Q. The solution of the differential equation $\frac{dy}{dx} = \frac{x - 2y + 1}{2x - 4y}$ is
A) $(x - 2y)^{2} + 2x = c$
B) $(x - 2y)^{2} + x = c$
C) $(x - 2y) + 2x^{2} = c$
D) $(x - 2y) + x^{2} = c$
Explanation
(a)
Given, $\frac{dy}{dx} = \frac{x - 2y + 1}{2x - 4y}$
Put $x - 2y = z \Rightarrow 1 - 2\ \frac{dy}{dx} = \frac{dz}{dx}$
$\therefore$ $\frac{1}{2}\left\lbrack - \frac{dz}{dx} + 1 \right\rbrack = \frac{z + 1}{2z}$
$\Rightarrow$ $zdz = - dx$
$\Rightarrow$ $\frac{z^{2}}{2} = - x + c_{1}$ [integrating]
$\Rightarrow$ $(x - 2y)^{2} + 2x = c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.