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anonymous
 one year ago
find the inverse laplace of 2s exp^s/(s1)^2
anonymous
 one year ago
find the inverse laplace of 2s exp^s/(s1)^2

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Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Could you rewrite that using the equation tool?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y u dont understand?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(2s \times e^{s})/(s1)^2\] ?

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Have you tried anything so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but im little bit confuse.

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Show us what you've tried so far and we'll see if we can help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok my ans is 2exp^t1.H(t1) but the book ans is different.

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0What's the book answer?

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Did you do this with partial fractions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no.i use unit step function.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2After applying time shifting, you'll need to take inverse laplace of 2s/(s1)^2 how did you find it? you say you got 2e^(t) for this?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2lets see :) Applying frequency shifting, \(\Large L^{1}[ 2s/ (s1)^2] = e^t L^{1} [2(s+1)/s^2 ]\) clear till here ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2because next steps are quite trivial !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the book ans is 2t.exp^t1H(t1).

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2lets see how we can get there :) you understood the properties i applied? Have the inverse laplace transform table with you ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2so from 2s exp^s/(s1)^2 we came to 2 (s+1)/s^2 by applying 2 properties. you understood both ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2for better clarity, do you want me to give you the steps till that step?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes give me details plz

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2give me a moment to type, by that time, please think of next steps :)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2Since \(\Large L^{1} [e^{as}F(s)] = f(ta) H(ta) \) we first find \(\Large L^{1} [\dfrac{2s }{(s1)^2}] \) and then use the above property Now \(\Large L^{1}f(sa) = e^{at}f(t) \) so \(\Large L^{1} [\dfrac{2s }{(s1)^2}] = e^t L^{1} [\dfrac{2(s+1) }{(s)^2}] \) see whether you get these steps first :)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2\(\Large L^{1}[f(sa)] = e^{at}L^{1} [f(s)]\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank u so much.and which softwaer u used to write this things clearly.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0last thing is tell me the the laplace of exp^t1 is what??

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2used \(Latex \), which is supported by this site. you can use it too :)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2why 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2and for the latex you can just start typing in `\( ..\)` try it type `\(Hi\)` and see what u get

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2so 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry exp ^tx laplace??

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2sorry, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand why u right 2(s+1) in numerater?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2glad that you ask doubts :) \(\Large L^{1}[f(sa)] = e^{at}L^{1} [f(s)]\) 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 <<Note how s1 on left became s>>

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2so applying that property to your function, the numerator s became (s+1)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2correction : If you want to remove e^at from inverse laplace , replace 's' by 's+a' in your function <<Note how sa on left became s>>

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u right the correct ans which i post in above comments??

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2if 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank u so much to give me precious time.nice to meet u.

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.1here \[F(s)=\frac{ 2s.e^{s} }{ (s1)^2 }\\s~In ~laplace~domain~means~\frac{ d }{ dt }~In ~time~domain\\f(t)=2.\frac{ d }{ dt }L^{1}[\frac{ e^{s} }{ (s1)^2 }]=2.\frac{ d }{ dt }[(t1)e^{t1}H(t1)]\\used~frequency~shifting~In~last~step\\f(t)=2[(t1)e^{t1}+e^{t1}]H(t1)=2t.e^{t1}H(t1)\] easy this way :) @Yasirist
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