Euler’s Method Follows an ODE One Step at a Time

Given $y'=f(t,y)$ and $y(t_0)=y_0$, Euler’s method advances by

$t_{n+1}=t_n+h,\qquad y_{n+1}=y_n+h f(t_n,y_n).$

Local approximation

The update comes from $y(t+h)\approx y(t)+hy'(t)$. For smooth solutions, one step has error proportional to $h^2$, while accumulated error over a fixed interval is typically proportional to $h$.

Step size

Smaller $h$ usually improves accuracy but increases work. A solution that changes rapidly may require short steps. Some stable differential equations can produce unstable Euler approximations when $h$ is too large.

Work the recurrence by hand

Approximate $y'=y$, $y(0)=1$, using $h=0.25$. Euler’s update is

$y_{n+1}=y_n+0.25y_n=1.25y_n.$

The first four approximations are $1$, $1.25$, $1.5625$, $1.953125$, and $2.44140625$. At $t=1$, the exact value is $e\approx2.71828$, so Euler underestimates the convex exponential.

Stability can be stricter than accuracy

Apply Euler to $y'=\lambda y$. The update is $y_{n+1}=(1+h\lambda)y_n$. When $\lambda<0$, the true solution decays. The numerical solution decays only if

$|1+h\lambda|<1.$

For $y'=-10y$, this requires $0<h<0.2$. A larger step makes a truly stable system oscillate or grow numerically.

Check your understanding

For $y'=-4y$ with $h=0.5$, what does Euler predict, and why is it misleading?

Show the reasoning

The multiplier is $1-4(0.5)=-1$, so the approximation alternates forever with constant magnitude. The exact solution $y_0e^{-4t}$ decays. The step lies on the stability boundary.