Question

$\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$

• A. ${\tan }^{-1}\left(2x+1\right)+{\tan }^{-1}\left(2y+1\right)=c$
• B. ${\tan }^{-1}\frac{2x+1}{\sqrt{3}}+{\tan }^{-1}\frac{2y+1}{\sqrt{3}}=c$
• C. ${\tan }^{-1}\frac{2x}{\sqrt{5}}+{\tan }^{-1}\frac{2y}{\sqrt{3}}=c$
• D. ${\tan }^{-1}\frac{2x-1}{\sqrt{3}}+{\tan }^{-1}\frac{2y-1}{\sqrt{3}}=c$

Answer 1

The correct option is B ${\tan }^{-1}\frac{2x+1}{\sqrt{3}}+{\tan }^{-1}\frac{2y+1}{\sqrt{3}}=c$

$\frac{dy}{dx}=-\frac{y^2+y+1}{x^2+x+1}$

$\Rightarrow \frac{dy}{y^2+y+1}=-\frac{dx}{x^2+x+1}$

On integrating, we get

$\int \frac{dy}{y^2+y+1}=-\int \frac{dx}{x^2+x+1}$

$\int \frac{dy}{y^2+2\times \frac{1}{2}y+\frac{1}{4}-\frac{1}{4}+1}=-\int \frac{dx}{x^2+2\times \frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+1}$

$\int \frac{dy}{{(y+\frac{1}{2})}^2+\frac{3}{4}}=-\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}$

$\int \frac{dy}{{(y+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=-\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+k$

or $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=k$

or ${\tan }^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=\frac{\sqrt{3}}{2}k$

or ${\tan }^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+{\tan }^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=c$ where $c=\frac{\sqrt{3}}{2}k$

or ${\tan }^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)+{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)=c$ where $c=\frac{\sqrt{3}}{2}k$