Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 17C / b / Solution

Expand the exact solution about $t_{n+1}=t$. The left side isSince $f(t,y(t))=y^{\prime }(t)$, the right side expands asThe $h{y}^{\prime }$ terms agree for every $\alpha ,\beta$, so the method is always consistent and is generically of order one.

The method is explicit exactly when $1+\alpha =0$. With $\alpha =-1$, its $h^2$ defect vanishes only for $\beta =3/2$, so a generic explicit member has order one; that exceptional explicit member has order two and cannot satisfy the next order condition.

For the unrestricted method to have order at least three, the $h^2$ and $h^3$ coefficients must agree:Solving givesFor these values, $1+2\alpha +\beta =1/2$, so the right-side $h^4$ coefficient is $1/12$, different from the left-side coefficient $1/24$. Hence the order is exactly three.