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

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!

• It's already in fully simplified form. That is, $A=0$ and $B=1$. – Simply Beautiful Art May 16 '17 at 23:48
• The decomposition would actually be $\dfrac{As + B}{s^2+16} + \dfrac{Cs + D}{(s^2+16)^2}$. But like SBA said, it's already fully simplified as originally given. – tilper May 16 '17 at 23:49
• @SimplyBeautifulArt r I am solving a differential equation using Laplace Transform. One of the fractions is the above. I need to make it look like one of the forms in the Laplace Transform Table. – socrates May 16 '17 at 23:50
• Sounds like you'll want to use convolutions. – tilper May 16 '17 at 23:52
• Oops, my derivative is wrong. But see the link @msm commented, it carries what I meant to say. – Simply Beautiful Art May 17 '17 at 0:02

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.