Given $\large\frac{dy}{dx}$$=2xy-y$
This can be written as
$\large\frac{dy}{dx}$$=y(2x-1)$
On seperating the variables we get,
$\large\frac{dy}{y}$$=(2x-1)dx$
Integrating on both sides we get,
$\int\large\frac{dy}{y}$$=\int (2x-1)dx$
$\log _e y=\bigg( \large\frac{2x^2}{2}-x\bigg)+c$
=>$\log _e y=x^2-x+c$
Converting this to exporential form we get,
$c\;e^{\large x-x^2}=y$
Hence the required equation is
$y=c\;e^{\large x-x^2}$