下册 7.3 微分方程的验证及变量替换 第11题

\section*{解题过程：} （1）在 $\displaystyle x=u v, y=\frac{1}{2}\left(u^{2}-v^{2}\right)$ 两边分别对 $x, y$ 求偏导数得 于是 $$ \begin{aligned} & 1=u_{x} v+u v_{x}, 0=u u_{x}-v v_{x}, 0=u_{y} v+u v_{y}, 1=u u_{y}-v v_{y} \\ & u_{x}=\frac{v}{u^{2}+v^{2}}, v_{x}=\frac{u}{u^{2}+v^{2}}, u_{y}=\frac{u}{u^{2}+v^{2}}, v_{y}=-\frac{v}{u^{2}+v^{2}} \\ & z_{x}=z_{u} \cdot u_{x}+z_{v} \cdot v_{x}=\frac{1}{u^{2}+v^{2}} \cdot\left(v z_{u}+u z_{v}\right) \\ & z_{y}=z_{u} \cdot u_{y}+z_{v} \cdot v_{y}=\frac{1}{u^{2}+v^{2}} \cdot\left(u z_{u}-v z_{v}\right) \\ & z_{x}^{2}+z_{y}^{2}=\frac{1}{\left(u^{2}+v^{2}\right)^{2}} \cdot\left[\left(v z_{u}+u z_{v}\right)^{2}+\left(u z_{u}-v z_{v}\right)^{2}\right] \end{aligned} $$ 而 $\displaystyle x^{2}+y^{2}=(u v)^{2}+\frac{1}{4}\left(u^{2}-v^{2}\right)^{2}=\frac{1}{4}\left(u^{2}+v^{2}\right)^{2}$ ，即 $\displaystyle \sqrt{x^{2}+y^{2}}=\frac{1}{2}\left(u^{2}+v^{2}\right)$ ，所以 $$ \frac{1}{\left(u^{2}+v^{2}\right)^{2}} \cdot\left[\left(v z_{u}+u z_{v}\right)^{2}+\left(u z_{u}-v z_{v}\right)^{2}\right]=\frac{2}{u^{2}+v^{2}} \text {, 即 } z_{u}^{2}+z_{v}^{2}=2 \text { 。 } $$ （2）已知 $\displaystyle z=z(u, v), u=\ln \sqrt{x^{2}+y^{2}}, v=\arctan \frac{x}{y}$ ，则 $$ \begin{aligned} & z_{x}=z_{u} u_{x}+z_{v} v_{x}=\frac{1}{x^{2}+y^{2}}\left(x \cdot z_{u}+y \cdot z_{v}\right), z_{y}=z_{u} u_{y}+z_{v} v_{y}=\frac{1}{x^{2}+y^{2}}\left(y \cdot z_{u}-x \cdot z_{v}\right) . \\ & (x+y) z_{x}-(x-y) z_{y}=\frac{x+y}{x^{2}+y^{2}}\left(x \cdot z_{u}+y \cdot z_{v}\right)-\frac{x-y}{x^{2}+y^{2}}\left(y \cdot z_{u}-x \cdot z_{v}\right)=z_{u}+z_{v} . \end{aligned} $$ 于是由 $(x+y) z_{x}-(x-y) z_{y}=0$ 得 $z_{u}+z_{v}=0$ ． （3）$\displaystyle z_{x}=z_{u} u_{x}+z_{v} v_{x}=y z_{u}+\frac{1}{y} \cdot z_{v}, z_{y}=z_{u} u_{y}+z_{v} v_{y}=x z_{u}-\frac{x}{y^{2}} \cdot z_{v}$ ． $$ \begin{aligned} z_{x x} & =\left(y z_{u}\right)_{x}+\left(\frac{1}{y} \cdot z_{v}\right)_{x}=y\left(z_{u u} u_{x}+z_{u v} v_{x}\right)+\frac{1}{y}\left(z_{v u} u_{x}+z_{v v} v_{x}\right)=y^{2} z_{u u}+2 z_{u v}+\frac{1}{y^{2}} z_{v v} . \\ z_{y y} & =\left(x z_{u}\right)_{y}-\left(\frac{x}{y^{2}} \cdot z_{v}\right)_{y}=x\left(z_{u u} u_{y}+z_{u v} v_{y}\right)+\frac{2 x}{y^{3}} \cdot z_{v}-\frac{x}{y^{2}}\left(z_{v u} u_{y}+z_{v v} v_{y}\right) \\ & =x^{2} z_{u u}-2 \frac{x^{2}}{y^{2}} z_{u v}+\frac{x^{2}}{y^{4}} z_{v v}+\frac{2 x}{y^{3}} \cdot z_{v} . \end{aligned} $$ 将上述 $z_{x x}, z_{y y}$ 代人原方程，并化简得 $2 x y z_{u v}=z_{v}$ ，即 $2 u z_{u v}=z_{v}$ ．