Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 8 / b / Solution

After normalization, $p=-6x^2/(1-x^3)$ and $q=-6x/(1-x^3)$. Both are analytic at zero, so $\boxed{x=0\text{ is ordinary}}$. At one,are analytic, whereas $p,q$ themselves have poles. Hence $\boxed{x=1\text{ is a regular singular point}}$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^n$. The coefficient of $x^0$ gives $a_2=0$; for $k\ge 1$ the coefficient of $x^k$ givesThus $a_n=a_{n-3}$ for $n\ge 3$. The initial data give $a_0=1$, $a_1=0$, soThe radius is one because the three cube roots of unity are poles of this solution. A direct structural check observes that the differential equation is exactly $[(1-x^3)y{]}^{\prime \prime }=0$. Thus $(1-x^3)y=A+Bx$, and the initial data select $A=1,B=0$. This is the product-derivative reduction of a second-order differential equation.