Question

use the differential of transforms theorem to find the inverse laplace transform of F(s)=ln [{(s+2)(s-5)]/{(s-4)(s+6)}] Thanks!

Answer 1

Could you write what the statement is for the "differential of transforms theorem" ? I am not sure I heard this before, although maybe it goes by a different name for me.

Answer 2

if f(0) = 0 then differentiation of f(t) corresponds to multiplication of its transform by s. differentiation of the transform F(s) corresponds to multiplication of the original function f(t) by -t. thanks for fast reply!

Answer 3

Ohh, okay. Not sure why I didn't think of that one!

F(s) = ln [{(s+2)(s-5)}/{(s-4)(s+6)}]
I feel like a really good action right here is to break this logarithm up by properties of logarithms. ln(AB) = ln A + ln B., ln(A/B) = ln A - ln B.
Because then we could take d/ds of both sides to get an easy to differentiate result.

Answer 4

The logarithm breaks up into four pieces, one of each factor. ln(s + #).
If you notice, the derivative of ln(s) with respect to s is actually a fairly well-known Laplace transform: 1/s. The d/ds of F(s) is equivalent to the Laplace transform of -t f(t). That should make it easy to apply an inverse Laplace transform of each side! :)

Answer 5

=\[= \ln (s-2)(s-5) - \ln (s-4)(s+6) \] and go from there? Thanks!

Answer 6

Just a minor change is that the original problem states ln(s + 2) rather than s -2.
Other than that, it looks good. We can still break up the factors into two more logarithms each (careful with the second because of the negative sign overall).

Answer 7

\[F'(s) = \frac{ 1 }{ (s+2) } +\frac{ 1 }{ (s-5) } - \frac{ 1 }{ s-4 } - \frac{ 1 }{ s+6 } = L ?\]

Answer 8

\( = \mathcal{L} \{ -t \ f(t) \} \)
Going from the original theorem of d/ds of F(s).

Answer 9

then look up the laplace stuffs and put them for (s+2), (s-5), (s-4) and (s+6)? Thanks for your help!

Answer 10

Yep.
Taking the inverse Laplace transform of both sides, you would have -t f(t) (the Laplace and inverse Laplace cancel). You can look up those inverse Laplace transforms of 1/(s - a). :)

Glad to help!

Answer 11

just curious....don't you have to multiple first to get \[\ln (s ^{2} -2s -10) / (s ^{2} +2s-24)\] before do anything else? Thanks!

Answer 12

You *could*, but your work becomes considerably more difficult when the factors are not completely separated. You should end up with the correct answer in any route that you choose, but it seems easier to make a simplification now than later...

If you left it right here and took the derivative:
F(s) = ln [{s^2 - 2s - 10}/{s^2 + 2s - 24}]
By derivative chain rule, we need to take the derivative of the inside. The inside is a quotient of two polynomials, so we need a considerable amount of effort to force our way through. Then, your end result will be something of the form 1/ ( A(s)B(s) ) x C(s) / D^2 (s). (A, B, C, and D are polynomials of variable s). That may leave you with a very awkward inverse Laplace transform.

We could break it into the two logarithms and it is a bit better:
F(s) = ln (s^2 - 2s - 10) - ln (s^2 + 2s - 24)
This is much easier to differentiate now, although it isn't quite the easiest because we still need chain rule.

Finally, we split it up into all four individual factors and the individual natural logarithms are of polynomials whose derivatives are 1.

Answer 13

I c...so I get \[f(t) = te ^{-2t} - te ^{4t} +te ^{5t} - te ^{6t}\] Thanks!

Answer 14

The t is multiplied or divided?
The equation would have ended up like this:
-t f(t) = inverse laplace transform found; solving by dividing by t...
f(t) = - 1/t * (inverse laplace transform found)
and I think there is one minor sign issue somewhere since we initially should have had two positives and two negatives, with no changes in sign except for the -t.

Answer 15

multiplied

Answer 16

laplace transforms from the table is e^at = 1/(s-a)

Answer 17

\[
\begin{align*}
F(s) & = \ln \left( \frac{ (s + 2)(s - 5) }{ (s - 4) (s + 6) } \right) \\
F(s) & = \ln \left( (s + 2) (s - 5) \right) - \ln \left( (s - 4)(s + 6) \right) \\
F(s) & = \ln \left( s + 2) \right) + \ln \left( s - 5 \right) - \ln \left( s - 4 \right) - \ln \left( s + 6 \right) \\
\dfrac{d}{ds} F(s) & = \dfrac{d}{ds} \left[ \ln \left( s + 2) \right) + \ln \left( s - 5 \right) - \ln \left( s - 4 \right) - \ln \left( s + 6 \right) \right] \\
\mathcal{L} \left\{ -t f(t) \right\} & = \dfrac{1}{s + 2} + \dfrac{1}{s - 5} - \dfrac{1}{s - 4} - \dfrac{1}{s + 6} \\
- t f(t) & = e^{-2t} + e^{5t} - e^{4t} - e^{-6t}
\end{align*} \]
This is where I ended up at. Can you check this against your work and see if there is anything that didn't match up here?
My end result is \( f(t) = - \dfrac{e^{-2t} + e^{5t} - e^{4t} - e^{-6t}}{t} \)

You posted: \( f(t) = te ^{-2t} - te ^{4t} +te ^{5t} - te ^{6t} \)

Answer 18

yours look good...thank you!

Answer 19

I provide the details in this case because I see your solution is very close but it seems to be some minor issues with signs and such rather than a fundamental flaw. Those little details are just to be looked out for in the future! :)

And, as I am looking, it seems there is a solution by Wolfram in terms of cosh(t). This would probably be the result of using ln (s^2 - 3s - 10) - ln (s^2 + 2s - 24). The answers are just in different forms yet meaning the same thing. (cosh t = e^t + e^-t / 2)

Again, I am glad to help you when I can!