Show that if y satisfies $y''+y=\sin^{2017} x \cdot \cos x$ then $y$ is a periodic function.
To deal with this problem, the first thing I do is solving $y''+y=0$. Then I get $y=c_1 \cos x +c_2 \sin x$, and I get stuck here.
Any suggestion? Thanks in advance!
Let $f(t)=(\sin(t))^{2017}\cos(t)$. Put $$g(x)=\sin(x)\int_0^xf(t)\cos(t)dt-\cos(x)\int_0^x f(t)\sin(t)dt$$ Then we compute that $g$ is a particular solution of our differential equation. We want to show that $g$ is periodic with period $2\pi$. This is equivalent , writing $g(x+2\pi)-g(x)=0$, to show that
$$\sin(x)\int_x^{x+2\pi}f(t)\cos(t)dt-\cos(x)\int_x^{x+2\pi}f(t)\sin(t)dt=0$$
It suffices to show that $\int_x^{x+2\pi}f(t)\cos(t)dt=\int_x^{x+2\pi}f(t)\sin(t)dt=0$. By computing the derivative of these functionss, we see that they are constant; so it suffices to show that they vanishes for a particular value of $x$. For the first, we take $x=-\pi$, and use the fact that $f(t)\cos(t)$ is odd to show that it is $0$. For the second, we take $x=0$, and use that a primitive of $f(t)\sin(t)$ is $\frac{(\sin(t)^{2019}}{2019}$ to show it is $0$ also, and we are done.