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find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))

Mathematics
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note
'note' sorry this is my first time using this what do you mean by that?
Im just bookmarking this question so that I can return to it later. :D Id want to know the answer too

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oh ok sounds good thanks
your question is \[\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]
??
@ragman you there?
hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )
uhh s^2 + 22 + 5??
dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)
so.. \[\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]
yup perfect
can u do partial fractions ??
@hartnn yeah thats what i was doing but for some reason i could figure out the other two variable maybe im solving it incorrectly but i got -3 for A and 5 for B
where's C?
those are correct, so u need only C right ? where u took C above s+1
your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]
so you have \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]
oh ok so thats where i messed up i thought that where C is it should be (Cs+D) so i was trying to solve for both of those variables
nopes only C, if it were s^2+1 u would have taken Cx+D
so i think u can get C now, right?
oh i see now and yeah i believe i can get C now trying it right now
i think it will change your A value too
so you best check
great tell us what u get C as....we all will verify that. A is correct.
nope nevermind..it stays the same
ok sounds good got B=5 finding A now
A=-3 and C=4 is that what you guys got?
yup, same.Good work.
A =-3?
yeah that's right
ok cool i was about to write out how i got it lol
everything's right
and now i just plug back all the variables right
yup, and use inverse laplace formulas.
yep find \[\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}\]
uhh that should be 5/s^2
ok sorry im just learning this but how do you go about doing that im trying to find the L^-1 formula but my notes arent really helping
here's a hint: \[\huge \mathcal L \{ a \} = \frac as\] \[\huge \mathcal L \{ t^n\} = \frac{n!}{s^{n+1}}\] \[\huge \mathcal L \{e^{at} \} = \frac{a}{s -a}\] does that help?
uhhh wait... wrong formula for last one
\[\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}\] better
yes thats very helpful but im getting confused with what n should be in the middle equations
n is the exponent
so it would be 2?
for example: \[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]
thats whats a lil confusing to me because on the denominator it would have to be n=1 so that it would equal s^2 but on the numerator 1! equals 1 right ?
yes thats true, so u will have 5t^1= 5t ....isn't it ? u can verify that u should get final answer as 5t-3+4e^(-t) did u get that?
yup thats what i got thanks
welcome :)

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