Euler’s Method and Local Slopes

The idea

Euler’s method follows the tangent line over a short time step. Starting from $(t_n,y_n)$ , it uses the slope $f(t_n,y_n)$ throughout the next step. The result is an approximation to the solution, not a new exact solution of the original equation.

Method and assumptions

With step $h>0$ , set $t_{n+1}=t_n+h$ and $y_{n+1}=y_n+h f(t_n,y_n)$ . Repeat to the desired time, shortening the final step if necessary. For sufficiently smooth, well-behaved problems over a fixed finite interval, the one-step defect is order $h^2$ and accumulated global error is order $h$ .

Worked example: one decay step

For $y'=-y$ , $y(0)=1$ , and $h=1/2$ , Euler gives $y_1=1-1/2=1/2$ . The exact value is $e^{-1/2}\approx0.6065$ , so the numerical step decays too far.

Worked example: refining at the same endpoint

At $t=1$ , two steps with $h=1/2$ give $(1/2)^2=0.25$ . Four steps with $h=1/4$ give $(3/4)^4\approx0.3164$ . The exact value is $e^{-1}\approx0.3679$ . Smaller steps reduce the error here, but require more evaluations.

Interpreting the result

Compare methods at the same final time. A smoother-looking polyline does not prove accuracy. Step refinement, exact checks when available and stability analysis provide stronger evidence.

Explore

Exact decay versus Euler steps

Try this. Keep the decay rate at 1 and compare 4, 8 and 16 steps. Then use rate 3 with one step: the numerical result fails to follow exact decay. Refine the steps to recover the shape.

Decay rate k1Initial value y(0)1Steps to t = 24

y′ = −1y, y(0) = 1. At t = 2: exact 0.135, Euler 0.063, signed error -0.073. Step h = 0.5; update factor = 0.5. Euler is in its decay stability range. Green: exact; orange: Euler; gray: slope field. Vertical scale adjusts to include the approximation.

This explorer compares Euler steps with the exact solution of $y'=-ky$ on $0\le t\le2$ . Change the decay rate, starting value or step count and compare error with the stability factor.

Practice

Take one Euler step for $y'=t+y$ from $(0,1)$ with $h=0.1$ .
What is the decay amplification factor for $y'=-ky$ ?
How many equal steps of $0.05$ reach $t=1$ from zero?

Show worked solutions

$y_1=1+0.1(0+1)=1.1$ .
$1-kh$ , so $y_{n+1}=(1-kh)y_n$ .
$20$ steps.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.