The solution of equation $(2y - 1) \, dx - (2x + 3) \, dy = 0$ is

$\frac{2x + 3}{2y - 1} = k$

The correct answer is Option (3) → $\frac{2x + 3}{2y - 1} = k$ ##

Given that, $(2y - 1) \, dx - (2x + 3) \, dy = 0$

$\Rightarrow (2y - 1) \, dx = (2x + 3) \, dy$

$\Rightarrow \frac{dx}{2x + 3} = \frac{dy}{2y - 1} \quad \text{[using variable separable method]}$

On integrating both sides, we get

$\frac{1}{2} \log (2x + 3) = \frac{1}{2} \log (2y - 1) + \log C$

$\Rightarrow \frac{1}{2} [\log (2x + 3) - \log (2y - 1)] = \log C$

$\Rightarrow \frac{1}{2} \log \left( \frac{2x + 3}{2y - 1} \right) = \log C$

$\Rightarrow \left( \frac{2x + 3}{2y - 1} \right)^{1/2} = C$

$\Rightarrow \frac{2x + 3}{2y - 1} = C^2$

$\Rightarrow \frac{2x + 3}{2y - 1} = k, \text{ where } k = C^2$