Question

Solve
$(1+e^{x/y})dx+e^{x/y}(1-\frac{x}{y})dy=0$

Answer 1

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

$(1+e^{x/y})dx=-e^{x/y}(1-\frac{x}{y})dy$

$\frac{dx}{dy}=\frac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$\frac{dx}{dy}=f(x,y)$

$f(x,y)=\frac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$f(\lambda x,\lambda y)=\frac{e^{-\lambda x/\lambda y}(1-\lambda x/\lambda y)}{1+e^{\lambda x/\lambda y}}$

$=\frac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$={\lambda }^{o}f(x,y)$

Let $x=vy$

Put

$\frac{dx}{dy}=v\frac{dy}{dy}+y\frac{dv}{dy}$

$\frac{dx}{dy}=v+y\frac{dv}{dy}$

As $\frac{dx}{dy}=\frac{e^{-x/y}(1-x/y)}{1+e^{x/y}}$

$V+y\frac{dv}{dy}=\frac{-e^{vy/y}(1-vy/y)}{1+e^{vy/y}}$

$y\frac{dv}{dy}=\frac{-e^v(1-v)}{1+e^v}-V$

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

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

Integrate Both sides

$\int \frac{1+e^v}{v+e^v}dv=-\log y+\log C$

$V+e^v=t\to (1+e^v)dv=dt$

$\int \frac{dt}{t}=-\log y+\log C$

now substitute $t$

$\log t=-\log y+\log c$

$\log (v+e^v)=-\log y+\log C$

$\log (v+e^v)+\log y=\log C$

$\log y(v+e^v)=\log C$

Put $v=\frac{x}{y}$

$\log y\left(\frac{x}{y}\right)+e^{\frac{x}{y}}$

$\log y(\frac{x}{y}+e^{\frac{x}{y}})=\log c$

$y(\frac{x}{y}+e^{x/y})=C$

$x+y{e}^{x/y}=C$.