Differential Equations — Class 12 Maths Solution

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

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

$\Rightarrow$ $\frac{{dx}}{{2x + 3}} = \frac{{dy}}{{2y - 1}}$

On integrating both sides,

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

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

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

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

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

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