Problem is from a past QR exam:
(1) Find the general solution to $$ y^{(5)}(t) + y''(t) = \cos t.$$ (2) Find an initial condition for which the solution to (1) remains bounded as $t \to +\infty$.
(3) Find an initial condition for which the solution to (1) remains unbounded as $t \to +\infty$.
(4) Find a function with the following property: If the function is entered in place of $\cos t$ on the right hand side of part (1), the solution $y \to -\infty$ as $t \to +\infty$ for all initial conditions.
The general solution I've found is
$$y(t) = C_1 + C_2 t + C_3 e^{-t} + C_4 e^{t/2}\cos(\sqrt{3}t/2) + C_5 e^{t/2}\sin(\sqrt{3}t/2) - \frac{1}{2} \cos t + \frac{1}{2} \sin t.$$
My question: Is there an easier way to answer (2) and (3) without differentiating this beast 4 times and picking conditions that make the proper coefficients vanish ($C_2 = C_4 = C_5 = 0$ for (1) and $C_3 = 0$ for (2))? For (4), I'm not sure how to proceed. My guess is that you want the inhomogeneous term to be in the span of the terms going that go to 0 as $t \to \infty$, which in this case is $e^{-t}$. But not sure how you ensure that $y \to -\infty$ for all conditions.
The differentiation would be much easier if you use the complex number representation instead: $$y(t) = C_1 + C_2 t + C_3 e^{-t} + C_4 e^{t\frac{2\pi i}3} + C_5 e^{t\frac{4\pi i}3} - \frac{1}{2} \cos t + \frac{1}{2} \sin t$$ The $C_4$ and $C_5$ coefficients are complex. For the last part, you would need to add a term that diverges to $-\infty$ as $t\to\infty$, so something like $-e^t$ in the particular solution. Taking the derivative two and 5 times, you will get the same thing, so your function will be $$f(t)=-2e^t$$ Notice that there is no term with $e^t$ in the general solution of the homogeneous equation, so you cannot cancel this term. And it diverges faster that any term in the homogeneous solution.