$\left( {1 + {e^{x/y}}} \right)dx + {e^{x/y}}\left( {1 - \cfrac{x}{y}} \right)dy = 0$

: $\left( {1 + {e^{x/y}}} \right)dx + {e^{x/y}}\left( {1 - \cfrac{x}{y}} \right)dy = 0$

$\Rightarrow \cfrac{{dx}}{{dy}} = \cfrac{{{e^{x/y}}\left( {\cfrac{x}{y} - 1} \right)}}{{{e^{x/y}} + 1}}$

…(1)
Since R.H.S. is of the $g(x/y)$ , so it is a homogeneous function of degree zero.

Therefore, equation is a homogeneous differential equation.

To solve this, we put $\cfrac{x}{y} = v \Rightarrow x = vy \Rightarrow \cfrac{{dy}}{{dx}} = v + y\cfrac{{dv}}{{dy}}$

Then, (1) becomes $v + y\cfrac{{dv}}{{dy}} = \cfrac{{{e^v}(v - 1)}}{{{e^\nu } + 1}}$

$\Rightarrow y\cfrac{{dv}}{{dy}} = \cfrac{{v{e^v} - {e^v} - v{e^v} - v}}{{{e^v} + 1}} \Rightarrow \cfrac{{{e^v} + i}}{{v + {e^v}}}dv = - \cfrac{{dy}}{y}$

…(2)
Integrating (2) both sides,

we get
$\int {\left( {\cfrac{{{e^v} + 1}}{{{e^v} + v}}} \right)} dv = - \int {\cfrac{{dy}}{y}} \Rightarrow \;\log |{e^v} + v| = - \log |y| + {C_1}$

$\Rightarrow \log |({e^v} + v)y| \cdot {C_1} \Rightarrow |({e^v} + v)y| = {e^{{C_1}}}$
$\Rightarrow ({e^v} + v)y = \pm {e^{{C_1}}} = C$

(say)
$\Rightarrow \left( {{e^{x/y}} + \cfrac{x}{y}} \right)y = C$

$\Rightarrow y{e^{x/y}} + x = C$,

which is the required general solution.

For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition: