问题及解答

设可微函数 $f(x,y)$ 满足方程 $(1+y)f(x,y)+xy\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=0$, 求 $f(x,y)$.

(猜测或尝试) 设 $f(x,y)=\frac{1}{xy}$, 则在其定义域内,

\[
\begin{aligned}
\frac{\partial f}{\partial x}&=\frac{1}{y}\cdot\frac{-1}{x^2}=-\frac{1}{x^2y},\\
\frac{\partial f}{\partial y}&=\frac{1}{x}\cdot\frac{-1}{y^2}=-\frac{1}{xy^2},\\
\end{aligned}
\]

于是

\[
\begin{split}
&(1+y)f(x,y)+xy\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}\\
=\ &(1+y)\cdot\frac{1}{xy}+xy\cdot\frac{-1}{x^2y}+y\cdot\frac{-1}{xy^2}\\
=\ &\frac{1}{xy}+\frac{1}{x}-\frac{1}{x}-\frac{1}{xy}\\
=\ &0.
\end{split}
\]