I am trying to solve the following exercise:
$y''+ ky=sin\pi x$. I need to find values of k such that this equation has no periodic solutions. I think that this is a harmonic oscillator equation and I must have resonance to achieve my goal. So if the general harmonic oscillator equation is : $x''+kx'+\omega^2x=f(t)$ then i need k=0 which I have and if f(t)=$C\sin\omega_0(t)$ then i need $\omega_0=\omega$ so in my exercise the correct answer is $k=^+_- \sqrt(\pi)$. Am I correct or is there something that I am missing?
You can ensure that the solution is not periodic by setting $k=\pi^2$.
This is so that the complementary function is $$y=A\cos\pi t+B\sin\pi t$$
and this means that, because of the repetition of $\sin \pi t$ on the right hand side, the particular integral must be of the form $$y=t(p\cos \pi t+q\sin \pi t)$$ thus ensuring that the solution is not periodic.