= Product-derivative reduction of a second-order differential equation
{title2=$p y''+2p'y'+p''y=(py)''$}
Recognizing the displayed product derivative reduces a homogeneous second-order equation to $(py)''=0$. On any interval where $p\ne0$, its general solution is $y=(A+Bx)/p$. For $p=1-x^3$, the equation becomes $(1-x^3)y''-6x^2y'-6xy=0$. Data $y(0)=1$, $y'(0)=0$ give $1/(1-x^3)=\sum_{j\geq0}x^{3j}$, 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.
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