Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 2A / Solution

Invert the equation and regard $x$ as a function of $y$:This is a first-order linear differential equation. Multiplication by the integrating factor $e^{-y}$ givesso $x=e^{2y}+C{e}^y$. The initial condition $x=1$ at $y=0$ gives $C=0$. Hence $x=e^{2y}$ and