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Math Help - Solve (x^2)*u_x + (y^2)*u_y = (x+y)u

  1. #1
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    Talking Solve (x^2)*u_x + (y^2)*u_y = (x+y)u

    Obtain the solution of

    (x^2)*u_x + (y^2)*u_y = (x+y)u

    (Hint: to obtain a general solution you may need to use the fact that dy/y^2 = (dx - dy)/ (x^2 - y^2)
    Last edited by mr fantastic; May 19th 2010 at 01:41 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by Waikato View Post
    Obtain the solution of

    (x^2)*u_x + (y^2)*u_y = (x+y)u

    (Hint: to obtain a general solution you may need to use the fact that dy/y^2 = (dx - dy)/ (x^2 - y^2)
    Characteristic equation

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

    so

    1)\;\;\;\frac{dx}{x^2} = \frac{dy}{y^2}\;\;\;\text{giving}\;\;\; I_1 = \frac{1}{x} - \frac{1}{y}

    2)\;\;\;\frac{dx}{x^2} - \frac{dy}{y^2} = \frac{du}{(x+y)u} or \frac{d(x-y)}{x-y} = \frac{du}{u} giving I_2 = \frac{u}{x+y}

    Solution: u = (x+y) f \left( \frac{1}{x} - \frac{1}{y}\right)
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  3. #3
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    Smile Differential Equation

    Are you able to solve this question?

    Consider u_t + (v.grad)u = 0 , where grad = upside down triangle, and v is a vector, dotted with the grad symbol.

    where v = (-ax, ay) represents 2D stagnation point flow and u = u(x,y,t). Obtain a solution of the Cauchy problem given that u = u0(x,y) at t = 0

    (u0 is read as "u naught")
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