So I'm working with a nonhomogeneous second order differential equation:
$$4y''-y=\sin(x)\cos(x/2).$$
I know that the general solution, $y$, equals $y_c + y_p$ where $y_c$ is the general solution to the complementary equation and $y_p$ is any particular solution to the nonhomogeneous equation. I'm struggling a little bit with $y_p$ because I'm not sure what form the particular solution should be.
I know (at least I think I do) that, for example, the general form of the particular solution for $\cos(x/2)$ is:
$$A\sin(x/2) + B\cos(x/2).$$
I also suspect that the general form of the particular solution for $\sin(x) + \cos(x/2)$ is:
$$A\sin(x) + B\cos(x) + C\sin(x/2) + D\cos(x/2).$$
However, I'm completely thrown off track with $\sin(x)\cdot\cos(x/2)$. I'd appreciate any insight on the matter, because frankly, the entire concept is still a little loose in my head.
Best Answer
There's a trig identity, $$\sin A\cos B=(\sin(A+B)+\sin(A-B))/2$$ which if you haven't seen it before you should be able to verify by expanding out $\sin(A+B)$ and $\sin(A-B)$. Now you can use that on your $\sin x\cos(x/2)$ to turn it into something you know how to handle.