1.2. Euler's Method

Consider the problem (1.1.1). Let $h= (a-b)/N$ be the step size. Let $X_i=a+ih, \,\, 1\leq i \leq n$, with $x_n=b$. Let $y_K$ be the approximate value of y at $x_k$. We define

$$y_{k+1}=y_k+hf_k, \,\,\, k=0,1,2...n-1$$ (1.2.1)

where $f_k=f(x_k, y_k)$. The method of determination of $y_k$ by (1.2.1) is called the Euler's method.
Remark : Euler's method is an one-step method. The Euler's method has many motivation. Firstly the derivative y' at $x = x_i$ can be approximated by

$$\frac{y(x_{i+1}) -y(x_i)}{h}$$

if h is sufficiently small. With this approximation in (1.1.1) we have

$$y'(x)\cong \frac{y(x_{i+1})-y(x_i)}{h}=f(x_i,y(x_i))$$

we can also look at (1.2.1) from the following point of View. Integration of (1.1.1) yields

$$y(x_{i+1})=y(x_i)+ \int\limits_{x_i}^{x_{i+1}}f(x,y(x))dx$$

The integral on the right hand side now is approximated by

$$\int\limits_{x_i}^{x_{i+1}}f(s,y(x))dx \cong f(x_i, y(x_i))h$$

for sufficiently small $h >0$.
Secondly if y is differentiable sufficient number of times, we use Taylor's expansion

$$y(x_{(i+1)})=y(x_i)+hf(x_i,y(x_i))+\frac{h^2}{2}(f(x_i,y(x_{i})))+...$$

neglecting terms containing with powers of $h \geq 2$, we arrive at (1.2.1). We illustrate the Euler's Algorithm with an example.