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Math Help - 2nd order differential equation - variation of parameters

  1. #1
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    2nd order differential equation - variation of parameters

    Hi,

    I would like to know how to solve the following:

    Solve the differential equation using the method of varaition of parameters.

    y" - 3y' + 2y = 1/(1+e^-x)

    What I have done till now (not really able to continue):

    - solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
    -y1 = e^x & y2 = e^2x
    -yp = u1e^x + u2e^2x

    However, I am not sure if I have u2 correctly:

    u1 = ln (1+e^-x)
    u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)

    can someone please check with me and help me solve the following please.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by zxcv View Post
    Hi,

    I would like to know how to solve the following:

    Solve the differential equation using the method of varaition of parameters.

    y" - 3y' + 2y = 1/(1+e^-x)

    What I have done till now (not really able to continue):

    - solution to y" - 3y' + 2y = 0 is c1e^x+c2e^2x
    -y1 = e^x & y2 = e^2x
    -yp = u1e^x + u2e^2x

    However, I am not sure if I have u2 correctly:

    u1 = ln (1+e^-x)
    u2 = -(1+e^-x) + ln (1+e^-x) where I have integral of (e^-2x)/(1+e^-x)

    can someone please check with me and help me solve the following please.

    Thanks in advance!
    When I did it, I got your u1 but for u2

    u_2 = -e^{-x} + \ln | 1 + e^{-x}|

    but I don't think it matters in the end b/c when you expand

    u_1 e^x + u_2 e^{2x}

    the first term in u2 (either yours or mine) can be absorbed into the complementary solution.
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