anonymous
  • anonymous
I'm struggling with an inverse laplace transform: 1/((s^2+1)^2) What I'd normally do is to use the convolution theorem, and get y(t)=sin^2(t), but I'm being a bit of a retard, and can't seem to get my answer to fit with the suggested solution.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
inverse laplace of \[1\div(s ^{2}+1)\] is sint using convolution we need to find \[\int\limits_{0}^{t} siny \sin(t-y) dy\]
anonymous
  • anonymous
sinAsinB = 1/2[cos(A - B) - cos(A + B)]
anonymous
  • anonymous
As I suspected, I'm a retard. I forgot to calculate the convolution after I had found the inverse transform. Thanks!

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anonymous
  • anonymous
welcome

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