Question

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

Answer 1

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

Separating the variables we get

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

On integration, we get

$\int \frac{dy}{(2y-1)}=\int \frac{dx}{2x+3}$

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

$\Rightarrow \log (2y-1)=\log 2c(2x+3)$

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

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

where $k=\frac{1}{2c}$

Hence,the solution of the equation is $\frac{2x+3}{2y-1}=k$