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I have the function $y=\cos(kt)$ and I want to find out for what values of k does that equation satisfy the differential equation $4y''=-25y$.

I differentiated $\cos(kt)$ twice and got $-k^2\cos(kt)$, then substituted that into the diff. eq.: $4(-k^2\cos(kt))=-25(\cos(kt))$. I'm not sure if I should just divide both terms by $\cos(kt)$ or add $25\cos(kt)$ to both terms; if I divide both terms by $cos(kt)$ I get the solutions $k=\pm\frac52$, and if I add $25\cos(kt)$ to both terms I also get the solution $k=\frac{\pi}{2t}+\frac{n\pi}t, n\in\mathbb Z$. Is the last solution simply useless because it depends on t? Am I just doing differential equations plain wrong?

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    k is a constant but when you write k in terms of t, it is no longer a constant. So the $\pm \frac{5}{2}$ looks good to me. – 2017-01-03

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You are trying to figure out which values of $k$ give a function which satisfies the differential equation for all $t$. In other words, you pick $k$ first, then you check the differential equation for all $t$. Since $k$ is chosen before $t$, then $k$ cannot depend on $t$.