## ranyai12 Group Title Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4)) one year ago one year ago

1. ranyai12 Group Title

@Spacelimbus

2. Spacelimbus Group Title

$\Large \frac{1}{s(s+4)^2}$?

3. ranyai12 Group Title

yes

4. Spacelimbus Group Title

Partial fraction decomposition $\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}$

5. ranyai12 Group Title

so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2

6. Spacelimbus Group Title

If you multiply through you are left with $\Large 1=A(s+4)^2+B(s+4)+Cs$

7. ranyai12 Group Title

I thought you are supposed to multiply A by the values under B and C?

8. Spacelimbus Group Title

I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression

9. Spacelimbus Group Title

$\Large 1=A(s+4)^2+Bs(s+4)+Cs$

10. Spacelimbus Group Title

better now it seems

11. ranyai12 Group Title

wouldnt you get A=0, B=0 and C=-1/4

12. Spacelimbus Group Title

Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.

13. Spacelimbus Group Title

$\Large 1=(A+B)s^2 +(8A+4B+C)s+16A$

14. ranyai12 Group Title

could you please explain how you go that

15. Spacelimbus Group Title

$A+B=0 \\ 8A+4B+C=0 \\ 16A=0$

16. Spacelimbus Group Title

Yes, I multiplied the above form by s(s+4)^2 (make sure you cancel on the right hand side right) and then you factor likewise terms. When you did that, you can set both sides of the equation equal, they are equal if and only if their coefficients match.

17. Spacelimbus Group Title

$\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}$ General setup of partial fraction decomposition, multiply through by $$s(s+4)^2$$, it will disappear entirely on the left hand side of the equation and piecewise on the righthandside of the equation.

18. Spacelimbus Group Title

Then expand the righthand side and factor likewise terms, set the coefficients equal to the coefficients on the lefthand side to make a valid equation out of it LHS=RHS, in this case a lot of the coefficients have to be zero, because there is only 1 on the LHS.

19. Spacelimbus Group Title

you will end up with a 3x3 system of linear equations. $A+B=0 \\ 8A+4B+C=0 \\ 16A=1$ *corrected the last line*

20. Spacelimbus Group Title

okay so far?

21. ranyai12 Group Title

sorry im still confused on how you got the equation

22. ranyai12 Group Title

can you please show me by the work?

23. Spacelimbus Group Title

The idea of partial fraction decomposition is that you find a different way of writing the quotient, so in the end there will be an easier form for it which is especially useful in integral calculus and for LaPlace Transformations. The idea is that you completely factor the denominator and then you split it up, so each new quotient becomes its own denominator

24. Spacelimbus Group Title

|dw:1344895643951:dw|

25. ranyai12 Group Title

i understand that part but after that I get confuesd on how you distributed the variables

26. Spacelimbus Group Title

$\Large 1=(A+B)s^2 +(8A+4B+C)s+16A$ This part?

27. ranyai12 Group Title

yea, I'm not sure how you got that

28. Spacelimbus Group Title

I multiplied both sides by s(s+4)^2 so the denominator will completely disappear on the LHS, on the right hand side I have: $\Large 1=A(s+4)^2+Bs(s+4)+Cs$

29. Spacelimbus Group Title

you want to solve for A, B, C in the end, but first you distribute all the binomials out and then factor likewise powers together to make a suitable system of equations.

30. Spacelimbus Group Title

$\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs$

31. Spacelimbus Group Title

now read the equation like this, on the righthand side you have a quadratic polynomial, but on the left hand side, you can understand it like this: $0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs$

32. Spacelimbus Group Title

okay?

33. Spacelimbus Group Title

so in order that the lefthand side is equal to the righthand side, the coefficients A,B,C on the righthand side have to match the coefficients of the same powers of s on the lefthand side.

34. Spacelimbus Group Title

$0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A$

35. ranyai12 Group Title

im sorry but i dont understand the LHS

36. ranyai12 Group Title

I mean RHS

37. Spacelimbus Group Title

$0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A$

38. ranyai12 Group Title

where did the numbers come in from

39. Spacelimbus Group Title

just a correction, was mistyping it.

40. ranyai12 Group Title

and why is A added to B

41. Spacelimbus Group Title

Well the numbers come from because you expand a polynomial, namely (s+4)^2

42. Spacelimbus Group Title

And $As^2+Bs^2 = (A+B)s^2$

43. Spacelimbus Group Title

just normal algebra.

44. ranyai12 Group Title

ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c

45. Spacelimbus Group Title

$\Large 1=A(s+4)^2+Bs(s+4)+Cs$ Could you follow to this step? You see that A, B, and C gets multiplied with all that stuff. A with the first entire term, B with the second entire term, and C with the third. To match the equation we have to write the LHS different and then factor terms together, not only the first expression will have an x^2, also the second one with the Bs in front of it!

46. Spacelimbus Group Title

$\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}}$

47. ranyai12 Group Title

oh ok! thanks! btw do we have to expand?

48. Spacelimbus Group Title

yes, in this case I suggest it, there are methods called cover up etc which can be applied to get at least some of the solutions easily, but I wouldn't recommend it.

49. Spacelimbus Group Title

For instance if you set $$s=-4$$ in the equation above, the first terms will all go to zero and you are left with $\Large 1=-4C$

50. Spacelimbus Group Title

but I don't recommend doing this, because this isn't an easy partial fraction decomposition.

51. ranyai12 Group Title

so then you get As^2+8As+16A+Bs^2+4Bs+Cs

52. Spacelimbus Group Title

yes, that's what you get. If you read your equation careful enough you will see that there are multiple s^2 and s in it, you want to collect them and then match them to the left hand side, this is a valid technique.

53. ranyai12 Group Title

ok now I understand that part! thank you! after that you solve for the variable right

54. Spacelimbus Group Title

$0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A$ See, the right hand side, you multiply A+B times s^2, if A+B are 0, then the right hand side becomes 0s^2, so it is equal to the left hand side and we are happy.

55. Spacelimbus Group Title

yes, match all the coefficients with the the coefficients on the LHS

56. ranyai12 Group Title

would you get C=1

57. Spacelimbus Group Title

$A+B=0 \\ 8A+4B+C=0 \\ 16A=1$ See how I got that 3x3 System?

58. ranyai12 Group Title

isnt it 16A=1

59. Spacelimbus Group Title

Sorry, C is not zero no.

60. Spacelimbus Group Title

yes, I got it right, confused myself with my silly notes (-:

61. Spacelimbus Group Title

I get the following (concentrating hard now) $A= \frac{1}{16}, \ B=- \frac{1}{16}, \ C= -\frac{1}{4}$

62. ranyai12 Group Title

so A=1/16 B=-1/16 C=-1/4

63. Spacelimbus Group Title

yes

64. ranyai12 Group Title

awesome! wat do we do now?

65. Spacelimbus Group Title

So we have just proved that the left hand side can also be written as $\Large \frac{1}{16s}- \frac{1}{16(s+1)}-\frac{1}{4(s+1)^2}$

66. ranyai12 Group Title

isnt it s+4

67. Spacelimbus Group Title

of course it is, I just typed it wrong. (-:

68. Spacelimbus Group Title

$\Large \frac{1}{16s}- \frac{1}{16(s+4)}-\frac{1}{4(s+4)^2}$

69. ranyai12 Group Title

ok awesome! whats the next step

70. Spacelimbus Group Title

well you know that the Laplace is a linear operator, so you can factor out constants and then grab a cheat sheet to see if you can find inverse laplace transforms

71. ranyai12 Group Title

ok but the there are two constaNTS

72. Spacelimbus Group Title

For the first one $\Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)$

73. ranyai12 Group Title

F(s)=(1/16)[(1/s)-(1/(s+4)]-(1/4)[1/(s+4)^2)]

74. Spacelimbus Group Title

The inverse of 1/s = 1

75. Spacelimbus Group Title
76. ranyai12 Group Title

1/16[e^0-e^-4t]

77. Spacelimbus Group Title

yes looks good.

78. ranyai12 Group Title

what do i do about 1/(s+4)^2

79. Spacelimbus Group Title

you forgot the last one though.

80. Spacelimbus Group Title

$\Large t^ne^{at}= \frac{n!}{(s-a)^{n+1}}$ with $$n=1, 2, 3 ...$$ in our case n=1

81. Spacelimbus Group Title

got that?

82. ranyai12 Group Title

kind of

83. Spacelimbus Group Title

You have to use a cheat sheet for these kind of problems, those formulas have been derived once, (some of them get ridiculously complicated, because of the long integration by parts problems)

84. ranyai12 Group Title

for 1/(s+4)-->n!/(s^(n+1))

85. Spacelimbus Group Title

I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

86. ranyai12 Group Title

i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n

87. Spacelimbus Group Title

$\Large e^{at}= \frac{1}{s-a}$ In our case $$a=-4$$

88. Spacelimbus Group Title

non related formulas.

89. ranyai12 Group Title

but we have s+4 not s-4

90. ranyai12 Group Title

oh nvr mind i confused t^n and e^at

91. Spacelimbus Group Title

$\Large t^1e^{-4t}= \frac{1!}{(s+4)^{2}}$

92. Spacelimbus Group Title

this is what I get for my substitution.

93. ranyai12 Group Title

ok so I ended up with (1/16)(1)-(1/16)[1/(s+4)]-(1/4)[1/(s+4)^5)]

94. Spacelimbus Group Title

No, that's not what you are doing.

95. Spacelimbus Group Title

You will use this for differential equations, so you compute the inverse, the inverse is a function.

96. ranyai12 Group Title

what do you do?

97. Spacelimbus Group Title

$\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t}$

98. Spacelimbus Group Title

That is what I get, if I didn't make any careless mistakes.

99. ranyai12 Group Title

so you dont use the transform equation

100. Spacelimbus Group Title

look at the cheat sheet again, right you have the Laplace Transform, and on the other side you have the inverse, you want to compute the inverse.

101. ranyai12 Group Title

oh ok

102. Spacelimbus Group Title

$\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace- \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....$

103. Spacelimbus Group Title

see that? I didn't carry it out entirely, but that above is a Laplace Transform, whenever you have something of such a form, you want to compute its inverse.

104. Spacelimbus Group Title

The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.

105. Spacelimbus Group Title

It basically turns differential equation problems into Algebra problems.

106. ranyai12 Group Title

oh i get it thank you sooo much!

107. Spacelimbus Group Title

welcome

108. ranyai12 Group Title

:)