# What is the value of k, so that the solution is unbounded

Consider the differential equation,$y''+2y=cos(kt)$, what is the values of k such that solutions to the differential equation are unbounded

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So many questions asked and you still do not know that one is supposed to show what one tried, why this failed, and so on? – Did May 7 '12 at 13:40
WolframAlpha can solve the differential equation for you. All that remains is to check for which values of $k$ this is unbounded. – Daan Michiels May 7 '12 at 13:43

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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This is the solution, found by Wolfram Alpha. Now you can try to answer by yourself.

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