Cfrgtkky

I'm trying to show that $x''+bx'+x=cos(t)$ has a unique bounded solution, for $b<0$, $binmathbb{R}$. 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:

$$x(t)=c_1e^{frac{-b+sqrt{b^2-4}}{2}t}+c_2e^{frac{-b-sqrt{b^2-4}}{2}t}+frac{sin(t)}{b},thinspacethinspace b^2-4>0 $$
$$x(t)=c_1e^t+c_2te^t-frac{sin(t)}{2},thinspacethinspace b^2-4=0 implies b=-2$$
$$x(t)=e^{frac{-b}{2}t}left(c_1cosleft(frac{sqrt{4-b^2}}{2}tright)+c_2sinleft(frac{sqrt{4-b^2}}{2}tright)right)+frac{sin(t)}{b},thinspacethinspace b^2-4<0$$

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!

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.