Consider the initial-value problem

Integrating both sides of the differential equation gives [latex]y={x}^{2}-3x+C[/latex], and solving for [latex]C[/latex] yields the particular solution [latex]y={x}^{2}-3x+3[/latex]. The solution for this initial-value problem appears as the parabola in Figure 10.

Figure 10. Euler’s Method for the initial-value problem [latex]{y}^{\prime }=2x - 3,y\left(0\right)=3[/latex].

The red graph consists of line segments that approximate the solution to the initial-value problem. The graph starts at the same initial value of [latex]\left(0,3\right)[/latex]. Then the slope of the solution at any point is determined by the right-hand side of the differential equation, and the length of the line segment is determined by increasing the [latex]x[/latex] value by [latex]0.5[/latex] each time (the step size). This approach is the basis of Euler’s Method.

Before we state Euler’s Method as a theorem, let’s consider another initial-value problem:

The idea behind direction fields can also be applied to this problem to study the behavior of its solution. For example, at the point [latex]\left(-1,2\right)[/latex], the slope of the solution is given by [latex]y^{\prime} ={\left(-1\right)}^{2}-{2}^{2}=-3[/latex], so the slope of the tangent line to the solution at that point is also equal to [latex]-3[/latex]. Now we define [latex]{x}_{0}=-1[/latex] and [latex]{y}_{0}=2[/latex]. Since the slope of the solution at this point is equal to [latex]-3[/latex], we can use the method of linear approximation to approximate [latex]y[/latex] near [latex]\left(-1,2\right)[/latex].

Here [latex]{x}_{0}=-1,{y}_{0}=2[/latex], and [latex]{f}^{\prime }\left({x}_{0}\right)=-3[/latex], so the linear approximation becomes

Now we choose a step size. The step size is a small value, typically [latex]0.1[/latex] or less, that serves as an increment for [latex]x[/latex]; it is represented by the variable [latex]h[/latex]. In our example, let [latex]h=0.1[/latex]. Incrementing [latex]{x}_{0}[/latex] by [latex]h[/latex] gives our next [latex]x[/latex] value:

We can substitute [latex]{x}_{1}=-0.9[/latex] into the linear approximation to calculate [latex]{y}_{1}[/latex].

Therefore the approximate [latex]y[/latex] value for the solution when [latex]x=-0.9[/latex] is [latex]y=1.7[/latex]. We can then repeat the process, using [latex]{x}_{1}=-0.9[/latex] and [latex]{y}_{1}=1.7[/latex] to calculate [latex]{x}_{2}[/latex] and [latex]{y}_{2}[/latex]. The new slope is given by [latex]y^{\prime} ={\left(-0.9\right)}^{2}-{\left(1.7\right)}^{2}=-2.08[/latex]. First, [latex]{x}_{2}={x}_{1}+h=-0.9+0.1=-0.8[/latex]. Using linear approximation gives

Finally, we substitute [latex]{x}_{2}=-0.8[/latex] into the linear approximation to calculate [latex]{y}_{2}[/latex].

Therefore the approximate value of the solution to the differential equation is [latex]y=1.492[/latex] when [latex]x=-0.8[/latex].

What we have just shown is the idea behind Euler’s Method. Repeating these steps gives a list of values for the solution. These values are shown in the table below, rounded off to four decimal places.

Typically [latex]h[/latex] is a small value, say [latex]0.1[/latex] or [latex]0.05[/latex]. The smaller the value of [latex]h[/latex], the more calculations are needed. The higher the value of [latex]h[/latex], the fewer calculations are needed. However, the tradeoff results in a lower degree of accuracy for larger step size, as illustrated in Figure 11.

Figure 11. Euler’s method for the initial-value problem [latex]{y}^{\prime }=2x - 3,y\left(0\right)=3[/latex] with (a) a step size of [latex]h=0.5[/latex]; and (b) a step size of [latex]h=0.25[/latex].

Watch the following videos to see the worked solution to Example: Using Euler’s Method

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You can view the transcript for this segmented clip of “4.2.3” here (opens in new window).