Fourier and differential equations

Question

Hey right now I'm practising Fourierseries and found this problem, just so you know it's my first time using Fourier to solve differential equations. $$ f''(x) + f(x) = 3\cos(2x) $$

Answer 1

$$ F[f''(x)+f(x)]=F[3\cos(2x)] $$

or

$$ ((i \omega)^2+1)F(\omega) = 3 \sqrt{\frac{\pi }{2}} \delta (w-2)+3 \sqrt{\frac{\pi }{2}} \delta (w+2) $$

and then

$$ F(\omega) = \frac{3 \sqrt{\frac{\pi }{2}} \delta (w-2)+3 \sqrt{\frac{\pi }{2}} \delta (w+2)}{((i \omega)^2+1)} $$

and finally

$$ F^{-1}\left(\frac{3 \sqrt{\frac{\pi }{2}} \delta (w-2)+3 \sqrt{\frac{\pi }{2}} \delta (w+2)}{((i \omega)^2+1)}\right) = -\cos(2 x) $$