Dear Saeb

Matlab often gives the inverse Laplace Transform in terms of sinhx and coshx. Using the following definition one can rewrite the hyperbolic expression as a function of exponentials:

• sinh(z)=(ez−e−z)/2
• cosh(z)=(ez+e−z)/2

then you use this command  ilaplace. One also needs to define the symbols t and s.

Lets calculate the inverse of function F(s),

F(s)=(s-5)/(s(s+2)2)

>> syms t s

>> F=(s-5)/(s*(s+2)^2);

>> ilaplace(F)

ans =

-5/4+(7/2*t+5/4)*exp(-2*t)

>> simplify(ans) ans =

-5/4+7/2*t*exp(-2*t)+5/4*exp(-2*t)

>> pretty(ans)

 - 5/4 + 7/2 t exp(-2 t) + 5/4 exp(-2 t)

 Which corresponds to f(t)

f (t) = −1.25 + 3.5te−2t +1.25e−2t

Alternatively one can write

>> ilaplace((s-5)/(s*(s+2)^2))

more information about ilaplace

https://www.mathworks.com/help/symbolic/mupad_ref/ilaplace.html

 Kind Regard

https://www.mathworks.com/help/symbolic/mupad_ref/ilaplace.html

Thank you for the answer. But I should add it that, Matlab and Maple don't surrender the inverse Laplace-transform result meaningfully. Eventhough using the exponential forms of hyperbolic functions, the result is out of sight.  

You can use next webpage:

https://www.symbolab.com/solver/inverse-laplace-calculator/inverse%20laplace%20sinh%5Cleft(s%5Cright)

and obtain the result 1/(s^2-1) which is totally FALSE. (be careful with online tools)

As for Maple, check why it does not give an answer in screenshot.

Mathematica works by the same way:

http://m.wolframalpha.com/input/?i=inverse+Laplace+transform+sinh%28s%29

Why that?

In order for an inverse Laplace transform to exist it is necessary the function F(s)-->0 when s-->infinity.

That is not the case for sinh(s), so they do not give answer, since inverse transform does not exist. (exception F(s)=1)

Look here for details:

http://home.iitk.ac.in/~sghorai/TEACHING/MTH203/ode17.pdf

Nice job Demetris. I have gotten the point. That was a good hint. Thank you. 

Thank you Fausto.

I think so, but my concern is to find an equivalent expression, better than Dirac delta, if it exists. In fact I want to calculate InvLaplace ( a× sinh(k1s)+b× cosh(k2s)).

Your inverse Laplace does not always exist, since your function is:

a*sinh(k1*s)+b*cosh(k2*s)=

(1/2)*a*exp(k1*s)-(1/2)*a*exp(-k1*s)+(1/2)*b*exp(k2*s)+(1/2)*b*exp(-k2*s)

Then only iff {b=-a,k2=k1} the inverse exists because it is the function:

a*sinh(k1*s)-a*cosh(k1*s)=-a*exp(-k1*s)

and gives inverse:

-a*Dirac(t-k1)

@Fausto

Thanks. 

https://www.researchgate.net/project/Analytic-solution-to-the-Blasius-equation

e^s is divergent, as was discussed already in this post.

I have added a pdf, just before your first comment under this thread.

Sech