= Solution
{c}
Integrate the <ordinary differential equation> across one step, using the <fundamental theorem of calculus>:
$$
y(x_{n+1})-y(x_n)=\int_{x_n}^{x_n+h}f(x,y(x))\,dx.
$$
The <midpoint quadrature rule> approximates the integral by $h f(x_n+h/2,y(x_n+h/2))$. Approximate the unknown midpoint state by the midpoint value of <linear interpolation> between the endpoint states:
$$
y(x_n+h/2)\approx\frac{y_n+y_{n+1}}2.
$$
Combining these approximations gives the <implicit midpoint rule>
$$
\boxed{y_{n+1}=y_n+h f\left(x_n+\frac h2,\frac{y_n+y_{n+1}}2\right),
\qquad y_0=\alpha.}
$$
It is an <implicit time-stepping method> because its new state occurs inside the evaluation of $f$. An equivalent derivation uses a straight-line approximation on the step and imposes the <ordinary differential equation> at its midpoint, a one-stage <collocation method>.
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