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Math Help - General Solution of a differential solution

  1. #1
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    General Solution of a differential solution

    y*y'+x=(x^2+y^2)^.5
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  2. #2
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    Quote Originally Posted by Ragingamer View Post
    y*y'+x=(x^2+y^2)^.5
    Your first thought should have been to divide both sides by x:

    \displaystyle \frac{y}{x} \frac{dy}{dx} + 1 = \sqrt{1 + \left( \frac{y}{x} \right)^2}.

    Now apply the usual technique for such DE's.
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  3. #3
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    I did that and got to y'=(x/y)((1+(y/x)^2)^.5-1) and then used a substitution of v=y/x.
    Reducing it to y'=(1/v)((1+v^2)^.5-1) and differentiating y=vx to get y'=v+x*dv/dx.

    Setting up an integral of v/((1+v^2)^.5-1-v^2)*dv=dx/x But I have attempted many times with this integral and have not gotten the correct solution.
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  4. #4
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    Quote Originally Posted by Ragingamer View Post
    I did that and got to y'=(x/y)((1+(y/x)^2)^.5-1) and then used a substitution of v=y/x.
    Reducing it to y'=(1/v)((1+v^2)^.5-1) and differentiating y=vx to get y'=v+x*dv/dx.

    Setting up an integral of v/((1+v^2)^.5-1-v^2)*dv=dx/x But I have attempted many times with this integral and have not gotten the correct solution.
    If people posted what they could do and said where they were stuck in the very first post, they would get the help they really needed a lot faster and other people wouldn't waste time doing things that had already been done.

    See here: integrate x/(Sqrt[x^2 + 1] - (x^2 + 1)) - Wolfram|Alpha (click on show steps).
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  5. #5
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     u^2=x^2+y^2
    uu'=x+yy'
    The initial equation becomes
    u'=1
     u=x+C
    y^2=2xC+C^2
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