Find the solution for the spring-mass problem $y′′+9y=\cos(3t)$. Solve with initial conditions $y(0) = 0$, $y′ (0) = 0$. Using Laplace transform

Question

I first took the Laplace transform of each part then getting $s^{2}Y+9Y=\frac{s}{s^{2}+9}$ then solving for Y, I got $Y=\frac{s}{(s^{2}+9)^{2}}$ but don't know how to simplify that to something that can be transformed back.

Answer 1

Hint: $\frac{s}{(s^2+9)^2} = \frac{d}{ds}\frac{-1}{2(s^2+9)}$

Answer 2

If you let k=3, then you can use that fact that L.T. t*sin(kt) = 2ks/(s^2+k^2)^2 or in your case, 2*3*s/(s^2+3^2)^2 -> 6s/(s^2+3^2)^2 which if you invert should lead to (1/6)t*sint(3t). Try that.