dy/dx + (2x^2 + 1)y + y^2 + (x^4 + x^2 + 2x) = 0
General solution
Let y = y1 + y2
where y1 = -x^2 is a solution
Then,
d(y1+y2)/dx + (2x^2 + 1)(y1+y2) + (y1+y2)^2 = -(x^4 + x^2 + 2x)
dy1/dx + dy2/dx + (2x^2 + 1)y1 + (2x^2 + 1)y2 + y1^2 + 2y1y2 + y2^2 = -(x^4 + x^2 + 2x)
[dy1/dx + (2x^2 + 1)y1 + y1^2 + (x^4 + x^2 + 2x)] + dy2/dx + (2x^2 + 1)y2 + 2y1y2 + y2^2 = 0
Since [dy1/dx + (2x^2 + 1)y1 + y1^2 + (x^4 + x^2 + 2x)] =0,
Then dy2/dx + (2x^2 + 1)y2 + 2y1y2 + y2^2 = 0
i.e. dy2/dx + (2x^2 + 1)y2 - 2x^2y2 + y2^2 = 0, using y1 = -x^2
dy2/dx + y2 + y2^2 = 0
dy2/(y2(1 + y2)) = -dx
Using partial fractions,
dy2/y2 - dy2/(1+y2) = -dx
integrating both sides,
ln(y2/[K(1 + y2)] = -x
y2/(1 + y2) = Ke^(-x)
(1 + y2)/y2 = Ce^x
1/y2 + 1 = Ce^x
1/y2 = Ce^x - 1
y2 = 1/(Ce^x - 1)
Therefore y = y1 + y2
i.e. y(x) = -x^2 + 1/(Ce^x - 1)