Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 9 / b / Solution

For $x\ne 0$, divide by $x$ and use the integrating factor $\mu (x)=e^x|x{|}^{\alpha }$, separately on each side of zero. The equation becomes $(\mu y{)}^{\prime }=|x{|}^{\alpha }/x$. ThusFor $x>0$ the absolute values may be omitted. The original equation is singular as a first-order evolution equation at zero, so these formulas describe regular intervals with $x\ne 0$; extension through zero needs a separate regularity condition.