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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!

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    $\begingroup$ It's already in fully simplified form. That is, $A=0$ and $B=1$. $\endgroup$ – Simply Beautiful Art May 16 '17 at 23:48
  • $\begingroup$ 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. $\endgroup$ – tilper May 16 '17 at 23:49
  • $\begingroup$ @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. $\endgroup$ – socrates May 16 '17 at 23:50
  • $\begingroup$ Sounds like you'll want to use convolutions. $\endgroup$ – tilper May 16 '17 at 23:52
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    $\begingroup$ Oops, my derivative is wrong. But see the link @msm commented, it carries what I meant to say. $\endgroup$ – Simply Beautiful Art May 17 '17 at 0:02
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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.

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    $\begingroup$ Easier than convolution is to use properties of LT here $\endgroup$ – msm May 16 '17 at 23:59
  • $\begingroup$ @msm, easier is largely subjective, and I saw you already linked this in the comments. Now OP has two ways to do it. :) Hope that downvote isn't yours, though. $\endgroup$ – tilper May 17 '17 at 0:07
  • $\begingroup$ what does op mean?? $\endgroup$ – socrates May 17 '17 at 0:08
  • $\begingroup$ @socrates, in this context, OP = original poster. $\endgroup$ – tilper May 17 '17 at 0:09
  • $\begingroup$ @tilper No it is not my downvote, convolution can also be used (+1). $\endgroup$ – msm May 17 '17 at 0:15

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