dy/dx=(4x∧2+y∧2-2y+1)/(2xy-2x) 感激不尽

∵dy/dx=(4x^2+y^2-2y+1)/(2xy-2x)
==>(2xy-2x)dy-(4x^2+y^2-2y+1)dx=0
==>2x(y-1)dy-(4x^2+(y-1)^2)dx=0
==>2(y-1)dy/x-(y-1)^2dx/x^2-4dx=0 (等式两端同除x^2)
==>d((y-1)^2)/x+(y-1)^2d(1/x)-4dx=0
==>d((y-1)^2/x)-4dx=0
==>(y-1)^2/x-4x=C (C是积分常数)
==>(y-1)^2=x(4x+C)
∴原方程的通解是(y-1)^2=x(4x+C).