Find the particular solution of the differential equation $\left( xe^{y/x} + y \right) dx = x \, dy$, given that $y = 1$ when $x = 1$.

$e^{-1} - e^{-y/x} = \log |x|$

The correct answer is Option (2) → $e^{-1} - e^{-y/x} = \log |x|$ ##

Given $\left( xe^{y/x} + y \right) dx = x \, dy$ and $y = 1, x = 1$

The given differential equation is a homogeneous function of degree zero.

To solve it, we make substitution $y = vx, \frac{dy}{dx} = v + x \frac{dv}{dx} \quad \dots(i)$

$\Rightarrow \frac{dy}{dx} = \frac{xe^{y/x} + y}{x}$

$\Rightarrow \frac{dy}{dx} = e^{y/x} + \frac{y}{x}$

from (i):

$v + x \frac{dv}{dx} = e^v + v$

$\Rightarrow \int \frac{dv}{e^v} = \int \frac{dx}{x}$

$\Rightarrow -e^{-v} = \log |x| + C$

$\Rightarrow -e^{-y/x} = \log |x| + C$

It is given that $x = 1, y = 1$

So, $-e^{-1} = \log(1) + C ⇒C = -e^{-1}$

Hence, required solution:

$e^{-1} - e^{-y/x} = \log |x|$