Finding the solution of dy / DX = 1 / (2x-y ^ 2)

Finding the solution of dy / DX = 1 / (2x-y ^ 2)

dx/dy=2x-y^2,
dx/dy-2x=-y^2,(1)
Let DX / dy-2x = 0,
dx/x=2dy,
lnx=2y+lnC1,
x=C1e^(2y),
The parameter variation method is used to replace C1 with V,
x=ve^(2y),(2)
dx/dy=e^(2y)dv/dy+2ve^(2y),(3)
Replace (2) (3) with (1),
e^(2y)dv/dy+2ve^(2y)-2ve^(2y)=-y^2,
dv=-y^2dy/e^(2y),
v=-∫y^2dy/e^(2y),
=（-1/2）[y^2e^(-2y)+ye^(-2y)+(1/2)e^(-2y)}+C2
Substituting (2),
x=(-1/2)e^(2y){[y^2e^(-2y)+ye^(-2y)+(1/2)e^(-2y)}+C2}
=(-1/2)[y^2+y+1/2+Ce^(2y)].

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