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

Question

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!

anon · Accepted Answer · 2013-01-25 14:29:06Z

up vote 5 down vote accepted

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 $t$.

edited Jan 25 at 14:29

answered Mar 26 '12 at 22:22

anon
39.7k244111

	Yes I can apply the inverse Laplace transform but I have one question: Which property of Laplace transforms do you use here $\mathcal{L}\{t f(t)\}=-F'(s)=$. ? – passenger Mar 26 '12 at 22:42
	@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 $s$. – anon Mar 26 '12 at 22:57
	It was obvious! Sorry it was a silly question! thank you for your time! – passenger Mar 26 '12 at 22:59
	@passenger: No problem. – anon Mar 26 '12 at 23:02
	Should that not be $F(s)=\mathcal{L}\{f(t)\}(s)=\log(1-s^2/a^2)$ ? – pbs Jan 25 at 13:25

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