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Sine Integral Function in differential equation

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Si(x)=[tex]\int(sint/t)dt[/tex] from 0 to x
    integrand is 1 at t=0

    express the solutiony(x) of the initial value problem x^3y'+2x^2y=10sinx, y(0)=0 in terms of Si(x)

    2. Relevant equations



    3. The attempt at a solution
    y'+2/xy=10x^-3sinx
    multiply by x^2
    x^2y'+2xy=10x^-2sinx
    (x^2y)'=10x^-2sinx
    Now I get stuck because I feel like this can't be integrated and I don't think i can just insert Si(x) because I need to account for 10x^-2.
     
  2. jcsd
  3. Nov 6, 2009 #2

    LCKurtz

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    Divide your original equation by x:

    [tex]x^2y'' + 2xy' = 10 \frac {\sin x}{x}[/tex]

    which will work for you. You just have an arithmetic mistake.
     
  4. Nov 7, 2009 #3
    So I have x^2y'+2xy=10sinx/x
    (x^2y)'=10sinx/x
    x^2y=10Si(x)
    y=10x^(-2)Si(x)
     
  5. Nov 7, 2009 #4

    LCKurtz

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    Your answer may be correct, but you need to show more explicitly the integral and how you used the initial condition y(0) = 1. Your second step isn't clear because the Si function is a definite integral.
     
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