Browse Questions

The solution of the equation $(2y-1)dx-(2x+3)dy=0$ is$(A)\;\frac{2x-1}{2y+3}=k \quad (B)\;\frac{2y+1}{2x-3}=k \quad (C)\;\frac{2x+3}{2y-1}=k \quad (D)\;\frac{2x-1}{2y+1}=k$

Toolbox:
• If the linear differential eqation is of the form $\large\frac{dy}{dx}$$=f(x), then it can be solved by seperaring the variable and integrating. • \int \large\frac{dx}{x+1}$$= \log |x+1|+c$
Given $(2y-1)dx-(2x+3)dy=0$
Seperating the variables we get,
$\large\frac{dy}{(2y-1)}=\frac{dx}{(2x+3)}$
On integrating we get
$\int \large\frac{dy}{(2y-1)}=\int \frac{dx}{(2x+3)}$
$=>\large\frac{1}{2} $$\log(2y-1)=\large\frac{1}{2}$$\log (2x+3)+\log c$
$=>\log(2y-1)=\log 2c (2x+3)$
$\large \frac{2y-1}{2x+3}$$=2c \large \frac{2x+3}{2y-1}=\frac{1}{2c}$$=k\qquad (where \;k=\large\frac{1}{2c})$