# How to solve this differential equation: x d y </mrow>

How to solve this differential equation:
$x\frac{dy}{dx}=y+x\frac{{e}^{x}}{{e}^{y}}?$
I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v=\frac{y}{x}$ to solve it.
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Tristan Ward
$x\frac{dy}{dx}=y+x\frac{{e}^{x}}{{e}^{y}}$
$x\frac{dy}{dx}=y+x{e}^{x-y}$
Let $u=y-x$,
Then $y=u+x$
$\frac{dy}{dx}=\frac{du}{dx}+1$
$\therefore x\left(\frac{du}{dx}+1\right)=u+x+x{e}^{-u}$
$x\frac{du}{dx}+x=u+x+x{e}^{-u}$
$x\frac{du}{dx}=x{e}^{-u}+u$
$\left(x{e}^{-u}+u\right)\frac{dx}{du}=x$
Let $v=x+u{e}^{u}$,
Then $x=v-u{e}^{u}$
$\frac{dx}{du}=\frac{dv}{du}-\left(u+1\right){e}^{u}$
$\therefore {e}^{-u}v\left(\frac{dv}{du}-\left(u+1\right){e}^{u}\right)=v-u{e}^{u}$
${e}^{-u}v\frac{dv}{du}-\left(u+1\right)v=v-u{e}^{u}$
${e}^{-u}v\frac{dv}{du}=\left(u+2\right)v-u{e}^{u}$
$v\frac{dv}{du}=\left(u+2\right){e}^{u}v-u{e}^{2u}$
This belongs to an Abel equation of the second kind.
Let $t=\left(u+1\right){e}^{u}$,
Then $u=W\left(et\right)-1$
$\frac{dv}{du}=\frac{dv}{dt}\frac{dt}{du}=\left(u+2\right){e}^{u}\frac{dv}{dt}$
$\therefore \left(u+2\right){e}^{u}v\frac{dv}{dt}=\left(u+2\right){e}^{u}v-u{e}^{2u}$
$v\frac{dv}{dt}=v-\frac{u{e}^{u}}{u+2}$
$v\frac{dv}{dt}-v=-\frac{t\left(W\left(et\right)-1\right)}{W\left(et\right)\left(W\left(et\right)+1\right)}$
This belongs to an Abel equation of the second kind in the canonical form.