Question

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

Answer 1

note

Answer 2

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

Answer 3

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

Answer 4

oh ok sounds good thanks

Answer 5

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

Answer 6

??

Answer 7

@ragman you there?

Answer 8

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

Answer 9

@lgbasallote

Answer 10

uhh s^2 + 22 + 5??

Answer 11

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

Answer 12

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

Answer 13

yup perfect

Answer 14

can u do partial fractions ??

Answer 15

@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

Answer 16

where's C?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

so i think u can get C now, right?

Answer 23

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

Answer 24

i think it will change your A value too

Answer 25

so you best check

Answer 26

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

Answer 27

nope nevermind..it stays the same

Answer 28

ok sounds good got B=5 finding A now

Answer 29

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

Answer 30

yup, same.Good work.

Answer 31

A =-3?

Answer 32

yeah that's right

Answer 33

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

Answer 34

everything's right

Answer 35

and now i just plug back all the variables right

Answer 36

yup, and use inverse laplace formulas.

Answer 37

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

Answer 38

uhh that should be 5/s^2

Answer 39

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

Answer 40

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?

Answer 41

uhhh wait... wrong formula for last one

Answer 42

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

Answer 43

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

Answer 44

n is the exponent

Answer 45

so it would be 2?

Answer 46

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

Answer 47

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 ?

Answer 48

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?

Answer 49

yup thats what i got thanks

Answer 50

welcome :)