Find the particular solution of the differential equation: xe^y/x - y…

The correct answer is Option (2) → $\left[ \sin\left(\frac{y}{x}\right) + \cos\left(\frac{y}{x}\right) \right] e^{-\frac{y}{x}} = \log x^2 + 1$ ##

Given differential equation is homogeneous.

$∴$ Putting $y = vx$ to get $\frac{dy}{dx} = v + x \frac{dv}{dx}$

$\frac{dy}{dx} = \frac{y \sin\left(\frac{y}{x}\right) - xe^{\frac{y}{x}}}{x \sin\left(\frac{y}{x}\right)}$

$\text{or } v + x \frac{dv}{dx} = \frac{v \sin v - e^v}{\sin v}$

$\text{or } v + x \frac{dv}{dx} = v - \frac{e^v}{\sin v}$

$\text{or } x \frac{dv}{dx} = -\frac{e^v}{\sin v}$

$∴\int \sin v e^{-v} dv = -\int \frac{dx}{x}$

$\text{or } I_1 = -\log x + C_1 \quad \dots(i)$

$\text{or } I_1 = -\sin v e^{-v} + \int \cos v e^{-v} dv$

$\text{or } I_1 = -\sin v e^{-v} - \cos v e^{-v} - \int \sin v e^{-v} dv$

$\text{or } I_1 = -\frac{1}{2}(\sin v + \cos v)e^{-v}$

Putting (i), $(\sin v + \cos v) e^{-v} = \log x^2 - 2C_1$

$\text{or } \left[ \sin\left(\frac{y}{x}\right) + \cos\left(\frac{y}{x}\right) \right] e^{-\frac{y}{x}} = \log x^2 + C_2$

For $x = 1, y = 0$ or $C_2 = 1$

Hence, solution is $\left[ \sin\left(\frac{y}{x}\right) + \cos\left(\frac{y}{x}\right) \right] e^{-\frac{y}{x}} = \log x^2 + 1$