# 在不进行积分运算的情况下，通过偏微分方程求解原函数

## 一、题目

$$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=\frac{2 y^{2}}{\sqrt{x^{2}+y^{2}}}$$

$$f(1) = ?$$

$$f^{\prime}(1) = ?$$

## 二、解析

$$z=\left(x^{2}+y^{2}\right)^{\frac{1}{2}} f\left(\frac{y}{x}\right)$$

$$\frac{\partial z}{\partial x}=\frac{1}{2} \cdot 2 x\left(x^{2}+y^{2}\right)^{\frac{-1}{2}} f\left(\frac{y}{x}\right)+$$

$$\left(x^{2}+y^{2}\right)^{\frac{1}{2}} f_{x}^{\prime}\left(\frac{y}{x}\right) \cdot \frac{-1}{x^{2}} \cdot y \Rightarrow$$

$$x \cdot \frac{\partial z}{\partial x}=\frac{x^{2} f\left(\frac{y}{x}\right)}{\sqrt{x^{2}+y^{2}}}+\frac{-x y \sqrt{x^{2}+y^{2}} \cdot f_{x}^{\prime}\left(\frac{y}{x}\right)}{x^{2}}.$$

$$\frac{\partial z}{\partial y}=\frac{1}{z} \cdot 2 y\left(x^{2}+y^{2}\right)^{\frac{-1}{2}} f\left(\frac{y^{x}}{x}\right)+$$

$$\left(x^{2}+y^{2}\right)^{\frac{1}{2}} f_{y}^{\prime}\left(\frac{y}{x}\right) \cdot \frac{1}{x} \Rightarrow$$

$$y \cdot \frac{\partial z}{\partial y}=\frac{y^{2} f\left(\frac{y}{x}\right)}{\sqrt{x^{2}+y^{2}}}+\frac{y \sqrt{x^{2}+y^{2}} f^{\prime} y\left(\frac{y}{x}\right)}{x}.$$

$$x \cdot \frac{\partial z}{\partial x}+y \cdot \frac{\partial z}{\partial y}=\frac{\left(x^{2}+y^{2}\right) f\left(\frac{y}{x}\right)}{\sqrt{x^{2}+y^{2}}}$$

$$\frac{y}{x}=1 \Rightarrow y=x \Rightarrow$$

$$x \cdot \frac{\partial z}{\partial x}+y \cdot \frac{\partial z}{\partial y}=\frac{2 y^{2} f\left(\frac{y}{x}\right)}{\sqrt{x^{2}+y^{2}}}. \quad ①$$

$$x \cdot \frac{\partial z}{\partial x}+y \cdot \frac{\partial z}{\partial y}=\frac{2 y^{2}}{\sqrt{x^{2}+y^{2}}}. \quad ②$$

$$f(1)=1$$

$$x \cdot \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=\sqrt{x^{2}+y^{2}} f\left(\frac{y}{x}\right) \Rightarrow$$

$$x \cdot \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=\frac{2 y^{2}}{\sqrt{x^{2}+y^{2}}} \Rightarrow$$

$$\sqrt{x^{2}+y^{2}} f\left(\frac{y}{x}\right)=\frac{2 y^{2}}{\sqrt{x^{2}+y^{2}}} \Rightarrow$$

$$\left(x^{2}+y^{2}\right) f\left(\frac{y}{x}\right)=2 y^{2} \frac{2 y^{2}}{x^{2}+y^{2}} \Rightarrow$$

$$f\left(\frac{y}{x}\right)=\frac{2 \frac{y^{2}}{x^{2}}}{1+\frac{y^{2}}{x^{2}}} \Rightarrow$$

$$f\left(u\right)=\frac{2 u^{2}}{1+u^{2}} \Rightarrow$$

$$f(1)=1$$

$$f^{\prime}(u)=\frac{4 u\left(1+u^{2}\right)-2 u^{2} \cdot 2 u}{\left(1+u^{2}\right)^{2}} \Rightarrow$$

$$f^{\prime}(1) = \frac{4 \times 2 – 4}{4} = 1.$$