If the function is rather well-behaved in the domain \( [a,b] \), we can use a fixed step size. If not, adaptive steps may be needed. Here we concentrate on fixed-step methods only. Let us try to generalize the above procedure by writing the step \( y_{i+1} \) in terms of the previous step \( y_i \) $$ \begin{equation} y_{i+1}=y(t=t_i+h)=y(t_i) + h\Delta(t_i,y_i(t_i)) + O(h^{p+1}), \tag{13} \end{equation} $$ where \( O(h^{p+1}) \) represents the truncation error. To determine \( \Delta \), we Taylor expand our function \( y \) $$ \begin{equation} y_{i+1}=y(t=t_i+h)=y(t_i) + h(y'(t_i)+\dots +y^{(p)}(t_i)\frac{h^{p-1}}{p!}) + O(h^{p+1}), \tag{14} \end{equation} $$ where we will associate the derivatives in the parenthesis with $$ \begin{equation} \Delta(t_i,y_i(t_i))=(y'(t_i)+\dots +y^{(p)}(t_i)\frac{h^{p-1}}{p!}). \tag{15} \end{equation} $$