$$\newcommand{\D}{\displaystyle} \renewcommand{\eqref}[1]{Eq.~(\ref{#1})}$$

## 2.6.5 Euler's method for a system

Euler's method may of course also be used for a system. Let's look at a simultaneous system of $$p$$ equations \begin{align} &y'_1=f_1(x,y_1,y_2,\dots y_p) \nonumber\\ &y'_2=f_2(x,y_1,y_2,\dots y_p) \nonumber\\ &. \tag{2.60}\\ &.\nonumber\\ &y'_p=f_p(x,y_1,y_2,\dots y_p)\nonumber \end{align} with initial values $$$$\tag{2.61} y_1(x_0)=a_1,\ y_2(x_0)=a_2,\dots,\ y_p(x_0)=a_p$$$$ Or, in vectorial format as follows, \begin{align} \tag{2.62} \mathbf{y'}&=\mathbf{f}(x,\mathbf{y})\\ \mathbf{y}&(x_0)=\mathbf{a} \nonumber \end{align} where $$\mathbf{y'}$$, $$\mathbf{f}$$, $$\mathbf{y}$$ and $$\mathbf{a}$$ are column vectors with $$p$$ components.

The Euler scheme (2.55) used on (2.62) gives $$$$\tag{2.63} \mathbf{y_{n+1}}=\mathbf{y_n}+h\cdot \mathbf{f}(x_n,\mathbf{y_n})$$$$ For a system of three equations we get \begin{align} y'_1=&y_2\nonumber\\ y'_2=&y_3 \tag{2.64}\\ y'_3=&-y_1y_3\nonumber \end{align} In this case (2.63) gives \begin{align} &(y_1)_{n+1}=(y_1)_n+h\cdot (y_2)_n\nonumber\\ &(y_2)_{n+1}=(y_2)_n+h\cdot (y_3)_n \tag{2.65}\\ &(y_3)_{n+1}=(y_3)_n-h\cdot (y_1)_n\cdot (y_3)_n\nonumber\\ \tag{2.66} \end{align} with $$y_1(x_0)=a_1,\ y_2(x_0)=a_2,\text{ and }y_3(x_0)=a_3$$

In the section 2.4 Reduction of Higher order Equations we have seen how we can reduce a higher order ODE to a set of first order ODEs. In (2.67) and (2.68) we have the equation $$\frac{d^2z}{dt^2}=g-\alpha\cdot \left(\frac{dz}{dt}\right)^2$$ which we have reduced to a system as \begin{align*} \frac{dz}{dt}= v&\\ \frac{dv}{dt}= g&-\alpha\cdot v^2 \end{align*} which gives an Euler scheme as follows, \begin{align*} &z_{n+1}=z_n+\Delta t\cdot v_n\\ &v_{n+1}=n_n+\Delta t\cdot [g-\alpha(v_n)^2]\\ &\text{med }z_0=0,\ v_0=0 \end{align*}