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Math Help - Solve y' = y-x-1+ 1/( x-y+2)

  1. #1
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    Solve y' = y-x-1+ 1/( x-y+2)

    Solve the ODE y' = y-x-1+ 1/( x-y+2) and give the solution in implicit form.
    Last edited by mr fantastic; July 7th 2011 at 02:20 PM. Reason: Title.
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    Re: Solve the equation y

    Quote Originally Posted by ryan1573 View Post
    Solve the ODE y' = y-x-1+ 1/( x-y+2) and give the solution in implicit form.
    Can we see some of your work on this problem?
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  3. #3
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    Re: Solve the equation y

    I really dont know how to start it . it does not fit the methods I learn in the class.
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    Quote Originally Posted by ryan1573 View Post
    Solve the ODE y' = y-x-1+ 1/( x-y+2) and give the solution in implicit form.
    Is it y' = y-x-1+ 1/( x-y+2) or y' = (y-x-1+ 1)/( x-y+2)? Big difference.
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    Try letting y = x + u.
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    Quote Originally Posted by mr fantastic View Post
    Is it y' = y-x-1+ 1/( x-y+2) or y' = (y-x-1+ 1)/( x-y+2)? Big difference.
    I doubt it would be the second, as there is no point in writing -1 + 1 in the numerator...
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    dy/dx = (y-x-1)+ 1/(x-y+2) is the correct equation. it is the first one.
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    I try y= x+u, dy/dx= 1+ (du/dx) dy/dx = (u-2)-[1/ (u-2)]. integrating u', I have u= 0.5 u^(2) - 2u- ln(u-2). I dont know how to do next.
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    Re: Solve y' = y-x-1+ 1/( x-y+2)

    Quote Originally Posted by ryan1573 View Post
    I try y= x+u, dy/dx= 1+ (du/dx) dy/dx = (u-2)-[1/ (u-2)]. integrating u', I have u= 0.5 u^(2) - 2u- ln(u-2). I dont know how to do next.
    Hi ryan1573,

    You have substituted incorrectly. It should be,

    1+\frac{du}{dx}=u-1-\frac{1}{u-2}

    \Rightarrow \frac{du}{dx}=(u-2)-\frac{1}{u-2}

    \Rightarrow \frac{du}{dx}=\frac{(u-1)(u-3)}{u-2}

    I hope you can continue from here.
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