# Solve differential equation1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0

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

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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
$$y=\sqrt{-\frac{2x}{x^2+1+2x}+\frac{c_{1}}{x^2+1+2x}}$$
$$y=\sqrt{-2x+\frac{c_1}{x^2+2x+1}}$$