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 The following family of Runge–Kutta methods is given: $$y_{j+1}=y_j+\frac{h}{2 \alpha}f(t_j, y_j) + h(1- \frac{1}{2 \alpha})f(t_j + \alpha h, y_j + \alpha h f(t_j, y_j))$$
We're asked to find all values of $\alpha$ such that the order of consistency $p$ is 2. For that, with $p=2$, we need to check that 
$$\tau_{j+1}= (\dot{y}(t_j)- \Phi(t_j, y(t_j), 0)) + \frac{h}{2}(\ddot{y}(t_j)-2 \dot{\Phi}(t_j, y(t_j), 0)) + O(h^{p=2})\overset{!}{=} O(h^{p=2}) $$ or 
$$ (\dot{y}(t_j)- \Phi(t_j, y(t_j), 0)) + \frac{h}{2}(\ddot{y}(t_j)-2 \dot{\Phi}(t_j, y(t_j), 0)) \overset{!}{=} 0$$
(we would go higher if $p > 2$). In our case, to hold, we need $\dot{y}(t_j)= \Phi(t_j, y(t_j), 0)$ and $\ddot{y}(t_j)= 2 \dot{\Phi}(t_j, y(t_j), 0)$
 A: 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 $\alpha = 1$.

So, to have consistency order $p=2$, there is only one possible value of $\alpha$, that is $\alpha = 1$.
A: 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 $β=\frac12\implies α=1$.
