A method to solve a difficult differential equation The equation is $\ddot u +4 u =sin^3t $ with the relative cauchy problem $u(0)=1; \dot u(0)=1 $.
Solve the homogeneous equation it's not difficult, but my problem is to find out the solutions of the sine.
I've tried with the method of variation of constants but it seems too long and there is too many calculations to do. 
Is there a simpler and faster method to solve this equation?
 A: One (somewhat nonstandard) way is to start by finding the fundamental solution. This is the solution to
$$u''+4u=0,u(0)=0,u'(0)=1.$$
In this case it is $u(t)=\frac{1}{2} \sin(2t)$. Now the solution to
$$z''+4z=f(t),z(0)=0,z'(0)=0$$
is given by $z=(u*f)(t)$, where $*$ denotes convolution. (If I were to prove this now, I would probably use distribution theory, but there is an ordinary calculus proof as well.) In other words, the solution to your problem (but with zero initial data) is
$$z(t) = \int_0^t \frac{1}{2} \sin(2s) \sin^3(t-s) ds.$$
Then you can add the solution to $v''+4v=0,v(0)=1,v'(0)=1$ to this in order to finish the problem. So
$$y(t) = \frac{1}{2} \left ( \cos(2t) + \sin(2t) + \int_0^t \sin(2s) \sin^3(t-s) ds \right )$$
solves your problem.
The reason I like this approach is that it puts the technical part of the problem in front of you automatically: you just have to compute an integral. One such technicality would arise from this problem if the $4$ were a $9$. Then you would want to work with $\sin(3t)$, but it wouldn't work because $\sin(3t)$ would be annihilated by the differential operator. With my suggested approach, that phenomenon would come out directly from the integration.
A: Hint: Expand $\sin{(3t)}$, you can find that 
$$\sin^3t=-\frac{1}{4}\sin{(3t)}+\frac{3}{4}\sin t$$
Then you can use the trial solutions $A\sin{(3t)}+B\cos{(3t)}$, $A\sin t+B\cos t$, together with the superposition rule to find the particular solutions.
