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Particular solutions - DE

  1. Apr 15, 2010 #1
    Find the particular solution of the differential equation

    [tex]xy' + 4y = -20xcos(x^5)[/tex]

    satisfying the initial condition y(pi) = 0.


    Solution

    [tex]y' + 4(y/x) = -20cos(x^5)[/tex]

    p(x) = [tex]4/x[/tex]
    q(x) = [tex]-20cos(x^5)[/tex]

    [tex]u(x) = int(4/x) = 4lnx[/tex]
    [tex]e^{u(x)} = x^4[/tex]

    [tex]1/e^{u(x)} int(e^{u(x)}*q(x)dx[/tex]
    [tex]1/(x^4) int(x^4*(-20cos(x^5)))dx[/tex]

    u substitution
    [tex]u = x^5[/tex]
    [tex]du = 5x^4dx[/tex]

    [tex]1/(x^4) int(x^4*(-20cos(u)))(du/(5x^4))[/tex]
    [tex]-4/(x^4) int(cosu)du[/tex]
    [tex]-4/(x^4) (sin(x^5)+c)[/tex]
    [tex](-4sin(x^5)/(x^4))-((4c)/(x^4))[/tex]

    [tex]y(pi) = 0 = (-4sin(pi^5)/(pi^4))-((4c)/(pi^4))[/tex]
    [tex]c = -sin(pi^5)[/tex]

    Final
    [tex](-4sin(x^5)/(x^4))+((4sin(pi^5))/(x^4))[/tex]

    So this is what I got, but it's a wrong answer. Can anybody tell me what I do wrong in this problem?
     
  2. jcsd
  3. Apr 15, 2010 #2

    tiny-tim

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    Hi shiri! :smile:

    (have a pi: π and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
    Haven't you lost the x in -20xcosx5? :redface:
     
  4. Apr 15, 2010 #3
    well I divide both sides by x, so...
     
  5. Apr 15, 2010 #4

    tiny-tim

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    Sorry, you're right :redface:

    xy' + 4y = -20xcosx5

    multiply by x3

    x4y' + 4x3y = -20x4cosx5 :wink:

    So (x4y)' = -4(sinx5)'

    So x4y = 4(sinπ5 - sinx5)

    So y = 4(sinπ5 - sinx5)/x4


    hmm :confused: … the condition y(π) = 0 is a bit strange for cos(x5) …

    is it possible the question should be (cosx)5 ?
     
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