\section{(Numerical Comparison)}
For this particular problem, where $f(x,y) = y$, each method can be
turned into an explicit method.  Euler's method and the 4-th-order
Runge-Kutta methods are already explicit, obviously.  For the others,
after plugging in the values for $f(x,y)$, we can solve for $y_{n+1}$ in
terms of $y_{n}$ (or solve for $y_{n+2}$ in terms of $y_{n+1}$ and $y_n$ in the
case of Simpson's rule).  This is what we obtain:
\begin{description}
\item[Euler's method]  $y_{n+1} = (h + 1) y_{n}$
\item[Trapezoidal rule] $y_{n+1} = (2 + h) / (2 - h) y_{n}$
\item[Simpson's rule] $y_{n+2} = (4hy_{n+1} + (h + 3) y_{n}) / (3 - h)$
\item[4th-order Runge-Kutta] $y_{n+1} = y_{n} + (h / 6) (K_{1} + 2 K_{2} + K_{4})$\\
 where $K_1 = y_n, K_2 = y_n + (h / 2)K_1, K_3 = y_n + (h / 2)K_2, K_4 = y_n + h K_3$
\end{description}
When we run each of these methods with step sizes $h = 2^{-s}$ for $s =
3,4,\ldots,15$, and plot the error (i.e., $e - y_n$) at the right endpoint
against $s$ we obtain the attached superimposed plots.  Euler's method
is shown in black, Trapezoidal rule in blue, Simpson's rule in red,
and the Runge-Kutta method in yellow.  It is interesting that Euler's
method performs better than all the others.
