PDE u_{xxy}=1

**dwsmith** · December 23rd 2010, 01:41 PM

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:

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

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

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

$\displaystyle\Rightarrow u(x,y)=\frac{yx^2}{2}+f(x)+xh(y)+g(y)$

**brouwer** · December 24th 2010, 04:27 AM

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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PDE u_{xxy}=1