In Improved Euler and Runge–Kutta Methods you met Euler's method, $y_{n+1} = y_n + hf(t_n, y_n)$, and the more accurate Heun and Runge–Kutta methods. Now we measure their error the way Truncation Error and Order of Accuracy taught: by how it shrinks with $h$. There are two different errors to keep apart.
The local error is the error of one step taken from an exact value: $y(t_n + h)$ minus the step's result when $y_n = y(t_n)$. The global error is what we actually care about: $y(T) - y_N$ at the final time, after all $N = T/h$ steps.
🔑 Key idea
For a method of order $p$, the local error is $O(h^{p+1})$ and the global error is $O(h^{p})$: summing about $1/h$ local errors costs one power of $h$. Euler: local $O(h^{2})$, global $O(h)$. Heun: global $O(h^{2})$. RK4: global $O(h^{4})$.
💡 Why it works
For Euler, Taylor's theorem gives $y(t + h) = y(t) + hy'(t) + \frac{h^{2}}{2}y''(\xi)$, and the step keeps only the first two terms: local error $\frac{h^{2}}{2}y''$. Over $[0, T]$ there are $N = T/h$ steps. Each adds an error of size about $Ch^{2}$, and earlier errors are carried along (grown or damped by the equation by at most a factor $e^{LT}$, where $L$ bounds $|\partial f/\partial y|$). So the global error is at most about $N \cdot Ch^{2}\,e^{LT} = CTe^{LT}h$: order $h$.
Euler's method follows the slope field in straight steps. Each step misses the curve a little (local error), and the misses add up (global error).
✏️ Worked example: One step: local error
For $y' = y$, $y(0) = 1$, take one Euler step with $h = 0.1$. Find the local error.
Euler: $y_1 = 1 + 0.1(1) = 1.1$.
Exact: $y(0.1) = e^{0.1} = 1.105171$.
Local error $\approx 0.00517$, close to $\frac{h^{2}}{2}y''(0) = 0.005$.
Answer: $\approx 0.0052$
✏️ Worked example: Many steps: global error
Solve the same problem to $t = 1$ (exact $y(1) = e \approx 2.71828$) with $h = 0.1$ and $h = 0.05$.
For $y' = y$, Euler multiplies by $1 + h$ each step: $y_N = (1 + h)^{N}$.
$h = 0.1$: $1.1^{10} = 2.59374$, global error $0.1245$.
$h = 0.05$: $1.05^{20} = 2.65330$, global error $0.0650$, about half: order 1.
The global error, $0.12$, is far bigger than one local error, $0.005$: ten of them added up and grew along the way.
Answer: $0.1245$ and $0.0650$
✏️ Worked example: A fourth-order method
RK4 on the same problem gives global errors $2.08 \times 10^{-6}$ ($h = 0.1$) and $1.36 \times 10^{-7}$ ($h = 0.05$). What is the order?
Ratio: $\frac{2.08 \times 10^{-6}}{1.36 \times 10^{-7}} \approx 15.3$, close to $2^{4} = 16$.
Order 4, and with $h = 0.1$ it is already 60,000 times more accurate than Euler, for 4 function values per step instead of 1.
Answer: $p = 4$
Global error at t = 1 against h for y′ = −2y. With "one step only", the plot shows the local error instead.
⚠️ Common mistake
Expecting the error at the end to be the size of one step's error. The local errors of all $N = T/h$ steps add up (and can grow along the way), which costs one power of $h$: a method with local error $O(h^{2})$ has global error only $O(h)$. Check: halve $h$ and compare final values; for Euler the error should halve, not quarter.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation (University of Delaware, free online edition): chapter "Initial-value problems for ODEs", sections "Euler's method" and "Runge–Kutta methods"
MIT OpenCourseWare 18.330, Introduction to Numerical Analysis: lecture notes on numerical methods for ODEs (local and global error)