Question

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

Answer 1

Given, differential equation is
$\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0\Rightarrow \frac{dy}{y^2+y+1}+\frac{dx}{x^2+x+1}=0$
On integrating both sides, we get
$\int \frac{dy}{y^2+y+1}+\int \frac{dx}{x^2+x+1}=C\Rightarrow \int \frac{dy}{y^2+y+1+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}+\int \frac{dx}{x^2+x+1+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}=C\Rightarrow \int \frac{dy}{{(y+\frac{1}{2})}^2+(1-\frac{1}{4})}+\int \frac{dx}{{(x+\frac{1}{2})}^2+(1-\frac{1}{4})}=C\Rightarrow \int \frac{dy}{{(y+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}+\int \frac{dx}{{(x+\frac{1}{2})}^2+(1-\frac{1}{4})}=C\Rightarrow \frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\frac{2}{\sqrt{3}}ta{n}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=C$
$[\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}ta{n}^{-1}\frac{x}{a}]$
$\Rightarrow ta{n}^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)+ta{n}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)=\frac{\sqrt{3}C}{2}=k$ (say)
$\Rightarrow ta{n}^{-1}\left[\frac{\frac{2y+1}{\sqrt{3}}+\frac{2x+1}{\sqrt{3}}}{1-\left(\frac{2y+1}{\sqrt{3}}\right)\left(\frac{2x+1}{\sqrt{3}}\right)}\right]=k$
$[\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)]$
$\Rightarrow ta{n}^{-1}\left[\frac{\frac{2y+1+2x+1}{\sqrt{3}}}{1-\left(\frac{4xy+2x+2y+1}{3}\right)}\right]=k\Rightarrow \frac{2\sqrt{3}(x+y+1)}{3-(4xy+2x+2y+1)}=tank\Rightarrow \frac{2\sqrt{3}(x+y+1)}{2(1-x-y-2xy)}=tank\Rightarrow x+y+1=\frac{1}{\sqrt{3}}tank(1-x-y-2xy)$
Let, $A=\frac{1}{\sqrt{3}}tank$ is an arbitrary constant.
$\Rightarrow x+y+1=A(1-x-y-2xy)$
Hence, the given result is proved.