Let *y* = *f* (*x*) be a solution to the IVP

What is the error when Euler's method is used to estimate *f* (2) with a step size of 0.5? With a step size of 0.25?

Answer

Since we're asked about error, we'll need to know the exact value of *f* (2). Thinking backwards, *f* (*x*) = 2*x*^{2} + *C*. Since *f* (1) = 3,

2(1)^{2} + *C* = 3

so *C* = 1 and *f* (*x*) = 2*x*^{2} + 1. The exact value in question is

*f* (2) = 2(2)^{2} + 1 = 9.

Euler's method with step size 0.5 gets us this table:

The error is 1, since the exact value of *f* (2) is 9 but Euler's method approximated *f* (2) ≅ 8. Euler's method with step size .25 gets us this table:

With step size 0.25 Euler's method approximates *f* (2) ≅ 8.5. The error in this case is 0.5.