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Math Help - Nonlinear Equation

  1. #1
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    Nonlinear Equation

    Solve the given DE by using the substitution u = y'
    (y + 1)y" = (y')^2
    (y + 1)u' = u^2

    u' = u*du/dy

    (y + 1)u*du/dy = u^2
    (y + 1)du/dy = u
    1/(y + 1) dy = (1/u) du
    ln(y + 1) = ln(u) + C
    y + 1 = Cu
    u = (y + 1)/C = dy/dx
    dx = C/(y + 1) dy
    x + C2 = C*ln(y + 1)

    I'd appreciate it if someone could look over this and tell me if I did it right, thanks in advance.
    Last edited by Naples; April 25th 2011 at 04:02 AM.
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  2. #2
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    Are you sure the substitution isn't meant to be u = y', not u = y''?
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  3. #3
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    Yeah, I just accidentally had that messed up in the first line but I did the problem by substituting u = y'
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  4. #4
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    (y + 1)(d^2y/dx^2) = (dy/dx)^2

    (y + 1)(d^2y/dx^2) = (dy/dx)(dy/dx)

    After the substitution you will have

    (y + 1)(du/dx) = u(dy/dx)

    (1/u)(du/dx) = [1/(y+1)](dy/dx)

    Now integrating both sides with respect to x gives

    int[(1/u)(du/dx)dx] = int{[1/(y+1)](dy/dx)dx}

    int[(1/u)du] = int{[1/(y+1)]dy}

    ln|u| + C_1 = ln|y + 1| + C_2

    ln|u| - ln|y + 1| = C where C = C_2 - C_1

    ln|u/(y + 1)| = C

    ln|(dy/dx)/(y + 1)| = C

    [1/(y + 1)](dy/dx) = A where A = (+-)e^C

    Integrating both sides with respect to x gives

    int{[1/(y+1)](dy/dx)dx} = int(A dx)

    int{[1/(y + 1)]dy} = Ax + C_3

    ln|y + 1| + C_4 = Ax + C_3

    ln|y + 1| = Ax + B where B = C_3 - C_4

    |y + 1| = e^(Ax + B)

    |y + 1| = (e^B)e^(Ax)

    y + 1 = De^(Ax) where D = (+-)e^B

    y = De^(Ax) - 1.
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  5. #5
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    Kudos. I have never seen a solution method like that before.

    -Dan
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