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Math Help - Semi Linear PDE

  1. #1
    Senior Member bugatti79's Avatar
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    Semi Linear PDE

    <br />
\displaystyle -2xy \frac{\partial u}{\partial x} +4x \frac{\partial u}{\partial y}=4xy-yu<br />
.

    u(x,0)=2 \sqrt x

    <br />
\displaystyle \frac{dy}{dx}=\frac{4x}{-2xy}=\frac{-2}{y} \implies \frac{y^2}{2}=-2x+k<br />

    <br />
\displaystyle \frac{du}{dx}=\frac{4xy-yu}{-2xy}=-2+\frac{u}{2x}<br />
. Solve using an integrating factor where p=\frac{-1}{2x} giving

    <br />
\displaystyle u x^{\frac{-1}{2}}= \int -2 x^{\frac{-1}{2}} dx=-4x^{\frac{1}{2}}+f(\frac{y^2}{2}+2x)<br />
Therefore

    <br />
\displaystyle u(x,y)=-4x+x^{\frac{1}{2}} f(\frac{y^2}{2}+2x)<br />

    Using intial conditions gives \displaystyle 2 \sqrt x=-4x+x^{\frac{1}{2}}f(2x) . Let \displaystyle t=2x \implies \frac{t}{2}=f(t) therefore

    <br />
\displaystyle f(t)=\{2\sqrt {\frac{t}{2}}+2t}\}({\frac{t}{2})^\frac{-1}{2}<br />
. Therefore

    <br />
\displaystyle u(x,y)=-4x+x^{\frac{1}{2}} \{2\sqrt {\frac{\frac{y^2}{2}+2x}{2}}+2(\frac{y^2}{2}+2x)\}  (({\frac{y^2}{2}+2x)/2)^{\frac{-1}{2}}<br />

    Could anyone check this is right so far? the above equation doesnt look correct.
    Thanks
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  2. #2
    MHF Contributor
    Jester's Avatar
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    It's correct but why not simple your f(t) a little. Such as

    f(t) = 2 + \sqrt{8t}.
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