Question

Find a particular solution to the nonhomogeneous differential equation y'' + 4 y = cos(2x) + sin(2x).

Answer 1

The characteristic solution follows from solving the characteristic equation,

$r^2+4=0\implies r=\pm2i$

so that

$y_c=C_1\cos2x+C_2\sin2x$

A guess for the particular solution may be $a\cos2x+b\sin2x$, but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to

$y_p=ax\cos2x+bx\sin2x$

which has second derivative

${y_p}''=(-4ax+4b)\cos2x+(-4bx-4a)\sin2x$

Substituting into the ODE, you have

$y''+4y=\cos2x+\sin2x$
$\implies4b\cos2x-4a\sin2x=\cos2x+\sin2x$
$\implies\begin{cases}4b=1\\-4a=1\end{cases}\implies a=-\dfrac14,b=\dfrac14$

Therefore the particular solution is

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

Note that you could have made a more precise guess of

$y_p=(a_1x+a_0)\cos2x+(b_1x+b_0)\sin2x$

but, of course, any solution of the form $a_0\cos2x+b_0\sin2x$ is already accounted for within $y_c$.