Question

What is B equal to ?

• A. -1
• B. 1
• C. 2
• D. None of the above

Answer 1

The correct option is A -1

$(x^2+x+1)dy+(y^2+y+1)dx=0$

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

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

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

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

For ease of solving let $c=\frac{2}{\sqrt{3}}{\tan }^{-1}c$ where c is a constant

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

We know that

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1+xy}$

On solving for c, we get

$c=\frac{2\sqrt{3}(x+y+1)}{-4xy-2y-2x+2}(x+y+1)=\frac{c}{\sqrt{3}}(1-x-y-2xy)$

Given general solution of the equation

$(x+y+1)=A(1+Bx+Cy+Dxy)$

Comparing, we get

$B=-1$