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=== Using step size equal to 1 ({{math|1=''h'' = 1}}) ===

[[Image:Numerical integration illustration, step=1.svg|right|thumb|(Figure 2) Illustration of numerical integration for the equation <math>y'=y, y(0)=1.</math> Blue is the Euler method; green, the [[midpoint method]]; red, the exact solution, <math>y=e^t.</math> The step size is <math>h=1.0.</math>]]
The Euler method is

:<math> y_{n+1} = y_n + hf(t_n,y_n). </math>

so first we must compute <math>f(t_0, y_0)</math>.  In this simple differential equation, the function <math>f</math> is defined by <math>f(t,y) = y</math>. We have

:<math> f(t_0,y_0) = f(0,1) = 1. </math>

By doing the above step, we have found the slope of the line that is tangent to the solution curve at the point <math>(0,1)</math>.  Recall that the slope is defined as the change in <math>y</math> divided by the change in <math>t</math>, or <math display="inline"> \frac{\Delta y} {\Delta t}</math>.

The next step is to multiply the above value by the step size <math>h</math>, which we take equal to one here:

:<math> h \cdot f(y_0) = 1 \cdot 1 = 1. </math>

Since the step size is the change in <math>t</math>, when we multiply the step size and the slope of the tangent, we get a change in <math>y</math> value.  This value is then added to the initial <math>y</math> value to obtain the next value to be used for computations.

:<math> y_0 + hf(y_0) = y_1 = 1 + 1 \cdot 1 = 2. </math>
 
The above steps should be repeated to find <math> y_2</math>, <math>y_3 </math> and <math>y_4</math>.

:<math> \begin{align}
y_2 &= y_1 + hf(y_1) = 2 + 1 \cdot 2 = 4, \\
y_3 &= y_2 + hf(y_2) = 4 + 1 \cdot 4 = 8, \\
y_4 &= y_3 + hf(y_3) = 8 + 1 \cdot 8 = 16.
\end{align} </math>

Due to the repetitive nature of this algorithm, it can be helpful to organize computations in a chart form, as seen below, to avoid making errors.

:{| class="wikitable"
|-
! <math>n</math> !! <math>y_n</math> !! <math>t_n</math> !!<math>f(t_n,y_n)</math> !! <math>h</math> !! <math>\Delta y</math> !! <math>y_{n+1}</math>
|-
| 0 || 1 || 0 || 1 || 1 || 1 || 2
|-
| 1 || 2 || 1 || 2 || 1 || 2 || 4
|-
| 2 || 4 || 2 || 4 || 1 || 4 || 8
|-
| 3 || 8 || 3 || 8 || 1 || 8 || 16
|}

The conclusion of this computation is that <math> y_4 = 16 </math>. The exact solution of the differential equation is <math> y(t) = e^t </math>, so <math> y(4) = e^4 \approx 54.598 </math>. Although the approximation of the Euler method was not very precise in this specific case, particularly due to a large value step size <math>h</math>, its behaviour is qualitatively correct as the figure shows.