= Solution
For real positive weight $w$, the <weighted inner product> is
$$
\boxed{\langle y_1,y_2\rangle_w=\int_a^b y_1(x)^*y_2(x)w(x)\,dx.}
$$
The functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $xe^{-x}$ gives
$$
(xe^{-x}y')'+ne^{-x}y=0.
$$
Thus the <Sturm-Liouville form> has $p=xe^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention define
$$
\boxed{\mathcal Ay=-\frac1w(py')'=-xy''+(x-1)y',\qquad\mathcal Ay=ny.}
$$
The existing <Laguerre differential operator on polynomials> uses the opposite operator $L=-\mathcal A$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.
For admissible $f,g$, <integration by parts> gives
$$
\langle f,\mathcal Ag\rangle_w-\langle\mathcal Af,g\rangle_w=\left[xe^{-x}(f'^*g-f^*g')\right]_0^\infty.
$$
The boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes $\lim_{x\downarrow0}xe^{-x}y'=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty)=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.
Self-adjointness yields $(n-m)\langle y_m,y_n\rangle_w=0$, giving <orthogonality of Sturm-Liouville eigenfunctions>:
$$
\boxed{\int_0^\infty e^{-x}y_m(x)^*y_n(x)\,dx=0\quad(m\ne n).}
$$
For the specified polynomials, direct differentiation gives $\mathcal Au=u$ and $\mathcal Av=3v$. Hence their eigenvalues in the displayed convention are \b[1 and 3], respectively, or \b[−1 and −3] for $L=-\mathcal A$.
To find the quadratic, put $h=ax^2+bx+c$. The equation $xh''+(1-x)h'+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,
$$
\boxed{h(x)=-x^2+4x-2.}
$$
These are scalar multiples of the <Laguerre polynomials> of degrees one, three and two.
For $n=0$, the first integral is $xe^{-x}y'=C$, so all solutions with the prescribed interior normalization are
$$
y(x)=1+C\int_{x_0}^x\frac{e^t}{t}dt.
$$
Two independent choices are
$$
\boxed{y_1(x)=1,\qquad y_2(x)=1+\int_{x_0}^x\frac{e^t}{t}dt.}
$$
They both equal one at $x_0$, and their Wronskian is $e^x/x\ne0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to\infty$. These are the <zero-eigenvalue solutions of the Laguerre differential equation>, with the second expressible through the <exponential integral>. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.
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