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The original PDF has a sign defect in the lower-left stage coefficient. Put $d=c_2-c_1$. With the printed negative sign, the row sums and the relevant Butcher order conditions areThus the claimed universal second order is false. For example $c_1=0,c_2=1$ gives, on $y^{\prime }=y$, two stages equal to $y$ and the update $Y=(1+h)y$. This is the first-order Forward Euler method. For the literal tableau, order at least two holds exactly when $c_1=1/2$ or $c_2=0$, subject to distinct allowed nodes.

For completeness the literal tableau's A-stability can also be classified algebraically. DefineThe stability function obtained from $R(z)=1+z{b}^T(I-zA{)}^{-1}e$ isIf $D\ge 0$, the denominator has no zero in the closed left half-plane: its reciprocal roots are the stage-matrix eigenvalues, whose positive sum and nonnegative product give positive real parts for nonzero eigenvalues. On the imaginary axis,Consequently for the literal tableau with uniquely solvable stages throughout the left half-plane, the complete parameter criterion isThese explicit inequalities in $c_1,c_2$ are necessary by the small and large imaginary parameters, and sufficient by the maximum modulus principle, including the bound at infinity. For $D=0$ the last inequality forces $r=0$, leaving the stable linear-fractional case. For $D<0$ there is a negative real stage pole. If one defines stability solely through an analytically continued scalar stability function, the only additional possibilities areIndeed these are precisely the cases where the left pole cancels and $R(z)=[1+(1-q)z]/[1-qz]$ is an A-stable theta method. The original stage equations are nevertheless singular at $z=-1/(q-a)$, so they do not give uniquely determined stages there. This distinction prevents a canceled rational factor from concealing an ill-defined stage solve.

The natural intended repair is to replace just the lower-left coefficient by $+c_2^2/(2d)$. This makes the method the collocation Runge-Kutta method on the two nodes. To derive the repaired entries, takeIntegration gives the other three printed entries and the weights unchanged, with the positive lower-left entry. Since $L_1+L_2=1$ and $c_1{L}_1+c_2{L}_2=t$, it follows that $Ae=c$, $b^{T}e=1$ and $b^{T}c=1/2$. Hence the corrected family is always of order at least two.

For the corrected family let $a=(c_1+c_2)/2$, $D=c_1{c}_2/2$ and $r=(1-c_1)(1-c_2)/2$. Its stability function has the same displayed rational form, now with $q=1/2$. The denominator again has no left-half-plane zero, and on the imaginary axis the squared-modulus difference is $(D^2-r^2)y^4$. Because $D,r\ge 0$, the necessary and sufficient inequality is $D\ge r$, orFor sufficiency, $|R|\le 1$ on the imaginary axis and at infinity; apply the maximum modulus principle on growing left half-discs to get the same bound inside. If $D=0$, the condition leaves the endpoint trapezoidal rule and its linear denominator. If $c_1+c_2<1$, every nonzero imaginary argument gives $|R|>1$. This is the two-node collocation A-stability criterion, explicitly distinguished from the literal defective tableau.