(x+ay)dx+ydy |
(x+y)2 |
df=
∂f |
∂x |
∂f |
∂y |
∂ |
∂x |
∂f |
∂y |
∂ |
∂y |
∂f |
∂x |
因此,
∂f |
∂x |
x+ay |
(x+y)2 |
∂f |
∂y |
y |
(x+y)2 |
∂ |
∂y |
∂f |
∂x |
∂ |
∂y |
x+ay |
(x+y)2 |
a |
(x+y)2 |
x+ay |
(x+y)3 |
(a−2)x−ay |
(x+y)3 |
∂ |
∂x |
∂f |
∂y |
∂ |
∂x |
y |
(x+y)2 |
y |
(x+y)3 |
所以,
(a−2)x−ay |
(x+y)3 |
−2y |
(x+y)3 |
所以,a=2
故选:D.
(x+ay)dx+ydy |
(x+y)2 |
(x+ay)dx+ydy |
(x+y)2 |
∂f |
∂x |
∂f |
∂y |
∂ |
∂x |
∂f |
∂y |
∂ |
∂y |
∂f |
∂x |
∂f |
∂x |
x+ay |
(x+y)2 |
∂f |
∂y |
y |
(x+y)2 |
∂ |
∂y |
∂f |
∂x |
∂ |
∂y |
x+ay |
(x+y)2 |
a |
(x+y)2 |
x+ay |
(x+y)3 |
(a−2)x−ay |
(x+y)3 |
∂ |
∂x |
∂f |
∂y |
∂ |
∂x |
y |
(x+y)2 |
y |
(x+y)3 |
(a−2)x−ay |
(x+y)3 |
−2y |
(x+y)3 |