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2xy dx + (x 2 + 1) dy = 0
(x 2 + y 2 − 2x) dx + 2xy dy = 0
y 2 dx + (xy − 1) dy = 0
R2
γ (0, 0) (x0, y0)
γ = γ1 ∪ γ2 γ1(x) = (x, 0) 0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y)
0 < y < y0 y0 ≤ y ≤ 0 ˙γ1(x) = (1, 0) ˙γ2(x) = (0, 1)
V (x0, y0) = ω = (2xy dx + (x 2
+ 1) dy) + (2xy dx + (x 2 + 1) dy)
γ γ1
γ2
= (2xy dx + (x
γ2
2 y0
+ 1) dy) = (x
0
2 0 + 1) dy = (x 2 0 + 1)y0.
V (x, y) = (x2 + 1)y (x2 + 1)y = c c ∈ R
1 + x2 = 0
y ′ (x) = dy 2xy
(x) = −
dx x2 + 1
y ′ 2x
= −
y x2 d
= −
+ 1 dx log(x2 + 1)
dy d
= −
y dx log(x2 + 1)
log |y| = − log(x 2 + 1) + d d ∈ R |y| = ed
x 2 +1
y = c
x 2 +1
c ∈ R c = ±e d
R 2
γ (0, 0) (x0, y0)
γ = γ1 ∪ γ2 γ1(x) = (x, 0) 0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y)
0 < y < y0 y0 ≤ y ≤ 0
V (x0, y0) = ω =
=
γ
x0
0
γ1
((x 2 + y 2 − 2x) dx + 2xy dy) +
(x 2 − 2x) dx +
y0
0
γ2
2x0y dy = x3 0
3 − x2 0 + x0y 2 0.
((x 2 + y 2 − 2x) dx + 2xy dy)
V (x, y) = x3
3 − x2 + xy 2 x(x 2 /3 − x + y 2 ) = c c ∈ R
R 2 \ {xy = 0} 2xy
y ′ (x) = dy
dx = −x2 + y2 − 2x
,
2xy
p(x, y) = y2 q(x, y) = xy − 1 ∂yp − ∂xq = 2y − y = y = 0 ω
∂yp − ∂xq = y = f(x)q(x, y) − g(y)p(x, y)
y f ≡ 0 g(y) = −1/y
(x0, y0) = (0, 1)
x y y
h(x, y) = exp f(t)dt + g(t)dt = exp −
1
1
t dt
= e log(1/y) = 1/y
x0
y0