Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 69 / 1 / b / Solution

For one stage, first-order consistency of a numerical method gives $b_1=1$. The condition $b^T{A}^{-1}1=1$ then gives $a_{11}=1$. Hence the scheme is the Backward Euler method: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 $p^{\prime }(t_n+h)=f(t_n+h,p(t_n+h))$. Its derivative is constant, so $p(t_n+h)-p(t_n)=h{p}^{\prime }(t_n+h)$, exactly the Backward Euler method update.