General solution of differential equation (2x-y ^ 2) * y '= 2Y? Thank you!

General solution of differential equation (2x-y ^ 2) * y '= 2Y? Thank you!

Obviously, y = 0 is the solution of the original equation
When y ≠ 0,
∵(2x-y^2)y'=2y
==>-2xdy+2ydx+y^2dy=0
==>-2xdy / y ^ 2 + 2DX / y + dy = 0 (both ends of the equation divide y ^ 2)
==>2xd(1/y)+2dx/y+dy=0
==>2d(x/y)+dy=0
==>2X / y + y = C (C is constant)
==>2x+y^2=Cy
The solution of the original equation is also 2x + y ^ 2 = cy
So the general solution of the original equation is y = 0 or 2x + y ^ 2 = cy