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Homework Help: Find the particular solution of the differential equation

  1. Mar 24, 2010 #1
    Find the particular solution of the differential equation

    dy/dx = (3x+42y)/7x

    satisfying the initial condition y(1) = 5.

    Attempt:
    dy/dx = 3/7 + 6y/x

    dy/dx - 6y/x = 3/7

    p(x) = -6/x
    q(x) = 3/7

    u(x) = -6 ∫(1/x)dx = -6ln|x|
    e^(u(x)) = -x^6


    1/e^(u(x)) ∫e^(u(x))q(x)dx

    = 1/(-x^6) ∫(-x^6)(3/7)dx

    = 3/(7(-x^6)) [(-x^7/7)+c]

    =3x/49 - 3c/(7x^6)


    y(1) = 5 = 3(1)/49 - 3c/(7(1)^6)

    5 = 3/49 - 3c/7

    242/49 = -3c/7

    c = -242/21


    So, what I got is

    y(x) = 3x/49 + 242/(49x^6)

    however, my professor said it's wrong. So, can anybody tell me what I did wrong? I'll be very appreciated. thanks.
     
  2. jcsd
  3. Mar 24, 2010 #2

    gabbagabbahey

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    [tex]e^u=e^{-6\ln|x|}=\left(e^{\ln|x|}\right)^{-6}=|x|^{-6}=x^{-6}\neq -x^6[/tex]

    :wink:

    It's always a good idea to check your answers before submitting them. Does this solution satisfy your original ODE?
     
  4. Mar 24, 2010 #3
    Just to be sure, the answer is

    [tex]y(x) = -3x/35 + 178/(35x^{6})[/tex]

    right?
     
    Last edited: Mar 24, 2010
  5. Mar 24, 2010 #4

    gabbagabbahey

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    You mean [tex]y(x) = -\frac{3}{35}x + \frac{178}{35}x^6[/tex]...right? If so, then yes.
     
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