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Math Help - Separation of variables..

  1. #1
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    Separation of variables..

    I get stuck at the integration step

    use the method of separation of variables to find the solution of the differential equation:


    dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1
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  2. #2
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    Quote Originally Posted by Raidan View Post
    I get stuck at the integration step

    use the method of separation of variables to find the solution of the differential equation:


    dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1
    So \int \frac{dy}{\sqrt{2y-y^2}} = \int \frac{dx}{x+1}
    then complete the square in the square root term

     <br />
\sqrt{2y-y^2} = \sqrt{1-(y-1)^2}<br />
Let u = y-1 and you should be able to integrate.
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    could someone help me to solve this question?
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  4. #4
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    Quote Originally Posted by Raidan View Post
    could someone help me to solve this question?
    Danny pointed it out:

    u = y-1 so du = dy

    Therefore your equation becomes

    \int{\frac{du}{\sqrt{1-u^2}}} = \int{\frac{dx}{x+1}}

    The left side is a standard integral and is equal to arcsin(u) +A
    wikipedia link (and don't forget to put y-1 wherever u appears)

    For the right side recall that \int{\frac{f'(x)}{f(x)}} = ln|f(x)| + B

    Where A and B are constants and so C = B-A which is also a constant (your constant of integration)
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