Question

Find the inverse Laplace transform of F(s)=4/(s-1)^3.

Answer 1

ok so this is inverse form of shifted function so we would have
\[\frac{ 4 }{ s^3 }|s->s-1\]
=>\[2\frac{ 2 }{ s^3 }|s->s-1\]
so laplace of that is \[2*t^{2}\]
and since we had s-1 we would have \[2t^{2}e^{t}\]

Answer 2

But what you wrote looks like limit, also, where did you get the e^t?

Answer 3

yeah it looks so messy , i cant write it good here dunno how to write the inverse laplace notation i will try explain better wait

Answer 4

\( \mathcal{L}^{-1} \) \mathcal{L}^{-1}
Also I think you were using -> for arrow, \( \to \) \to

Answer 5

yeah i didnt know how to write them -> is arrow

Answer 6

Thanks.

Answer 7

\( \displaystyle \mathcal{L}^{-1} \left(\frac{4}{(s - 1)^3}\right) \)
This s-variable function is shifted to the right by 1 unit, observing the \(s - 1\).
We can use the concept stating that a \(s - n\) shift will correspond to the t-variable function's exponential \(e^{nt}\). Also we can quickly pull out the coefficent 4 because the operation is linear. So, once we pull the exponential, \( s - 1 \to s\) and we have...
\( \displaystyle \mathcal{L}^{-1} \left(\frac{4}{(s - 1)^3}\right) = 4e^{t} \mathcal{L}^{-1} \left(\frac{1}{s^3}\right) \)

As stated, our inverse Laplace transform of this s-variable function is of the form
\( \displaystyle \mathcal{L}^{-1} \left(\frac{n!}{s^{n+1}}\right) = t^n \) For n=2.
We pull in a factor of 2 and we can simply convert that inverse Laplace transform to the t-domain.
\[ \begin{align}
4e^{t} \mathcal{L}^{-1} \left(\frac{1}{s^3}\right) &= 2e^{t} \mathcal{L}^{-1} \left(\frac{2}{s^3}\right) \\
&= 2e^{t} t^2 \\
\end{align} \]
As we were hoping for. :)

Answer 8

http://www.youtube.com/watch?v=FxoQvGONMj0

heres a yt video has laplace and inverse laplace of shifted function example should help understand better