## ragman 3 years ago find inverse laplace transform of F(s)= (s^2+2s+5/(s^2(s+1))

1. Skaematik

note

2. ragman

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

3. Skaematik

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

4. ragman

oh ok sounds good thanks

5. lgbasallote

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

6. lgbasallote

??

7. lgbasallote

@ragman you there?

8. ragman

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

9. ragman

@lgbasallote

10. lgbasallote

uhh s^2 + 22 + 5??

11. ragman

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

12. lgbasallote

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

13. ragman

yup perfect

14. hartnn

can u do partial fractions ??

15. ragman

@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

16. lgbasallote

where's C?

17. hartnn

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

18. lgbasallote

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

19. lgbasallote

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

20. ragman

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

21. hartnn

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

22. hartnn

so i think u can get C now, right?

23. ragman

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

24. lgbasallote

i think it will change your A value too

25. lgbasallote

so you best check

26. hartnn

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

27. lgbasallote

nope nevermind..it stays the same

28. ragman

ok sounds good got B=5 finding A now

29. ragman

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

30. hartnn

yup, same.Good work.

31. lgbasallote

A =-3?

32. lgbasallote

yeah that's right

33. ragman

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

34. lgbasallote

everything's right

35. ragman

and now i just plug back all the variables right

36. hartnn

yup, and use inverse laplace formulas.

37. lgbasallote

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

38. lgbasallote

uhh that should be 5/s^2

39. ragman

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

40. 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?

41. lgbasallote

uhhh wait... wrong formula for last one

42. lgbasallote

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

43. ragman

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

44. lgbasallote

n is the exponent

45. ragman

so it would be 2?

46. lgbasallote

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

47. ragman

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 ?

48. 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?

49. ragman

yup thats what i got thanks

50. hartnn

welcome :)