由pg(x)=2xy→q(x)=x2+c 王由q0)=0,知e=0→ p(r)=x2 1,1) 故「xy2dx+pp(x) (0,0) =∫yt=1 2 上页由(0) = 0，知c = 0 2  (x) = x . 故  +  (1,1) (0,0) 2 xy dx y (x)dy 由y(x) = 2xy   x = x + c 2 ( )   = + 1 0 1 0 0dx ydy . 2 1 =