Find a solution to the PDE for $u(x,y)$ Solve the PDE for $u(x,y)$ $$\frac{\partial^2 u}{\partial x \, \partial y} = 0$$
I was thinking to integrated both sides in respect to $x$ first to get $$x= c(x)$$ then i will have $$c(x)-x=0$$ then i will integrate in respect to y but i think this wrong because it does not making any sense to me. 
 A: The functions $a(x,y)$ for which
$$ \frac{\partial}{\partial x} a(x,y) = 0 $$
everywhere are precisely those that are constant in $x$, i.e. functions of $y$ alone. If we set $a = \partial u/\partial y$, the original equation becomes after one integration
$$ \frac{\partial u}{\partial y} = A(y), $$
for $A$ an arbitrary function of $y$. Integrating with respect to $y$ (and holding $x$ constant, in the same way we get
$$ u(x,y) = B(y)+C(x), $$
where $B(y) = \int_{y_0}^y A(t) \, dt  $. It is easy to check that this function does satisfy the original equation. 
A: If you integrate both sides w.r.t. x, you should get   $$\frac{\partial u}{ \partial y} = c(y)$$ for some arbitrary function c. Integrating again  w.r.t. y now gives $$u=a(y)+b(x)$$ where $b$ is an arbitrary function of $x$ and $a$ is the function you get when integrating $c$ which is again arbitrary since $c$ was. So the general solution is simply the sum of 2 differentiable functions, one only dependent on $x$ and the other one only dependent on $y$. For e.g $u(x,y)=x+e^y$ is one of the infinite possible solutions.
A: After one integration w.r.t. $x$ you have $u_y=k(y),$ and then integrating that w.r.t. $y$ arrive at $u=\int(k(y))+h(x).$ So it looks like you can say $u(x,y)=f(x)+g(y)$ for two single variable functions $f,g.$
