I had trouble to prove the following:
If f is a twice continuously differentiable function on $\mathbb{R}$ which is a solution of the equation: $$f^{"}(t)+c^{2}f(t)=0,$$
then there exist constants $a$ and $b$ such that:
$$f(t)=acos(ct)+bsin(ct).$$
This can be done by differentiating the two functions $$g(t)=f(t)cos(ct)-c^{-1}f'(t)sin(ct)$$
and
$$h(t)=f(t)sin(ct)+c^{-1}f'(t)cos(ct).$$
On
I think you are supposed to prove the result without using any result from ODE. If you differentiate $g$ and $h$ using the formula $ (\phi \psi)'=(\phi) (\psi)'+(\phi') (\psi)$ you will see that $g'=0$ and $h'=0$. Hence $g$ and $h$ are both constants. Now write down the equations $g(x)=A$ and $h(x)=B$ (where $A$ and $B$ are constants) and eliminate $f'$ to find $f$.
I guess $c\neq 0$, otherwise it's not true. Consider the following equation, over $\mathbb{R}$ :
$$f''+c^2f=0 \quad (E) $$
It's a homogeneous linear differential equation with constant coefficient. Its characteristic polynomial, $X^2+c^2$, has (simple) roots : $\pm ic$ (this is where $c\neq 0$ is important).
Therefore, $f:\mathbb{R}\to\mathbb{C}$ is a solution to $(E)$ iff there exists $\lambda,\mu\in\mathbb{C}$ s.t. for all $t\in\mathbb{R}$,
$$f(t)=\lambda\exp(ict)+\mu\exp(-ict).$$
Now, $\forall t\in\mathbb{R}, \exp(it)=\cos(t)+i\sin(t)$. Plugging this into the previous expression gives the desired result.