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

1. anonymous

@Spacelimbus

2. anonymous

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

3. anonymous

yes

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

better now it seems

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

could you please explain how you go that

15. anonymous

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

16. anonymous

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. anonymous

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

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. anonymous

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. anonymous

okay so far?

21. anonymous

sorry im still confused on how you got the equation

22. anonymous

can you please show me by the work?

23. anonymous

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. anonymous

|dw:1344895643951:dw|

25. anonymous

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

26. anonymous

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

27. anonymous

yea, I'm not sure how you got that

28. anonymous

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. anonymous

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. anonymous

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

31. anonymous

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. anonymous

okay?

33. anonymous

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. anonymous

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

35. anonymous

im sorry but i dont understand the LHS

36. anonymous

I mean RHS

37. anonymous

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

38. anonymous

where did the numbers come in from

39. anonymous

just a correction, was mistyping it.

40. anonymous

and why is A added to B

41. anonymous

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

42. anonymous

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

43. anonymous

just normal algebra.

44. anonymous

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

45. anonymous

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

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

47. anonymous

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

48. anonymous

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. anonymous

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. anonymous

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

51. anonymous

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

52. anonymous

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. anonymous

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

54. anonymous

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

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

56. anonymous

would you get C=1

57. anonymous

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

58. anonymous

isnt it 16A=1

59. anonymous

Sorry, C is not zero no.

60. anonymous

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

61. anonymous

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

62. anonymous

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

63. anonymous

yes

64. anonymous

awesome! wat do we do now?

65. anonymous

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. anonymous

isnt it s+4

67. anonymous

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

68. anonymous

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

69. anonymous

ok awesome! whats the next step

70. anonymous

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. anonymous

ok but the there are two constaNTS

72. anonymous

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

73. anonymous

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

74. anonymous

The inverse of 1/s = 1

75. anonymous
76. anonymous

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

77. anonymous

yes looks good.

78. anonymous

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

79. anonymous

you forgot the last one though.

80. anonymous

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

81. anonymous

got that?

82. anonymous

kind of

83. anonymous

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. anonymous

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

85. anonymous

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

86. anonymous

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. anonymous

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

88. anonymous

non related formulas.

89. anonymous

but we have s+4 not s-4

90. anonymous

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

91. anonymous

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

92. anonymous

this is what I get for my substitution.

93. anonymous

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

94. anonymous

No, that's not what you are doing.

95. anonymous

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

96. anonymous

what do you do?

97. anonymous

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

98. anonymous

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

99. anonymous

so you dont use the transform equation

100. anonymous

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. anonymous

oh ok

102. anonymous

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

103. anonymous

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. anonymous

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

105. anonymous

It basically turns differential equation problems into Algebra problems.

106. anonymous

oh i get it thank you sooo much!

107. anonymous

welcome

108. anonymous

:)