Q. The general solution of $y^{2}dx + \left( x^{2} - xy + y^{2} \right)dy = 0$ is
A) $\tan^{- 1}\left( \frac{x}{y} \right) + \log y + c = 0$
B) $2\tan^{- 1}\left( \frac{x}{y} \right) + \log x + c = 0\$
C) $\log\left( y + \sqrt{x^{2} + y^{2}} \right) + \log y + c = 0$
D) $\sin h^{- 1}\left( \frac{x}{y} \right) + \log y + c = 0$
Explanation
(a)
Given, $\frac{dy}{dx} + \left( \frac{x}{y} \right)^{2} - \left( \frac{x}{y} \right) + 1 = 0$
Put $v = \frac{x}{y}\ \ \ \Rightarrow x = vy$
$\Rightarrow$ $\frac{dx}{dy} = v + y\ \frac{dv}{dy}$
$\therefore$ $v + y\ \frac{dv}{dy} + v^{2} - v + 1 = 0$
$\Rightarrow$ $\frac{dv}{v^{2} + 1} + \frac{dy}{y} = 0$
$\Rightarrow$ $\tan^{- 1}v + \log y + c = 0$ [integrating]
$\Rightarrow$ $\tan^{- 1}\frac{x}{y} + \log y + c = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.