Verify that the functions $y_1$ and $y_2$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
$$
y^{\prime \prime}+4 y=0, y(t)=\cos (2 t), y_{2}(t)=\sin (2 t)
$$
$$
\begin{aligned}
w=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|&=\left|\begin{array}{cc}{\cos (2 t)} & {\sin (2 t)} \\ {-2 \sin (2 t)} & {2 \cos (2 t)}\end{array}\right|\\
&=2 \cos ^{2}(2 t)+2 \sin ^{2}(2 t) \neq 0
\end{aligned}
$$
$$
\begin{aligned} y_{1}=\cos (2 t) \quad& y_{1}^{\prime}=-2 \sin 2 t \quad y_{1}^{\prime \prime}=-4 \cos 2 t \\ y_{1}^{\prime \prime}+4 y_{1}=&(-4 \cos 2 t)+4 \cos (2 t)=0 \\ y_{2}=\sin (2 t)\quad & y_{2}^{\prime}=2 \cos (2 t) \quad y_{2}^{\prime \prime}=-4 \sin (2 t) \\ y_{2}^{\prime \prime}+4 y_{2} &=-4 \sin (2 t)+4 \sin (2 t)=0 \end{aligned}
$$
Hence, they constitute a fundamental set of solutions.
