设\( z=f(u)\)，二元函数\(u=u(x,y) \)是由方程 \( u=\varphi(u) + \int^x_y p(t) dt \)确定的。其中\(f(u)\), \( \varphi(u)\)可微，\(p(t)\), \(\varphi'(u)\)连续，且\(\varphi'(u) \neq 1 \),求\(p(y) \frac{\partial z}{\partial x} + p(x) \frac{\partial z}{\partial y}\)

[collapse title=”解答”]

\[\frac{\partial z}{\partial x} = f'(u) \frac{\partial u}{\partial x},\frac{\partial z}{\partial y} = f'(u) \frac{\partial u}{\partial y}\]

又由\( u=\varphi(u) + \int^x_y p(t) dt \)，两边分别对 \(x\)与\(y\)求偏导，有

\[ \begin{cases}
\frac{\partial u}{\partial x} =& \varphi'(u) \frac{\partial u}{\partial x} +p(x)\\
\frac{\partial u}{\partial y} =& \varphi'(u) \frac{\partial u}{\partial y} -p(y) \\
\end{cases} \Rightarrow
\begin{cases}
\frac{\partial u}{\partial x} =& \frac{p(x)}{1-\varphi'(u)}\\
\frac{\partial u}{\partial y} =&- \frac{p(y)}{1-\varphi'(u)}
\end{cases} \]

所以 \begin{align} ~ & p(y) \frac{\partial z}{\partial x} + p(x) \frac{\partial z}{\partial y} \\
=& f'(u)\left(p(y)\frac{\partial u}{\partial x}+p(x)\frac{\partial u}{\partial y}\right) \\
=& f'(u)\left(p(y)\frac{p(x)}{1-\varphi'(u)} – p(x)\frac{p(y)}{1-\varphi'(u)}\right) \\
=&0
\end{align}

[/collapse]