Inverse Laplace

**gconfused** · June 1st 2009, 01:29 PM

Can someone please help me figure out the inverse Laplace of:

$ \frac {2x^{2s-1}}{s^3(2s-1)} $

Well i know what the inverse of
$ \frac {2}{s^3(2s-1)} $ is but i Dont know what to do with the $x^{2s-1}$

Any guidance would be very appreciated
Thank you

**TheEmptySet** · June 1st 2009, 07:55 PM

Originally Posted by gconfused

Can someone please help me figure out the inverse Laplace of:

$ \frac {2x^{2s-1}}{s^3(2s-1)} $

Well i know what the inverse of
$ \frac {2}{s^3(2s-1)} $ is but i Dont know what to do with the $x^{2s-1}$

Any guidance would be very appreciated
Thank you

I will take a shot.

Assume that $x \ne 0$ and x is NOT a function of s then

$ \frac{2}{x} \cdot \frac {x^{2s}}{s^3(2s-1))}=\frac{1}{x} \cdot \frac{e^{s(2\ln(x))}}{s^3(2s-1)} $

Now this is in the form of a Heaviside function

I hope this helps.

$\frac{1}{x}\mathcal{U}(t+2\ln(x) \mathcal{L}^{-1}\left( \frac{1}{s^3(2s-1)}\right) \bigg|_{t \to t+2\ln(x)}$

Where

$\mathcal{U}(t-a)=\begin{cases} 0, t < a \\ 1,t>a \end{cases}$

**gconfused** · June 1st 2009, 08:54 PM

Oh wow thanks, i get it now, you helped me heaps

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