= Solution
The nodes are $c=(0,1/2,1)^T$. Their <Lagrange interpolation> basis on the normalized step is
$$
\ell_1(s)=2s^2-3s+1,\quad \ell_2(s)=4s-4s^2,\quad \ell_3(s)=2s^2-s.
$$
Integrating these polynomials from zero to each node gives $a_{ij}=\int_0^{c_i}\ell_j(s)ds$. At $c_1=0$ the row vanishes; at $c_2=1/2$ the integrals are $(5/24,1/3,-1/24)$; at $c_3=1$ they are $(1/6,2/3,1/6)$. The endpoint integrals also give $b=(1/6,2/3,1/6)^T$. These are exactly the coefficients in the original PDF.
Indeed, the degree-three polynomial $q$ determined by $q(0)=y_n$ and $q'(hs)=\sum_j\ell_j(s)f(q(hc_j))$ satisfies $q'(hc_i)=f(q(hc_i))$. Integrating it to the nodes gives precisely the stage equations, and integrating to the endpoint gives the update. This proves it is a <collocation Runge-Kutta method>, specifically the three-stage <Lobatto IIIA method>.
For an explicit order proof, the collocation integrals imply $A\mathbf1=c$, $Ac=c^2/2$ and $Ac^2=c^3/3$, with powers componentwise. The endpoint weights are Simpson weights, so $b^Tc^j=1/(j+1)$ for $j=0,1,2,3$. Therefore the eight <Butcher order conditions> through order four are satisfied:
$$
\begin{gathered}
b^T\mathbf1=1,\quad b^Tc=\tfrac12,\quad b^Tc^2=\tfrac13,\quad b^TAc=\tfrac16,\\
b^Tc^3=\tfrac14,\quad b^T\operatorname{diag}(c)Ac=\tfrac18,\quad
b^TAc^2=\tfrac1{12},\quad b^TA^2c=\tfrac1{24}.
\end{gathered}
$$
For example the last expression is $b^Tc^3/6$, because $A^2c=Ac^2/2=c^3/6$. These conditions are obtained by matching all elementary differential terms through degree four in the exact and numerical <Taylor expansions>, not just by testing a scalar linear equation. Nonautonomous equations are included by adjoining $t'=1$.
To exclude higher order, the scalar <stability function> computed below satisfies $R(z)-e^z=-z^5/720+O(z^6)$. The local defect is therefore nonzero at degree five even on the scalar test equation. Hence \b[$\boxed{\text{the classical order is exactly }4}$].
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