Help with inverse laplace

**jdk022** · Jul 15th 2010, 02:31 PM

Hi, i have attempted to solve the following:

L^-1: 3/(4s^2-9)

and got

(1/8)sinh(3t/2).

I first moved a (1/4) outside of the laplace in order to get the bottom closer to the laplace of sinh(kt) = k/(s^2-k^2). I then pulled a (1/2) out to get the 3 on the top to be 3/2 so that it would fit the equation. Is this correct? Thanks for any help!

**chiph588@** · Jul 15th 2010, 03:00 PM

$\frac{3}{4s^2-9}=\frac14\cdot\frac{3}{s^2-9/4} = \frac{3}{4}\cdot\frac23\cdot\frac{3/2}{s^2-(3/2)^2} = \frac12\cdot\frac{3/2}{s^2-(3/2)^2}$

Thus $\mathcal{L}^{-1}\left(\frac{3}{4s^2-9}\right) = \frac12\cdot\sinh\left(\frac{3t}{2}\right)$

I believe your mistake is your last step where you pulled out a $\frac12$ when you should've pulled out a $2$ .

**jdk022** · Jul 15th 2010, 03:11 PM

ok, that makes sense. Thanks for the help!

Thread: Help with inverse laplace

LinkBack

Thread Tools

Display

Help with inverse laplace

Similar Math Help Forum Discussions

Laplace Inverse

laplace inverse

Laplace/Inverse Laplace Questions

inverse laplace

Inverse Laplace

Search Tags