Question

The solution of $\frac{dy}{dx}=1+y+y^2+x+xy+x{y}^2$ is

• A. ${\tan }^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)=x+x^2+C$
• B. $4{\tan }^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)=\sqrt{3}(2x+x^2)+C$
• C. $\sqrt{3}{\tan }^{-1}\left(\frac{3y+1}{3}\right)=4(1+x+x^2)+C$
• D. ${\tan }^{-1}\left(\frac{2y+1}{3}\right)=4(2x+x^2)+C$

Answer 1

The correct option is B $4{\tan }^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)=\sqrt{3}(2x+x^2)+C$

Given, $\frac{dy}{dx}=1+y+y^2+x+xy+x{y}^2$

$⟹\frac{dy}{1+y+y^2}=(1+x)dx$

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

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

$4{\tan }^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)=\sqrt{3}(2x+x^2)+C$