# 一个简单偏微分方程的求解

$z$是一个关于$x$和$y$的函数，并且有$x$$\cfrac{dz}{dy}$$+$$y$$\cfrac{dz}{dx}$$=z$，求$z$的解析式。

$\\ y\frac{\partial z}{\partial x} + x\frac{\partial z}{\partial y}=z \\ m=x+y,n=x-y \\ \frac{m-n}{2}\left ( \frac{\partial z}{\partial m} + \frac{\partial z}{\partial n} \right )+\frac{m+n}{2}\left ( \frac{\partial z}{\partial m}-\frac{\partial z}{\partial n} \right )=z \\ m \frac{\partial z}{\partial m}-n\frac{\partial z}{\partial n}=z \\ t = mn \\ m \left ( \frac{\partial z}{\partial m} + n\frac{\partial z}{\partial t}\right )-n\left ( m \frac{\partial z}{\partial t} \right )=z \\\therefore z=Cm=mf(t)=(x+y)f(x^2-y^2)$

\begin{align} & x \frac{\partial z}{\partial y}+y \frac{\partial z}{\partial x}=z \\ & \text{令} y = y(x),那么\frac{d}{dx}z(x,y(x))=\frac{\partial z}{\partial x} + \frac{dy}{dx}\frac{\partial z}{\partial y} \\ & x \frac{\partial z}{\partial y} +y (\frac{dz}{dx}-\frac{dy}{dx}\frac{\partial z}{\partial y})=z \\ & \text{选取合适的y(x)，可以使得}\frac{\partial z}{\partial y}\text{的系数为0，此时可以得到下面的方程组} \\ & \left\{\begin{matrix} y \frac{dz}{dx}=z (1) \\ x=y\frac{dy}{dx} (2) \end{matrix}\right. \\ &\text{解方程组(2)可以得到}x^2-y^2=C_1 \\ &\text{代人方程组(1)可以解出}z=(x+\sqrt{x^2-C_1})C_2\\ &\text{由于}C_2\text{和}C_1\text{都是与x，y无关的常数，所以}C_2\text{可以写成}x^2-y^2\text{的函数} \\ &\therefore z=(x+y)f(x^2-y^2) \end{align}