4.2 Direction fields and numerical methods  (Page 4/14)

Euler’s method

Consider the initial-value problem

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

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 $\left(0,3\right).$ 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 $x$ value by $0.5$ 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 $\left(\mathrm{-1},2\right),$ the slope of the solution is given by $y\prime ={\left(\mathrm{-1}\right)}^2-2^2=\mathrm{-3},$ so the slope of the tangent line to the solution at that point is also equal to $\mathrm{-3}.$ Now we define $x_0=\mathrm{-1}$ and $y_0=2.$ Since the slope of the solution at this point is equal to $\mathrm{-3},$ we can use the method of linear approximation to approximate $y$ near $\left(\mathrm{-1},2\right).$

Here $x_0=\mathrm{-1},y_0=2,$ and $f^{\prime }\left(x_0\right)=\mathrm{-3},$ so the linear approximation becomes

Now we choose a step size    . The step size is a small value, typically $0.1$ or less, that serves as an increment for $x;$ it is represented by the variable $h.$ In our example, let $h=0.1.$ Incrementing $x_0$ by $h$ gives our next $x$ value:

We can substitute $x_1=\mathrm{-0.9}$ into the linear approximation to calculate $y_1.$

Therefore the approximate $y$ value for the solution when $x=\mathrm{-0.9}$ is $y=1.7.$ We can then repeat the process, using $x_1=\mathrm{-0.9}$ and $y_1=1.7$ to calculate $x_2$ and $y_2.$ The new slope is given by $y\prime ={\left(\mathrm{-0.9}\right)}^2-{\left(1.7\right)}^2=\mathrm{-2.08}.$ First, $x_2=x_1+h=\mathrm{-0.9}+0.1=\mathrm{-0.8}.$ Using linear approximation gives

Finally, we substitute $x_2=\mathrm{-0.8}$ into the linear approximation to calculate $y_2.$

Therefore the approximate value of the solution to the differential equation is $y=1.492$ when $x=\mathrm{-0.8}.$

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 [link] , rounded off to four decimal places.