Answer
$${\text{The function is a solution of the initial value problem}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& y\left( t \right) = \frac{1}{4}\left( {{e^{2x}} - {e^{ - 2x}}} \right);{\text{ }}y''\left( x \right) - 4y = 0 \cr
& {\text{Initial conditions: }}y\left( 0 \right) = 0{\text{ and }}y'\left( 0 \right) = 1 \cr
& y\left( t \right) = - 3\cos 3t \cr
& {\text{Calculate the first and second derivative}} \cr
& y'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{4}\left( {{e^{2x}} - {e^{ - 2x}}} \right)} \right] \cr
& y'\left( t \right) = \frac{1}{4}\left( {2{e^{2x}} + 2{e^{ - 2x}}} \right) \cr
& y'\left( t \right) = \frac{1}{2}\left( {{e^{2x}} + {e^{ - 2x}}} \right) \cr
& \cr
& y''\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{2}\left( {{e^{2x}} + {e^{ - 2x}}} \right)} \right] \cr
& y''\left( t \right) = \frac{1}{2}\left( {2{e^{2x}} - 2{e^{ - 2x}}} \right) \cr
& y''\left( t \right) = {e^{2x}} - {e^{ - 2x}} \cr
& {\text{Substitute the derivatives into the given differential equation }} \cr
& y''\left( x \right) - 4y = 0 \cr
& {e^{2x}} - {e^{ - 2x}} - 4\left( {\frac{1}{4}\left( {{e^{2x}} - {e^{ - 2x}}} \right)} \right) = 0 \cr
& {e^{2x}} - {e^{ - 2x}} - \left( {{e^{2x}} - {e^{ - 2x}}} \right) = 0 \cr
& {e^{2x}} - {e^{ - 2x}} - {e^{2x}} + {e^{ - 2x}} = 0 \cr
& 0 = 0{\text{ Proved}} \cr
& \cr
& {\text{Verifying the initial conditions}} \cr
& y\left( t \right) = \frac{1}{4}\left( {{e^{2x}} - {e^{ - 2x}}} \right),{\text{ }}y\left( 0 \right) = 0 \cr
& y\left( 0 \right) = \frac{1}{4}\left( {{e^0} - {e^0}} \right) \cr
& y\left( 0 \right) = 0 \cr
& \cr
& and \cr
& y'\left( t \right) = \frac{1}{2}\left( {{e^{2x}} + {e^{ - 2x}}} \right),{\text{ }}y'\left( 0 \right) = 1 \cr
& y'\left( 0 \right) = \frac{1}{2}\left( {{e^0} + {e^0}} \right) \cr
& y'\left( 0 \right) = \frac{1}{2}\left( 2 \right) \cr
& y'\left( 0 \right) = 1 \cr
& \cr
& {\text{The function is a solution of the initial value problem}}{\text{.}} \cr} $$
