Question

Y′′−3y′+2y=3e^(-x)-10cos(3x)

Answer 1

First, the characteristic solution. The homogeneous ODE

$y''-3y'+2y=0$

has characteristic equation

$r^2-3r+2=(r-2)(r-1)=0$

which has roots at $r=2$ and $r=1$, which means the characteristic solution is

$y_c=C_1e^{2x}+C_2e^x$

To find the particular solution, the method of undetermined coefficients will probably be the easiest to use. Suppose the particular solution and its derivative are

$y_p=ae^{-x}+b\cos3x+c\sin3x$
${y_p}'=-ae^{-x}-3b\sin3x+3c\cos3x$
${y_p}''=ae^{-x}-9b\cos3x-9c\sin3x$

Substituting into the ODE gives

$\left(ae^{-x}-9b\cos3x-9c\sin3x\right)-3\left(-ae^{-x}-3b\sin3x+3c\cos3x\right)+2\left(ae^{-x}+b\cos3x+c\sin3x\right)=3e^{-x}-10\cos3x$
$6ae^{-x}+(-9b-9c+2b)\cos3x+(-9c+9b+2c)\sin3x=3e^{-x}-10\cos3x$
$6ae^{-x}+(-7b-9c)\cos3x+(-7c+9b)\sin3x=3e^{-x}-10\cos3x$

It follows that

$\begin{cases}6a=3\\-7b-9c=-10\\-7c+9b=0\end{cases}\implies a=\dfrac12,b=\dfrac7{13},c=\dfrac9{13}$

So the solution is

$y=y_c+y_p$
$y=C_1e^{2x}+C_2e^x+\dfrac12e^{-x}+\dfrac7{13}\cos3x+\dfrac9{13}\sin3x$