Show that the general solution of the differentia equation `(dy)/(dx)+(y^2+y+1)/(x^2+x+1)=0` is given by `x+y+1=A(1-x-y-2x y)` where A is a parameter.

Explanation

$$x+y+1=A(1-x-y-2 x y)$$
Differentiate both sides w.r.t. `x`
$$\Rightarrow 1+y^{\prime}=A\left(-1-y^{\prime}-2 y-2 x y\right)$$
$$\Rightarrow(1+A+2 A x) y^{\prime}=-1+A(-1-2 y)$$
$$\Rightarrow y^{\prime}=-\frac{1+A(1+2 y)}{1+A(1+2 x)} \ldots \ldots(1)$$
$$\Rightarrow(1+A+2 A x) y^{\prime}=-1+A(-1-2 y)$$
$$\Rightarrow y^{\prime}=-\frac{1+A(1+2 y)}{1+A(1+2 x)}$$
Now, from the given equation, $$A=\frac{x+y+1}{1-x-y-2 x y}$$.
...