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Math Help - A differential equation

  1. #1
    Tau
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    A differential equation

    Hello

    xy' + y^2 = 1, y(1) = \frac{1}{3}

    My thinking:

    Rearrange the equation and you will get this:

    \frac{1}{1 - y^{2}}\delta y = \frac{1}{x} \delta x

    Integrate it...

    \frac{1}{2}(\ln|1 - y| + \ln|1 + y|) = \ln|x| + C

    I think that the absolutes can be ignored, in that case I get:

    y = \pm \sqrt{C - x^2}

    But the correct f(x) is something very different.

    Thanks for your help.
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  2. #2
    Math Engineering Student
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    Quote Originally Posted by Tau View Post

    But the correct f(x) is something very different.
    Probably 'cause it was set C=\ln|C| for being constant.
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  3. #3
    Tau
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    The "answer" to this problem is:

    y = \frac{2x^{2} - 1}{2x^{2} + 1}

    According to the book.

    Don't ask me what it means or in what context it is a "answer".
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  4. #4
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    Quote Originally Posted by Tau View Post
    Hello

    xy' + y^2 = 1, y(1) = \frac{1}{3}

    My thinking:

    Rearrange the equation and you will get this:

    \frac{1}{1 - y^{2}}\delta y = \frac{1}{x} \delta x

    Integrate it...

    \frac{1}{2}({\color{red} - } \ln|1 - y| + \ln|1 + y|) = \ln|x| + C Mr F says: The mistake is in this line. I've added the correction (in red).

    I think that the absolutes can be ignored, in that case I get:

    y = \pm \sqrt{C - x^2}

    But the correct f(x) is something very different.

    Thanks for your help.
    After the correction it follows that

    \ln \left| \frac{1 + y}{1 - y} \right| = \ln (x^2) + 2C


    \Rightarrow \frac{1 + y}{1 - y} = B x^2

    where B = e^{2C} is also completely arbitrary.

    At this stage I'd substitute y(1) = \frac{1}{3} to get the value of B. Then I'd substitute the value for B and make y the subject.

    The book's answer is correct.
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