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

\((1+y^2+xy^2)+(x^2y+y+2xy)dy/dx=0\)

Put \(dy/dx= y'\)

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

Now converting the above equation (1) in the form

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

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

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

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

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

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

\(1/(x+1)y+y'= -1/(x+1)^2 y^-1\)

\(y'+1/(x+1)y= - 1/(x+1)^2 y^-1\)

\(p(x)= 1/(x+1)\)

\(q(x)= -1/(x+1)^2\)

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

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

\(y= v^(1/(1-n))\)

\(1/(1-n)v'+p(x)v= q(x)\)

Solving by putting all the values we get

\((v')/2+v/(x+1)= −1/(x+1)^2\)

Put \(v= y^2

and then solve for v

\(v= - 2x/(x^2+1+2x)+c_1/(x^2+1+2x)\)

Substituting back the value of y

\(y= sqrt(-(2x)/(x^2+1+2x)+c_1/(x^2+1+2x))\)

\(y= sqrt((-2x+c_1)/(x^2+2x+1))\)

\((1+y^2+xy^2)+(x^2y+y+2xy)dy/dx=0\)

Put \(dy/dx= y'\)

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

Now converting the above equation (1) in the form

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

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

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

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

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

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

\(1/(x+1)y+y'= -1/(x+1)^2 y^-1\)

\(y'+1/(x+1)y= - 1/(x+1)^2 y^-1\)

\(p(x)= 1/(x+1)\)

\(q(x)= -1/(x+1)^2\)

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

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

\(y= v^(1/(1-n))\)

\(1/(1-n)v'+p(x)v= q(x)\)

Solving by putting all the values we get

\((v')/2+v/(x+1)= −1/(x+1)^2\)

Put \(v= y^2

and then solve for v

\(v= - 2x/(x^2+1+2x)+c_1/(x^2+1+2x)\)

Substituting back the value of y

\(y= sqrt(-(2x)/(x^2+1+2x)+c_1/(x^2+1+2x))\)

\(y= sqrt((-2x+c_1)/(x^2+2x+1))\)