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Kelumptus

  • 3 years ago

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

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  1. Kelumptus
    • 3 years ago
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    \[\frac{ d ^{2}y }{ dx ^{2} }-6\frac{ dy }{ dx }+8y=8e^{4x}\]

  2. Kelumptus
    • 3 years ago
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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.

  3. exraven
    • 3 years ago
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    use the undetermined coefficients method

  4. Kelumptus
    • 3 years ago
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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?

  5. Kelumptus
    • 3 years ago
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    What is the undetermined coefficients method?

  6. Kelumptus
    • 3 years ago
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    Oh, i believe that is what i am doing

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

  8. Kelumptus
    • 3 years ago
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    thats what i tried, but i get C = 0

  9. Kelumptus
    • 3 years ago
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    I believe 'k' is actually 4 in this instance.

  10. Kelumptus
    • 3 years ago
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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?

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

  12. Kelumptus
    • 3 years ago
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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.

  13. exraven
    • 3 years ago
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    try\[y = Cxe^{4x}\]it works

  14. Kelumptus
    • 3 years ago
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    Just trying that now...

  15. Kelumptus
    • 3 years ago
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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 =)

  16. Kelumptus
    • 3 years ago
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    Oh and thanks heaps... =)

  17. exraven
    • 3 years ago
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    lol, you are welcome

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