First Order Bernoulli differential equation is of the form \(y'+p(x)y=q(x)y^n\)

\((1+y^2+xy^2)+(x^2y+y+2xy)\ \frac{dy}{dx}=0\)

Put \(\frac{dy}{dx}= y'\)

\((1+y^2+xy^2)+(x^2y+y+2xy)\ y'=0\) (1)

Now converting the above equation (1) in the form

\(y'+p(x)y=q(x)y^n\)

Dividing the whole equation (1) by \((x^2y+y+2xy)\)

\((\frac{1+y^2+xy^2}{x^2y+y+2xy}+\frac{x^2y+y+2xy}{x^2y+y+2xy})y'=0\)

\((\frac{1+y^2(x+1)}{y(x^2+1+2x)})+y'=0\)

\(\frac{1+y^2(x+1)}{y(x+1)^2}+y'=0\)

\(\frac{1}{y(x+1)^2}+\frac{y^2(x+1)}{y(x+1)^2}+y'=0\)

\(\frac{1}{x+1}y+y'= -\frac{1}{x+1}^2 y^{-1}\)

\(y'+\frac{1}{x+1}y= - \frac{1}{x+1}^2 y^{-1}\)

\(p(x)= \frac{1}{x+1}\)

\(q(x)= -\frac{1}{x+1}^2\)

\(n= -1\) Now, substitute \(v=y^{1−n}\) in the Bernoulli equation, we get

\(v= y^{1-n}\)

\(y= v^{\frac{1}{1}-n}\)

\(\frac{1}{1-n}v'+p(x)v= q(x)\)

Solving by putting all the values we get

\(\frac{v'}{2}+\frac{v}{x+1}= −\frac{1}{x+1}^2\)

Put \(v= y^{2}\)

and then solve for v

\(v= -\frac{ 2x}{x^2+1+2x}+\frac{c_1}{x^2+1+2x}\)

Substituting back the value of y

\(y=\sqrt{-\frac{2x}{x^2+1+2x}+\frac{c_{1}}{x^2+1+2x}}\)

\(y=\sqrt{-2x+\frac{c_1}{x^2+2x+1}}\)