How to solve this partial fraction decomposition for taking the Inverse Laplace Transform?

Question

I am trying to solve the differential equation $y''+16y=2\sin(4x)$ with initial conditions of $y(0)=\frac{-1}{2}$ and $y'(0)=0$. Upon solving for $\mathcal{L}\{y\}$, I obtained the below fraction. $$\frac{1}{(s^2+16)^2} $$ And I need to solve this fraction using partial fraction decomposition to make it look like one of the forms in the Laplace Transform Table so I can take the Inverse Laplace Transform to solve the differential equation. Please help me!

Answer 1

From the comments we see that what you actually want is the inverse Laplace transform of $\dfrac1{(s^2+16)^2}$. You'll need to use the convolution theorem, which essentially (details withheld) says: $$ f \ast g = \mathcal L^{-1}\left\{ \mathcal L\{f\} \cdot \mathcal L\{g\}\right\}$$

In this case, you'll want to take $\mathcal L\{f\} = \mathcal L\{g\} = \dfrac1{s^2+16}$. This means that $f = g$. Find the $f$ that gives you $\mathcal L\{f\} = \dfrac1{s^2+16}$ and then find $f \ast f$, where $\ast$ is the convolution.