The general solution of differential equation log_e≤ft(dy/dx)=2x-y,y(…

$\log\left(\frac{dy}{dx}\right)=2x-y$

so $\frac{dy}{dx}=\frac{e^{2x}}{e^y}⇒\int e^ydy=\int e^{2x}dx$

$e^y=\frac{e^{2x}}{2}+c$ at (0, 0)

$e^0=\frac{e^0}{2}+c⇒c=\frac{1}{2}$

$e^y=\frac{e^{2x}}{2}+\frac{1}{2}$

so $2e^y=e^{2x}+1$