Question

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

Answer 1

We have,

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

$\frac{dx}{dy}=-\frac{(y-2x{e}^{\frac{x}{y}})}{2y{e}^{\frac{x}{y}}}$ ……. (1)

Put $x=vy$

On differentiating w.r.t $y$, we get

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

From equation (1), we get

$v+y\frac{dv}{dy}=-\frac{(y-2vy{e}^v)}{2y{e}^v}$

$y\frac{dv}{dy}=-\frac{(y-2vy{e}^v)}{2y{e}^v}-v$

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

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

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

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

On integration both sides, we get

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

$2e^v=-\ln y+C$

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

Hence, this is the answer.