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RolyPoly

  • 3 years ago

(again)Solve by Laplace transform: \[\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0\] y(0)=6, y'(0) = 0

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  1. RolyPoly
    • 3 years ago
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    \[\frac{d^2y}{dx^2} + a \frac{dy}{dx} - 2a^2y=0\]\[sY'-0 +a(sY-6)-2a^2Y=0\]\[s(sY-6) +a(sY-6)-2a^2Y=0\]\[(s^2+as-2a^2)Y=6a+6s\]\[Y=\frac{6a+6s}{(s+2a)(s-a)}\]So far so good? Or...??

  2. experimentX
    • 3 years ago
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    |dw:1355844666804:dw|

  3. RolyPoly
    • 3 years ago
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    y(0)=6, y'(0) = 0

  4. experimentX
    • 3 years ago
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    sorry ... i didn't see, looks like the transform of d^2y/dx^2 is incorrect.

  5. experimentX
    • 3 years ago
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    http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf no 37

  6. RolyPoly
    • 3 years ago
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    I guess so.. \[L(y'') = s^2Y - s(6) -0=s^2Y-6s\]What's wrong?

  7. experimentX
    • 3 years ago
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    \[ sY'-0 +a(sY-6)-2a^2Y=0 \] this should be like \[ s^2 Y - 6s + a(sY - 0) -2a^2 Y = 0 \]

  8. RolyPoly
    • 3 years ago
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    \[L(\frac{dy}{dx}) = sY - f(0)\]Right?

  9. experimentX
    • 3 years ago
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    yep!!

  10. experimentX
    • 3 years ago
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    woops!! sorry I misread ... i put opposite!!

  11. experimentX
    • 3 years ago
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    \[ s^2 Y - 0 - 6 + a(sY - 6) - 2a^2 Y = 0\]

  12. RolyPoly
    • 3 years ago
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    It's okay! But what's wrong with my workings?

  13. RolyPoly
    • 3 years ago
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    \[L(y'') = s^2Y - s(6) -0=s^2Y-6s\]

  14. experimentX
    • 3 years ago
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    the second step!! lol ... i again put it wrong!!

  15. RolyPoly
    • 3 years ago
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    The second step is right. You just get the third immediately.. \[L(y'') = sL(y')-f'(0) = s^2 Y-sf(0) - f'(0)\] Anyway.. I still get the third step..?!

  16. experimentX
    • 3 years ago
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    Y' ... well i thought you differentiated it.

  17. experimentX
    • 3 years ago
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    the answer should be according to mathematica \[ \left\{\left\{y[x]\to 2 e^{-2 a x} \left(1+2 e^{3 a x}\right)\right\}\right\} \]

  18. RolyPoly
    • 3 years ago
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    Sorry :(

  19. RolyPoly
    • 3 years ago
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    Yes... But I can't get it :(

  20. experimentX
    • 3 years ago
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    the final answer is right ... you get \[ F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}\] looks lkie you factorized it.

  21. experimentX
    • 3 years ago
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    now just take the inverse transform.

  22. RolyPoly
    • 3 years ago
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    Factorize it -> partial fraction -> inverse Right?

  23. experimentX
    • 3 years ago
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    yeah!! that's the way

  24. RolyPoly
    • 3 years ago
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    \[F(s) = \frac{6(a+s)}{s^2 + as - 2a^2}=\frac{6(a+s)}{(a+2a)(s-a)}\]I suppose..

  25. RolyPoly
    • 3 years ago
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    s+2a typo

  26. RolyPoly
    • 3 years ago
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    \[Y = \frac{6a+6s}{(s+2a)(s-a)}=\frac{A}{s+2a}+\frac{B}{s-a}\] A(s-a) + B(s+2a) =6a+6s A+B = 6 2B-A = 6 => B=4 , A=2 \[Y =\frac{2}{s+2a}+\frac{4}{s-a}\]\[y=2e^{-2at}+4e^{at}\]Why couldn't I get this answer... :\ Thanks!!

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