= Solution
For one stage, first-order <consistency of a numerical method> gives $b_1=1$. The condition $b^TA^{-1}\mathbf1=1$ then gives $a_{11}=1$. Hence the scheme is \b[the <Backward Euler method>]:
$$
y_{n+1}=y_n+h f(t_n+h,y_{n+1}).
$$
It is a one-node <collocation Runge-Kutta method> with $c_1=1$. Specifically, choose a linear <polynomial> $p(t)$ on $[t_n,t_n+h]$ with $p(t_n)=y_n$ and impose $p'(t_n+h)=f(t_n+h,p(t_n+h))$. Its derivative is constant, so $p(t_n+h)-p(t_n)=hp'(t_n+h)$, exactly the <Backward Euler method> update.
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