Q. The solution of $\frac{dy}{dx} = 1 + y + y^{2} + x + xy + xy^{2}$ is
A) $\tan^{- 1}\left( \frac{2y + 1}{\sqrt{3}} \right) = x + x^{2} + c$
B) $4\tan^{- 1}\left( \frac{4y + 1}{\sqrt{3}} \right) = \sqrt{3}\left( 2x + x^{2} \right) + c$
C) $\sqrt{3}\tan^{- 1}{\left( \frac{3y + 1}{3} \right) = 4\ \left( 1 + x + x^{2} \right) + c}$
D) $4\tan^{- 1}\left( \frac{2y + 1}{\sqrt{3}} \right) = \sqrt{3}\left( 2x + x^{2} \right) + c$
Explanation
(d)
Given, $\frac{dy}{1 + y + y^{2}} = (1 + x)dx$
$\Rightarrow$ $\int_{}^{}\frac{dy}{\left( y + \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}} = \int_{}^{}{(1 + x)}dx$
$\Rightarrow$ $\frac{1}{\frac{\sqrt{3}}{2}}\tan^{- 1}\left( \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = x + \frac{x^{2}}{2} + \frac{c}{2}$
$\Rightarrow$ $4\tan^{- 1}\left( \frac{2y + 1}{\sqrt{3}} \right) = \sqrt{3}\left( 2x + x^{2} \right) + c\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.