Forcing function If the forcing function on the right-hand side of a linear $n^{th}$ order differential equation is nonconstant and periodic, can the solution of the equation be a nonperiodic function?
 A: Yes.  Consider the equation
$\ddot y + \omega^2 y = A \cos \omega t, \; \omega \ne 0; \tag{1}$
with initial conditions
$y(0) = 0, \tag{2}$
$\dot y(0) = 0; \tag{3}$
then the unique solution is
$y(t) = \dfrac{A}{2\omega}t\sin \omega t; \tag{4}$
we see that  with $y(t)$ as in (4) we have 
$\dot y(t) = \dfrac{A}{2\omega}\sin \omega t + \dfrac{A}{2} t \cos \omega t, \tag{5}$
$\ddot y(t) = \dfrac{A}{2} \cos \omega t +  \dfrac{A}{2} \cos \omega t - \dfrac{A\omega}{2} t \sin \omega t = A \cos \omega t - \dfrac{A\omega}{2} t \sin \omega t, \tag{6}$
and furthermore (4) and (5) satisfy (2) and (3); since
$\omega^2 y(t) = \dfrac{A \omega}{2}t \sin \omega t, \tag{7}$
we see that (4) satisfies (1).  But the solution (4) is not periodic; in fact, it is unbounded as $t \to \pm \infty$.  This gives an affirmative example in the case $n = 2$, 
We generalize the above to construct examples for all $n > 2$ as follows:  
A.)  differentiate (1) $n - 2$ times to obtain the equation:
$\dfrac{d^n y}{dt^n} + \omega^2 \dfrac{d^{n - 2} y}{dt^{n - 2}} = \dfrac{d^{n - 2}}{dt^{n - 2}}(A \cos \omega t); \tag{7}$
note the right-hand side of (7) is periodic.
B.)  Evaluate $\dfrac{d^j}{dt^j}(\dfrac{A}{2\omega} t \sin \omega t)$, $2 \le j \le n - 1$, at $t = 0$ to obtain a complete set of initial conditions $\dfrac{d^j y}{dt^j}(0)$ for (7), using $y(0)$ and $\dot y(0)$ as in (2), (3);
C.)  Observe that then $y(t)$ as in (4) is the unique solution to (7) satisfying the initial conditions obtained under item (B.) above.  
I leave the case $n = 1$ to the interested members of my readership; it is not difficult, but if I am not mistaken the sought-for solution must be of the form $y(t) = t(C\sin \omega t + D\cos \omega t)$ if (1) is replaced by $\dot y + \omega y = A\cos \omega t$.  And, of course the value of $y(0)$ must be specified.  Quite likely we must have $y(0) = 0$. ;-)
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
