The general solution of differential equation \(\log_{e}\left(\frac{dy}{dx}\right)=2x-y,y(0)=0\) is

\(e^y=2e^{2x}+1\) \(e^y=2e^{2x}+1\) \(e^y=2e^{2x}+3\) \(2e^{y}=e^{2x}+1\)

\(2e^{y}=e^{2x}+1\)

$\log\left(\frac{dy}{dx}\right)=2x-y$ 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 $2e^y=e^{2x}+1$