# Inverse Laplace

• Jun 1st 2009, 01:29 PM
gconfused
Inverse Laplace
Can someone please help me figure out the inverse Laplace of:

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

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

Any guidance would be very appreciated
Thank you
• Jun 1st 2009, 07:55 PM
TheEmptySet
Quote:

Originally Posted by gconfused
Can someone please help me figure out the inverse Laplace of:

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

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

Any guidance would be very appreciated
Thank you

I will take a shot.

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

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

Now this is in the form of a Heaviside function

I hope this helps.

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

Where

$\displaystyle \mathcal{U}(t-a)=\begin{cases} 0, t < a \\ 1,t>a \end{cases}$
• Jun 1st 2009, 08:54 PM
gconfused
Oh wow thanks, i get it now, you helped me heaps :D