= Solution
Multiplication by $x^2$ puts the <eigenvalue problem> into <Sturm-Liouville form>, $-(x^2y')'=\lambda x^2y$, with weight $x^2$. If $u,v$ correspond to distinct <eigenvalues> $\lambda,\mu$, subtract their multiplied equations and integrate to get
$$
(\lambda-\mu)\int_0^\pi x^2uv\,dx=\left[x^2(uv'-vu')\right]_0^\pi=0.
$$
The upper boundary term vanishes by the <Dirichlet boundary condition>. For a bounded solution near zero, $x^2y'$ has a limit because $(x^2y')'=-\lambda x^2y$ is integrable. A nonzero limit would force a $1/x$ divergence in $y$, so that limit is zero; integration then gives $y'=O(x)$. The lower boundary term therefore also vanishes, proving the weighted <orthogonality>.
Set $z=xy$. For $x>0$, $y''+2y'/x=z''/x$, and finiteness of $y(0)$ gives $z(0)=0$. Thus $z''+\lambda z=0$ with zero values at zero and $\pi$. The <radial Dirichlet spectrum in three dimensions> gives
$$
\boxed{\lambda_n=n^2,\qquad y_n(x)=\frac{\sin nx}{x},\qquad n=1,2,\ldots,\quad y_n(0)=n.}
$$
The zero and negative <eigenvalues> yield only the zero solution under these <boundary conditions>. Also $\int_0^\pi x^2y_n^2dx=\pi/2$.
For the forced equation, the same substitution gives $z''-\beta z=x$. The <Fourier sine series> of $x$ on $(0,\pi)$ has coefficients
$$
b_n=\frac2\pi\int_0^\pi x\sin nx\,dx=\frac{2(-1)^{n+1}}n.
$$
Expanding $z=\sum c_n\sin nx$ gives $-(n^2+\beta)c_n=b_n$, hence
$$
\boxed{y(x)=2\sum_{n=1}^\infty\frac{(-1)^n}{n(n^2+\beta)}\frac{\sin nx}{x}.}
$$
The value at zero is defined by a <limit>. In fact $|\sin nx/x|\leq n$, so <uniform convergence> holds even there, by comparison with $2\sum(n^2+\beta)^{-1}$. An independent check comes from solving the constant-coefficient equation directly:
$$
y(x)=-\frac1\beta+\frac{\pi\sinh(\sqrt\beta x)}{\beta x\sinh(\pi\sqrt\beta)}.
$$
It satisfies both endpoint conditions and the forcing for $x>0$, and its finite origin limit agrees with the series. For $\beta>0$ the homogeneous boundary value problem has no nonzero solution, so the solution is unique.
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