Answer
$y(x)=c_1e^{-x}\cos 2x+c_2e^{-x}\sin 2x-\frac{4}{5}\cos 2x-\frac{1}{5}\sin 2x$
Work Step by Step
Given: $y''+2y'+5y=3\sin 2x$
Solve the auxiliary equation for the differential equation.
$r^2+2r+5=0$
Factor and solve for the roots.
$(r+1)(r-2)=0$
Roots are: $r_1=-1+2i$, as a multiplicity of 1 and $r_2=-1-2i$ as a multiplicity of 1.
The general solution is
$y(x)=c_1e^{-x}\cos 2x+c_2e^{-x}\sin 2x$
We have $F(x)=3\sin 2x$
Obtain:
$(D^2+2D+5)(D^2+4)y_p(x)=3\sin 2x$
Therefore, the general solution for $(D^2+2D+5)(D^2+4)y_p(x)=0$ is:
$y(x)=c_1e^{-x}\cos 2x+c_2e^{-x}\sin 2x+A_0\cos 2x+B_0\sin 2x$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0\cos 2x+B_0\sin 2x$
So, we have:
$(D^2+2D+5)(D^2+4)y_p(x)=3\sin 2x\\
(D^2+2D+5)(D^2+4)(A_0\cos 2x+B_0\sin 2x)=3\sin 2x$
On comparing coefficients, we get:
$A_0=-\frac{4}{5}\\
B_0=-\frac{1}{5}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1e^{-x}\cos 2x+c_2e^{-x}\sin 2x-\frac{4}{5}\cos 2x-\frac{1}{5}\sin 2x$
