
PDE u_{xxy}=1
Find the general solution of $\displaystyle u_{xxy}=1$
I know how to find the general solution of $\displaystyle u_{xx}=1$
Since this is $\displaystyle u_{xxy}$, does this mean integrate with respect to y, than x, and x again?
Thanks.
Would this be the general solution:
$\displaystyle \displaystyle \int u_{xxy}dy=\int dy\Rightarrow u_{xx}(x,y)=y+f(x)\Rightarrow\int u_{xx}(x,y)dx=\int (y+f(x))dx$
$\displaystyle \displaystyle\Rightarrow u_{x}(x,y)=yx+f(x)+h(y)$
$\displaystyle \displaystyle\Rightarrow \int u_{x}(x,y)dx=\int (yx+f(x)+h(y))dx$
$\displaystyle \displaystyle\Rightarrow u(x,y)=\frac{yx^2}{2}+f(x)+xh(y)+g(y)$

This seems correct. Differentiating back again for your found solution: $\displaystyle \frac{\partial}{\partial y}\frac{\partial}{\partial x}\frac{\partial}{\partial x}u(x,y)$ yields 1 as well.