下册 7.4 隐函数的微分 第13题

\section*{解题过程：} （1）令 $\displaystyle u=\frac{x}{z}, v=\frac{y}{z}$ ．方程两边对 $x$ 求偏导得 $$ F_{u}\left(\frac{1}{z}-\frac{x}{z^{2}} z_{x}\right)+F_{v}\left(-\frac{y}{z^{2}}\right) z_{x}=0 . $$ 于是 $\displaystyle z_{x}=\frac{z F_{u}}{x F_{u}+y F_{v}}$ ． 同理可得：$\displaystyle z_{y}=\frac{z F_{v}}{x F_{u}+y F_{v}}$ ． 从而 $x z_{x}+y z_{y}=z$ 。 （2）由 $z=f(u)$ 得 $\displaystyle \frac{\partial z}{\partial x}=f^{\prime}(u) \cdot \frac{\partial u}{\partial x}, \frac{\partial z}{\partial y}=f^{\prime}(u) \cdot \frac{\partial u}{\partial y}$ ． 方程 $u=\varphi(u)+\int_{y}^{x} p(t) \mathrm{d} t$ 两边对 $x$ 求偏导，$u_{x}=\varphi^{\prime}(u) \cdot u_{x}+p(x)$ ，即 $\displaystyle u_{x}=\frac{p(x)}{1-\varphi^{\prime}(u)}$ ； 方程 $u=\varphi(u)+\int_{y}^{x} p(t) \mathrm{d} t$ 两边对 $y$ 求偏导，$u_{y}=\varphi^{\prime}(u) \cdot u_{y}-p(y)$ ，即 $\displaystyle u_{y}=-\frac{p(y)}{1-\varphi^{\prime}(u)}$ ． 故 $p(y) z_{x}+p(x) z_{y}=0$ 。 （3）用 $G_{i}$ 表示 $G$ 对第 $i$ 个中间变量的偏导数． 在方程 $G(c x-a z, c y-b z)=0$ 两边分别对 $x$ 和 $y$ 求偏导数得 $$ G_{1}\left(c-a z_{x}\right)+G_{2}\left(-b z_{x}\right)=0, G_{1}\left(-a z_{y}\right)+G_{2}\left(c-b z_{y}\right)=0 . $$ 于是 $\displaystyle z_{x}=\frac{c G_{1}}{a G_{1}+b G_{2}}, z_{y}=\frac{c G_{2}}{a G_{1}+b G_{2}}$ ，从而 $a z_{x}+b z_{y}=c$ 。 再在 $G_{1}\left(c-a z_{x}\right)+G_{2}\left(-b z_{x}\right)=0$ 两边对 $y$ 求偏导数得 $$ \left[G_{11}\left(-a z_{y}\right)+G_{12}\left(c-b z_{y}\right)\right]\left(c-a z_{x}\right)+G_{1}\left(-a z_{x y}\right)+\left[G_{21}\left(-a z_{y}\right)+G_{22}\left(c-b z_{y}\right)\right]\left(-b z_{x}\right)+G_{2}\left(-b z_{x y}\right)=0 . $$ 于是 $$ \begin{aligned} z_{x y} & =\frac{1}{a G_{1}+b G_{2}}\left\{\left[c G_{12}-\left(a G_{11}+b G_{12}\right) z_{y}\right]\left(c-a z_{x}\right)-b z_{x}\left[c G_{22}-\left(a G_{21}+b G_{22}\right) z_{y}\right]\right\} \\ & =\frac{a b c^{2}}{\left(a G_{1}+b G_{2}\right)^{3}}\left(2 G_{1} G_{2} G_{12}-G_{2}^{2} G_{11}-G_{1}^{2} G_{22}\right) \end{aligned} $$ （4）由 $u=x+a y, v=x-a y(a \neq 0)$ 得 $\displaystyle x=\frac{1}{2}(u+v), y=\frac{1}{2 a}(u-v)$ ． $$ z_{u}=z_{x} x_{u}+z_{y} y_{u}=\frac{1}{2} z_{x}+\frac{1}{2 a} z_{y} $$ $$ z_{u v}=\left(\frac{1}{2} z_{x}+\frac{1}{2 a} z_{y}\right)_{v}=\frac{1}{2}\left(z_{x x} x_{v}+z_{x y} y_{v}\right)+\frac{1}{2 a}\left(z_{y x} x_{v}+z_{y y} y_{v}\right)=\frac{1}{4} z_{x x}-\frac{1}{4 a^{2}} z_{y y} . $$ （5）由 $F(x+a z, y+b z)=0$ 知 $F_{u}\left(1+a z_{x}\right)+F_{v} \cdot b z_{x}=0$ 。所以 $\displaystyle z_{x}=-\frac{F_{u}}{a F_{u}+b F_{v}}$ ． 由 $F(x+a z, y+b z)=0$ 知 $F_{u} \cdot a z_{y}+F_{v}\left(1+b \cdot z_{y}\right)=0$ ．所以 $\displaystyle z_{y}=-\frac{F_{v}}{a F_{u}+b F_{v}}$ ． 从而 $a z_{x}+b z_{y}=-1$ ．于是 $$ \begin{aligned} \iint_{x^{2}+y^{2}<1} \mathrm{e}^{-\left(x^{2}+y^{2}\right)}\left(a z_{x}+b z_{y}\right) \mathrm{d} x \mathrm{~d} y & =\iint_{x^{2}+y^{2}<1} \mathrm{e}^{-\left(x^{2}+y^{2}\right)}(-1) \mathrm{d} x \mathrm{~d} y=-\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} \mathrm{e}^{-r^{2}} r \mathrm{~d} r \\ & =-2 \pi \int_{0}^{1}\left(-\frac{1}{2} \mathrm{e}^{-r^{2}}\right)^{\prime} \mathrm{d} r=\pi\left(\mathrm{e}^{-1}-1\right) \end{aligned} $$