Answer
$${\text{The function is a solution of the initial value problem}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& y\left( t \right) = - 3\cos 3t;{\text{ }}y''\left( t \right) + 9y = 0 \cr
& {\text{Initial conditions: }}y\left( 0 \right) = - 3{\text{ and }}y'\left( 0 \right) = 0 \cr
& y\left( t \right) = - 3\cos 3t \cr
& {\text{Calculate the first and second derivative}} \cr
& y'\left( t \right) = \frac{d}{{dt}}\left[ { - 3\cos 3t} \right] \cr
& y'\left( t \right) = - 3\left( { - 3\sin 3t} \right) \cr
& y'\left( t \right) = 9\sin 3t \cr
& y''\left( t \right) = \frac{d}{{dt}}\left[ {9\sin 3t} \right] \cr
& y''\left( t \right) = 9\left( {3\cos 3t} \right) \cr
& y''\left( t \right) = 27\cos 3t \cr
& {\text{Substitute the derivatives into the given differential equation }} \cr
& y''\left( t \right) + 9y = 0 \cr
& 27\cos 3t + 9\left( { - 3\cos 3t} \right) = 0 \cr
& 27\cos 3t - 27\cos 3t = 0 \cr
& 0 = 0{\text{ Proved}} \cr
& \cr
& {\text{Verifying the initial conditions}} \cr
& y\left( t \right) = - 3\cos 3t,{\text{ }}y\left( 0 \right) = - 3 \cr
& y\left( 0 \right) = - 3\cos 3\left( 0 \right) \cr
& y\left( 0 \right) = - 3 \cr
& \cr
& and \cr
& y'\left( t \right) = 9\sin 3t,{\text{ }}y'\left( 0 \right) = 0 \cr
& y'\left( 0 \right) = 9\sin 3\left( 0 \right) \cr
& y\left( 0 \right) = 0 \cr
& \cr
& {\text{The function is a solution of the initial value problem}}{\text{.}} \cr} $$
