If xdy/dx=y(log y-log x + 1) then the solution of the equation is:

$x\,dy-y\,dx=y\log(\frac{y}{x})dx⇒\frac{x\,dy-y\,dx}{x^2}=\frac{y}{x}.\log(\frac{y}{x}).\frac{dx}{x}⇒\frac{d(\frac{y}{x})}{(\frac{y}{x})\log(\frac{y}{x})}=\frac{dy}{x}$

Substitute $\log (\frac{y}{x})=t⇒\log (\frac{y}{x})=cx$