= Solution
For $f'=\Phi(x,f)$, the standard explicit <midpoint method> is
$$
\boxed{\widetilde y_{i+1/2}=y_i+\frac h2\Phi(x_i,y_i),\qquad
y_{i+1}=y_i+h\Phi(x_i+h/2,\widetilde y_{i+1/2}).}
$$
The <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 $y_{i+1}=y_i+hf'(x_i+h/2)$.
The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\tfrac h2[\Phi(x_i,y_i)+\Phi(x_{i+1},y_{i+1})]$. 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.
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