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anonymous

  • one year ago

Hi fellas! Differential Equation SOS!!!!

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  1. anonymous
    • one year ago
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    \[y''+6y'-7y=x^2e^x+7e^(-7x)+e^(-7x)sinx\]

  2. anonymous
    • one year ago
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    I need to write down the general and the private solution of the DE @Loser66

  3. anonymous
    • one year ago
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    @freckles

  4. anonymous
    • one year ago
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    and THANKS A LOT guys I truly appreciate that!

  5. anonymous
    • one year ago
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    genuinely

  6. anonymous
    • one year ago
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    I know how to get to the general solution, but the private one is killing me

  7. anonymous
    • one year ago
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    \[yh=C1e ^{-7x}+C2e ^{x}+yp\]

  8. anonymous
    • one year ago
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    can you help me out with the solution of the private solution?

  9. Loser66
    • one year ago
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    The homogenious part give you the solutions, right? what are they?

  10. Loser66
    • one year ago
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    \(y_h= C_1 e^{-7x}+C_2e^x\), right?

  11. anonymous
    • one year ago
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    yes

  12. Loser66
    • one year ago
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    hence \(e^x\) is available. Now private part, the first term is \(x^2e^x\) , that is we have \(y_{p1}= (Ax^2+Bx+C)*x(e^x)\) \(y'_{p1}= you~~ do\) \(y"_{p1} =.....\) then apply to the equation y"+6y'+7y to solve for \(y_{p1}\)

  13. anonymous
    • one year ago
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    not sure I'm following, let me have a look though

  14. anonymous
    • one year ago
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    wow it's so long

  15. anonymous
    • one year ago
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    a couple of minutes I need to sort it out

  16. Loser66
    • one year ago
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    The second term is \(7e^{-7x}\) , I do it for you as a sample We need find \(y_{p2}= Dte^{-7x}\\y'_{p2}= -7De^{-7x}\\y"_{p2}= 49 De^{-7x}\) Now combine \(y"_{p2}+6y'_{p2}-7y_{p2}= 49De^{-7x} -42De^{-7x}+7Dte^{-7x}= 7e^{-7x}\)

  17. Loser66
    • one year ago
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    from that you have \(7De^{-7x}+7Dxe^{-7x}= 7e^{-7x}\) Make a comparison, you have D =1 hence the solution for that is \(y_{p2}= xe^{-7x}\) Sorry for t at the beginning, it is x.

  18. anonymous
    • one year ago
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    sorry but what does D refer to?

  19. Loser66
    • one year ago
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    Do the same with the last term\(e^{-x} sinx\)

  20. Loser66
    • one year ago
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    |dw:1437757184146:dw|

  21. Loser66
    • one year ago
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    and you apply the ODE from the left to solve for D

  22. anonymous
    • one year ago
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    wow thanks a lot @Loser66 !

  23. anonymous
    • one year ago
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    but please just for clarification

  24. anonymous
    • one year ago
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    D is the function?

  25. Loser66
    • one year ago
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    D is a constant, like A, B, C for \(y_{p1}\). I use D to make sure that you are not confused and differentiate them.

  26. Loser66
    • one year ago
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    when finding private solution for \(e^x sinx\), you need use E, F, G.... to know that those constants are not the constants from \(y_{p1}~~or~~y_{p2}\)

  27. anonymous
    • one year ago
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    thanks a lot!

  28. anonymous
    • one year ago
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    I hope I'll nail this now

  29. Loser66
    • one year ago
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    ok, good luck. gtg.

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