
The steady state oscillation is $y_{p}$ from part (a). Its period is $\frac{2 \pi}{3}$ and its amplitude is
$\sqrt{(-14)^{2}+18^{2}}=2 \sqrt{130} .$
Note: You can find the amplitude from the "complexified" solution without working out its real part. From the beginning of part (a), the complexified oscillation is
$z=\frac{270}{-7+9 i} e^{3 i t}$
so the amplitude is
$\left|\frac{270}{-7+9 i}\right|=\frac{270}{|-7+9 i|}=\frac{270}{\sqrt{130}}$
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Complexifying gives
$z^{\prime \prime}+3 z^{\prime}+2 z=260 e^{3 i t}$
Look for a particular solution of the form $z=c e^{3 i t}$. Substituting into the equation gives
$(-9+9 i+2) c=260$
so
$c=\frac{260}{-7+9 i}=\frac{260(-7-9 i)}{130}=-14-18 i$
Therefore
\begin{aligned} z &=(-14-18 i) e^{3 i t} \\ &=(-14-18 i)(\cos 3 t+i \sin 3 t) \\ &=-14 \cos 3 t+18 \sin 3 t+i(\cdots) \end{aligned}
and a particular solution to the original equation is
$y_{p}=\Re z=-14 \cos 3 t+18 \sin 3 t$
The general solution is
$y=y_{p}+y_{h}=(-14 \cos 3 t+18 \sin 3 t)+c_{1} e^{-t}+c_{2} e^{-2 t}$
by Platinum (93,389 points)

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