# 1st order Runge-Kutta method and Euler's method

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.

• 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.
– John
Apr 17, 2020 at 7:00
• I transferred my comments to an answer. Apr 17, 2020 at 8:36

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}