Question

use the differential of transforms theorem to find the inverse laplace transform of F(s)=ln(((s-2)/(s+4))*((s+5)/(s-6))

Answer 1

\[\ln (\frac{s-2}{s+4}*\frac{s+5}{s-6})\]

Answer 2

First, split up the big logarithm into four easier ones, and then take the derivative, giving you F'(s). Is there a theorem you can use to easily transform from a derivative of F into the original function?

Answer 3

the derivative would just flip the whole thing, right? ie derivative of ln(x)=1/x ?

Answer 4

Yep.
I'll give you a hint: the relevant theorem begins on page 6 of this document
http://www.math.uah.edu/howell/DEtext/Part4/Laplace_derivatives.pdf

Answer 5

They do a more thorough job of explaining it than I could here, anyway. The main statement is this:
\[L(t∗f(t))=-F′(s).\] Now, you've got several F'(s) terms. Work backwards!

Answer 6

so F'(s)= \[\frac{s+4}{s-2}*\frac{s-6}{s+5}\]?

Answer 7

No, you need to split up the logarithm first, like so:\[F(s) = \ln(s-2) + \ln(s+5) - \ln(s+4) - \ln(s-6) \\ F'(s) = \frac{1}{s-2} + \frac{1}{s+5} -\frac{1}{s+4} - \frac{1}{s-6}\]

Answer 8

so then When I get -F'(s), what do I do to get it to the next step? would the laplace transforms cancel out?

Answer 9

Well, let's just take the first term as an example:\[F'(s) = \frac{1}{s-2} \\ -t*f(t) = L^{-1}(\frac{1}{s-2})\] after taking the inverse Laplace transform of both sides, and we know that the right-hand side gives us e^{2t}. Hence, f(t) for that term must be -e^{2t}/t.

Answer 10

So they would all be like that, wouldn't they?

Answer 11

They would all be similar, yep.

Answer 12

so the answer would be \[\frac{e^(-4t)+e^(6t)-e^(-5t)-e^(2t)}{t}\]

Answer 13

or can that be simplified?

Answer 14

Looks good to me.

Answer 15

Sweet thank you very much. I appreciate it