Solution
= Solution
Invert the equation and regard $x$ as a function of $y$:
$$
\frac{dx}{dy}-x=e^{2y}.
$$
This is a <first-order linear differential equation>. Multiplication by the <integrating factor> $e^{-y}$ gives
$$
\frac d{dy}(xe^{-y})=e^y,
$$
so $x=e^{2y}+Ce^y$. The initial condition $x=1$ at $y=0$ gives $C=0$. Hence $x=e^{2y}$ and
$$
\boxed{y(x)=\frac12\log x},
\qquad x>0.
$$