= Solution
\b[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> are
$$
Ae=\begin{pmatrix}c_1\\-c_1c_2/d\end{pmatrix},\qquad
b^Te=1,\quad b^Tc=\tfrac12,\quad
q:=b^TAe=\tfrac12-\frac{(\tfrac12-c_1)c_2^2}{d^2}.
$$
Thus the claimed universal second order is false. For example $c_1=0,c_2=1$ gives, on $y'=y$, 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. Define
$$
a=\frac{c_1+c_2}{2}>0,\qquad
D=\frac{c_1c_2}{2}\left(1-\frac{c_1c_2}{d^2}\right),\qquad
r=D+q-a.
$$
The <stability function> obtained from $R(z)=1+zb^T(I-zA)^{-1}e$ is
$$
R(z)=\frac{1+(1-a)z+rz^2}{1-az+Dz^2}.
$$
If $D\geq0$, 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,
$$
|1-aiy+D(iy)^2|^2-|1+(1-a)iy+r(iy)^2|^2
=(2q-1)y^2+(D^2-r^2)y^4.
$$
Consequently \b[for the literal tableau with uniquely solvable stages throughout the left half-plane, the complete parameter criterion is]
$$
\boxed{D\geq0,\qquad q\geq\tfrac12,\qquad D^2\geq(D+q-a)^2.}
$$
These 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 are
$$
D=q(a-q),\qquad q>a,\qquad q\geq\tfrac12.
$$
Indeed 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, take
$$
L_1(t)=\frac{c_2-t}{d},\qquad L_2(t)=\frac{t-c_1}{d},\qquad
a_{ij}=\int_0^{c_i}L_j(t)\,dt,\quad b_j=\int_0^1L_j(t)\,dt.
$$
Integration gives the other three printed entries and the weights unchanged, with the positive lower-left entry. Since $L_1+L_2=1$ and $c_1L_1+c_2L_2=t$, it follows that $Ae=c$, $b^Te=1$ and $b^Tc=1/2$. Hence \b[the corrected family is always of order at least two].
For the corrected family let $a=(c_1+c_2)/2$, $D=c_1c_2/2$ 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 $D,r\geq0$, the necessary and sufficient inequality is $D\geq r$, or
$$
\boxed{c_1+c_2\geq1\quad\text{for the corrected two-node collocation family}.}
$$
For sufficiency, $|R|\leq1$ 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.
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