Find $\alpha$ such that $y_{j+1}=y_j+\frac{h}{2 \alpha}k_1 + h(1- \frac{1}{2 \alpha})k_2$ has order of consistency 2

Solution 1:

Assuming that $y(x)$ is an exact solution of $y'(x)=f(x,y(x))$, you have that \begin{align} y(x)+αy'(x)h&=y(x+αh)+O(h^2) \\[1em] y(x)+\beta y'(x)h&+(1-β)f(x+αh, y(x+αh)+O(h^2))h \\ &=y(x)+\beta y'(x)h+(1-β)y'(x+αh)h+O(h^3) \\ &=y(x)+y'(x)h+α(1-β)y''(x)h^2+O(h^3) \end{align} so that you get an order 2 method if and only if $α(1-β)=\frac12$. This means that you have the coefficients of the outer convex combination of the $k_1,k_2$ in the wrong order, which only is not relevant in the symmetric case $β=\frac12\implies α=1$.

Solution 2:

The Butcher Tableau of this Runge-Kutta method is, with $k_1= f(t_j, y_j)$,
$ k_2= f(t_j + \alpha h, y_j + \alpha k_1)$,
$ y_{j+1}=y_j+\frac{h}{2 \alpha}k_1 + h(1- \frac{1}{2 \alpha})k_2$ :
$$ \ \renewcommand\arraystretch{1.2} \begin{array} {c|cc} 0\\ \alpha & \alpha\\ \hline & \frac{1}{2 \alpha} &1- \frac{1}{2 \alpha} \end{array} \ $$

(Which is by the way very similar to https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods#Second-order_methods_with_two_stages)

If $$\Phi(h)= \frac{1}{2 \alpha}f(y(t_j)) + (1- \frac{1}{2 \alpha})f(y(t_j) + \alpha hf(y(t_j)))$$ then $$\dot{\Phi}(h) = 0 + (1- \frac{1}{2 \alpha}) \frac{df}{dy}(y(t_j)+ \alpha hf(y(t_j))) \cdot (\alpha f(y(t_j))) = (\alpha - \frac{1}{2})f(y(t_j)) \frac{df}{dy}(y(t_j)+ \alpha hf(y(t_j)))$$

Also, if $$\dot{y}(t)= f(y(t))$$ then $$\ddot{y}(t) = \frac{d}{dt}f(y(t))= \frac{df}{dy}(y(t))f(y(t))$$

Now, in fact, if we compare: $$\Phi(t_j, y(t_j), h=0) = \frac{1}{2 \alpha}f(y(t_j)) + (1- \frac{1}{2 \alpha})f(y(t_j) + 0)= 1 \cdot f(y(t_j)) + 0 = \dot{y}(t_j)$$ And $$2 \dot{\Phi}(t_j, y(t_j), h=0) = 2(\alpha - \frac{1}{2})f(y(t_j)) \frac{df}{dy}(y(t_j)) \overset{!}{=} \frac{df}{dy}(y(t_j))f(y(t_j))$$

This holds for $\alpha = 1$.