# Solve differential equation 1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0

Question
Solve differential equation $$1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0$$

2021-03-08
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))$$

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