题目
已知 ((x+ay)dy-ydx)/((x+y)^2) 为某二元函数 u(x,y) 的全微分,则 a 等于:A. -1B. 1C. 2D. 0
已知 $\frac{(x+ay)dy-ydx}{(x+y)^2}$ 为某二元函数 $u(x,y)$ 的全微分,则 $a$ 等于:
A. -1
B. 1
C. 2
D. 0
题目解答
答案
D. 0
解析
步骤 1：将给定微分形式重写为标准形式
将给定的微分形式重写为：\[ P(x, y)dx + Q(x, y)dy = -\frac{y}{(x+y)^2}dx + \frac{x+ay}{(x+y)^2}dy \] 其中，$P(x, y) = -\frac{y}{(x+y)^2}$，$Q(x, y) = \frac{x+ay}{(x+y)^2}$。
步骤 2：计算偏导数
为了使该形式为全微分，需满足：\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \] 计算得：\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{(x+y)^2}\right) = \frac{-x^2 + y^2}{(x+y)^4} \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x+ay}{(x+y)^2}\right) = \frac{-x^2 + y^2 - 2axy - 2ay^2}{(x+y)^4} \]
步骤 3：令偏导数相等并求解 $a$
令两式相等，消去相同项后得：\[ 0 = -2axy - 2ay^2 \implies 0 = -2ay(x + y) \] 由于 $x + y \neq 0$，解得：\[ a = 0 \]
将给定的微分形式重写为：\[ P(x, y)dx + Q(x, y)dy = -\frac{y}{(x+y)^2}dx + \frac{x+ay}{(x+y)^2}dy \] 其中，$P(x, y) = -\frac{y}{(x+y)^2}$，$Q(x, y) = \frac{x+ay}{(x+y)^2}$。
步骤 2：计算偏导数
为了使该形式为全微分，需满足：\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \] 计算得：\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{(x+y)^2}\right) = \frac{-x^2 + y^2}{(x+y)^4} \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x+ay}{(x+y)^2}\right) = \frac{-x^2 + y^2 - 2axy - 2ay^2}{(x+y)^4} \]
步骤 3：令偏导数相等并求解 $a$
令两式相等，消去相同项后得：\[ 0 = -2axy - 2ay^2 \implies 0 = -2ay(x + y) \] 由于 $x + y \neq 0$，解得：\[ a = 0 \]