= Solution
Write $L_n(t)=\sum_{k=0}^na_kt^k$. Matching coefficients in the differential equation gives
$$
(k+1)^2a_{k+1}+(n-k)a_k=0.
$$
Taking $a_0=1$ yields
$$
\boxed{L_n(t)=\sum_{k=0}^n(-1)^k\binom nk\frac{t^k}{k!}.}
$$
The top coefficient is $(-1)^n/n!\ne0$, so this is a real <polynomial> of degree exactly $n$.
Put $D_n=L_n'-L_n$. <Integration by parts>, with vanishing endpoint terms because of the factor $t$ at zero and exponential decay at infinity, gives
$$
\begin{aligned}
\int_0^\infty tD_nD_me^{-t}dt
&=-\int_0^\infty[(tD_ne^{-t})'+tD_ne^{-t}]L_mdt\\
&=-\int_0^\infty[D_n+tD_n']L_me^{-t}dt\\
&=(n+1)\int_0^\infty L_nL_me^{-t}dt.
\end{aligned}
$$
The last step uses $D_n+tD_n'=tL_n''+(1-t)L_n'-L_n=-(n+1)L_n$. The left side is symmetric in $m,n$. Interchanging them therefore gives $(n-m)\int L_nL_me^{-t}=0$, proving <orthogonality> for distinct indices. These integrals are exactly $\langle\varphi_n,\varphi_m\rangle$, and each $\varphi_n$ is square-integrable and nonzero. Thus they form an <orthogonal> sequence in $L^2(0,\infty)$.
Multiplying the equations by $z^n$ and summing gives
$$
\boxed{tG_{tt}+(1-t)G_t+zG_z=0.}
$$
Conversely equality of power-series coefficients recovers every <polynomial> equation. For the proposed expression,
$$
\frac{G_t}G=-\frac z{1-z},\qquad\frac{G_{tt}}G=\frac{z^2}{(1-z)^2},\qquad\frac{G_z}G=\frac1{1-z}-\frac t{(1-z)^2}.
$$
Substitution makes the PDE left side zero. Its coefficient of $z^n$ has value one at $t=0$, so it corresponds to the normalization $L_n(0)=1$ chosen above.
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