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I missed a class this week in maths and been a bit lost since with Inverse Laplace, how do I go about finding the Inverse laplace of: $$\frac{(s+1)^3}{s^4}$$

Do I simply expand the numerator? then inverse laplace? or is there a quicker way of doing it, expanding the brackets seems like a lengthy process...if someone could help me I'd be grateful as I have a test on this Friday!

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  • $\begingroup$ Doesn't look that onerous to me...and the inverse transforms of the $1/s^n$ are straightforward. $\endgroup$ – Simon S Nov 4 '15 at 18:48
  • $\begingroup$ I thought I could handle it but surpisingly I'm stuck here, maybe my brain is fried from this weeks exams, I'm confused to as why there's an "s" in the numerator...if you can give me another hint at least I'd be grateful. $\endgroup$ – Modrisco Nov 4 '15 at 18:50
  • $\begingroup$ So I expand it? I thought they're would be a quicker way which why I'm skeptical. $\endgroup$ – Modrisco Nov 4 '15 at 18:52
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$$\frac{(s+1)^3}{s^4} = \frac 1s + \frac 3{s^2} + \frac 3{s^3} + \frac 1{s^4}$$ and the inverse Laplace transform of each of those terms should be standard to you. After you've found it, it may be possible to simplify the answer!

(If the inverse transform of these terms are not in your head, go back to your notes, text or this nice MIT lecture on the topic: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/ )

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  • $\begingroup$ I didn't see the simplicity of it, how embaressing. Thanks! $\endgroup$ – Modrisco Nov 4 '15 at 19:06

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