In PLA1, p26,
Prove that if $f$ is a twice continuously differentiable function on
$\mathbb{R}$ which is a solution of the equation

f”(t) + c^2f(t) = 0,

then there exist constants $a$ and $b$ such that

f(t) = a\cos{ct} + b\sin{ct}.

This can be done by differentiating the two functions $g(t) = f(t)\cos{ct} – c^{-1}f'(t)\sin{ct}$ and $h(t) = f(t)\sin{ct} + c^{-1}f'(t)\cos{ct}$.

Obviously we need that c \neq 0

Smart! But how to get the trick?

See

f(t) = a\cos{ct} + b\sin{ct},

then by differentiating the equation, we get

f'(t) = -ac\sin{ct} + bc\cos{ct}

Now we get equations with two variables $a$ and $b$,

\left\{
\begin{array}{rl}
a\cos{ct} + b\sin{ct} &= f(t)\\
-ac\sin{ct} + bc\cos{ct} &= f'(t).
\end{array}
\right.

Solving it by simple linear algebra theorem,

\left\{
\begin{array}{l}
a = f(t)\cos{ct} – c^{-1}f'(t)\sin{ct}\\
b = f(t)\sin{ct} + c^{-1}f'(t)\cos{ct}.
\end{array}
\right.

and let $g(t) = a, h(t) = b$ . Bingo.