I am trying to understand analytical theory of RK method. I am following the book A first course in Numerical Analysis by Ralston and Rabinowitz(2nd Edition) Page 225.
We start with the solution of $\frac{\mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = f(x,y)$ to be $y_{n+1} = y_{n} + \sum_{i=1}^{m}(w_{i}k_{i})$ where $k_{i} = h_{n}f(x_{n}+\alpha_{i}h_{n}, y_{n}+\sum_{j=1}^{i-1}\beta_{ij}k_{j})$.
I have derived upto
\begin{align} y_{n+1}-y_{n} &= \sum_{t = 1}^{\infty}(h_{n})^{t}\frac{y_{n}^{(t)}}{t!}= \sum_{t = 1}^{\infty}h_{n}^{t+1}\cdot \frac{(\frac{\partial }{\partial x} +f\frac{\partial }{\partial y})^{t}}{(t+1)!}\cdot f(x_{n},y_{n})\\ &=\bigg[hf + \frac{h^{2}}{2!}Df + \frac{h^{3}}{3!}\bigg(D^{2}f+f_{y}Df\bigg)+ \frac{h^{4}}{4!}\bigg(D^{3}f + f_{y}D^{2}f + f_{y}^{2}Df + 3DfDf_{y}\bigg) \\ &+ \frac{h^{5}}{5!}\bigg(D^{4}f+6DfD^{2}f_{y} + f_{y}^{2}D^{2}f + f_{y}^{3}Df+ 3f_{yy}(Df)^{2} + f_{y}D^{3}f + 7f_{y}DfDf_{y}\bigg)\bigg] + O(h^{6}) \end{align}
Also
\begin{align} k_{i} &= h_{n}f(x_{n}+\alpha_{i}h_{n}, y_{n}+\sum_{j=1}^{i-1}\beta_{ij}k_{j})\\ &= h_{n}f\bigg[x_{n}+\alpha_{4}h_{n}, y_{n}+(\sum_{j=1}^{i-1}\beta_{ij})h_{n}f_{n} + \bigg(\sum_{j=2}^{i-1}\beta_{ij}(k_{j}-h_{n}f_{n})\bigg)\bigg] \\ &= h_{n}\sum_{t=0}^{\infty}\frac{\bigg(h_{n}D_{i}+\sum_{j=2}^{i-1}\beta_{ij}(k_{j}-h_{n}f_{n})\frac{\partial}{\partial y}\bigg)^{t}}{t!}f(x_{n},y_{n}) \end{align} where $$D_{i} = \alpha_{i}\frac{\partial}{\partial x} + \sum (\beta_{ij})f_{n}\frac{\partial}{\partial y}$$
Now i have to find $k_{i}$ for $i = 1,2,3,4$
Prove that,
\begin{align*} \omega_{1} + \omega_{2} + \omega_{3} + \omega_{4} &= 1\\ \omega_{2}\alpha_{2} + \omega_{3}\alpha_{3} + \omega_{4}\alpha_{4} &= \frac{1}{2}\\ \omega_{2}\alpha_{2}^{2} + \omega_{3}\alpha_{3}^{2} + \omega_{4}\alpha_{4}^{2} &= \frac{1}{3}\\ \omega_{3}\beta_{32}\alpha_{2} + \omega_{4}\bigg(\beta_{42}\alpha_{2} + \beta_{43}\alpha_{3}\bigg) &= \frac{1}{6}\\ \omega_{2}\alpha_{2}^{3} + \omega_{3}\alpha_{3}^{3} + \omega_{4}\alpha_{4}^{3} &= \frac{1}{4}\\ \omega_{3}\beta_{32}\alpha_{2}^{2} + \omega_{4}\bigg(\beta_{42}\alpha_{2}^{2} + \beta_{43}\alpha_{3}^{2}\bigg) &= \frac{1}{12}\\ \omega_{4}\beta_{32}\beta_{43}\alpha_{2} &= \frac{1}{24}\\ \omega_{3}\beta_{32}\alpha_{2}\alpha_{3}+ \omega_{4}\bigg(\beta_{42}\alpha_{2}\alpha_{4} + \beta_{43}\alpha_{3}\alpha_{4}\bigg) &= \frac{1}{8} \end{align*}
Edit: I finally got the proof done.