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Math Help - PDE u_{xxy}=1

  1. #1
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    PDE u_{xxy}=1

    Find the general solution of u_{xxy}=1

    I know how to find the general solution of u_{xx}=1

    Since this is u_{xxy}, does this mean integrate with respect to y, than x, and x again?

    Thanks.

    Would this be the general solution:

    \displaystyle \int u_{xxy}dy=\int dy\Rightarrow u_{xx}(x,y)=y+f(x)\Rightarrow\int u_{xx}(x,y)dx=\int (y+f(x))dx

    \displaystyle\Rightarrow u_{x}(x,y)=yx+f(x)+h(y)

    \displaystyle\Rightarrow \int u_{x}(x,y)dx=\int (yx+f(x)+h(y))dx

    \displaystyle\Rightarrow u(x,y)=\frac{yx^2}{2}+f(x)+xh(y)+g(y)
    Last edited by dwsmith; December 23rd 2010 at 02:11 PM.
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  2. #2
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    This seems correct. Differentiating back again for your found solution: \frac{\partial}{\partial y}\frac{\partial}{\partial x}\frac{\partial}{\partial x}u(x,y) yields 1 as well.
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