$xy \, dx - (x + 2) \, dy = 0$
$\dfrac{xy \, dx}{y(x + 2)} - \dfrac{(x + 2) \, dy}{y(x + 2)} = 0$
$\dfrac{x \, dx}{x + 2} - \dfrac{dy}{y} = 0$
$\left( 1 - \dfrac{2}{x + 2} \right) \, dx - \dfrac{dy}{y} = 0$
$\displaystyle \int dx - 2 \int \dfrac{dx}{x + 2} - \int \dfrac{dy}{y} = 0$
$x - 2 \ln (x + 2) - \ln y = \ln c$
$x = \ln c + \ln y + 2 \ln (x + 2)$
$x = \ln c + \ln y + \ln (x + 2)^2$
$x = \ln cy(x + 2)^2$
$\ln e^x = \ln cy(x + 2)^2$
$e^x = cy(x + 2)^2$ answer