# Solve the differential equation (x y'-y)y=(x+y)^2 e^(-y/x) ?

Oct 31, 2016

${\log}_{e} x + C = {e}^{\frac{y}{x}} / {\left(1 + \frac{y}{x}\right)}^{2}$

#### Explanation:

Making $y \left(x\right) = x \lambda \left(x\right)$ and substituting into the differential equation we have

${x}^{2} \left(x \lambda \left(x\right) \lambda ' \left(x\right) - {\left(1 + \lambda \left(x\right)\right)}^{2} {e}^{- \lambda \left(x\right)}\right) = 0$

then we have

$\frac{\mathrm{dl} a m b \mathrm{da}}{\mathrm{dx}} = \frac{{\left(1 + \lambda\right)}^{2}}{x \lambda} {e}^{- \lambda}$ which is separable then

$\frac{\mathrm{dx}}{x} = \frac{\lambda \mathrm{dl} a m b \mathrm{da}}{1 + \lambda} ^ 2 {e}^{\lambda}$

integrating we have

${\log}_{e} x + C = {e}^{\lambda} / \left(1 + \lambda\right)$ or

${\log}_{e} x + C = {e}^{\frac{y}{x}} / {\left(1 + \frac{y}{x}\right)}^{2}$