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 $u(x, y)$ $=$ $v(x, y) \mathrm{e}^{ax + by}$ 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 $u(x, y)$ $=$ $v(x, y) \mathrm{e}^{ax + by}$ 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 $u(x, y)$ $=$ $v(x, y) \mathrm{e}^{ax + by}$ 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 $u(x, y)$ $=$ $v(x, y) \mathrm{e}^{ax + by}$ 可得：
$$
\begin{aligned}
\textcolor{orange}{ \frac{\partial u}{\partial x} } & = \textcolor{orange}{ \frac{\partial v}{\partial x} \mathrm{e}^{a x + b y} + a v \mathrm{e}^{a x + b y} } \\ \\
\textcolor{lightgreen}{ \frac {\partial u}{\partial y} } & = \textcolor{lightgreen}{ \frac{\partial v}{\partial y} \mathrm{e}^{a x + b y} + b v \mathrm{e}^{a x + b y} } \\ \\
\textcolor{magenta}{ \frac{\partial^{2} u}{\partial x^{2}} } & = \textcolor{magenta}{ \frac{\partial^{2} v}{\partial x^{2}} \mathrm{e}^{a x + b y} + 2 a \frac{\partial v}{\partial x} \mathrm{e}^{a x + b y} + a^{2} v \mathrm{e}^{a x + b y} } \\ \\
\textcolor{lightblue}{ \frac{\partial^{2} u}{\partial y^{2}} } & = \textcolor{lightblue}{ \frac{\partial^{2} v }{\partial y^{2}} \mathrm{e}^{a x + b y} + 2 b \frac{\partial v}{\partial y} \mathrm{e}^{a x + b y} + b^{2} v \mathrm{e}^{a x + b y} }
\end{aligned}
$$
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 \textcolor{magenta}{ \frac{\partial^{2} u}{\partial x^{2}} } – 2 \times \textcolor{lightblue}{ \frac{\partial^{2} u}{\partial y^{2}} } + 3 \times \textcolor{orange}{ \frac{\partial u}{\partial x} } + 3 \times \textcolor{lightgreen}{ \frac{\partial u}{\partial y} } = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & 2 \times \left( \textcolor{magenta}{ \frac{\partial^{2} v}{\partial x^{2}} \mathrm{e}^{a x + b y} + 2 a \frac{\partial v}{\partial x} \mathrm{e}^{a x + b y} + a^{2} v \mathrm{e}^{a x + b y} } \right) – \\
& 2 \times \left( \textcolor{lightblue}{ \frac{\partial^{2} v }{\partial y^{2}} \mathrm{e}^{a x + b y} + 2 b \frac{\partial v}{\partial y} \mathrm{e}^{a x + b y} + b^{2} v \mathrm{e}^{a x + b y} } \right) + \\
& 3 \times \left( \textcolor{orange}{ \frac{\partial v}{\partial x} \mathrm{e}^{a x + b y} + a v \mathrm{e}^{a x + b y} } \right) + \\
& 3 \times \left( \textcolor{lightgreen}{ \frac{\partial v}{\partial y} \mathrm{e}^{a x + b y} + b v \mathrm{e}^{a x + b y} } \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & 2 \times \left( \textcolor{magenta}{ \frac{\partial^{2} v}{\partial x^{2}} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{magenta}{ 2 a \frac{\partial v}{\partial x} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{magenta}{ a^{2} v } \textcolor{gray}{ \mathrm{e}^{a x + b y} } \right) – \\
& 2 \times \left( \textcolor{lightblue}{ \frac{\partial^{2} v }{\partial y^{2}} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{lightblue}{ 2 b \frac{\partial v}{\partial y} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{lightblue}{ b^{2} v } \textcolor{gray}{ \mathrm{e}^{a x + b y} } \right) + \\
& 3 \times \left( \textcolor{orange}{ \frac{\partial v}{\partial x} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{orange}{ a v } \textcolor{gray}{ \mathrm{e}^{a x + b y} } \right) + \\
& 3 \times \left( \textcolor{lightgreen}{ \frac{\partial v}{\partial y} } \textcolor{gray}{ \mathrm{e}^{a x + b y} } + \textcolor{lightgreen}{ b v } \textcolor{gray}{ \mathrm{e}^{a x + b y} } \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & 2 \times \left( \textcolor{magenta}{ \frac{\partial^{2} v}{\partial x^{2}} } + \textcolor{magenta}{ 2 a \frac{\partial v}{\partial x} } + \textcolor{magenta}{ a^{2} v } \right) – \\
& 2 \times \left( \textcolor{lightblue}{ \frac{\partial^{2} v }{\partial y^{2}} } + \textcolor{lightblue}{ 2 b \frac{\partial v}{\partial y} } + \textcolor{lightblue}{ b^{2} v } \right) + \\
& 3 \times \left( \textcolor{orange}{ \frac{\partial v}{\partial x} } + \textcolor{orange}{ a v } \right) + \\
& 3 \times \left( \textcolor{lightgreen}{ \frac{\partial v}{\partial y} } + \textcolor{lightgreen}{ b v } \right) = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & 2 \times \textcolor{magenta}{ \frac{\partial^{2} v }{\partial x^{2}} } – 2 \times \textcolor{lightblue}{ \frac{\partial^{2} v}{\partial y^{2}} } + \left( 4a + 3 \right) \times \textcolor{orange}{ \frac{\partial v}{\partial x} } + \left( 3 – 4b \right) \times \textcolor{lightgreen}{ \frac{\partial v}{\partial y} } + \\
& \left(2a^{2} – 2b^{2} + 3a + 3b \right) \times v = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & 2 \times \frac{\partial^{2} v }{\partial x^{2}} – 2 \times \frac{\partial^{2} v}{\partial y^{2}} + \textcolor{red}{ \cancel{\textcolor{yellow}{\left( 4a + 3 \right) } \textcolor{white}{\times} \textcolor{white}{ \frac{\partial v}{\partial x}}} } + \textcolor{red}{\cancel{\textcolor{yellow}{\left( 3 – 4b \right) } \textcolor{white}{\times} \textcolor{white}{ \frac{\partial v}{\partial y}}}} + \\
& \left(2a^{2} – 2b^{2} + 3a + 3b \right) \times v = 0 \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \begin{cases}
\textcolor{yellow}{4a + 3} = \textcolor{yellow}{0} \\
\textcolor{yellow}{3 – 4b} = \textcolor{yellow}{0}
\end{cases} \\ \\
\textcolor{lightgreen}{ \leadsto } \ & \begin{cases}
\boldsymbol{\textcolor{green}{a} = \textcolor{green}{\frac{-3}{4}}} \\
\boldsymbol{\textcolor{green}{b} = \textcolor{green}{\frac{3}{4}}}
\end{cases}
\end{aligned}
$$
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