 Solution

For $f^{\prime }=\Phi (x,f)$, the standard explicit midpoint method isThe implicit midpoint rule instead uses $y_{i+1}=y_i+h\Phi (x_i+h/2,(y_i+y_{i+1})/2)$. When the derivative depends only on $x$, both give $y_{i+1}=y_i+h{f}^{\prime }(x_i+h/2)$.

The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\frac{h}{2}[\Phi (x_i,y_i)+\Phi (x_{i+1},y_{i+1})]$. This is the trapezoidal rule, and generally differs from the implicit midpoint rule. A further use of “midpoint” is the two-step explicit midpoint linear multistep method, $y_{i+1}=y_{i-1}+2h\Phi (x_i,y_i)$, which needs a second starting value. The printed question supplies no convention to distinguish these meanings; (e) spells out the resulting alternatives.