Thread: how to find inverse laplace of : 1/(s^3-s)

1. how to find inverse laplace of : 1/(s^3-s)

and also laplace trans form of sin^2t

2. Re: how to find inverse laplace of : 1/(s^3-s)

use partial fractions for the inverse transform

$\sin^2(t)=\frac{1}{2} (1-\cos (2 t))$

you should be able to finish from there.

3. Re: how to find inverse laplace of : 1/(s^3-s)

Part A)$\displaystyle \frac{1}{s^3-s} = \frac{1}{(s)(s^2-1)}$ This is the product of two Laplace transforms (Hint: Convolution Theorem is of interest here) :
$\displaystyle \mathcal{L}^{-1}\lbrace F(s)G(s) \rbrace = f(t)*g(t) = \int_{t'=0}^{t'=t} f(t')g(t'-t)dt'$

What are $\displaystyle F(s)$ and $\displaystyle G(s)$? What are the corresponding $\displaystyle f(t)$ and $\displaystyle g(t)$?

Part B) Laplace transform of $\displaystyle sin^2(t)$ Recall that $\displaystyle sin^2(t) = \frac{1}{2}-\frac{cos(2t)}{2}$ and we also have the definition of the Laplace transform: $\displaystyle \mathcal{L}\lbrace f(t) \rbrace = \int_{t=0}^{t\rightarrow \infty} e^{-st}f(t)dt$. What do you get when you substitute the above trig identity in the definition of the laplace transform (Hint: you should notice the answer will be a combination of two simpler Laplace transforms)

4. Re: how to find inverse laplace of : 1/(s^3-s)

Another method:

$\displaystyle \dfrac{1}{s^3-s} = \dfrac{s^2-(s^2-1)}{s^3-s} = \dfrac{s^2}{s(s^2-1)} - \dfrac{s^2-1}{s(s^2-1)} = \dfrac{s}{s^2-1} - \dfrac{1}{s}$

Those two have easy inverse Laplace Transforms.

5. Re: how to find inverse laplace of : 1/(s^3-s)

Originally Posted by SlipEternal
Another method:

$\displaystyle \dfrac{1}{s^3-s} = \dfrac{s^2-(s^2-1)}{s^3-s} = \dfrac{s^2}{s(s^2-1)} - \dfrac{s^2-1}{s(s^2-1)} = \dfrac{s}{s^2-1} - \dfrac{1}{s}$

Those two have easy inverse Laplace Transforms.
Or even easier transforms still if you split up \displaystyle \begin{align*} \frac{s}{s^2 - 1} \end{align*} using partial fractions too.

6. Re: how to find inverse laplace of : 1/(s^3-s)

Originally Posted by Prove It
Or even easier transforms still if you split up \displaystyle \begin{align*} \frac{s}{s^2 - 1} \end{align*} using partial fractions too.
Inverse Laplace of $\displaystyle \dfrac{s}{s^2-1}$ is $\displaystyle \cosh t$. No need to use partial fractions.