# 2023年考研数二第13题解析：偏导数的特解

## 二、解析

$$\begin{cases} x=1 \\ y=1 \end{cases} \Rightarrow$$

$$e^{z}+x z=2 x-y \Rightarrow e^{z}+z=1 \Rightarrow z=0$$

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

$$\begin{cases} x=1 \\ y=1 \\ z=0 \end{cases} \Rightarrow$$

$$\frac{\partial z}{\partial x}+\frac{\partial z}{\partial x}=2 \Rightarrow \frac{\partial z}{\partial x}=1$$

$$\frac{\partial^{2} z}{\partial x^{2}} \cdot e^{z}+\frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial x} \cdot e^{z}+\frac{\partial z}{\partial x}+\frac{\partial z}{\partial x}+x \cdot \frac{\partial^{2} z}{\partial x^{2}}=0 \Rightarrow$$

$$\begin{cases} x=1 \\ y=1 \\ z=0 \\ \frac{\partial z}{\partial x} = 1 \end{cases} \Rightarrow$$

$$\frac{\partial^{2} z}{\partial x^{2}}+1+1+1+\frac{\partial^{2} z}{\partial x^{2}}=0 \Rightarrow$$

$$2 \frac{\partial^{2} z}{\partial x^{2}}=-3 \Rightarrow$$

$$\frac{\partial^{2} z}{\partial x^{2}}=\frac{-3}{2}$$

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