Solve the differential equation 2ye^x/y dx + ≤ft( y - 2xe^x/y ) dy =…

The correct answer is Option (2) → $e^{\frac{x}{y}} = \log \sqrt{\frac{c}{y}}$ ##

The given differential equation is

$2ye^{x/y} dx + \left( y - 2xe^{x/y} \right) dy = 0$

Rewrite the differential equation:

$2ye^{x/y} \frac{dx}{dy} = 2xe^{x/y} - y$

$\frac{dx}{dy} = \frac{2xe^{x/y}}{2ye^{x/y}} - \frac{y}{2ye^{x/y}}$

$\frac{dx}{dy} = \frac{x}{y} - \frac{1}{2e^{x/y}}$

Assume that, $x = yv$, this implies $\frac{dx}{dy} = y \frac{dv}{dy} + v$

$y \frac{dv}{dy} + v - v = -\frac{1}{2e^v}$

$y \frac{dv}{dy} = -\frac{1}{2e^v}$

$2e^v dv + \frac{dy}{y} = 0$

Integrating both sides:

$\int 2e^v \, dv + \int \frac{dy}{y} = 0$

$2e^v + \log y = \log c$

$e^{\frac{x}{y}} = \log \sqrt{\frac{c}{y}}$

The solution of the differential equation is $e^{\frac{x}{y}} = \log_e \sqrt{\frac{c}{y}}$.