8
$\begingroup$

I am styding Laplace transforms and for some reason I have stuck in the followning exercise.

Find the inverse Laplace Transform $ L^{-1} \{\log \frac{s^2 - a^2}{s^2} \}$.

Any help?

Thank's in advance!

1 Answers 1

9

If

$$F(s)=\mathcal{L}\{f(t)\}(s)=\log(1-s^2/a^2)$$

then

$$\mathcal{L}\{t f(t)\}=-F'(s)=-\frac{d}{ds}\log(1-a^2/s^2)=\frac{2}{s}-\frac{1}{s+a}-\frac{1}{s-a}.$$

Now, can you apply the inverse Laplace transform to both sides here? Then just divide by $t$.

  • 0
    Yes I can apply the inverse Laplace transform but I have one question: Which property of Laplace transforms do you use here $\mathcal{L}\{t f(t)\}=-F'(s)=$. ?2012-03-26
  • 0
    @passenger: $\mathcal{L}\{tf(t)\}=-F'(s)$ ***is*** the property I used there; it is standard. It can be proven by taking $\mathcal{L}\{f\}=F(s)$ and differentiating both sides with respect to $s$.2012-03-26
  • 0
    It was obvious! Sorry it was a silly question! thank you for your time!2012-03-26
  • 0
    @passenger: No problem.2012-03-26
  • 0
    Should that not be $F(s)=\mathcal{L}\{f(t)\}(s)=\log(1-s^2/a^2)$ ?2013-01-25
  • 0
    @pbs Right, thanks.2013-01-25