# Solution to Second order PDE with elementary method

Solve $UU_{xy}-U_xU_y=0$ I know how to solve $UU_{xy}+U_xU_y=0$ But i cant find any trick to do with the differential operators .This is an exercise and im not interested in that specific pde but i want to know how to handle the differential operators and when can i "partial integrate" things like in the second $$UU_{xy}+U_xU_y=0$$ i said that this is $$\frac {\partial U U_x}{\partial y}=0$$ So then i partial integrated by $y$ so $UU_x=g(x)$ which is $$\frac{U \partial U}{\partial x}=g(x)$$ then integrated again so I got $\frac{U^2} {2} =G(x)$ Did the same by integrating first with $x$ so I got $U=\pm\sqrt{G(x)+F(y)}$ Cant find a way to do the same for the other one.

## 1 Answer

In this case $\dfrac{\partial}{\partial x} \dfrac{U_y}{U} = 0$, so $U_y = g(y) U$ for some function $g(y)$. Now $u'(y) = g(y) u(y)$ is a linear homogeneous ODE whose solutions are of the form $u(y) = c G(y)$ where $G(y)$ is one solution. Thus $U(x,y) = F(x) G(y)$ for some functions $F$, $G$. On the other hand, for any differentiable $F$ and $G$, $U(x,y) = F(x) G(y)$ satisfies your PDE.

• One could also contract to $(U^2)_{xy}=0$ for the first case and $(\ln|U|)_{xy}=0$ for the second case, both leading to $U^2=f(x)+g(y)$ resp. $\ln|U|=f(x)+g(y)$. – LutzL Mar 9 '16 at 17:22