Solving system of linear second order differential equations Solve this system of differential equations:
$$y_1'' + y_2' + 3y_1 = e^{-x}$$
$$y_2''-4y_1' + 3y_2 = \sin(2x)$$
My try: Using $y_1' = p$ and $y_2'= q$ , we get:
$$\begin{pmatrix} p' \\q'\\y_1'\\y_2'\end{pmatrix} =\begin{pmatrix}
 0 & -1 & -3 &  0 \\
 4 & 0 & 0 &  -3 \\
1 & 0 & 0 &  0 \\
 0 & 1 & 0  & 0    
 \end{pmatrix}\begin{pmatrix} p \\q\\y_1\\y_2\end{pmatrix} + \begin{pmatrix} e^{-x} \\\sin(2x)\\0\\0\end{pmatrix}$$
I solved the homogeneous part and find :
$$ y_h = c_1e^{-ix}\begin{pmatrix} 1/2 \\-i\\i/2\\1\end{pmatrix} + c_2e^{ix}\begin{pmatrix} 1/2 \\i\\-i/2\\1\end{pmatrix} + c_3e^{3ix}\begin{pmatrix} -3/2 \\3i\\i/2\\1\end{pmatrix} + c_4e^{-3ix}\begin{pmatrix} -3/2 \\-3i\\-i/2\\1\end{pmatrix}$$ 
I don't know how to solve non-homogeneous part of the equation.
 A: I am going to rewrite the system as
$$\tag 1 x'' + y' + 3 x = e^{-t}  \\ y'' - 4 x' + 3 y = \sin 2t $$
Using the first equation, we have
$$\tag 2 y' = -x'' - 3 x + e^{-t} \\ y'' = -x''' - 3 x' - e^{-t} \\ y''' = - x'''' - 3 x'' + e^{-t}$$
Taking the derivative of the second equation
$$\tag 3 y''' - 4 x'' + 3 y' = 2 \cos 2 t$$
Substituting $(2)$ into $(3)$
$$(- x'''' - 3 x'' + e^{-t}) - 4 x'' + 3 (-x'' - 3 x + e^{-t}) = 2 \cos 2t $$
This simplifies to
$$-x'''' - 10 x'' - 9x = 2 \cos 2 t - 4 e^{-t} $$
Solving this using your favorite method
$$x(t)= c_1 \cos 3t +c_2 \sin 3t+c_3 \cos t + c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
Now, using the first equation and $x(t)$, we need to solve
$$\tag 4 y' = -x'' - 3 x + e^{-t} = 6 c_1 \cos 3t+6 c_2 \sin 3t-2 c_3 \cos t -2 c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
By integrating $(4)$, we get 
$$y(t) = 2 c_1 \sin 3t-2 c_2 \cos 3t-2 c_3 \sin t+2 c_4 \cos t-\dfrac{e^{-t}}{5}+\dfrac{1}{15} \sin 2t$$
You can substitute $(1)$ and $(2)$ into the original system and verify they are solutions.
