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Math Help - Diff Eq w/ Initial Value problem

  1. #1
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    Diff Eq w/ Initial Value problem

    Here is the problem: {(dy/dx)+(x(y+1)^2)/(y(x+1)^2)=0, y(0)=0}

    I have separated and integrated using partial fractions, which gives:
    ln|y+1| + 1/(y+1) = ln|x+1| + 1/(x+1) + Constant

    Maybe I've been at this for too long, but I can't figure out how to solve for y to determine my constant...
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  2. #2
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    Re: Diff Eq w/ Initial Value problem

    Quote Originally Posted by shane18 View Post
    Here is the problem: {(dy/dx)+(x(y+1)^2)/(y(x+1)^2)=0, y(0)=0}

    I have separated and integrated using partial fractions, which gives:
    ln|y+1| + 1/(y+1) = ln|x+1| + 1/(x+1) + Constant

    Maybe I've been at this for too long, but I can't figure out how to solve for y to determine my constant...
    First up, you have a sign error on the RHS...the whole RHS is negative. Second, you can't solve for y explicitly. Finally, to get the constant simply plug x = 0 and y = 0 into your expression.

    -Dan
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  3. #3
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    Re: Diff Eq w/ Initial Value problem

    Thanks! Sorry for the simple error. Correcting the signs gives:

    ln|y+1| + 1/(y+1) = - ln|x+1| - 1/(x+1) + C

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

    for y(0)=0, ln|1| + 1 + ln|1| + 1 = C

    1 + 1 = C = 2

    so, ln|y+1| + 1/(y+1) + ln|x+1| + 1/(x+1) = 2

    Thanks again, Dan.

    -Shane
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  4. #4
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    Re: Diff Eq w/ Initial Value problem

    The simplest expression of the resut is on the form of the equation already given.
    The closed form of y(x) requires a special function : the Lambert W function :
    y =-1 -1/W(X) where X = -|x+1|*exp(1/(x+1))/eČ
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