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Math Help - Solve y^2 + xy dy/dx = 0 by integrating factor?

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    Solve y^2 + xy dy/dx = 0 by integrating factor?

    a) Given (dN/dx - dM/dy) = Q, which is a function of only y. Prove that

    e^ \int Q dy is an integrating factor of the equation
    M dx + N dy = 0, where M and N are functions of x and y.

    I have already answered this.

    b) Solve
    y^2 + x y dy/dx = 0
    by finding an integrating factor to make it exact.

    This is how I did:
    M = y^2
    N = x y
    dN/dx = y
    dM/dy = 2y

    dN/dx - dM/dy = y - 2y = -y
    (dN/dx - dM/dy)/M = -y/y^2 = -1/y, which is a function of y
    Therefore, from part a,
    an integrating factor is

    e^[ \int -1/y dy] = e^[-ln |y|]

    Multiplying both sides of the given equation by the integrating factor,
    y^2 e^[-ln |y|] + x y e^[-ln |y|] dy/dx = 0
    d/dx (x y^2 e^[-ln|y|]) = 0
    x y^2 e^[-ln|y|] = C ( a constant)
    y^2 e^[-ln|y|] = C/xIs this correct?


    c) Solve the above differential equation by finding am integrating factor to make it separable.
    I think the integrating factor in part b also makes it separable. So can I use the same answer?
    But I think the equation can be solved much easier as follows:

    y^2 + x y dy/dx = 0
    y (y + x dy/dx) = 0
    y = 0 Or
    y + x dy/dx = 0
    Or y = -x dy/dx
    Or, dy/y = -1/x dx
    Or,dy/dy + dx/x = 0
    ln y + ln x = C (a constant)
    ln(xy) = C
    xy = e^C = K, where K is a constant
    y = K/x, if X is not zero

    If x is zero, then from the given equation
    y^2 + 0 = 0
    Or, y = 0

    Is this correct? Can I use this in the answer?
    Last edited by visharad; February 3rd 2013 at 10:58 PM.
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