How i can find the general solution of the following partial differential equation:
$$\frac{{{\partial }^{2}}I}{\partial x\partial y}=\frac{1}{2}\left( \frac{\partial I}{\partial x}+\frac{\partial I}{\partial y} \right)$$ I made a guess $I\left( x,y \right)={{e}^{x+y}}$
Hint:
Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,
Then $\dfrac{\partial I}{\partial x}=\dfrac{\partial I}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial I}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial I}{\partial p}+\dfrac{\partial I}{\partial q}$
$\dfrac{\partial I}{\partial y}=\dfrac{\partial I}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial I}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{\partial I}{\partial p}-\dfrac{\partial I}{\partial q}$
$\dfrac{\partial^2I}{\partial x\partial y}=\dfrac{\partial I}{\partial y}\left(\dfrac{\partial I}{\partial p}+\dfrac{\partial I}{\partial q}\right)=\dfrac{\partial I}{\partial p}\left(\dfrac{\partial I}{\partial p}+\dfrac{\partial I}{\partial q}\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial I}{\partial q}\left(\dfrac{\partial I}{\partial p}+\dfrac{\partial I}{\partial q}\right)\dfrac{\partial q}{\partial y}=\dfrac{\partial^2I}{\partial p^2}+\dfrac{\partial^2I}{\partial p\partial q}-\dfrac{\partial^2I}{\partial p\partial q}-\dfrac{\partial^2I}{\partial q^2}=\dfrac{\partial^2I}{\partial p^2}-\dfrac{\partial^2I}{\partial q^2}$
$\therefore\dfrac{\partial^2I}{\partial p^2}-\dfrac{\partial^2I}{\partial q^2}=\dfrac{1}{2}\left(\dfrac{\partial I}{\partial p}+\dfrac{\partial I}{\partial q}+\dfrac{\partial I}{\partial p}-\dfrac{\partial I}{\partial q}\right)$
$\dfrac{\partial^2I}{\partial p^2}-\dfrac{\partial^2I}{\partial q^2}=\dfrac{\partial I}{\partial p}$
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