Find the general solution of $y^2 dx + (x^2 - xy + y^2) dy = 0$.

$\tan^{-1}(x/y) + \ln|y| = C$

The correct answer is Option (2) → $\tan^{-1}(x/y) + \ln|y| = C$ ##

Given, differential equation is

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

$\Rightarrow y^2 dx = -(x^2 - xy + y^2) dy$

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

$\Rightarrow \frac{dx}{dy} = -\left( \frac{x^2}{y^2} - \frac{x}{y} + 1 \right) \quad \dots (i)$

which is a homogeneous differential equation.

Put $\frac{x}{y} = v$ or $x = vy$

$\Rightarrow \frac{dx}{dy} = v + y \frac{dv}{dy}$

On substituting these values in Eq. (i), we get

$v + y \frac{dv}{dy} = -[v^2 - v + 1]$

$\Rightarrow y \frac{dv}{dy} = -v^2 + v - 1 - v$

$\Rightarrow y \frac{dv}{dy} = -v^2 - 1 \Rightarrow \int \frac{dv}{v^2 + 1} = -\int \frac{dy}{y}$

$\text{[applying variable separable method]}$

On integrating both sides, we get

$\tan^{-1}(v) = -\log y + C$

$\Rightarrow \tan^{-1} \left( \frac{x}{y} \right) + \log y = C \quad \left[ ∵v = \frac{x}{y} \right]$