Question

Solve the differential equation $y{e}^{\frac{x}{y}}dx=(x{e}^{\frac{x}{y}}+y^2)dy(y\ne 0)$.

Answer 1

The differential equation is
$y{e}^{\frac{x}{y}}dx=(x{e}^{\frac{x}{y}}+y^2)dy$
let's take p as $p=e^{\frac{x}{y}}$
so differential equation will become
$ypdx=(xp+y^2)dy$
differentiating p with respect to x

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

$⟹y^2\frac{dp}{dx}=p(y-x\frac{dy}{dx})$

$⟹pydx=y^{2}dp+pxdy$
apply this on above differential equation

$xpdy+y^{2}dp=xpdy+y^{2}dy$

$⟹dy=dp,$

$⟹y=p+C,$
hence the solution is :
$y=e^{\frac{x}{y}}+C$