20.(本题满分12分)-|||-已知函数 =u(x,y)(e)^ax+by, 且 dfrac ({a)^2u}(partial xpartial y)=0, 确定常数a,b,使函数 z=z(x,y) 满足方程-|||-dfrac ({partial )^2z}(partial xpartial y)-dfrac (partial z)(partial x)-dfrac (partial z)(partial y)+z=0.

步骤 1：计算 $\dfrac {\partial z}{\partial x}$
根据 $z=u(x,y){e}^{ax+by}$，我们有 $\dfrac {\partial z}{\partial x} = \dfrac {\partial u}{\partial x}{e}^{ax+by} + u(x,y)a{e}^{ax+by}$。

步骤 2：计算 $\dfrac {\partial z}{\partial y}$
同样地，$\dfrac {\partial z}{\partial y} = \dfrac {\partial u}{\partial y}{e}^{ax+by} + u(x,y)b{e}^{ax+by}$。

步骤 3：计算 $\dfrac {{\partial }^{2}z}{\partial x\partial y}$
$\dfrac {{\partial }^{2}z}{\partial x\partial y} = \dfrac {\partial }{\partial y}(\dfrac {\partial z}{\partial x}) = \dfrac {\partial }{\partial y}(\dfrac {\partial u}{\partial x}{e}^{ax+by} + u(x,y)a{e}^{ax+by}) = \dfrac {{\partial }^{2}u}{\partial x\partial y}{e}^{ax+by} + \dfrac {\partial u}{\partial x}b{e}^{ax+by} + u(x,y)ab{e}^{ax+by}$。
由于 $\dfrac {{\partial }^{2}u}{\partial {x}^{\partial }y}=0$，所以 $\dfrac {{\partial }^{2}z}{\partial x\partial y} = \dfrac {\partial u}{\partial x}b{e}^{ax+by} + u(x,y)ab{e}^{ax+by}$。

步骤 4：代入方程 $\dfrac {{\partial }^{2}z}{\partial x\partial y}-\dfrac {\partial z}{\partial x}-\dfrac {\partial z}{\partial y}+z=0$
将上述结果代入方程，我们得到：
$\dfrac {\partial u}{\partial x}b{e}^{ax+by} + u(x,y)ab{e}^{ax+by} - (\dfrac {\partial u}{\partial x}{e}^{ax+by} + u(x,y)a{e}^{ax+by}) - (\dfrac {\partial u}{\partial y}{e}^{ax+by} + u(x,y)b{e}^{ax+by}) + u(x,y){e}^{ax+by} = 0$。
简化后得到：
$(b-1)\dfrac {\partial u}{\partial x}{e}^{ax+by} + (ab-a-b+1)u(x,y){e}^{ax+by} = 0$。
由于 $u(x,y)$ 和 ${e}^{ax+by}$ 都不是零，所以必须有：
$(b-1)\dfrac {\partial u}{\partial x} + (ab-a-b+1)u(x,y) = 0$。
由于 $\dfrac {{\partial }^{2}u}{\partial {x}^{\partial }y}=0$，则 $\dfrac {\partial u}{\partial x}$ 和 $u(x,y)$ 都不是零，因此必须有：
$b-1=0$ 和 $ab-a-b+1=0$。
解得 $b=1$ 和 $a=1$。