∵
=f′(u)?z ?x
,?u ?x
=f′(u)?z ?y
?u ?y
而u=φ(u)+
p(t)dt
∫
两边对x求偏导得:
=φ′(u)?u ?x
+p(x)?u ?x
两边对y求偏导得:
=φ′(u)?u ?y
?p(y)?u ?y
∴
=?u ?x
p(x) 1?φ′(u)
=?u ?y
?p(y) 1?φ′(u)
从而
p(y)
+p(x)?z ?x
=f′(u)[?z ?y
+p(x)p(y) 1?φ′(u)
]=0?p(x)p(y) 1?φ′(u)