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Math Help - Help with Homogeneous/Seperable equation

  1. #1
    cYn
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    Help with Homogeneous/Seperable equation

    OK so here is my problem:

    \frac{dy}{dx} = \frac{(x+y)^2}{2x}

    My steps:

    y = vx
    y' = v + v'x
    y' = \frac{(x+y)^2}{2x}
    v + v'x = \frac{(x+y)^2}{2x}
    v + \frac{dv}{dx}x = \frac{(x+y)^2}{2x}
    v + \frac{dv}{dx}x = \frac{(x+vx)^2}{2x}
    v + \frac{dv}{dx}x = \frac{x^2 + 2vx^2 +v^2x^2}{2x}
    v + \frac{dv}{dx}x = \frac{x^2}{2x} + \frac{2vx^2}{2x} + \frac{v^2x^2}{2x}
    v + \frac{dv}{dx}x = \frac{x}{2} + vx + \frac{v^2x}{2}

    If I divide both sides by x I get:

    \frac{v}{x} + \frac{dv}{dx} = \frac{1}{2} + v + \frac{v^2}{2}
    \frac{v}{x} + dv = (\frac{1}{2} + v + \frac{v^2}{2})dx

    This is the part that I'm having trouble with. I haven't done homogeneous/seperable in a long time. Can someone help me with the next step? And possibly a solution?
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  2. #2
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    Quote Originally Posted by cYn View Post
    OK so here is my problem:

    \frac{dy}{dx} = \frac{(x+y)^2}{2x}

    My steps:

    y = vx
    y' = v + v'x
    y' = \frac{(x+y)^2}{2x}
    v + v'x = \frac{(x+y)^2}{2x}
    v + \frac{dv}{dx}x = \frac{(x+y)^2}{2x}
    v + \frac{dv}{dx}x = \frac{(x+vx)^2}{2x}
    v + \frac{dv}{dx}x = \frac{x^2 + 2vx^2 +v^2x^2}{2x}
    v + \frac{dv}{dx}x = \frac{x^2}{2x} + \frac{2vx^2}{2x} + \frac{v^2x^2}{2x}
    v + \frac{dv}{dx}x = \frac{x}{2} + vx + \frac{v^2x}{2}

    If I divide both sides by x I get:

    \frac{v}{x} + \frac{dv}{dx} = \frac{1}{2} + v + \frac{v^2}{2}
    \frac{v}{x} + dv = (\frac{1}{2} + v + \frac{v^2}{2})dx

    This is the part that I'm having trouble with. I haven't done homogeneous/seperable in a long time. Can someone help me with the next step? And possibly a solution?
    If you expand out the right hand side you will get an equation of the form.

    \frac{dy}{dx}=P(x)+Q(x)y+R(x)y^2

    This is know as Ricatti's equation see here

    Riccati equation - Wikipedia, the free encyclopedia

    The page has a method to solve it.

    Good luck.
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