Answer
See below
Work Step by Step
The general solution for the given differential equation is: $y(x)=c_1\cos 3x+c_2 \sin 3x+A_0\sin 2x+B_0\cos 2x$
The trial solution for $y_p=A_0\sin 2x+B_0\cos 2x$ can be computed as by plugging back into the given differential equation.
So, we have: $y''+9y=5\cos 2x\\(D^2+9)y_p(x)=5\cos 2x\\ 5A_0\sin 2x +5B_0\cos 2x=5\cos 2x$
On comparing coefficients, we get: $A_0=0,B_0=1$
Therefore, the general solution for the given differential equation is: $y(x)=c_1\cos 3x+c_2 \sin 3x+\cos 2x$
Substitute $y(0)=2\\y'(0)=3$
Then we have $c_1=2\\
c_2=1$
Hence, $y=2\cos 3x+\sin 3x+\cos 2x$
