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

1. anonymous

note

2. anonymous

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

3. anonymous

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

4. anonymous

oh ok sounds good thanks

5. anonymous

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

6. anonymous

??

7. anonymous

@ragman you there?

8. 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 )

9. anonymous

@lgbasallote

10. anonymous

uhh s^2 + 22 + 5??

11. anonymous

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

12. anonymous

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

13. anonymous

yup perfect

14. hartnn

can u do partial fractions ??

15. 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

16. anonymous

where's C?

17. hartnn

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

18. anonymous

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

19. anonymous

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

20. 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

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. anonymous

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

24. anonymous

i think it will change your A value too

25. anonymous

so you best check

26. hartnn

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

27. anonymous

nope nevermind..it stays the same

28. anonymous

ok sounds good got B=5 finding A now

29. anonymous

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

30. hartnn

yup, same.Good work.

31. anonymous

A =-3?

32. anonymous

yeah that's right

33. anonymous

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

34. anonymous

everything's right

35. anonymous

and now i just plug back all the variables right

36. hartnn

yup, and use inverse laplace formulas.

37. anonymous

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

38. anonymous

uhh that should be 5/s^2

39. 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

40. anonymous

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. anonymous

uhhh wait... wrong formula for last one

42. anonymous

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

43. anonymous

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

44. anonymous

n is the exponent

45. anonymous

so it would be 2?

46. anonymous

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

47. 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 ?

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. anonymous

yup thats what i got thanks

50. hartnn

welcome :)