Past exam of the mathematical methods I course of the University of Oxford / 2017 / 2 / Solution

An integrating factor for $u^{\prime }+P(x)u=Q(x)$ is a nonzero function $\mu (x)=\exp (\int P(x)dx)$, chosen so that $(\mu u{)}^{\prime }=\mu Q$. Integrating this total derivative reduces the first-order linear ordinary differential equation to one quadrature.

For the given nonlinear equation, let $u=y^2$, so $u^{\prime }=2y{y}^{\prime }$. Multiplication by $2y$ gives $u^{\prime }+u=x^2+2x$. Using the integrating factor $e^x$, or recognizing the particular solution $u=x^2$, yields $u=x^2+C{e}^{-x}$. The initial value gives $C=1$ and selects the positive branch:The radicand is strictly positive for every real $x$. This solution never reaches the singular denominator $y=0$ of the original equation, so multiplication by $y$ has introduced no forbidden crossing. The negative branch would violate $y(0)=1$.