Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 5 / b

On the interval $x>-1$ containing the initial condition, division by $1+x$ gives a linear ordinary differential equation with integrating factor $(1+x{)}^{-3}$. ConsequentlyThe initial value gives $C=1$. ThusThis polynomial also satisfies the original equation at $x=-1$ and gives an extension to all real $x$. The initial value problem is uniquely determined on $x>-1$; the vanishing leading coefficient at $-1$ does not ensure unique continuation to its left.