= Solution
With $t=1-x^2$ the <chain rule> gives $y'=-2xu'$ and $y''=-2u'+4x^2u''$. Substitution, initially for $x>0$, gives
$$
\boxed{t(1-t)u''+(1-t)u'+\frac14u=0.}
$$
The transformed coefficients are $P(t)=1/t$ and $Q(t)=1/[4t(1-t)]$. At $t=0$, both $tP$ and $t^2Q$ are analytic; at $t=1$, both $(t-1)P$ and $(t-1)^2Q$ are analytic. Each point is singular because a normalized coefficient has a pole, but satisfies the <regular singular point criterion for a second-order equation>. Thus \b[both points are regular singular points].
For a <Frobenius method> substitution $u=t^\rho\sum_{n\ge0}a_nt^n$, the lowest power gives the <indicial equation>
$$
\rho(\rho-1)+\rho=\rho^2=0.
$$
The repeated <indicial root> is zero. Setting $\rho=0$ and equating the coefficient of $t^n$ gives
$$
(n+1)^2a_{n+1}+\left(\frac14-n^2\right)a_n=0,
\qquad
\boxed{a_{n+1}=\frac{n^2-\tfrac14}{(n+1)^2}a_n.}
$$
Consequently the analytic <power-series solution of a differential equation> is
$$
u_1(t)=a_0\left(1-\frac t4-\frac{3t^2}{64}+\cdots\right).
$$
Taking $a_0=2\pi$ reproduces the <ellipse> perimeter expansion.
For a second independent solution, normalize $u_1(0)=1$. <Reduction of order> and $P=1/t$ give
$$
u_2(t)=u_1(t)\int^t\frac{ds}{s\,u_1(s)^2}.
$$
Since $u_1(s)^{-2}=1+O(s)$, this has the <Logarithmic Frobenius solution> form
$$
\boxed{u_2(t)=u_1(t)\log t+v(t),\qquad v\ \text{analytic near }0.}
$$
The logarithm is $\log|t|$ on the real negative side, or a chosen <complex logarithm> on a slit complex neighborhood. Its nonzero logarithmic coefficient establishes independence. The physical perimeter is finite at $t=0$, so it selects the analytic branch rather than this logarithmic one.
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