## ranyai12 Group Title Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4)) 2 years ago 2 years ago

1. ranyai12

@Spacelimbus

2. Spacelimbus

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

3. ranyai12

yes

4. Spacelimbus

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

5. ranyai12

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

6. Spacelimbus

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

7. ranyai12

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

8. Spacelimbus

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

9. Spacelimbus

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

10. Spacelimbus

better now it seems

11. ranyai12

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

12. Spacelimbus

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

13. Spacelimbus

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

14. ranyai12

could you please explain how you go that

15. Spacelimbus

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

16. Spacelimbus

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

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

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

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

okay so far?

21. ranyai12

sorry im still confused on how you got the equation

22. ranyai12

can you please show me by the work?

23. Spacelimbus

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

|dw:1344895643951:dw|

25. ranyai12

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

26. Spacelimbus

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

27. ranyai12

yea, I'm not sure how you got that

28. Spacelimbus

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

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

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

31. Spacelimbus

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

okay?

33. Spacelimbus

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

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

35. ranyai12

im sorry but i dont understand the LHS

36. ranyai12

I mean RHS

37. Spacelimbus

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

38. ranyai12

where did the numbers come in from

39. Spacelimbus

just a correction, was mistyping it.

40. ranyai12

and why is A added to B

41. Spacelimbus

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

42. Spacelimbus

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

43. Spacelimbus

just normal algebra.

44. ranyai12

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

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

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

47. ranyai12

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

48. Spacelimbus

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

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

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

51. ranyai12

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

52. Spacelimbus

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

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

54. Spacelimbus

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

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

56. ranyai12

would you get C=1

57. Spacelimbus

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

58. ranyai12

isnt it 16A=1

59. Spacelimbus

Sorry, C is not zero no.

60. Spacelimbus

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

61. Spacelimbus

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

62. ranyai12

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

63. Spacelimbus

yes

64. ranyai12

awesome! wat do we do now?

65. Spacelimbus

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

isnt it s+4

67. Spacelimbus

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

68. Spacelimbus

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

69. ranyai12

ok awesome! whats the next step

70. Spacelimbus

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

ok but the there are two constaNTS

72. Spacelimbus

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

73. ranyai12

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

74. Spacelimbus

The inverse of 1/s = 1

75. Spacelimbus
76. ranyai12

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

77. Spacelimbus

yes looks good.

78. ranyai12

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

79. Spacelimbus

you forgot the last one though.

80. Spacelimbus

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

81. Spacelimbus

got that?

82. ranyai12

kind of

83. Spacelimbus

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

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

85. Spacelimbus

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

86. ranyai12

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

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

88. Spacelimbus

non related formulas.

89. ranyai12

but we have s+4 not s-4

90. ranyai12

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

91. Spacelimbus

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

92. Spacelimbus

this is what I get for my substitution.

93. ranyai12

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

94. Spacelimbus

No, that's not what you are doing.

95. Spacelimbus

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

96. ranyai12

what do you do?

97. Spacelimbus

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

98. Spacelimbus

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

99. ranyai12

so you dont use the transform equation

100. Spacelimbus

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

oh ok

102. Spacelimbus

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

103. Spacelimbus

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

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

105. Spacelimbus

It basically turns differential equation problems into Algebra problems.

106. ranyai12

oh i get it thank you sooo much!

107. Spacelimbus

welcome

108. ranyai12

:)