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Math Help - separable differential equation

  1. #1
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    separable differential equation

    I can't seem to get the solution in my text book, the problem is:

    xy' + y = y^{2}

    I began by reorganizing to:

    \frac{1}{y(y-1)}dy = \frac{1}{x}dx

    expanding the y term and integrating:

    \int{\frac{1}{y}dy} - \int{\frac{1}{y-1}dy} = \int{\frac{1}{x}dx}

    giving the following:

    ln|y| - ln|y-1| = ln|x| + K

    where K is the combined constant of integration from both sides
    now when I take the exponential, I think I have to raise e to the entire
    right side and that I can't do it term by term which simplifies to a division
    using log properties and then canceling out the e and ln

    |\frac{y}{y-1}| = |x|e^{K}

    now I was thinking I could drop all the absolute values if I throw a +\-
    in front of the remaining exponential, so we'll say C=\pm e^{K}

    \frac{y}{y-1} = Cx

    solving for y yields:

    y = \frac{-Cx}{1-Cx}

    however my textbook gives:

    y = \frac{1}{1-Cx}

    Please let me know if you can see my mistake, thanks
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  2. #2
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    Quote Originally Posted by stevedave View Post
    I can't seem to get the solution in my text book, the problem is:

    xy' + y = y^{2}

    I began by reorganizing to:

    \frac{1}{y(y-1)}dy = \frac{1}{x}dx

    expanding the y term and integrating:

    \int{\frac{1}{y}dy} - \int{\frac{1}{y-1}dy} = \int{\frac{1}{x}dx}

    Wrong sign! Left side should be \color{red}\int{\frac{1}{y-1}dy} - \int{\frac{1}{y}dy}
    ..
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  3. #3
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    Quote Originally Posted by stevedave View Post
    I can't seem to get the solution in my text book, the problem is:

    xy' + y = y^{2}

    I began by reorganizing to:

    \frac{1}{y(y-1)}dy = \frac{1}{x}dx

    expanding the y term and integrating:

    \int{\frac{1}{y}dy} - \int{\frac{1}{y-1}dy} = \int{\frac{1}{x}dx}
    \frac{1}{y}- \frac{1}{y-1}= \frac{y-1}{y(y-1)}- \frac{y}{y(y-1)}= -\frac{1}{y(y-1)}. You have the sign wrong. You should have
    \int{\frac{1}{y-1}dy} - \int{\frac{1}{y}dy} = \int{\frac{1}{x}dx}
    giving the following:

    ln|y| - ln|y-1| = ln|x| + K

    where K is the combined constant of integration from both sides
    now when I take the exponential, I think I have to raise e to the entire
    right side and that I can't do it term by term which simplifies to a division
    using log properties and then canceling out the e and ln

    |\frac{y}{y-1}| = |x|e^{K}

    now I was thinking I could drop all the absolute values if I throw a +\-
    in front of the remaining exponential, so we'll say C=\pm e^{K}

    \frac{y}{y-1} = Cx
    The sign error I pointed out above means this should be \frac{y-1}{y}= Cx so y-1= Cxy, Cxy- y= (Cx-1)y= -1. That gives what your textbook says.

    solving for y yields:

    y = \frac{-Cx}{1-Cx}

    however my textbook gives:

    y = \frac{1}{1-Cx}

    Please let me know if you can see my mistake, thanks
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  4. #4
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    bah, I always do stuff like that . I'll rework it and see what happens. Thanks
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  5. #5
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    skeeter's Avatar
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    correction ...

    \frac{1}{y(y-1)} = -\frac{1}{y} + \frac{1}{y-1}
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