Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:33:08 AM

Find the general solution for equation
\begin{equation*}
y''(t)+6y'(t)+10y(t)=5e^{3t} +13\cos(t) .
\end{equation*}

Here is my solution.

For the homogeneous part, I think $r=3\pm i$ . Therefore, $$y_c(t)=C_1e^{3t}cost+C_2e^{3t}sint$$
I will post my answer below.

First, we consider $y''+6y'+10y=0$
$$r^2+6r+10=0$$
$$∴r=3\pm i$$
$$∴y_c(t)=C_1e^{3t}\cos t+C_2e^{3t}\sin t$$
Second, we consider $y''+6y'+10y=5e^{3t}$, let $$y_{p1}(t)= Ae^{3t}, y'_{p1}(t)=3Ae^{3t}, y''_{p1}(t)=9Ae^{t}$$
$$9Ae^{3t} 18Ae^{3t}+10Ae^{3t} =5e^{3t}$$
$$∴A=5$$
$$∴y_{p1}(t)= 5e^{3t}$$
Third, we consider $y''+6y'+10y=13\cos t$ , let $$y_{p2}(t)= B\cos t+C\sin t, y'_{p2}(t)= B\sin t+C\cos t,y''_{p2}(t)= B\cos tC\sin t$$
$$∴B\cos tC\sin t+(B\sin t+C\cos t)+10(B\cos t+C\sin t)=13\cos t$$
$$∴9C6B=0, 9B+6C=13 $$
$$∴ B=1, C=\frac{2}{3}$$
$$∴y_{p2}(t)= \cos t+\frac{2}{3}\sin t$$
Therefore,
$$y(t)=y_c(t)+y_{p1}(t)+y_{p2}(t) =C_1e^{3t}\cos t+C_2e^{3t}\sin t+5e^{3t}+\cos t+\frac{2}{3}\sin t$$

Jialu found a correct particular solution for inhomogeneous equation, but made a mistake in the general solution for homogeneous equation. Shengying did everything right (and the LaTeX is good).