I was asked to work out a differential equation using the Euler method and then followed by the Runge-Kutta method. Based on the theory I have come across it says that the Euler method agrees with the Taylor series solution up to the term $h$ and therefore the Euler method is the Runge-Kutta method of first order. Now based on the formula for both the Euler method as well as the first order Runge-Kutta method:

$$y_{n+1} = y_n + hy'_n$$

This means both methods are the same and if I have to use the Runge-Kutta method to work out the question it would make sense to use some higher order for the Runge-Kutta method to arrive at my answer. I am hoping that I haven't not confused both the R-K method of first order and the Euler's method of being the same.

  • $\begingroup$ what was my confusion was the question stated to use the runge kutta method but previous to that it said to work the question using the eulers method so i wanted to make sure that the first order runge kutta method and the eulers method is the same and i was not mixing up anyhting @Lutz Lehmann. Therefore i decided to work the question using a higher oder R-K method. $\endgroup$
    – John
    Apr 17, 2020 at 7:00
  • $\begingroup$ I transferred my comments to an answer. $\endgroup$ Apr 17, 2020 at 8:36

1 Answer 1


What you say about the Euler method is correct in the context of explicit methods.

Runge-Kutta methods are a class of methods by Martin Wilhelm Kutta. The Euler method is one of them. Then there is THE Runge-Kutta method of 4th order, or classical RK4, that Kutta constructed to simultaneously fit the type of methods of Karl Heun.

There are lots of 1-stage first order RK methods $$ k=f(x_n+αh,y_n+αhk), \\ y_{n+1}=y_n+hk, $$ but the only explicit one is the explicit Euler method. Additionally, for $α=\frac12$ (and only that) you get the implicit midpoint method which has order $2$.

$\newcommand{\D}{\mathit{\Delta}}$ If told to use the Runge-Kutta method without further specification, what is meant is the method \begin{align} y_{n+1}&=y_n+\frac{\D y+2\D'y+2\D''y+\D'''y}6\\ \D y&=f(x_n+\D x,y_n+\D'y)\D x\\ \D'y&=f(x_n+\tfrac12\D x,y_n+\tfrac12\D''y)\D x\\ \D''y&=f(x_n+\tfrac12\D x_n,y_n+\tfrac12\D'''y)\D x\\ \D'''y&=f(x_n,y_n)\D x \end{align}


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