2019年考研数二第20题解析：波纹函数、定积分累加求和、等比数列

一、题目

已知函数 $u (x, y)$ 满足 $2 \frac{\partial^{2} u}{\partial x^{2}}$ $-$ $2 \frac{\partial^{2} u}{\partial y^{2}}$ $+$ $3 \frac{\partial u}{\partial x}$ $+$ $3 \frac{\partial u}{\partial y}$ $=$ $0$ , 求 $a$ , $b$ 的值，使得在变换 $u (x, y)$ $=$ $v (x, y) e^{a x + b y}$ 下，上述等式可化为 $v (x, y)$ 不含一阶偏导数的等式.

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二、解析

题目中所说的“ 使得在变换 $u (x, y)$ $=$ $v (x, y) e^{a x + b y}$ 下，上述等式可化为 $v (x, y)$ 不含一阶偏导数的等式 ”，意思就是对式子 $u (x, y)$ $=$ $v (x, y) e^{a x + b y}$ 求解一阶偏导数 $\frac{\partial u}{\partial x}$ , $\frac{\partial u}{\partial y}$ 和二阶偏导数 $\frac{\partial^{2} u}{\partial x^{2}}$ , $\frac{\partial^{2} u}{\partial y^{2}}$ , 并将求得的结果代入到原来的式子 $2 \frac{\partial^{2} u}{\partial x^{2}}$ $-$ $2 \frac{\partial^{2} u}{\partial y^{2}}$ $+$ $3 \frac{\partial u}{\partial x}$ $+$ $3 \frac{\partial u}{\partial y}$ $=$ $0$ 中，之后根据题目条件，求解出 $a$ 和 $b$ 的值即可。

首先，由 $u (x, y)$ $=$ $v (x, y) e^{a x + b y}$ 可得：

$\begin{aligned} \frac{\partial u}{\partial x} & = \frac{\partial v}{\partial x} e^{a x + b y} + a v e^{a x + b y} \\ \frac{\partial u}{\partial y} & = \frac{\partial v}{\partial y} e^{a x + b y} + b v e^{a x + b y} \\ \frac{\partial^{2} u}{\partial x^{2}} & = \frac{\partial^{2} v}{\partial x^{2}} e^{a x + b y} + 2 a \frac{\partial v}{\partial x} e^{a x + b y} + a^{2} v e^{a x + b y} \\ \frac{\partial^{2} u}{\partial y^{2}} & = \frac{\partial^{2} v}{\partial y^{2}} e^{a x + b y} + 2 b \frac{\partial v}{\partial y} e^{a x + b y} + b^{2} v e^{a x + b y} \end{aligned}$

将上面的计算结果，代入 $2 \frac{\partial^{2} u}{\partial x^{2}}$ $-$ $2 \frac{\partial^{2} u}{\partial y^{2}}$ $+$ $3 \frac{\partial u}{\partial x}$ $+$ $3 \frac{\partial u}{\partial y}$ $=$ $0$ , 可得：

$\begin{aligned} 2 \times \frac{\partial^{2} u}{\partial x^{2}} - 2 \times \frac{\partial^{2} u}{\partial y^{2}} + 3 \times \frac{\partial u}{\partial x} + 3 \times \frac{\partial u}{\partial y} = 0 \\ ⇝ & 2 \times (\frac{\partial^{2} v}{\partial x^{2}} e^{a x + b y} + 2 a \frac{\partial v}{\partial x} e^{a x + b y} + a^{2} v e^{a x + b y}) - \\ 2 \times (\frac{\partial^{2} v}{\partial y^{2}} e^{a x + b y} + 2 b \frac{\partial v}{\partial y} e^{a x + b y} + b^{2} v e^{a x + b y}) + \\ 3 \times (\frac{\partial v}{\partial x} e^{a x + b y} + a v e^{a x + b y}) + \\ 3 \times (\frac{\partial v}{\partial y} e^{a x + b y} + b v e^{a x + b y}) = 0 \\ ⇝ & 2 \times (\frac{\partial^{2} v}{\partial x^{2}} e^{a x + b y} + 2 a \frac{\partial v}{\partial x} e^{a x + b y} + a^{2} v e^{a x + b y}) - \\ 2 \times (\frac{\partial^{2} v}{\partial y^{2}} e^{a x + b y} + 2 b \frac{\partial v}{\partial y} e^{a x + b y} + b^{2} v e^{a x + b y}) + \\ 3 \times (\frac{\partial v}{\partial x} e^{a x + b y} + a v e^{a x + b y}) + \\ 3 \times (\frac{\partial v}{\partial y} e^{a x + b y} + b v e^{a x + b y}) = 0 \\ ⇝ & 2 \times (\frac{\partial^{2} v}{\partial x^{2}} + 2 a \frac{\partial v}{\partial x} + a^{2} v) - \\ 2 \times (\frac{\partial^{2} v}{\partial y^{2}} + 2 b \frac{\partial v}{\partial y} + b^{2} v) + \\ 3 \times (\frac{\partial v}{\partial x} + a v) + \\ 3 \times (\frac{\partial v}{\partial y} + b v) = 0 \\ ⇝ & 2 \times \frac{\partial^{2} v}{\partial x^{2}} - 2 \times \frac{\partial^{2} v}{\partial y^{2}} + (4 a + 3) \times \frac{\partial v}{\partial x} + (3 - 4 b) \times \frac{\partial v}{\partial y} + \\ (2 a^{2} - 2 b^{2} + 3 a + 3 b) \times v = 0 \\ ⇝ & 2 \times \frac{\partial^{2} v}{\partial x^{2}} - 2 \times \frac{\partial^{2} v}{\partial y^{2}} + (4 a + 3) \times \frac{\partial v}{\partial x} + (3 - 4 b) \times \frac{\partial v}{\partial y} + \\ (2 a^{2} - 2 b^{2} + 3 a + 3 b) \times v = 0 \\ ⇝ & {\begin{cases} 4 a + 3 = 0 \\ 3 - 4 b = 0 \end{cases} \\ ⇝ & {\begin{cases} a = \frac{- 3}{4} \\ b = \frac{3}{4} \end{cases} \end{aligned}$

2019年考研数二第20题解析：波纹函数、定积分累加求和、等比数列

一、题目

二、解析

高等数学

线性代数

特别专题

一、题目

二、解析

高等数学

线性代数

特别专题

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