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suzi20

y" + 3y' + 2y = 6 y(0)=0 y'(0) = 2 with laplace

  • 2 years ago
  • 2 years ago

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  1. elecengineer
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    y'' = sy^2 - sy(0) -y'(0)

    • 2 years ago
  2. elecengineer
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    y' = sy - y(0)

    • 2 years ago
  3. elecengineer
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    (sy^2 -2) +3(sy) +2y = 6

    • 2 years ago
  4. elecengineer
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    solve the quadratic

    • 2 years ago
  5. suzi20
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    yes, im stuck in integral (6+2s) / s(s+2)(s+1)

    • 2 years ago
  6. math1234
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    First write everything in terms of the laplace transform. THen solve the equation by converting back.

    • 2 years ago
  7. elecengineer
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    no, so you got y to be that

    • 2 years ago
  8. math1234
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    Use particla fractions.

    • 2 years ago
  9. elecengineer
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    then you must use partial fractions

    • 2 years ago
  10. suzi20
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    6+2s = As + B(s+2)+C(s+1) then?

    • 2 years ago
  11. elecengineer
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    = 3/s -4/(s+1) +1/(s+2)

    • 2 years ago
  12. elecengineer
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    I created and went straight to wolframa for the partial fractions

    • 2 years ago
  13. elecengineer
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    cheated

    • 2 years ago
  14. elecengineer
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    partial fractions are very standard

    • 2 years ago
  15. elecengineer
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    then you convert them back

    • 2 years ago
  16. elecengineer
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    the 3/s goes to 3 from memory

    • 2 years ago
  17. elecengineer
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    and the other two are time shifted exponentials

    • 2 years ago
  18. math1234
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    Laplace is all about matching and partial fractions, at least in solving simple ODE systems.

    • 2 years ago
  19. suzi20
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    6+2s = As + B(s+2)+C(s+1) s=-1 --> 4 = -A+B s=-2 --> 2 = -2A-C for the last s what number should i choose?

    • 2 years ago
  20. elecengineer
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    1/(s-a) = e^(at) ( I googled this lol )

    • 2 years ago
  21. elecengineer
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    s=0 lol

    • 2 years ago
  22. elecengineer
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    remember you can pick any value for s, just that some values will make the simultaneous eqns alot easierto solve

    • 2 years ago
  23. suzi20
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    6+2s = As + B(s+2)+C(s+1) s=-1 --> 4 = -A+B s=-2 --> 2 = -2A-C s=0 --> 6 = 2B+C ---------------------- 4+A = B 6 = 2(4+A)+C 6 = 8 +2A+C elimination -2 = 2A+C 2 = -2A -C infinity?

    • 2 years ago
  24. elecengineer
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    y= 3-4e^(-t) + e^(-2t)

    • 2 years ago
  25. elecengineer
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    you make no sense at all lol

    • 2 years ago
  26. elecengineer
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    this is why people need to pay attention in high school and first year uni maths course , so they absolutely hammer in the basics

    • 2 years ago
  27. suzi20
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    could u help me?

    • 2 years ago
  28. elecengineer
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    its just simultaneous eqns , takes for ever, you need to set up a matrix etc.

    • 2 years ago
  29. elecengineer
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    it will take me like 10mins to type it up , I aint doing it lol

    • 2 years ago
  30. suzi20
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    matrix? really?

    • 2 years ago
  31. AnwarA
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    I can see that you reach the point \(Y(s)=\frac{2s+6}{s(s+1)(s+2)}\). Now we should use partial fractions to write the expression in a form that can be easily to find its inverse laplace transform. That's \[{2s+6 \over s(s+1)(s+2)}={a \over s}+{b \over s+1}+{c \over s+2}\]. Multiplying both sides by \(s(s+1)(s+2)\) gives: \[2s+6=a(s+1)(s+2)+bs(s+2)+cs(s+1)\] Plugging \(s=0\) gives \(2a=6 \implies a=3\); \(s=-1\) gives \(-b=4 \implies b=-4\), and \(s=-2\) gives \(2c=2 \implies c=1\). So, \(Y(s)=3\frac{1}{s}-\frac{4}{s+1}+\frac{1}{s+2}\). Hence \(y(t)=3-4e^{-t}+e^{-2t}\).

    • 2 years ago
  32. AnwarA
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    Hello suzi!! Does the answer make sense to you?

    • 2 years ago
  33. suzi20
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    thank you anwara

    • 2 years ago
  34. AnwarA
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    You're welcome!

    • 2 years ago
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