Question

I'm trying to find the inverse Laplace of some given argument and I think I know how to do it, but I'm curious if there is a better way. (Comments section)

Answer 1

I'm trying to find \[L^{-1} \left\{ \vphantom{} \frac {(s+1)^{3}}{s^4} \right\}\] The way I'm imagining approaching this is by expanding the numerator and then just evaluating them as individual inverse laplaces, but that sounds like a pain. Is there a better, quicker way to do this?

Answer 2

I don't know of a faster way

Answer 3

None that I can remember, but fast is how your brain interprets it. It's gonna be different speed for everyone.

Answer 4

\[L^{-1} \left\{ \vphantom{} \frac{(s+1)^3}{s^4} \right\} = L^{-1} \left\{\vphantom{} \frac{(s^2+2s+1)(s+1)}{s^4} \right\}=L^{-1} \left\{\vphantom{} \frac{s^3+2s^2 +s+s^2+2s+1}{s^4} \right\}\]
(Simplify)

Answer 5

\[L^{-1} \left\{\vphantom{} \frac{s^3+3s^2+3s+1}{s^4} \right\}\]

Answer 6

\[= L^{-1} \left\{\vphantom{} \frac{1}{s} \right\} +L^{-1} \left\{\vphantom{} \frac{3}{s^2} \right\}+L^{-1} \left\{\vphantom{} \frac {3}{s^3}\right\}+L^{-1} \left\{\vphantom{} \frac{1}{s^{4}} \right\}\]

Answer 7

\[= 1 + 3t +\frac{3}{2}t^2+\frac{1}{6}t^3\]
Does this look correct?

Answer 8

No, I don't think you can do inverse of Laplace for \(1/s^4\).

Answer 9

I believe you must do partial fraction decomposition for this.

Answer 10

One way to go around the expanding is by adding/subtracting the right terms. It's not exactly any faster than simply expanding, nor is it really that different from binomial expansion (more like factorization into a binomial).

First substitute \(x=s+1\), then you have
\[\frac{(s+1)^3}{s^4}=\frac{x^3}{(x-1)^4}\]
What we want is a power of \(x-1\) in the numerator, so what we can do is add/subtract the terms you'd see in the expansion of \((x-1)^3\):
\[\begin{align*}\frac{x^3}{(x-1)^4}&=\frac{\color{red}{x^3}+(\color{red}{-3x^2+3x-1})+(\color{blue}{3x^2-3x}+1)}{(x-1)^4}\\\\
&=\frac{(x-1)^3+3x(x-1)+1}{(x-1)^4}\\\\
&=\frac{1}{x-1}+\frac{3x}{(x-1)^3}+\frac{1}{(x-1)^4}
\end{align*}\]
Do something similar for that second term:
\[\begin{align*}\cdots&=\frac{1}{x-1}+\frac{\color{red}{3x}+(\color{red}{-3}+3)}{(x-1)^3}+\frac{1}{(x-1)^4}\\\\
&=\frac{1}{x-1}+\frac{3(x-1)+3}{(x-1)^3}+\frac{1}{(x-1)^4}\\\\
&=\frac{1}{x-1}+\frac{3}{(x-1)^2}+\frac{3}{(x-1)^3}+\frac{1}{(x-1)^4}\\\\
&=\frac{1}{s}+\frac{3}{s^2}+\frac{3}{s^3}+\frac{1}{s^4}\end{align*}\]
which is the same result as you got before, with what's really a fundamentally identical approach. It's one way to go around expanding, but no more efficient.

Answer 11

Wait, so can you take the inverse Laplace of \(s^{-4}\)? I figured it would be an issue due to \(\Gamma (-4+1)\) being an issue.

Answer 12

Yes,
\[\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}~~\iff~~t^n=\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}\]