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note

'note' sorry this is my first time using this what do you mean by that?

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?

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 ??

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 \[\implies As(s+1) + B(s+1) + Cs^2 = s^2 + 2s + 5\]

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

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:
\[\huge \mathcal L \{ t^2 \} = \frac{2!}{s^{2+1}} \implies \frac{2}{s^3}\]

yup thats what i got thanks

welcome :)