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zmey [24]
3 years ago
14

Inverse laplace of [(1/s^2)-(48/s^5)]

Mathematics
1 answer:
Katen [24]3 years ago
7 0
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
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