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Kelumptus Group Title

I am having a problem in arriving at the correct answer to a second order inhomogeneous differential equation (below):

  • one year ago
  • one year ago

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  1. Kelumptus Group Title
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    \[\frac{ d ^{2}y }{ dx ^{2} }-6\frac{ dy }{ dx }+8y=8e^{4x}\]

    • one year ago
  2. Kelumptus Group Title
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    I know that the complimentary function is \[Ae^{4x}+Be^{2x}\] however i seem to arrive at an incorrect answer for the particular integral.

    • one year ago
  3. exraven Group Title
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    use the undetermined coefficients method

    • one year ago
  4. Kelumptus Group Title
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    For the particular integral i get \[16Ce^{4x}-24Ce^{4x}+8Ce^{4x}=8e^{4x}\] Based on derivatives of the general function \[Ce^{4x}\] I therefore get \[e^{4x}(16C-24C+8C)=8e^{4x}\] Therefore \[0=8\] Which means there is no solution. The answer in my book for the particular integral however is \[4e^{4x}\]. Where did i go wrong?

    • one year ago
  5. Kelumptus Group Title
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    What is the undetermined coefficients method?

    • one year ago
  6. Kelumptus Group Title
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    Oh, i believe that is what i am doing

    • one year ago
  7. exraven Group Title
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    your should assume the particular\[y = Ce^{kx}\]then you solve for C and k

    • one year ago
  8. Kelumptus Group Title
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    thats what i tried, but i get C = 0

    • one year ago
  9. Kelumptus Group Title
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    I believe 'k' is actually 4 in this instance.

    • one year ago
  10. Kelumptus Group Title
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    Can you see where i went wrong in working through the answer? Or do you think it is an error in my book?

    • one year ago
  11. exraven Group Title
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    wait I am working on it too, if Ce^(kx) won't work, try Cxe^(kx)

    • one year ago
  12. Kelumptus Group Title
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    Oh... i'll try that only i thought you only did that when there was already a \[e^{kx}\] on the LHS.

    • one year ago
  13. exraven Group Title
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    try\[y = Cxe^{4x}\]it works

    • one year ago
  14. Kelumptus Group Title
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    Just trying that now...

    • one year ago
  15. Kelumptus Group Title
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    Awesome, as you said, that works. I didn't realize that you could use that method when there is no solution of a coefficient. I thought it was only when an element of the general solution appeared in the LHS. Good to know, i've learned a new trick =)

    • one year ago
  16. Kelumptus Group Title
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    Oh and thanks heaps... =)

    • one year ago
  17. exraven Group Title
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    lol, you are welcome

    • one year ago
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