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Math Help - Variation of parameters Diff EQ

  1. #1
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    Variation of parameters Diff EQ

    y" + 2y' + y = 4(e^-x)*ln(x)

    Here's my work:

    1. r^2 + 2r + 1 = 0
    2. (r + 1)^2 = 0; r= -1, -1
    3. yc = c1e^(-x) + c2xe^(-x)
    4. yp = f1e^(-x) + f2xe^(-x)

    So:

    f1'e^(-x) + f2'xe^(-x) = 0
    -f1'e^(-x) + f2'(e^(-x) - xe^(-x)) = 4(e^-x)*ln(x)

    This system simplifies to:

    f1' + f2'x = 0
    -f1' + f2'(1 - x) = 4*ln(x)

    Then I used Cramer's Rule to solve for f1' and f2'. I end up with:

    f1' = 4x*ln(x)
    f2' = 4*ln(x)

    Then I integrate these to find f1 and f2 and plug back into yp = f1e^(-x) + f2xe^(-x). Then the answer is y = yc + yp. However, I don't get the answer in the back of the book. My yc is correct; so my error has to lie in my calculation of the yp. Are my f1' and f2' correct? Any suggestions?

    Thanks a bunch,

    Jim
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    I used maple to check your work.
    I agree you are very close but have lost a minus sign.

    I hope you can follow the maple screen shot below.

    f_1'=-4x\ln(x)

    I hope this helps.
    Good luck.

    Variation of parameters Diff EQ-capture.jpg
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  3. #3
    Junior Member
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    Thanks Empty Set! Are you a math teacher?

    JN
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