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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

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  1. ranyai12 Group Title
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    @Spacelimbus

    • 2 years ago
  2. Spacelimbus Group Title
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    \[\Large \frac{1}{s(s+4)^2} \]?

    • 2 years ago
  3. ranyai12 Group Title
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    yes

    • 2 years ago
  4. Spacelimbus Group Title
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    Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]

    • 2 years ago
  5. ranyai12 Group Title
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    so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2

    • 2 years ago
  6. Spacelimbus Group Title
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    If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]

    • 2 years ago
  7. ranyai12 Group Title
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    I thought you are supposed to multiply A by the values under B and C?

    • 2 years ago
  8. Spacelimbus Group Title
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    I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression

    • 2 years ago
  9. Spacelimbus Group Title
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    \[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]

    • 2 years ago
  10. Spacelimbus Group Title
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    better now it seems

    • 2 years ago
  11. ranyai12 Group Title
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    wouldnt you get A=0, B=0 and C=-1/4

    • 2 years ago
  12. Spacelimbus Group Title
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    Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.

    • 2 years ago
  13. Spacelimbus Group Title
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    \[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]

    • 2 years ago
  14. ranyai12 Group Title
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    could you please explain how you go that

    • 2 years ago
  15. Spacelimbus Group Title
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    \[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]

    • 2 years ago
  16. Spacelimbus Group Title
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    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.

    • 2 years ago
  17. Spacelimbus Group Title
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    \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] 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.

    • 2 years ago
  18. Spacelimbus Group Title
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    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.

    • 2 years ago
  19. Spacelimbus Group Title
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    you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*

    • 2 years ago
  20. Spacelimbus Group Title
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    okay so far?

    • 2 years ago
  21. ranyai12 Group Title
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    sorry im still confused on how you got the equation

    • 2 years ago
  22. ranyai12 Group Title
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    can you please show me by the work?

    • 2 years ago
  23. Spacelimbus Group Title
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    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

    • 2 years ago
  24. Spacelimbus Group Title
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    |dw:1344895643951:dw|

    • 2 years ago
  25. ranyai12 Group Title
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    i understand that part but after that I get confuesd on how you distributed the variables

    • 2 years ago
  26. Spacelimbus Group Title
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    \[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?

    • 2 years ago
  27. ranyai12 Group Title
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    yea, I'm not sure how you got that

    • 2 years ago
  28. Spacelimbus Group Title
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    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: \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]

    • 2 years ago
  29. Spacelimbus Group Title
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    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.

    • 2 years ago
  30. Spacelimbus Group Title
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    \[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

    • 2 years ago
  31. Spacelimbus Group Title
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    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: \[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

    • 2 years ago
  32. Spacelimbus Group Title
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    okay?

    • 2 years ago
  33. Spacelimbus Group Title
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    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.

    • 2 years ago
  34. Spacelimbus Group Title
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    \[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]

    • 2 years ago
  35. ranyai12 Group Title
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    im sorry but i dont understand the LHS

    • 2 years ago
  36. ranyai12 Group Title
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    I mean RHS

    • 2 years ago
  37. Spacelimbus Group Title
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    \[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]

    • 2 years ago
  38. ranyai12 Group Title
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    where did the numbers come in from

    • 2 years ago
  39. Spacelimbus Group Title
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    just a correction, was mistyping it.

    • 2 years ago
  40. ranyai12 Group Title
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    and why is A added to B

    • 2 years ago
  41. Spacelimbus Group Title
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    Well the numbers come from because you expand a polynomial, namely (s+4)^2

    • 2 years ago
  42. Spacelimbus Group Title
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    And \[ As^2+Bs^2 = (A+B)s^2\]

    • 2 years ago
  43. Spacelimbus Group Title
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    just normal algebra.

    • 2 years ago
  44. ranyai12 Group Title
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    ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c

    • 2 years ago
  45. Spacelimbus Group Title
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    \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] 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!

    • 2 years ago
  46. Spacelimbus Group Title
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    \[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]

    • 2 years ago
  47. ranyai12 Group Title
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    oh ok! thanks! btw do we have to expand?

    • 2 years ago
  48. Spacelimbus Group Title
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    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.

    • 2 years ago
  49. Spacelimbus Group Title
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    For instance if you set \( s=-4\) in the equation above, the first terms will all go to zero and you are left with \[ \Large 1=-4C\]

    • 2 years ago
  50. Spacelimbus Group Title
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    but I don't recommend doing this, because this isn't an easy partial fraction decomposition.

    • 2 years ago
  51. ranyai12 Group Title
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    so then you get As^2+8As+16A+Bs^2+4Bs+Cs

    • 2 years ago
  52. Spacelimbus Group Title
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    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.

    • 2 years ago
  53. ranyai12 Group Title
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    ok now I understand that part! thank you! after that you solve for the variable right

    • 2 years ago
  54. Spacelimbus Group Title
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    \[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \] 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.

    • 2 years ago
  55. Spacelimbus Group Title
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    yes, match all the coefficients with the the coefficients on the LHS

    • 2 years ago
  56. ranyai12 Group Title
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    would you get C=1

    • 2 years ago
  57. Spacelimbus Group Title
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    \[ A+B=0 \\ 8A+4B+C=0 \\ 16A=1\] See how I got that 3x3 System?

    • 2 years ago
  58. ranyai12 Group Title
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    isnt it 16A=1

    • 2 years ago
  59. Spacelimbus Group Title
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    Sorry, C is not zero no.

    • 2 years ago
  60. Spacelimbus Group Title
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    yes, I got it right, confused myself with my silly notes (-:

    • 2 years ago
  61. Spacelimbus Group Title
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    I get the following (concentrating hard now) \[A= \frac{1}{16}, \ B=- \frac{1}{16}, \ C= -\frac{1}{4} \]

    • 2 years ago
  62. ranyai12 Group Title
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    so A=1/16 B=-1/16 C=-1/4

    • 2 years ago
  63. Spacelimbus Group Title
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    yes

    • 2 years ago
  64. ranyai12 Group Title
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    awesome! wat do we do now?

    • 2 years ago
  65. Spacelimbus Group Title
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    So we have just proved that the left hand side can also be written as \[\Large \frac{1}{16s}- \frac{1}{16(s+1)}-\frac{1}{4(s+1)^2}\]

    • 2 years ago
  66. ranyai12 Group Title
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    isnt it s+4

    • 2 years ago
  67. Spacelimbus Group Title
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    of course it is, I just typed it wrong. (-:

    • 2 years ago
  68. Spacelimbus Group Title
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    \[\Large \frac{1}{16s}- \frac{1}{16(s+4)}-\frac{1}{4(s+4)^2} \]

    • 2 years ago
  69. ranyai12 Group Title
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    ok awesome! whats the next step

    • 2 years ago
  70. Spacelimbus Group Title
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    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

    • 2 years ago
  71. ranyai12 Group Title
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    ok but the there are two constaNTS

    • 2 years ago
  72. Spacelimbus Group Title
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    For the first one \[ \Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)\]

    • 2 years ago
  73. ranyai12 Group Title
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    F(s)=(1/16)[(1/s)-(1/(s+4)]-(1/4)[1/(s+4)^2)]

    • 2 years ago
  74. Spacelimbus Group Title
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    The inverse of 1/s = 1

    • 2 years ago
  75. Spacelimbus Group Title
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    http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

    • 2 years ago
  76. ranyai12 Group Title
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    1/16[e^0-e^-4t]

    • 2 years ago
  77. Spacelimbus Group Title
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    yes looks good.

    • 2 years ago
  78. ranyai12 Group Title
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    what do i do about 1/(s+4)^2

    • 2 years ago
  79. Spacelimbus Group Title
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    you forgot the last one though.

    • 2 years ago
  80. Spacelimbus Group Title
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    \[\Large t^ne^{at}= \frac{n!}{(s-a)^{n+1}} \] with \(n=1, 2, 3 ...\) in our case n=1

    • 2 years ago
  81. Spacelimbus Group Title
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    got that?

    • 2 years ago
  82. ranyai12 Group Title
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    kind of

    • 2 years ago
  83. Spacelimbus Group Title
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    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)

    • 2 years ago
  84. ranyai12 Group Title
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    for 1/(s+4)-->n!/(s^(n+1))

    • 2 years ago
  85. Spacelimbus Group Title
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    I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

    • 2 years ago
  86. ranyai12 Group Title
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    i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n

    • 2 years ago
  87. Spacelimbus Group Title
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    \[\Large e^{at}= \frac{1}{s-a} \] In our case \( a=-4\)

    • 2 years ago
  88. Spacelimbus Group Title
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    non related formulas.

    • 2 years ago
  89. ranyai12 Group Title
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    but we have s+4 not s-4

    • 2 years ago
  90. ranyai12 Group Title
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    oh nvr mind i confused t^n and e^at

    • 2 years ago
  91. Spacelimbus Group Title
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    \[ \Large t^1e^{-4t}= \frac{1!}{(s+4)^{2}} \]

    • 2 years ago
  92. Spacelimbus Group Title
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    this is what I get for my substitution.

    • 2 years ago
  93. ranyai12 Group Title
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    ok so I ended up with (1/16)(1)-(1/16)[1/(s+4)]-(1/4)[1/(s+4)^5)]

    • 2 years ago
  94. Spacelimbus Group Title
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    No, that's not what you are doing.

    • 2 years ago
  95. Spacelimbus Group Title
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    You will use this for differential equations, so you compute the inverse, the inverse is a function.

    • 2 years ago
  96. ranyai12 Group Title
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    what do you do?

    • 2 years ago
  97. Spacelimbus Group Title
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    \[\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t} \]

    • 2 years ago
  98. Spacelimbus Group Title
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    That is what I get, if I didn't make any careless mistakes.

    • 2 years ago
  99. ranyai12 Group Title
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    so you dont use the transform equation

    • 2 years ago
  100. Spacelimbus Group Title
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    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.

    • 2 years ago
  101. ranyai12 Group Title
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    oh ok

    • 2 years ago
  102. Spacelimbus Group Title
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    \[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace- \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]

    • 2 years ago
  103. Spacelimbus Group Title
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    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.

    • 2 years ago
  104. Spacelimbus Group Title
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    The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.

    • 2 years ago
  105. Spacelimbus Group Title
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    It basically turns differential equation problems into Algebra problems.

    • 2 years ago
  106. ranyai12 Group Title
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    oh i get it thank you sooo much!

    • 2 years ago
  107. Spacelimbus Group Title
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    welcome

    • 2 years ago
  108. ranyai12 Group Title
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    :)

    • 2 years ago
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