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Find the inverse Laplace transform of \(\frac{1}{s(s^2 + \omega ^2)}\)

Mathematics
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\[L(1*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\] But then, I don't know how to continue...
i would rather do partial fractions. \(\huge \frac{1}{s(s^2 + \omega ^2)}=\frac{Ax+B}{(s^2 + \omega ^2)}+\frac{C}{s}\)
find A,B,C

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Other answers:

Ax+B ?! As+B?!
yes, As+B
i am getting C=1/w^2 , B=0 , A =-1/w^2 are u getting the same ?
\[(As+B)s + C(s^2 + \omega^2) = 1\] \[A+C = 0\]\[B =0\]\[C = \frac{1}{\omega^2}\] \[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-1}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\]Hmm.. How to do the inverse of Laplace tramsform of \(\frac{-1}{\omega^2(s^2 + \omega ^2)}\) ??
did u forget s ?
As+B
\(L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\)
w^2 is constant
Oh!!! Sorry!!!
\[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})=\frac{1}{\omega^2}-\frac{cos\omega t}{\omega^2} = \frac{1-cos\omega t}{\omega^2}\]No wonder why I couldn't get the answer!!! Thanks!! May I know if how to get the answer using the convolution one?
i was trying exactly that...using convolution
\(\int_0^t u(t-u)sin (\omega u)/\omega \:\:du\) did u reach here ?
and i am getting same answer with convolution also.
also its , \(\huge L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\)
if and after u get this : \(\int \limits_0^t u(t-u) \frac{\sin (\omega u)}{\omega} \:\:du\) u just put t-u = something and solve the integral, yo'll get the same answer.....
doubts ? i assume you are trying...
Why is it \( L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\) ???
because L[u(t)] = 1/s and NOT L[1] = 1/s ...
!!!!! How come...
do u want me to prove that ? i suggest you try first...using definition of Laplace, with u(t), u get 1/s with 1, you won't get 1/s..
Of course I try first :) Sorry to keep you waiting again!
in the definition its 0 to infinity or -infinity to infinity ??
0 to infinity. That's what I've learnt..
okk....then u get same answer for 1 and u(t) because when t>=0, u(t) = 1 then u'll get same answer for both u(t)*... and 1*....
so suit yourself...i have habit of taking u(t) because for some transforms, the definition is -infinity to infinity where u(t)and 1 are not same...
I haven't learnt \(L (u(t)) = \frac{1}{s}\)....
then take it as 1...
\(\int_0^t 1.sin (\omega (t-u))/\omega \:\:du\) u got this?
I... think I got that..
then its just a normal definite integration problem
didn't get it ?
\[f*g =\int_0^tf(\tau) g(t-\tau) d\tau\] \[\int_0^t 1\cdot \frac{sin (\omega (t-u))}{\omega} du\]\[=\frac{1}{\omega^2}cos (\omega t-\omega u)|_0^t\]\[=\frac{1}{\omega^2}[cos(0) - cos\omega t]=\frac{1-cos\omega t}{\omega^2}\] Wow!!! Thanks a ton!!!!
welcome ^_^

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