[This is the picture Lutz Sir] 1For single differential equation With y'=f(x,y) and y(x_0)=y_0 ,it is x_(k+1)=x_(k)+(1/6)(m1+2m2+2m3+m4)
For higher order of differential equations it is,when converted to system of differential equations
x'=f(t,x,y) and y'=g(t,x,y)=x(in case of higher de being converted to system of differential equations) and x(t_0)=x_0 and y(t_0)=y_0
X_(k+1)=x_(k)+(1/6)(mk1+mk2+mk3+mk4) And for y_(k+1)=y_(k)+(1/6)(nk1+nk2+nk3+nk4)
Why so?
Where did the 2 of the 2nd and 3rd term disappear and why?
All the mk's and nk's and M's denote the usual terms in runge kutta method???
All the well-known methods are designed to work for scalars as well as for vector-valued ODE or ODE systems. That is, for the application it makes no difference if the $x$ in $\dot x=f(x)$ is a scalar or a vector.
There are methods where one gets different error orders, the scalar variant having a higher order than the vector variant. This is due to the fact that in the scalar case all derivatives are scalar valued and thus commute. In the system case the derivatives are tensors, vector-valued symmetric multi-linear forms where commutativity does not make much sense. For example, $f'(x)f''(x)[f(x),f(x)]$ is different from $f''(x)[f'(x)f(x),f(x)]$. Expressed in partial derivatives with summation convention $$ \frac{\partial f_i}{\partial x_j} \frac{\partial^2f_j}{\partial x_k\partial x_l} f_kf_l ~~\text{ vs. }~~ \frac{\partial^2f_i}{\partial x_j\partial x_k} \frac{\partial f_j}{\partial x_l} f_lf_k $$
Or one order higher, $$ f'f''[f'f,f],~~ f''[f'f'f,f]~~\text{and}~~ f''[f'f,f'f] $$ are all different in the non-scalar case. This means that certain order conditions for the method coefficients that are different for the vector case combine into one if derived strictly for the scalar case. The most famous example it the 5th order 5-stage method that Kutta gave in his original paper. Such a method is impossible in the system case.
In summary:
Any change in the coefficients gives a different method, any contrary claim is wrong, it is an error either in the naming of the method or in the coefficients presented.
Any method has to be at least zero-order consistent. Meaning that the integration of $\dot x=f(x)=1$ has to result in $x(t)=t+x_0-t_0$ without any errors. This implies that the sum of the combination coefficients for the result of the step has to be $1$, and $\frac16(1+1+1+1)=\frac23$ as per the presented source is not one.