Q. The solution of the differential equation $\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition $y(1) = 1$ is
A) $y = x\log x + x$
B) $y = \log x + x$
C) $y = x\log x + x^{2}$
D) $y = xe^{(x - 1)}$
Explanation
(a)
Given equation can be rewritten as
$\frac{dy}{dx} - \frac{1}{x}.y = 1$
$\therefore$ $IF = e^{- \int_{}^{}{\frac{1}{x}dx}} = e^{- \log x} = \frac{1}{x}$
$\therefore$ Required solution is
$y\left( \frac{1}{x} \right) = \int_{}^{}\frac{1}{x}\ dx = \log x + c$
Since, $y(1) = 1$
$\Rightarrow$ $c = 1$
$\therefore$ $y = x\log x + x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.