I'm trying to show that $x''+bx'+x=\cos(t)$ has a unique bounded solution, for $b<0$, $b\in\mathbb{R}$. I believe I understand that it has a unique solution as all coefficients and the non-homogeneous term are all continuous, but the problem I'm having is the bounded part. I have that the three general solutions are:
$$x(t)=c_1e^{\frac{-b+\sqrt{b^2-4}}{2}t}+c_2e^{\frac{-b-\sqrt{b^2-4}}{2}t}+\frac{\sin(t)}{b},\thinspace\thinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-\frac{\sin(t)}{2},\thinspace\thinspace b^2-4=0 \implies b=-2$$
$$x(t)=e^{\frac{-b}{2}t}\left(c_1\cos\left(\frac{\sqrt{4-b^2}}{2}t\right)+c_2\sin\left(\frac{\sqrt{4-b^2}}{2}t\right)\right)+\frac{\sin(t)}{b},\thinspace\thinspace b^2-4<0$$
The only thing that I can see is that if for all cases, the solution is only bounded if $c_1=c_2=0$, seeing as the exponential term is always growing. Am I missing something about boundedness? Thank you for your time!
Best Answer
You are correct.
Since $b$ is the damping coefficient we get boundedness if $b>0$
If $b>0$ then it is easy to see that the solutions are bounded.
There is a typo in $b<0$ part.