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So I want to obtain a inverse Laplace transform from mathematica but I get this:

enter image description here

How can I get the solution?

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  • $\begingroup$ I'm solving a heat transfer problem, so there is a solution for that inverse, I think. I only wrote this: InverseLaplaceTransform[Sinh[a sqrt(s)]/(s^2 Sinh[b sqrt(s)), s, t] $\endgroup$ – Francesco Camussoni Mar 1 at 13:58
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    $\begingroup$ That means Mathematica doesn't know the answer. $\endgroup$ – Mariusz Iwaniuk Mar 1 at 14:07
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From here we have:

Sinh[a Sqrt[sn]]/(sn^2 Cosh[b*Sqrt[sn]])*2*Sqrt[sn]*Exp[sn t] /. sn -> -n^2*Pi^2 // FullSimplify
(* -((2 E^(-n^2 \[Pi]^2 t) Sec[b n \[Pi]] Sin[a n \[Pi]])/(n^3 \[Pi]^3))*)

And

Limit[D[Sinh[a Sqrt[s]]/Sinh[b Sqrt[s]]*Exp[s t], s], s -> 0, Assumptions -> t > 0] // Expand
(*a^3/(6 b) - (a b)/6 + (a t)/b*)

Then:

HoldForm[InverseLaplaceTransform[Sinh[a Sqrt[s]]/(s^2*Sinh[b Sqrt[s]]), s, t] == 
a^3/(6 b) - (a b)/6 + (a t)/b - Sum[(2 E^(-n^2 \[Pi]^2 t) Sec[b n \[Pi]] Sin[a n \[Pi]])/(n^3 \[Pi]^3), {n, 1, Infinity}]] // TeXForm

$$\mathcal{L}_s^{-1}\left[\frac{\sinh \left(a \sqrt{s}\right)}{s^2 \sinh \left(b \sqrt{s}\right)}\right](t)=\frac{a^3}{6 b}-\frac{a b}{6}+\frac{a t}{b}-\sum _{n=1}^{\infty } \frac{2 e^{-n^2 \pi ^2 t} \sec (b n \pi ) \sin (a n \pi )}{n^3 \pi ^3}$$

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