Non bounded ODE solution Find the value of $w \ge0$ so that the ODE below doesn't have a bounded solution
$$y''+y=cos(wt)$$
My attempt:
The solution should be in the form
$y(t)=Asin(wt)+Bcos(wt)$
$y'(t)=Awcos(wt)-Bwsin(wt)$
$y''(t)=-Aw^2sin(wt)-Bw^2cos(wt)$
So $(1): -Aw^2sin(wt)-Bw^2cos(wt)+Asin(wt)+Bcos(wt)=cos(wt)$
$Asin(wt)(1-w^2)+Bcos(wt)(1-w^2)=cos(wt)$ $\Rightarrow A=0$ and $B=\frac{1}{1-w^2}$
Thus $y(t)=\frac{1}{1-w^2}cos(wt)$
Since $cosx$ is bounded for every $x\in \mathbb{R}$ does that mean that $w=1$ so that the solution is not bounded? Does not bounded means not defined in this case? Or am I very wrong?
 A: Your solution gives a particular solution, but not the whole solution.  You have to add the solution to the corresponding homogeneous equation $y''+y=0$, which is easily seen to be $c_1 \cos(t) + c_2 \sin(t)$.  Hence, $$y(t)=c_1 \cos(t) + c_2 \sin(t) + \frac{1}{1-w^2}\cos(wt)$$
But this is still bounded for all $t$.  However, notice that this solution is valid for $w \neq 1$, but not for $w=1$ since you would obtain $0=\cos(wt)$ when you plug derivatives back into the differential equation if $w=1$.
For $w=1$, you may guess the solution to be of the form $A \cos(t) + B \sin(t)$, but it can't be since these are already included in the solution to the homogeneous equation.  The strategy here is to assume the solution has the form $A t \cos(t) + B t \sin(t)$. Then,
$$\begin{cases} y=At \sin(t)+Bt \cos(t) \\ y'=At \cos(t)+B \cos(t)-Bt \sin(t) +A \sin(t) \\ y''= -Bt \cos(t)+2A \cos(t)-At\sin(t)-2B \sin(t) \end{cases}$$
$$\implies \left( -Bt \cos(t)+2A \cos(t)-At\sin(t)-2B \sin(t) \right) + \left( At \sin(t)+Bt \cos(t) \right) = \cos(t)$$
$$\implies 2A \cos(t) - 2B \sin(t) = \cos(t) \implies \begin{cases} A=\frac{1}{2} \\ B =0 \end{cases}$$
So we obtain the full solution for when $w=1$ as
$$y(t)=c_1 \cos(t) + c_2 \sin(t) + \frac12t\sin(t)$$
which you can see is unbounded.
A: Compare the solution provided to that obtained by use of the Laplace transform. Using $y(t) \to y(s)$ and $y''(t) \to s^2 \, y(s) - y'(0) - y(0) \, s$ then
\begin{align}
s^2 \, y(s) - y'(0) - y(0) \, s + y(s) &= \frac{s}{s^2 + \omega^2} \\
y(s) &= \frac{y'(0)}{s^2 + 1} + \frac{y(0) \, s}{s^2 + 1} + \frac{s}{(s^2 +1)(s^2 + \omega^2)} \\
y(t) &= y(0) \, \cos(t) + y'(0) \, \sin(t) - \frac{\cos(t) - \cos(\omega t)}{1-\omega^2} \\
&= \left( y(0) -\frac{1}{1-\omega^2} \right) \, \cos(t) + y'(0) \, \sin(t) + \frac{\cos(\omega t)}{1- \omega^2}.
\end{align}
This solution is in agreement with that of the proposer and the Laplace transform used is
$$y(s) = \int_{0}^{\infty} e^{- s t} \, y(t) \, dt.$$
