## 3 Construction of the New Scheme  
In order to construct new single-step methods to solve the IVP Equation (1), we rely on a variant of the standard form of the Runge-Kutta method given by  
\[y_{n+1}=y_{n}+h\psi(x_{n},y_{n};h)\]  
where  
\[\psi(x_{n},y_{n};h)=\sum_{i=1}^{2m}w_{i}k_{i+1}\]  
and  
\[k_{1}=f(x_{n},y_{n}),\;k_{i}=f\left(x_{n}+c_{i}h,y_{n}+h\overset{j=1}{\underset {j\
eq i}{\sum}}a_{ij}k_{j}\right),\;i=2,\;3,\ldots,\;2m+1\]  
and  
\[\sum_{j=1}^{i-1}a_{ij}=c_{i},\;i=2,\;3,\;\ldots,\;2m+1\]  
By using the Taylor series expansion of any arbitrary function \(\psi(x_{n},y_{n};h)\) for the non-autonomous Equation (1), we have  
\[\psi_{T}(x_{n},y_{n};h)=\sum_{r=0}^{\infty}\frac{1}{(r+1)!}\left(\frac{\partial }{\partial x}+f\frac{\partial}{\partial y}\right)^{r}f(x,y)\]  
and in the autonomous case Equation (2), \(\psi_{T}(x_{n},y_{n};h)\) becomes  
\[\psi_{T}(y_{n};h)=\sum_{r=0}^{\infty}\frac{1}{(r+1)!}\left(f\frac{\partial}{ \partial y}\right)^{r}f(y)\]  
We now start with the numerical method by using the family of explicit Runge-Kutta methods listed below to solve the mentioned problem in (2). \[y_{n+1}=y_{n}+h(w_{1}k_{2}+w_{2}k_{3}) \tag{9}\]  
with  
\[\left. \begin{array}{c}k_{1}=f(x_{n},y_{n})\\ k_{2}=f(x_{n}+c_{2}h,y_{n}+a_{1}hk_{1})\\ k_{3}=f(x_{n}+c_{3}h,y_{n}+h(a_{2}k_{1}+a_{3}k_{2}))\end{array} \right\} \tag{10}\]where \(a_{1}=a_{21},\ a_{2}=a_{31}\), and \(a_{3}=a_{32}\). We must first calculate the unknowns \(a_{1},\ \ a_{2}\), \(a_{3}\), \(w_{2}\) and \(w_{3}\). Using Taylor series expansion around \(f(y)\), we obtain  
\[k_{1}=f(y_{n}) \tag{11}\]  
\[k_{2}=\sum_{r=0}^{\infty}\frac{1}{r!}\bigg{(}ha_{1}k_{1}\frac{d}{dy}\bigg{)}^{r }f(y_{n}) \tag{12}\]  
\[k_{3}=\sum_{r=0}^{\infty}\frac{1}{r!}\bigg{(}h(a_{2}k_{1}+a_{3}k_{2})\frac{d}{ dy}\bigg{)}^{r}f(y_{n}) \tag{13}\]  
By expanding \(k_{1}\), \(k_{2}\), and \(k_{3}\) in Equations (11)-(13), we have  
\[\left. \begin{array}{c}k_{1}=f\\ k_{2}=f+ha_{1}ffy+\frac{h^{2}}{2!}\left(f^{2}a_{1}{}^{2}f_{yy}\right)+\frac{h^ {2}}{3!}\left(a_{1}{}^{3}f^{3}f_{yyyy}\right)+\cdots\\ k_{3}=f+h(a_{2}+a_{3})ffy+\frac{h^{2}}{2!}\left(\left(a_{2}{}^{2}+2a_{2}a_{3}+ a_{3}{}^{2}\right)f^{2}f_{yy}+2a_{1}a_{3}ffy^{2}\right)\\ +h^{3}\Big{(}\left(\frac{1}{2}a_{2}a_{3}{}^{3}+a_{2}{}^{2}a_{3}+\frac{1}{6}a_{ 3}{}^{3}+\frac{1}{6}a_{3}{}^{3}+\frac{1}{6}a_{3}{}^{2}\right)f^{3}f_{yyyy}+a_{ 1}a_{2}a_{3}f^{2}f_{y}f_{yy}+\frac{1}{2}a_{1}{}^{2}a_{3}f^{2}f_{y}f_{yy}\Big{)} +\cdots\end{array} \right\} \tag{14}\]  
Substituting Equation (14) into Equation (9) yields  
\[y_{n+1}=y_{n}+(w_{1}+w_{2})f\ h+((a_{2}+a_{3})\ w_{2}+a_{1}\ w_{1})ff_{y}\ h^{2}+\frac{1}{2} \Big{(}\big{(}(a_{2}+a_{3})^{2}\ w_{2}+a_{1}{}^{2}\ w_{1}\big{)}ff_{yy}+2\ w_{2}a_{ 1}a_{3}\ f^{2}\Big{)}f\ h^{3}+\cdots \tag{15}\]  
The Taylor series expansion of an exact solution \(y(x_{n+1})\) is given by  
\[y(x_{n+1})=y(x_{n})+fh+\bigg{(}\frac{1}{2}ff_{y}\bigg{)}h^{2}+\frac{1}{6} \Big{(}f^{2}f_{yy}+ffy^{2}\Big{)}h^{3}+\frac{1}{24}\Big{(}f^{3}f_{yy}+4f^{2}f _{y}f_{yy}+ffy^{3}\Big{)}h^{4}+\ldots \tag{16}\]  
The following system of equations is obtained by expressing \(k_{1}\), \(k_{2}\), and \(k_{3}\) in the Taylor expansion, ignoring terms with powers of \(h\) higher than 3, and then substituting them into formula (15) and comparing them to Equation (16):  
\[hf:\ w_{1}+w_{2}=1\]  
\[h^{2}ffy:(a_{2}+a_{3})w_{2}+w_{1}a_{1}=\frac{1}{2}\]  
\[h^{3}f^{2}f_{yy}:\ (a_{2}+a_{3})^{2}w_{2}+a_{1}{}^{2}w_{1}=\frac{1}{3}\]  
\[h^{3}ffy^{2}:\ w_{2}a_{1}a_{3}=\frac{1}{6}\]  
This is a system with an infinite number of solutions comprising four equations and five unknowns. Assuming that \(w_{1}=\frac{3}{7}\), we obtain the optimal solution listed below:  
\[w_{1}=\frac{3}{7}\,\ w_{2}=\frac{4}{7},\ a_{1}=\frac{1}{6}\,\ a_{2}=-1,\ a_{3}= \frac{7}{4}\]  
Additionally, from Equation (6), we obtain \(c_{2}=\frac{1}{6}\) and \(c_{3}=\frac{3}{4}\).