Consider the differential equation,$y''+2y=cos(kt)$, what is the values of k such that solutions to the differential equation are unbounded
Hint: the solution of the associated homogeneous equation $y''+2y=0$ is $$y_h=C_1\cos\sqrt2t+C_2\sin\sqrt2 t.$$
If you've studied the method of undetermined coefficients, you should know that a particular solution of $y''+2y=\cos kt$ has the form
$(1)\ \ \ \ \ y_p=A\cos kt +B\sin kt$, if $k\ne \sqrt 2$
$(2)\ \ \ \ \ y_p=At\cos kt +Bt\sin kt $, if $k=\sqrt2$
(note that when $k=\sqrt2$, the guess $(1)$ for a particular solution is already a solution of the homogeneous equation, hence the multiplication by $t$ in $(2)$ for the corrrect guess of $y_p$).
The general solution of your equation is $y_h+y_p$.
Find the solution in each of the two cases $k=\sqrt2$ and $k\ne\sqrt2$, and determine if you have an unbounded solution.
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