I have the following equality:
$$\ddot{x} = -\omega^2 x \tag{1}$$
and I know that this must give me :
$$x= A cos(\omega t) + B sin(\omega t) \tag{2}$$
How do I reach this result? What happens if I add an extra term which depends on the variable $(t)$ , eg. $x^2 = -\omega^2 x + bt$?
You can rewrite as
$$x''+\omega^2 x. = 0$$
Multiply both sides by $2x'$
$$2x'x'' + 2\omega^2xx' = 0$$
$$(x'^2 + \omega^2x^2)' = 0$$
Can you proceed from here?