= Solution
{c}
First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y'$ has finite endpoint limits. A nonzero limit would make $y'$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequently
$$
\boxed{\int_{-1}^1f(x)\,dx=0}.
$$
This is <orthogonality> to the constant <eigenfunction> $P_0$, hence the <Fredholm alternative> condition. It is also sufficient under the stated endpoint regularity: put
$$
F(x)=\int_{-1}^xf(t)\,dt,\qquad
y(x)=C+\int_0^x\frac{F(s)}{1-s^2}\,ds.
$$
If $M=\max|f|$, the zero-mean condition gives $|F(s)|\leq M\min(1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z')'=0$, whose other solution is $\tfrac12\log((1+x)/(1-x))$.
For the requested <eigenfunction expansion>, set
$$
N_k=\int_{-1}^1P_k(t)^2\,dt,\qquad
f_k=\frac{\int_{-1}^1f(t)P_k(t)\,dt}{N_k}.
$$
The <Legendre polynomials> span all <polynomials>, so their completeness follows from the <Weierstrass approximation theorem>. Projecting $\mathcal Ly=f$ onto each <eigenfunction> gives $-k(k+1)y_k=f_k$ for $k\geq1$. The zero mode is free, while $f_0=0$. Thus the full family is
$$
\boxed{y(x)=C-\sum_{k=1}^\infty
\frac{P_k(x)}{k(k+1)}
\frac{\int_{-1}^1f(t)P_k(t)\,dt}{\int_{-1}^1P_k(t)^2\,dt},\qquad C\in\mathbb R}.
$$
The formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\leq1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the <Cauchy-Schwarz inequality>, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second <differentiation>.
There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, let
$$
H(x)=\log\frac{e}{1-x^2},\qquad
F(x)=\frac1{H(x)},\qquad
f(x)=F'(x)=\frac{-2x}{(1-x^2)H(x)^2}.
$$
This $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y'$ has
$$
y'(x)=\frac1{(1-x^2)H(x)},
$$
whose <integral> diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makes
$$
\int_0^x\frac{B+\int_0^sf(t)\,dt}{1-s^2}\,ds
$$
bounded as $x$ approaches both endpoints. The <Bounded solvability for the Legendre operator> criterion above covers the intended regular forcing and makes this endpoint limitation explicit.
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