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 $\ddot{y}(t_j)= 2 \dot{\Phi}(t_j, y(t_j), 0)$
Best Answer
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 $β=\frac12\implies α=1$.