For $y ≠ 0$, the particular solution of the differential equation $2ye^{x/y} dx + (y - 2xe^{x/y})dy = 0$ at the point (1, 1) is

$ln |y|+ 2(e^{x/y}-e) = 0$

The correct answer is Option (1) → $ln |y|+ 2(e^{x/y}-e) = 0$

Given differential equation:

$2y e^{\frac{x}{y}} dx + \left(y - 2x e^{\frac{x}{y}}\right) dy = 0$

Rewriting:

$2y e^{\frac{x}{y}} \, dx + \left(y - 2x e^{\frac{x}{y}}\right) \, dy = 0$

Divide the whole equation by $y$ (since $y \ne 0$):

$2 e^{\frac{x}{y}} \, dx + \left(1 - \frac{2x}{y} e^{\frac{x}{y}}\right) \, dy = 0$

Let $u = \frac{x}{y} \Rightarrow x = uy$

Differentiate: $dx = u \, dy + y \, du$

Substitute into the original equation:

$2 e^{u}(u \, dy + y \, du) + \left(1 - 2u e^{u} \right) dy = 0$

Distribute:

$2u e^u \, dy + 2y e^u \, du + \left(1 - 2u e^u \right) dy = 0$

Combine $dy$ terms:

$\left(2u e^u + 1 - 2u e^u\right) dy + 2y e^u \, du = 0$

$dy + 2y e^u \, du = 0$

$\Rightarrow dy = -2y e^u \, du$

Separate variables:

$\frac{dy}{y} = -2 e^u \, du$

Integrate both sides:

$\int \frac{1}{y} dy = -2 \int e^u \, du$

$\ln |y| = -2 e^u + C$

Recall $u = \frac{x}{y}$:

$\ln |y| = -2 e^{\frac{x}{y}} + C$

Use point $(1, 1)$ to find $C$:

$\ln 1 = -2 e^{1} + C \Rightarrow 0 = -2e + C \Rightarrow C = 2e$

Final answer: ${\ln y = -2 e^{\frac{x}{y}} + 2e}$