I have the differential equation $y''+by=\cos(t)$. The general solution is
$y(t)=\begin{cases}c_1\cos(\sqrt{b}t)+c_2\sin(\sqrt{b}t)+\frac{1}{b-1}\cos(t), b>0, b\neq 1 \\ c_1\cos(t)+c_2\sin(t)+\frac{1}{2}t\sin(t), b =1 \\ c_1t+c_2+\frac{1}{b-1}\cos(t), b=0 \\ c_1 e^{\sqrt{|b|t}}+c_2e^{-\sqrt{|b|t}}+\frac{1}{b-1}\cos(t), b<0\end{cases}$
How do I determine if the solutions are bounded on the interval $[0,\infty)$?
Best Answer
This is a very informal answer, but you can expand on it if you wish.
First, let's identify which functions are bounded and which are not.
A constant solution like $f(t)=c$ is obviously always bounded
A linear combination of trig functions is always bounded, since $|\sin t|, |\cos t| \le 1$.
A decaying exponential is bounded on $[0,\infty)$ since $0< e^{-t} \le 1 $
A growing exponential is unbounded on $[0,\infty)$ since $\lim_\limits{t\to\infty} e^t = \infty$
Polynomials of degree $1$ and higher are never bounded, since $\lim_\limits{t\to \infty} t^n = \infty$
Let's look at the cases
$b > 0 \ne 1$ is a linear combination of bounded functions, so is it always bounded. More formally $$ \big|y(t)\big| \le |c_1|\big|\cos(\sqrt{b}t)\big| + |c_2|\big|\cos(\sqrt{b}t)\big| + \frac{1}{|b-1|}|\cos t| \le |c_1| + |c_2| + \frac{1}{|b-1|} $$
$b=1$ grows without bounds, since it contains the term $ \frac{1}{2}t\sin t $ which has no bound over $[0,\infty)$
For the remaining cases, the particular solution is always bounded, while the general solutions both contain one unbounded term. Therefore, they will either be bounded or unbounded under certain conditions.
$b=0$ contains a linear term, so it will be bounded if $c_1 = 0$. For example, when $y'(0) = 0$
Similarly, $b < 0$ will be unbounded if $c_1 = 0$. For example, when $\lim_\limits{t\to\infty}y(t)=0$