下册 7.3 微分方程的验证及变量替换 第1题

解题过程： （1） $$ \begin{aligned} & \frac{\partial z}{\partial x}=-\frac{y}{f^{2}\left(x^{2}-y^{2}\right)} \frac{\partial f\left(x^{2}-y^{2}\right)}{\partial x}=-\frac{2 x y f^{\prime}\left(x^{2}-y^{2}\right)}{f^{2}\left(x^{2}-y^{2}\right)} . \\ & \frac{\partial z}{\partial y}=\frac{1}{f\left(x^{2}-y^{2}\right)}-\frac{y}{f^{2}\left(x^{2}-y^{2}\right)} \frac{\partial f\left(x^{2}-y^{2}\right)}{\partial y}=\frac{1}{f\left(x^{2}-y^{2}\right)}+\frac{2 y^{2} f^{\prime}\left(x^{2}-y^{2}\right)}{f^{2}\left(x^{2}-y^{2}\right)} . \end{aligned} $$ 直接计算可得 $\displaystyle \frac{1}{x} \frac{\partial z}{\partial x}+\frac{1}{y} \frac{\partial z}{\partial y}=\frac{1}{y f\left(x^{2}-y^{2}\right)}=\frac{z}{y^{2}}$ ． （2）记 $\displaystyle u=\frac{z}{y}$ ．方程 $x^{2}+y^{2}+z^{2}=y f(u)$ 两边分别对 $x, y$ 求偏导数得 $$ 2 x+2 z \frac{\partial z}{\partial x}=y \frac{1}{y} f^{\prime}(u) \frac{\partial z}{\partial x}, 2 y+2 z \frac{\partial z}{\partial y}=f(u)+y f^{\prime}(u) \frac{1}{y^{2}}\left(\frac{\partial z}{\partial y} y-z\right) . $$ 所以 $\displaystyle \frac{\partial z}{\partial x}=\frac{2 x}{f^{\prime}(u)-2 z}, \frac{\partial z}{\partial y}=\frac{2 y+\frac{z}{y} f^{\prime}(u)-f(u)}{f^{\prime}(u)-2 z}$ ．于是 $\displaystyle \left(x^{2}-y^{2}-z^{2}\right) \frac{\partial z}{\partial x}+2 x y \frac{\partial z}{\partial y}=\left(x^{2}-y^{2}-z^{2}\right) \frac{2 x}{f^{\prime}(u)-2 z}+2 x y \frac{2 y+\frac{z}{y} f^{\prime}(u)-f(u)}{f^{\prime}(u)-2 z}=\frac{2 x z f^{\prime}(u)-4 x z^{2}}{f^{\prime}(u)-2 z}=2 x z$. （3）设 $\displaystyle t=\frac{y}{x^{2}}$ ，则 $\displaystyle \frac{\partial z}{\partial x}=n x^{n-1} f(t)-2 x^{n-3} y f^{\prime}(t), \frac{\partial z}{\partial y}=x^{n-2} f^{\prime}(t)$ 。于是 $$ x \frac{\partial z}{\partial x}+2 y \frac{\partial z}{\partial y}=n x^{n} f(t)-2 x^{n-2} y f^{\prime}(t)+2 x^{n-2} y f^{\prime}(t)=n x^{n} f(t)=n z $$ （4）在方程 $a x+b y+c z=\varphi\left(x^{2}+y^{2}+z^{2}\right)$ 两边对 $x$ 求偏导数， $$ a+c z_{x}=2\left(x+z z_{x}\right) \varphi^{\prime} \text {, 即 } \frac{\partial z}{\partial x}=\frac{2 x \varphi^{\prime}-a}{c-2 z \varphi^{\prime}} \text {. } $$ 在方程 $a x+b y+c z=\varphi\left(x^{2}+y^{2}+z^{2}\right)$ 两边对 $y$ 求偏导数， $$ b+c z_{y}=2\left(y+z z_{y}\right) \varphi^{\prime} \text {, 即 } \frac{\partial z}{\partial y}=\frac{2 y \varphi^{\prime}-b}{c-2 z \varphi^{\prime}} \text {. } $$ 于是 $$ (c y-b z) \frac{\partial z}{\partial x}+(a z-c x) \frac{\partial z}{\partial y}=(c y-b z) \frac{2 x \varphi^{\prime}-a}{c-2 z \varphi^{\prime}}+(a z-c x) \frac{2 y \varphi^{\prime}-b}{c-2 z \varphi^{\prime}}=b x-a y $$