Question:

Xd2y/dx2 - 2dy/dx + x =0?

by Guest63499  |  earlier

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dy/dx = v

please help.

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  1. EDIT: My mistake...apparently this is a nonlinear equation.

    xy'' - 2y' = -x

    xv' - 2v = -x

    v' - (2/x)v = -1, x≠0

    This is now a linear 1st-order diffeq.  It can be written in a generic form:

    v' + p(x)v = q(x)

    The general solution to this is:

    v(x) = [∫u(x)q(x)dx + C] / u(x)

    Where

    u(x) = e^(∫p(x)dx)

    For your problem:

    p(x) = -2/x

    q(x) = -1...(I don't know if this works)

    u(x) = e^(-2∫dx/x) = e^(-2(ln(x)+A))

    = e^(-2ln(x))∙e^(-2A)

    = e^(-2A)∙[e^(ln(x))]^(-2)

    = e^(-2A) / x²

    ∫u(x)q(x)dx = -e^(-2A) ∫dx/x²

    = (e^(-2A) / x) + B

    v(x) = [∫u(x)q(x)dx + C] / u(x)

    = [(e^(-2A) / x) + B + C] / [e^(-2A) / x²]

    = [(e^(-2A) + (B+C)x) / x] / [e^(-2A) / x²]

    = x∙[(e^(-2A) + (B+C)x)] / e^(-2A)

    = x∙[1 + x∙(B+C)e^(2A)]

    = x + Kx²

    Where K = (B+C)e^(2A)

    Now integrate this to find y(x):

    y(x) = ∫v(x)dx = x²/2 + Kx³/3 + C

    Check:

    y'(x) = x + Kx²

    y''(x) = 1 + 2Kx

    xy'' - 2y' + x = x(1+2Kx) - 2(x+Kx²) + x

    = x + 2Kx² - 2x - 2Kx² + x

    = 0


  2. with your sub the problem reduces to dv / dx - [2/x]  v = -1 with integrating factor 1/x².....problem is now easily solvable...you do the work

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