find the inverse laplace of 2s exp^-s/(s-1)^2

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find the inverse laplace of 2s exp^-s/(s-1)^2

Mathematics
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Could you rewrite that using the equation tool?
y u dont understand?
\[(2s \times e^{-s})/(s-1)^2\] ?

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yes this is right.
Have you tried anything so far?
yes but im little bit confuse.
Show us what you've tried so far and we'll see if we can help
ok my ans is 2exp^t-1.H(t-1) but the book ans is different.
What's the book answer?
2t.exp^t-1H(t-1).
Did you do this with partial fractions?
no.i use unit step function.
After applying time shifting, you'll need to take inverse laplace of 2s/(s-1)^2 how did you find it? you say you got 2e^(t) for this?
lets see :) Applying frequency shifting, \(\Large L^{-1}[ 2s/ (s-1)^2] = e^t L^{-1} [2(s+1)/s^2 ]\) clear till here ?
because next steps are quite trivial !
the book ans is 2t.exp^t-1H(t-1).
i need this answer.
lets see how we can get there :) you understood the properties i applied? Have the inverse laplace transform table with you ?
yes i have.
so from 2s exp^-s/(s-1)^2 we came to 2 (s+1)/s^2 by applying 2 properties. you understood both ?
for better clarity, do you want me to give you the steps till that step?
yes give me details plz
give me a moment to type, by that time, please think of next steps :)
ok.thank u.
Since \(\Large L^{-1} [e^{-as}F(s)] = f(t-a) H(t-a) \) we first find \(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] \) and then use the above property Now \(\Large L^{-1}f(s-a) = e^{at}f(t) \) so \(\Large L^{-1} [\dfrac{2s }{(s-1)^2}] = e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}] \) see whether you get these steps first :)
\(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\)
ok thank u so much.and which softwaer u used to write this things clearly.
last thing is tell me the the laplace of exp^t-1 is what??
used \(Latex \), which is supported by this site. you can use it too :)
can i download it?/
why would you need that ? we're here : \(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\) you'll get the same answer which is in you book, be patient lol we're just half way done
and for the latex you can just start typing in `\( ..\)` try it type `\(Hi\)` and see what u get
ok
so we need to solve \(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\) `\(e^t L^{-1} [\dfrac{2(s+1) }{(s)^2}]\)` any ideas how to proceed?
sorry exp ^t-x laplace??
sorry, didn't get your question. we're solving an inverse laplace question, and I just asked you whether you know how we can proceed with \( L^{-1} [\dfrac{2(s+1) }{(s)^2}]\) are we on a same page?
i dont understand why u right 2(s+1) in numerater?
glad that you ask doubts :) \(\Large L^{-1}[f(s-a)] = e^{at}L^{-1} [f(s)]\) This is a standard property what does it say? If you want to remove e^at from inverse laplace , replace 's' by 's+1' in your function <>
so applying that property to your function, the numerator s became (s+1)
correction : If you want to remove e^at from inverse laplace , replace 's' by 's+a' in your function <>
can u right the correct ans which i post in above comments??
if I just post all the steps, you'll understand none of them. I don't suppose you want that... thats why I am posting few steps and asking you whether you got them... if this approach doesn't suit you, I am really not the right person to help you...
ok thank u so much to give me precious time.nice to meet u.
ah, you too :)
here \[F(s)=\frac{ 2s.e^{-s} }{ (s-1)^2 }\\s~In ~laplace~domain~means~\frac{ d }{ dt }~In ~time~domain\\f(t)=2.\frac{ d }{ dt }L^{-1}[\frac{ e^{-s} }{ (s-1)^2 }]=2.\frac{ d }{ dt }[(t-1)e^{t-1}H(t-1)]\\used~frequency~shifting~In~last~step\\f(t)=2[(t-1)e^{t-1}+e^{t-1}]H(t-1)=2t.e^{t-1}H(t-1)\] easy this way :) @Yasirist

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