题目
14.已知 dfrac ((x+ay)dy-ydx)({(x+y))^2} 为某个函数的全微分,则 a= () .-|||-(A) -1 ; (B)0; (C)1; (D)2
题目解答
答案
解析
步骤 1：确定全微分条件
全微分的条件是：对于函数 $f(x,y)$，其微分形式 $df = P(x,y)dx + Q(x,y)dy$ 必须满足 $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$。这里，$P(x,y) = -\dfrac{y}{(x+y)^2}$，$Q(x,y) = \dfrac{x+ay}{(x+y)^2}$。
步骤 2：计算偏导数
计算 $\dfrac{\partial P}{\partial y}$ 和 $\dfrac{\partial Q}{\partial x}$。
$\dfrac{\partial P}{\partial y} = \dfrac{\partial}{\partial y} \left(-\dfrac{y}{(x+y)^2}\right) = -\dfrac{(x+y)^2 - 2y(x+y)}{(x+y)^4} = -\dfrac{x+y-2y}{(x+y)^3} = -\dfrac{x-y}{(x+y)^3}$。
$\dfrac{\partial Q}{\partial x} = \dfrac{\partial}{\partial x} \left(\dfrac{x+ay}{(x+y)^2}\right) = \dfrac{(x+y)^2 - 2(x+ay)(x+y)}{(x+y)^4} = \dfrac{x+y-2(x+ay)}{(x+y)^3} = \dfrac{x+y-2x-2ay}{(x+y)^3} = \dfrac{-x+y-2ay}{(x+y)^3}$。
步骤 3：设置等式并求解
根据全微分条件，$\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$，即 $-\dfrac{x-y}{(x+y)^3} = \dfrac{-x+y-2ay}{(x+y)^3}$。化简得 $x-y = x-y+2ay$，从而 $2ay = 0$。由于 $y$ 不恒为零，因此 $a = 0$。
全微分的条件是：对于函数 $f(x,y)$，其微分形式 $df = P(x,y)dx + Q(x,y)dy$ 必须满足 $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$。这里，$P(x,y) = -\dfrac{y}{(x+y)^2}$，$Q(x,y) = \dfrac{x+ay}{(x+y)^2}$。
步骤 2：计算偏导数
计算 $\dfrac{\partial P}{\partial y}$ 和 $\dfrac{\partial Q}{\partial x}$。
$\dfrac{\partial P}{\partial y} = \dfrac{\partial}{\partial y} \left(-\dfrac{y}{(x+y)^2}\right) = -\dfrac{(x+y)^2 - 2y(x+y)}{(x+y)^4} = -\dfrac{x+y-2y}{(x+y)^3} = -\dfrac{x-y}{(x+y)^3}$。
$\dfrac{\partial Q}{\partial x} = \dfrac{\partial}{\partial x} \left(\dfrac{x+ay}{(x+y)^2}\right) = \dfrac{(x+y)^2 - 2(x+ay)(x+y)}{(x+y)^4} = \dfrac{x+y-2(x+ay)}{(x+y)^3} = \dfrac{x+y-2x-2ay}{(x+y)^3} = \dfrac{-x+y-2ay}{(x+y)^3}$。
步骤 3：设置等式并求解
根据全微分条件，$\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$，即 $-\dfrac{x-y}{(x+y)^3} = \dfrac{-x+y-2ay}{(x+y)^3}$。化简得 $x-y = x-y+2ay$，从而 $2ay = 0$。由于 $y$ 不恒为零，因此 $a = 0$。