Division by zero when putting an ODE in standard form for analyzing singular points

Question

Suppose we want to find the singular points of the ODE $\sin(x^2)y''+xy'+y=0$, where $0 \leq x \leq \pi/2$. We first seek to put the ODE in standard form $y''+a(x)y'+b(x) = 0$ so we can analyze $a(x)$ and $b(x)$ to find the singular points of the ODE. In this case, we divide through by $\sin(x^2)$. Since the ODE is specified for $0 \leq x \leq \pi/2$, wouldn't dividing by $\sin(x^2)$ be dividing by zero when $x = 0$? What is the best explanation for why doing this is valid? Is it because we have the trivial solution when $x = 0$?

Answer 1

Yes, it is exactly because whenever $\sin(x^2)=0$ on this interval, $x=0$ and $y\equiv 0$.

To be formal, I would say, either $\sin(x^2)=0$ and $$ y\equiv 0,x\in[0,\pi/2] $$ or we can divide through by $\sin(x^2)=0$.....

Answer 2

It's really not much different from any other case where $a(x)$ and $b(x)$ have singularities. What matters is not the value at the point where $\sin(x^2) = 0$ but the limiting behaviour as $x$ approaches such a point. As $x \to 0$, $\sin(x^2) \sim x^2$ so $x a(x) = x^2/\sin(x^2) \to 1$ while $x^2 b(x) = x^2/\sin(x^2) \to 1$. This makes $x=0$ a regular singular point.