Euler's method steps forward along the slope at the start of each step: $y_{n+1} = y_n + hf(t_n, y_n)$. It is simple, but its error shrinks only in proportion to $h$. Better methods sample the slope at more than one place in each step.
Idea clave
Improved Euler (Heun): $k_1 = f(t_n, y_n)$, $k_2 = f(t_n + h,\ y_n + hk_1)$, $y_{n+1} = y_n + \frac{h}{2}(k_1 + k_2)$.
Runge–Kutta (RK4): $k_1 = f(t_n, y_n)$, $k_2 = f(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_1)$, $k_3 = f(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_2)$, $k_4 = f(t_n + h, y_n + hk_3)$, $$y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4).$$
Runge–Kutta (RK4): $k_1 = f(t_n, y_n)$, $k_2 = f(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_1)$, $k_3 = f(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_2)$, $k_4 = f(t_n + h, y_n + hk_3)$, $$y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4).$$
Por qué funciona
Taylor's theorem gives $y(t + h)$ $= y + hy' + \frac{h^{2}}{2}y'' + \dots$. Euler keeps only $hy'$, so each step is off by about $\frac{h^{2}}{2}y''$; after $T/h$ steps the errors add up to something proportional to $h$. Improved Euler averages the slope at both ends of the step, like the trapezoid rule, and that average matches the $h^{2}$ term too: the total error is proportional to $h^{2}$. RK4 weights four slopes like Simpson's rule and reaches $h^{4}$.
| Method | Slopes per step | Global error | Halve $h$: error shrinks by |
|---|---|---|---|
| Euler | 1 | $\approx Ch$ | 2 |
| Improved Euler | 2 | $\approx Ch^{2}$ | 4 |
| Runge–Kutta (RK4) | 4 | $\approx Ch^{4}$ | 16 |
Ejemplo resuelto: One step of improved Euler
For $y' = t + y$, $y(0) = 1$, take one step with $h = 0.1$. The exact solution is $y = 2e^{t} - t - 1$.
- $k_1 = f(0, 1) = 1$.
- Euler prediction: $1 + 0.1(1) = 1.1$ at $t = 0.1$, so $k_2 = f(0.1, 1.1) = 1.2$.
- $y_1 = 1 + \frac{0.1}{2}(1 + 1.2)$ $= 1.11$.
- Exact: $y(0.1) = 2e^{0.1} - 1.1$ $\approx 1.110342$. Euler alone gives $1.1$ (error $0.0103$); improved Euler's error is $0.00034$.
Respuesta: $y(0.1) \approx 1.11$
Ejemplo resuelto: One step of RK4
For $y' = y$, $y(0) = 1$, take one RK4 step with $h = 0.5$.
- $k_1 = 1$; $k_2 = f(0.25, 1.25) = 1.25$; $k_3 = f(0.25, 1.3125)$ $= 1.3125$; $k_4 = f(0.5, 1.65625)$ $= 1.65625$.
- $y_1$ $= 1$ $+ \frac{0.5}{6}(1 + 2.5 + 2.625 + 1.65625)$ $= 1 + \frac{0.5}{6}(7.78125)$ $\approx 1.648438$.
- Exact: $e^{0.5} \approx 1.648721$, an error of about $0.0003$ with a single large step.
Respuesta: $y(0.5) \approx 1.6484$
Error común
In improved Euler, evaluating $k_2$ at the old value $y_n$ instead of at the Euler prediction $y_n + hk_1$, or using $h$ instead of $\frac{h}{2}$ in the average. In RK4, the middle slopes use half a step, $t_n + \frac{h}{2}$, but the final update uses the whole $h$. A quick check: your answer should be close to plain Euler's, and closer to the exact value when you know it.