Find the general solution to \[ y^{\prime \prime}+3 y^{\prime}+2 y=260 \cos (3 t) \]

Question

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
\[
\left|\frac{270}{-7+9 i}\right|=\frac{270}{|-7+9 i|}=\frac{270}{\sqrt{130}}
\]

Answer 1

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
\[
y=y_{p}+y_{h}=(-14 \cos 3 t+18 \sin 3 t)+c_{1} e^{-t}+c_{2} e^{-2 t}
\]