I'm having issues trying to solve this differential equation
$x''(t)\ + \ x(t) \ = \ \frac{1}{6} \cdot \cos^3(t),\ x(0)=0 \ ,\ x'(0)=0$
I've been trying to convert $\frac{1}{6} \cdot \cos^3(t)$ to $\frac{1}{48} \cdot (e^{it}+e^{-it})^3$ and then guessing a solution in the form $A\cdot(e^{it}+e^{-it})^3$ but that doesn't work.
I then tried to guess a solution in the form $A\cdot t\cdot(e^{it}+e^{-it})^3$ but it does not seem to work either...
I know the solution to the differential equation is $x(t)=\frac{1}{192} \cdot \cos(t)\ -\frac{1}{192} \cdot \cos(3t) \ +\frac{1}{16}\cdot t\cdot \sin(t)$
Any help would be appreciated!
hint
Replace $\cos^3 (t) $ by
$$\frac {(e^{it}+e^{-it})^3}{8}=$$
$$\frac {1}{8}(2\cos (3t)+6\cos (t)) $$
and use superposition principle.
The general solution will have the form $$y_g=y_h+y_1+y_2$$ with $$y_h=A\cos (t)+B\sin (t) $$ $$y_1=C\cos (3t)+D\sin (3t) $$ and $$y_2=t (E\cos(t)+F\sin (t)) $$