OK, here is my current thinking on Pietro's approach.
We use the differential equation
$$
(x^4-34x^3+x^2)U'''(x) + 3(2x^3-51x^2+x)U''(x)+(7x^2-112x+1)U'(x)+(x-5)U(x)=0
\tag{ODE3}$$
Multiply by $x^k$, $k \ge 0$, then integrate by parts as much as possible. The result is
$$
\int \!{x}^{k-1} \left( {k}^{3}{x}^{2}-34\,{k}^{3}x
+3\,{k}^{2}{x}^{2}+{k}^{3}-51\,{k}^{2}x+3\,k{x}^{2}-27\,kx+{x}^{2}-5\,
x \right) U \left( x \right) {dx}
=
\left( {x}^{k+2}{k}^{2}-34\,{x}^{1+k}{k}^{2}+{x}^{k+2}k+{x}^{k}{k}^{2
}-17\,{x}^{1+k}k+{x}^{k+2}-10\,{x}^{1+k} \right) U \left( x \right)
+
\left( -{x}^{1+k}k-{x}^{3+k}k+34\,{x}^{k+2}k+{x}^{1+k}+2\,{x}^{3+k}-
51\,{x}^{k+2} \right) U' \left( x \right)
+
\left( {x}^{k+4}-34\,{x}^{3+k}+{x}^{k+2} \right) U'' \left( x \right)
\tag{A}$$
In particular, for $k=0$,
$$
\int (x-5)U(x)\;dx
=
\left( {x}^{2}-10\,x \right) U \left( x \right) + \left( 2\,{x}^{3}-
51\,{x}^{2}+x \right) U' \left( x \right) + \left( {x}^{
4}-34\,{x}^{3}+{x}^{2} \right) U'' \left( x
\right)
\tag{A0}$$
Write $Q_k(x)$ for the right-hand-side of (A) and $\int R_k(x)U(x)\;dx$ for the left side. We will be doing differences like
$$
\big[Q_k(x)\big]_a^b := Q_k(b)-Q_k(a)
$$
since that will equal the integral on the left $\int_a^b R_k(x)U(x)\;dx$.
We want to arrange a solution $U$ so that $\int_0^c R_k(x)U(x)\;dx$. As Pietro noted, this will give us the recurrence we want for the moments $M(k):=\int_0^c x^kU(x)\;dx$. And in particular from (A0) we would have $\int_0^c (x-5)U(x)\;dx = 0$ so that $M(1)=5M(0)$.
Now consider (ODE3).
TO BE CONTINUED