由p(x)=2xy→q(x)=x2+c 由φ(0)=0,知c=0→q(x)=x2 (1,1) 故「。x+p(x)d (0,0) a+y=1由y(x) = 2xy   x = x + c 2 ( ) 由(0) = 0，知c = 0 2  (x) = x . 故  +  (1,1) (0,0) 2 xy dx y (x)dy   = + 1 0 1 0 0dx ydy . 2 1 =