Q. The differential equation representing the family of curves $y = xe^{cx}$ ($c$ is a constant) is
A) $\frac{dy}{dx} = \frac{y}{x}\left( 1 - \log\frac{y}{x} \right)$
B) $\frac{dy}{dx} = \frac{y}{x}\log\left( \frac{y}{x} \right) + 1$
C) $\frac{dy}{dx} = \frac{y}{x}\left( 1 + \log\frac{y}{x} \right)$
D) $\frac{dy}{dx} + 1 = \frac{y}{x}\log\left( \frac{y}{x} \right)$
Explanation
(c)
Given, $y = xe^{cx}$ …(i)
$\Rightarrow$ $\frac{dy}{dx} = e^{cx} + xe^{cx}.c = \frac{y}{x} + y.c$ …(ii)
From Eq. (ii),
Log $y = \log{x + cx}$
$\Rightarrow c\frac{1}{x}\log\frac{y}{x}$
$\frac{dy}{dx} = \frac{y}{x} + \frac{y}{x}\log\frac{y}{x}$
$= \frac{y}{x}\left( 1 + \log\frac{y}{x} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.