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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

note

- anonymous

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

- anonymous

Im just bookmarking this question so that I can return to it later. :D
Id want to know the answer too

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

oh ok sounds good thanks

- lgbasallote

your question is \[\huge \mathcal L ^{-1} \left \{ s^2 + 2s + \frac 5{s^2(s+1)}\right \}\]

- lgbasallote

??

- lgbasallote

@ragman you there?

- anonymous

hey sorry about the late reply but my question is actually (s^2+22+5) / (s^2(s+1) sorry i forgot the other )

- anonymous

@lgbasallote

- lgbasallote

uhh s^2 + 22 + 5??

- anonymous

dang i keep doing typos 2s not 22 so yeah its (s^2+2s+5)

- lgbasallote

so..
\[\huge \mathcal L ^{-1} \left \{ \frac{s^2 + 2s + 5}{s^2(s+1)} \right \}\]

- anonymous

yup perfect

- hartnn

can u do partial fractions ??

- anonymous

@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

- lgbasallote

where's C?

- hartnn

those are correct, so u need only C right ?
where u took C above s+1

- lgbasallote

your P.F. should be \[\huge \frac As + \frac B{s^2} + \frac C{s+1}\]

- lgbasallote

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

- anonymous

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

- hartnn

nopes only C, if it were s^2+1 u would have taken Cx+D

- hartnn

so i think u can get C now, right?

- anonymous

oh i see now and yeah i believe i can get C now trying it right now

- lgbasallote

i think it will change your A value too

- lgbasallote

so you best check

- hartnn

great tell us what u get C as....we all will verify that.
A is correct.

- lgbasallote

nope nevermind..it stays the same

- anonymous

ok sounds good got B=5 finding A now

- anonymous

A=-3 and C=4 is that what you guys got?

- hartnn

yup, same.Good work.

- lgbasallote

A =-3?

- lgbasallote

yeah that's right

- anonymous

ok cool i was about to write out how i got it lol

- lgbasallote

everything's right

- anonymous

and now i just plug back all the variables right

- hartnn

yup, and use inverse laplace formulas.

- lgbasallote

yep find \[\huge \mathcal L ^{-1} \left \{ -\frac 3s + \frac 5s + \frac 4{s+1} \right \}\]

- lgbasallote

uhh that should be 5/s^2

- anonymous

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

- lgbasallote

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?

- lgbasallote

uhhh wait... wrong formula for last one

- lgbasallote

\[\huge \mathcal L \{e^{at} \} = \frac{1}{s-a}\]
better

- anonymous

yes thats very helpful but im getting confused with what n should be in the middle equations

- lgbasallote

n is the exponent

- anonymous

so it would be 2?

- lgbasallote

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

- anonymous

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 ?

- hartnn

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?

- anonymous

yup thats what i got thanks

- hartnn

welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.