= Solution
Write $t=1-x$. The analytic solution has $y_1=a_0+O(t)$ with $a_0\ne0$, so the integrand in the <reduction of order> formula satisfies
$$
\frac1{(1-x)y_1(x)^2}=\frac1{a_0^2(1-x)}+O(1).
$$
Integrating with respect to $x$ gives $-a_0^{-2}\log(1-x)+O(1)$. Multiplication by $y_1$ therefore gives
$$
y_2(x)=-\frac1{a_0}\log(1-x)+O(1).
$$
Any solution independent of $y_1$ has a nonzero coefficient of this second solution. Hence
$$
\boxed{y(x)=K\log(1-x)+O(1),\qquad K\ne0,\quad x\to1^-.}
$$
An added multiple of $y_1$ stays bounded and cannot cancel this logarithm. On the real side $x>1$, the corresponding statement uses $\log|1-x|$.
Back to article page
