= Solution
{c}
Use $t=x+1$. The <Legendre differential equation> becomes
$$
t(2-t)y_{tt}+2(1-t)y_t+Ny=0.
$$
Substituting a <Frobenius series> $y=t^r\sum_{k\geq0}a_kt^k$, $a_0\ne0$, gives the <indicial equation> $2r^2=0$: the <indicial root> zero is repeated. For $r=0$, the coefficient of $t^k$ yields
$$
2(k+1)^2a_{k+1}+[n(n+1)-k(k+1)]a_k=0,
$$
or
$$
\boxed{a_{k+1}=\frac{(k-n)(k+n+1)}{2(k+1)^2}a_k}.
$$
For $0\leq k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a <polynomial> of exact <polynomial degree> $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.
The <linear recurrence relation> determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, <reduction of order> gives the second solution locally as
$$
P_n(x)\int^x\frac{ds}{(1-s^2)P_n(s)^2}.
$$
Near $t=0$, the integrand is $1/(2a_0^2t)$ plus an <analytic function>. After rescaling the second solution, its form is therefore
$$
\boxed{y_2(x)=P_n(x)\log(x+1)+\sum_{k=0}^\infty b_k(x+1)^k}.
$$
The analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ <linearly independent> of $P_n$. This <Frobenius expansion of the Legendre equation at an endpoint> gives a <Logarithmic Frobenius solution>; the same endpoint calculation at $x=1$ explains why the independent <Legendre function of the second kind> is excluded when boundedness at both ends is required.
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