I'm asked to find the conditions on the coefficients $a_1,a_2,b_1,b_2$ in the RK method $$y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n + b_1h, y_n + b_2hf(t_n,y_n))$$ such that is consistent of (a) first order and (b) second order.
The first assumption I make is that $b_1 = b_2 := b$ otherwise I have no idea how to solve it. I know it should be like this, but why is this reasonable? I know it follows if you look at the butcher tablaeu, but why is a method with $b_1 \neq b_2$ "impossible"?
I taylor expand $y(t_{n+1})$, $$y(t_{n+1}) = y(t_n) + hy'(t_n) + O(h^2).$$ The order of consistency is $p$ if $l = O(h^{p+1})$ where $l$ is the local residual.
The local residual takes the form $$y(t_n) + hy'(t_n) + O(h^2) - y(t_n) - ha_1f(t_n,y_n) - ha_2f(t_{n+b}, y_{n+b}) $$ $$= hy'(t_n) - ha_1y'(t_n) - ha_2y'(t_{n+b}) + O(h^2).$$ So I need to find $a_1,a_2,b$ so that every term except the $O(h^2)$ remains. I get for (a) $$a_1 + a_2 = 1, b=0.$$
For (b) I expand one term more $$y(t_{n+1}) = y(t_n) + hy'(t_n) + \frac{h^2y''(t_n)}{2} + O(h^3)$$ The local residual becomes $$y(t_n) + hy'(t_n) + \frac{h^2y''(t_n)}{2} + O(h^3) - y(t_n) - ha_1y(t_n) - ha_2y(t_{n+b}) $$
But here I am stuck. How am I supposed to choose the coefficients as to remove the $\frac{h^2y''(t_n)}{2}$ term?
How do I solve (b)? And is my answer on (a) correct?
Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get \begin{align} f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &\implies& a_1+a_2=1\\ f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &\implies& 2a_2b_1=1\\ f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&\implies& a_2b_2=\frac12 \end{align} so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=\frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.