下册 7.3 微分方程的验证及变量替换 第18题

\section*{解题过程：} （1）由 $\displaystyle z=\frac{w+y}{x}$ 得 $$ z_{y}=\frac{1}{x}\left(1+w_{u} u_{y}+w_{v} v_{y}\right)=\frac{1}{x}\left(1-\frac{x}{y^{2}} w_{u}\right), z_{y y}=\frac{2}{y^{3}} w_{u}-\frac{1}{y^{2}}\left(w_{u u} u_{y}+w_{u v} v_{y}\right)=\frac{2}{y^{3}} w_{u}+\frac{x}{y^{4}} w_{u u} . $$ 于是 $\displaystyle y z_{y y}+2 z_{y}-\frac{2}{x}=\frac{2}{y^{2}} w_{u}+\frac{x}{y^{3}} w_{u u}+\frac{2}{x}-\frac{2}{y^{2}} w_{u}-\frac{2}{x}=\frac{x}{y^{3}} w_{u u}=0$ ． 所以 $w_{u u}=0$ ． $$ \begin{aligned} & \text { (2) } f_{x}=-\frac{y}{x^{2}} f_{u}, f_{y}=f_{u} \frac{1}{x}+f_{v}, \\ & f_{x x}=\frac{y^{2}}{x^{4}} f_{u u}+\frac{2 y}{x^{3}} f_{u}, f_{x y}=-\left(f_{u u} \frac{1}{x}+f_{u v}\right) \frac{y}{x^{2}}-f_{u} \frac{1}{x^{2}}, f_{y y}=\left(f_{u u} \frac{1}{x}+f_{u v}\right) \frac{1}{x}+f_{v u} \frac{1}{x}+f_{v v} . \\ & x^{2} f_{x x}=f_{u u} \frac{y^{2}}{x^{2}}+2 f_{u} \frac{y}{x}, 2 x y f_{x y}=-2\left(f_{u u} \frac{1}{x}+f_{u v}\right) \frac{y^{2}}{x}-2 f_{u} \frac{y}{x}, y^{2} f_{y y}=f_{u u} \frac{y^{2}}{x^{2}}+2 f_{u v} \frac{y^{2}}{x}+y^{2} f_{v v} . \end{aligned} $$ 以上各式相加可得 $f_{v v}=0$ 。