Proof

To show that the modified Euler method builds in the curvature term, we start from the Taylor series expansion of (5) and combine it with a backward Taylor series expansion about $y(t+h)$:

$$y(t) = y(t+h) - h \frac{dy(t+h)}{dt} + h^2/2 \frac{d^2y(t+h)}{dt^2} + {\cal O}(h^3).$$

The difference between the second derivative at $t+h$ and at $t$ is of order $h$. So if we evaluate the second derivative term in the expansion above at $t$ instead of $t+h$, namely, $d^2y(t)/dt^2$, its contribution changes by an amount of net order $h^3$, which may be absorbed in the last term. Solving this equation for $y(t+h)$ then gives

$$y(t+h) = y(t) + h \frac{dy(t+h)}{dt} - h^2/2 \frac{d^2y(t)}{dt^2} + {\cal O}(h^3).$$

Averaging this equation with (5) then gives

$$y(t+h) = y(t) + \frac{h}{2}\left[\frac{dy(t)}{dt} + \frac{dy(t+h)}{dt}\right] + {\cal O}(h^3).$$

Notice that the second derivative term is gone. Finally, substituting the definitions of the approximate values $w_i$ and of the discrete time $t_i$ and using the differential equation (3) leads to (6).