Answer
$y(x)=C_1\cos 2x+C_2\sin 2x+C_3\cos 3x+C_4\sin 3x$
Work Step by Step
Given $y^{(iv)}+13y''+36y=0$
Solve the auxiliary equation for the differential equation.
$r^4+13r^2+36=0$
Factor and solve for the roots.
$(r^2+4)(r^2+9)=0$
Roots are: $r_1=\pm 2i$, as a multiplicity of 2 and $r_2=\pm 3i$ as a multiplicity of 2.
This implies that there are two independent solutions to the differential equation
$y_1(x)=\sin 2x$
$y_2=\cos 2x$ and
$y_3(x)=\sin 3x\\
y_4(x)=\cos 3x$
Therefore, the general equation is equal to $y(x)=C_1\cos 2x+C_2\sin 2x+C_3\cos 3x+C_4\sin 3x$
