 Product-derivative reduction of a second-order differential equation

Recognizing the displayed product derivative reduces a homogeneous second-order equation to $(py{)}^{\prime \prime }=0$. On any interval where $p\ne 0$, its general solution is $y=(A+Bx)/p$. For $p=1-x^3$, the equation becomes $(1-x^3)y^{\prime \prime }-6x^2{y}^{\prime }-6xy=0$. Data $y(0)=1$, $y^{\prime }(0)=0$ give $1/(1-x^3)={\sum }_{j\ge 0}{x}^{3j}$, with radius of convergence one. The poles occur at the three cube roots of unity. A solution singularity can disappear for a special numerator that cancels the corresponding factor, so an equation's singular point need not be a singularity of every solution.