Answer
$$y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + 8$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 3\cos 2x + 2\sin 3x \cr
& {\text{separate the variables}} \cr
& dy = \left( {3\cos 2x + 2\sin 3x} \right)dx \cr
& {\text{integrate both sides }} \cr
& \int {dy} = \int {\left( {3\cos 2x + 2\sin 3x} \right)dt} \cr
& y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + C \cr
& {\text{the initial condition }}y\left( {\pi /2} \right) = 8{\text{ implies that}} \cr
& 8 = \frac{3}{2}\sin 2\left( {\pi /2} \right) - \frac{2}{3}\cos 3\left( {\pi /2} \right) + C \cr
& 8 = \frac{3}{2}\sin \left( \pi \right) - \frac{2}{3}\cos \left( {3\pi /2} \right) + C \cr
& 8 = C \cr
& {\text{so the solution of the initial value problem is}} \cr
& y\left( x \right) = \frac{3}{2}\sin 2x - \frac{2}{3}\cos 3x + 8 \cr} $$
