[题目]-|||-设变换 ^2)+dfrac ({partial )^2z}(partial xpartial y)-dfrac ({partial )^2z}(partial {y)^2}=0-|||-简化为 dfrac ({a)^2z}(3upartial v)=0, 求常数a.

本题主要考察利用复合函数求导法则，将含$x,y$的二阶偏微分方程转化为含$u,v$的形式，进而求解常数$a$。核心思路是通过变量替换$u=x-2y$、$v=x+ay$，计算$z$对$x,y$的二阶偏导数，代入原方程后根据简化结果的条件确定$a$。

关键步骤1：计算一阶偏导数

$z$是$u,v$的函数，而$u,v$是$x,y$的函数，根据复合函数求导法则：
$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v}$
$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial y} = -2\frac{\partial z}{\partial u} + a\frac{\partial z}{\partial v}$

关键步骤2：计算二阶偏导数

对$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$分别求$x$和$y$的偏导，得到二阶偏导数：
$\frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial u^2} + 2\frac{\partial^2 z}{\partial u\partial v} + \frac{\partial^2 z}{\partial v^2}$
$\frac{\partial^2 z}{\partial x\partial y} = -2\frac{\partial^2 z}{\partial u^2} + (a-2)\frac{\partial^2 z}{\partial u\partial v} + a\frac{\partial^2 z}{\partial v^2}$
$\frac{\partial^2 z}{\partial y^2} = 4\frac{\partial^2 z}{\partial u^2} - 4a\frac{\partial^2 z}{\partial u\partial v} + a^2\frac{\partial^2 z}{\partial v^2}$

关键步骤3：代入原方程化简

将二阶偏导数代入原方程$6\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial x\partial y} - \frac{\partial^2 z}{\partial y^2} = 0$，整理得：
$(10 + 5a)\frac{\partial^2 z}{\partial u^2} + (6 + a - a^2)\frac{\partial^2 z}{\partial v^2} = 0$
要使方程简化为$\frac{\partial^2 z}{\partial u\partial v} = 0$，需$\frac{\partial^2 z}{\partial v^2}$的系数为0，且$\frac{\partial^2 z}{\partial u^2}$的系数不为0：
$6 + a - a^2 = 0 \quad \text{和} \quad 10 + 5a \neq 0$
解得$a=3$（$a=-2$舍去，因$10+5a=0$）。