# Show that the general solution of the differential equation$$\large\frac{dy}{dx}+\large\frac{y^2+y+1}{x^2+x+1}=0$$ is given by $$(x+y+1)\;=\;A(1-x-y-2xy)$$, where A is parameter

Toolbox:
• A first order degree differential equation is of the form $\large\frac{dy}{dx} = $$F(x,y), and if F(x,y) can be expressed as a product of g(x).h(y), where g(x) is a function of x and h(y) is a function of y, then it is said to be of variables seperable type. Step 1: Given :\large\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}$$=0$
$\large\frac{dy}{dx}=-\large\frac{y^2+y+1}{x^2+x+1}$
Using the information in the tool box we identify the equation as variables seperable type.
Seperating the variables we get,
$\large\frac{dy}{y^2+y+1}=-\large\frac{dx}{x^2+x+1}$
Step 2:
Integrating on both sides
$\int \large\frac{dy}{y^2+y+1}=\int\large\frac{-dx}{x^2+x+1}$
$\int \large\frac{dy}{(y+\large\frac{1}{2})^2+(\sqrt{\large\frac{3}{2}})^2}=-\int \large\frac{dx}{(x+\large\frac{1}{2})^2+(\sqrt{\large\frac{3}{2}})^2}$
$\large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2y+1)}{\sqrt 3}=-\large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2x+1)}{\sqrt 3}$$+C \large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2y+1)}{\sqrt 3}+\large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2x+1)}{\sqrt 3}$$=C$
Step 3:
Apply the formula $\tan^{-1}x +\ tan^{-1}y = tan^{-1}\large\frac{(x+y)}{1-xy}$
$\large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2y+1)}{\sqrt 3}+\large\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{(2x+1)}{\sqrt 3}=\large\frac{2}{\sqrt 3}\tan^{-1}\large\frac{(2x+2y+1)}{\sqrt 3}\times \large\frac{3}{(4-4xy+2x+2y)}=C$
Step 4:
On simplifying we get,
$\tan^{-1}\large\frac{(x+y+1)}{1-x-y-2xy)}=\sqrt{\large\frac{3}{2}}$$+C \large\frac{(x+y+1)}{1-x-y-2xy}$$=\tan\sqrt{\large\frac{3C}{2}}$
Let $\tan\sqrt {\large\frac{3C}{2}}=$$A \large\frac{(x+y+1)}{1-x-y-2xy}$$=A$
$x+y+1=A(1-x-y-2xy)$
Hence proved.