5.4 — Runge-Kutta Methods

The workhorse of numerical differential equations

Learning Objectives

Understand the RK4 algorithm and why it uses four slope evaluations
Compute all four slopes ( $k_{1}$ , $k_{2}$ , $k_{3}$ , $k_{4}$ ) in a single step
Compare the accuracy of RK4 with Euler's method
Understand the error order and practical implications
Apply RK4 to systems of differential equations

Motivation

Why Runge-Kutta? The RK4 method is considered the workhorse of numerical methods in science and engineering. It provides exceptional accuracy with manageable computational cost. You will encounter RK4 in:

Engineering simulations (structural mechanics, control systems)
Physics simulations (orbital mechanics, fluid dynamics)
Biology and medicine (pharmacokinetics, population dynamics)
Climate modeling and weather prediction

From Euler to Runge-Kutta

Euler's method uses a single slope evaluation at the beginning of each step, giving $O (h)$ accuracy. Runge-Kutta methods use multiple slope evaluations at strategic points to achieve higher-order accuracy. The 4th-order method (RK4) uses four slope evaluations to achieve $O (h^{4})$ accuracy.

The RK4 Formulas

Definition: Fourth-Order Runge-Kutta (RK4)

Given $y' = f (x, y)$ with initial condition $y (x_{0}) = y_{0}$ and step size $h$ , the RK4 step is:

k_1 = f(x_n, y_n)$$ $$k_2 = f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_1\right)$$ $$k_3 = f\!\left(x_n + \tfrac{h}{2},\; y_n + \tfrac{h}{2}k_2\right)$$ $$k_4 = f(x_n + h,\; y_n + h\,k_3)$$ $$y_{n+1} = y_n + \tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)

Intuition Behind RK4

$k_{1}$ : Slope at the start of the interval: $f (x_{n}, y_{n})$

$k_{2}$ : Slope at the midpoint using $k_{1}$ to estimate the midpoint value

$k_{3}$ : Slope at the midpoint using $k_{2}$ (a refined estimate)

$k_{4}$ : Slope at the end of the interval using $k_{3}$

The final update uses a weighted average of these four slopes:

\text{weight} = \frac{1}{6}(1 + 2 + 2 + 1)

This weighting (Simpson's rule-like) gives extraordinary accuracy by sampling the slope in a strategic pattern.

Error Analysis

Theorem: RK4 Error

Local Truncation Error: $O (h^{5})$

Global Error: $O (h^{4})$

RK4 is a fourth-order method. To reduce error by a factor of 16, halve the step size $h$ .

Comparison of Methods

Quick Reference: Comparing Numerical Methods

Method	Order	$f$ Evaluations per Step	Typical Use
Euler	$O (h)$	1	Quick sketches, educational purposes
Improved Euler (RK2)	$O (h^{2})$	2	Better accuracy, simple code
RK4	$O (h^{4})$	4	Standard default for production code

Step-by-Step RK4 Algorithm

Set up: Define the differential equation $y' = f (x, y)$ , initial condition $(x_{0}, y_{0})$ , and step size $h$ .
Compute $k_{1}$ : $k_{1} = f (x_{n}, y_{n})$
Compute $k_{2}$ : $k_{2} = f (x_{n} + h / 2, y_{n} + (h / 2) k_{1})$
Compute $k_{3}$ : $k_{3} = f (x_{n} + h / 2, y_{n} + (h / 2) k_{2})$
Compute $k_{4}$ : $k_{4} = f (x_{n} + h, y_{n} + h k_{3})$
Update: $y_{n + 1} = y_{n} + (h / 6) (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})$
Advance: $x_{n + 1} = x_{n} + h$
Repeat: Continue until you reach the desired endpoint.

Worked Examples

Problem: Solve $y' = y$ , $y (0) = 1$ , with $h = 0.1$ . Find $y (0.1)$ .

Solution:

Setup: $f (x, y) = y$ , $x_{0} = 0$ , $y_{0} = 1$ , $h = 0.1$

Step	Calculation	Value
$k_{1}$	$f (0, 1) = 1$	1
$k_{2}$	$f (0.05, 1 + 0.05) = f (0.05, 1.05) = 1.05$	1.05
$k_{3}$	$f (0.05, 1 + 0.05 (1.05)) = 1.0525$	1.0525
$k_{4}$	$f (0.1, 1 + 0.1 (1.0525)) = 1.10525$	1.10525

RK4 Update:

y_1 = 1 + \frac{0.1}{6}(1 + 2(1.05) + 2(1.0525) + 1.10525)$$ $$y_1 = 1 + \frac{0.1}{6}(6.31075) = 1 + 0.10517083... = 1.10517083...

Exact solution: $y (x) = e^{x}$ , so $y (0.1) = e^{0.1} = 1.10517092...$

Error: $| 1.10517083 - 1.10517092 | \approx 8.5 \times 10^{- 8}$ (microscopic!)

Compare with Euler: Euler gives $y_1 = 1 + 0.1(1) = 1.1$ with error $\approx 0.005$ — about 59,000 times larger!

Problem: Solve $y' = x + y$ , $y (0) = 1$ , with $h = 0.2$ . Find $y (1)$ .

Solution:

We compute 5 steps, each with the full RK4 formula. Here's a summary table:

$n$	$x_{n}$	$y_{n}$	$k_{1}$	$k_{2}$	$k_{3}$	$k_{4}$
0	0.0	1.0000	1.0000	1.2000	1.2200	1.4440
1	0.2	1.2428	1.4428	1.6871	1.7115	1.9851
2	0.4	1.5836	1.9836	2.2820	2.3118	2.6460
3	0.6	2.0442	2.6442	3.0086	3.0451	3.4532
4	0.8	2.6510	3.4510	3.8961	3.9407	4.4392
5	1.0	3.4365	—	—	—	—

RK4 Result: $y (1) \approx 3.4365$

Exact solution: $y = 2 e^{x} - x - 1$ , so $y (1) = 2 e^{1} - 1 - 1 = 2 e - 2 \approx 3.4366$

Error: $| 3.4365 - 3.4366 | \approx 0.0001$ (excellent accuracy for 5 steps with $h = 0.2$ )

Problem: Solve $y' = - 2 x y$ , $y (0) = 1$ , with $h = 0.25$ . Find $y (1)$ .

Solution:

This problem has an exact solution: $y (x) = e^{- x^{2}}$ , so $y (1) = e^{- 1} \approx 0.36788$

Using RK4 with only 4 steps ( $h = 0.25$ ):

Step	$x$	RK4 $y$	Exact $y$	Error
0	0.0	1.00000	1.00000	0.00000
1	0.25	0.93941	0.93941	0.00000
2	0.50	0.77848	0.77880	0.00032
3	0.75	0.55383	0.55432	0.00049
4	1.0	0.36779	0.36788	0.00009

Result: RK4 with just 4 steps gives $y (1) \approx 0.36779$ , accurate to 4 decimal places! The error is only $0.00009$ .

Problem: Solve $y' = y$ , $y (0) = 1$ to $x = 1$ with $h = 0.5$ . Compare three methods.

Solution:

Exact solution: $y (1) = e \approx 2.71828$

Method	$y (1)$	Error	Functions Called
Euler	2.2500	0.4683	2
Improved Euler (RK2)	2.6406	0.0777	4
RK4	2.71735	0.00093	8
Exact	2.71828	—	—

Observation: RK4 with 8 function evaluations gives error $0.00093$ , while Euler with 2 evaluations gives error $0.4683$ — RK4 is 500 times more accurate for the same computational work!

Problem: Solve the system:

x' = -y, \quad y' = x, \quad x(0) = 1, \quad y(0) = 0

with $h = 0.1$ . Compute one step.

Solution:

For systems, RK4 uses vector slopes. Let $f_1 (x, y) = - y$ and $f_2 (x, y) = x$ .

At step 0: $x_{0} = 1$ , $y_{0} = 0$

Slope	$k$ (for $x$ )	$l$ (for $y$ )
$k_{1}, l_{1}$	$- 0 = 0$	$1$
$k_{2}, l_{2}$	$- (0 + 0.05) = - 0.05$	$1$
$k_{3}, l_{3}$	$- (0 + 0.05 (1)) = - 0.05$	$1 + 0.05 (- 0.05) = 0.9975$
$k_{4}, l_{4}$	$- (0 + 0.1 (0.9975)) = - 0.09975$	$1 + 0.1 (- 0.05) = 0.995$

RK4 Updates:

x_1 = 1 + \frac{0.1}{6}\big(0 + 2(-0.05) + 2(-0.05) + (-0.09975)\big) = 1 - 0.00499583 = 0.99500417$$ $$y_1 = 0 + \frac{0.1}{6}\big(1 + 2(1) + 2(0.9975) + 0.995\big) = \frac{0.1}{6}(5.99) = 0.09983333

Result: $(x_{1}, y_{1}) \approx (0.99500417, 0.09983333)$

Exact solution: $x (t) = \cos t$ , $y (t) = \sin t$ (unit circle)

At $t = 0.1$ : $(0.99500, 0.09983)$

Error: $| Δ x | \approx 1.4 \times 10^{- 9}$ and $| Δ y | \approx 8.3 \times 10^{- 8}$ — tiny! RK4 tracks circular motion almost exactly. Note that each stage must update both components before the next stage is evaluated; freezing $l$ at its $l_{1}$ value is the classic mistake and costs six digits of accuracy.

📝 Exam-Style Practice Problems

These are the kind of RK4 questions that appear on exams: one or two steps by hand, carried to full precision, with an accuracy comparison at the end. Keep at least eight decimal places in the intermediate slopes — RK4 is accurate enough that premature rounding, not the method, becomes your error. Click a problem to reveal the full solution.

Practice 1: Take one RK4 step for $y' = x + y$, $y(0) = 1$ with $h = 0.2$, and compare against Euler and the exact solution ▼

Solution

Step 1: Write down the four slopes

$$k_1 = f(x_0, y_0),\quad k_2 = f\!\left(x_0 + \tfrac{h}{2},\, y_0 + \tfrac{h}{2}k_1\right),\quad k_3 = f\!\left(x_0 + \tfrac{h}{2},\, y_0 + \tfrac{h}{2}k_2\right),\quad k_4 = f(x_0 + h,\, y_0 + h k_3)$$

With $f(x,y) = x + y$, $(x_0,y_0) = (0,1)$, $h = 0.2$ and $\tfrac{h}{2} = 0.1$:

Step 2: Evaluate them in order

$$k_1 = f(0, 1) = 0 + 1 = 1$$ $$k_2 = f(0.1,\; 1 + 0.1(1)) = f(0.1,\, 1.1) = 0.1 + 1.1 = 1.2$$ $$k_3 = f(0.1,\; 1 + 0.1(1.2)) = f(0.1,\, 1.12) = 0.1 + 1.12 = 1.22$$ $$k_4 = f(0.2,\; 1 + 0.2(1.22)) = f(0.2,\, 1.244) = 0.2 + 1.244 = 1.444$$

Notice each slope feeds the next: $k_2$ uses $k_1$, $k_3$ uses $k_2$, $k_4$ uses $k_3$.

Step 3: Weighted average and update

$$k_1 + 2k_2 + 2k_3 + k_4 = 1 + 2.4 + 2.44 + 1.444 = 7.284$$ $$y_1 = 1 + \frac{0.2}{6}(7.284) = 1 + 0.2428 = 1.2428$$

Step 4: Compare

The exact solution is $y = 2e^{x} - x - 1$, so $y(0.2) = 1.24280552$.

Method ($h = 0.2$, one step)	$y(0.2)$	Error	$f$-evaluations
Euler	1.2000000	$2.4\times10^{-2}$	1
Improved Euler	1.2400000	$2.8\times10^{-3}$	2
RK4	1.2428000	$5.5\times10^{-6}$	4
Exact	1.2428055	—	—

$$\boxed{y(0.2) \approx 1.2428,\qquad \text{error} \approx 5.5\times10^{-6}}$$

Exam tip: four times the work of Euler buys roughly four thousand times the accuracy here. That ratio only improves as $h$ shrinks, because the errors scale as $h$ and $h^4$ respectively.

Practice 2: Take two RK4 steps for $y' = y - x^2 + 1$, $y(0) = 0.5$ with $h = 0.2$, and verify the fourth-order accuracy ▼

Solution

Step 1: Record the exact solution first

$$y(x) = (x+1)^2 - \tfrac{1}{2}e^{x}$$

Check: $y(0) = 1 - 0.5 = 0.5$ ✓, and $y' = 2(x+1) - \tfrac12 e^x$, while $y - x^2 + 1 = (x^2+2x+1) - \tfrac12 e^x - x^2 + 1 = 2x + 2 - \tfrac12 e^x$ ✓

Step 2: First step, $x_0 = 0 \to x_1 = 0.2$

$$k_1 = f(0,\,0.5) = 0.5 - 0 + 1 = 1.5$$ $$k_2 = f(0.1,\; 0.5 + 0.1(1.5)) = f(0.1,\, 0.65) = 0.65 - 0.01 + 1 = 1.64$$ $$k_3 = f(0.1,\; 0.5 + 0.1(1.64)) = f(0.1,\, 0.664) = 0.664 - 0.01 + 1 = 1.654$$ $$k_4 = f(0.2,\; 0.5 + 0.2(1.654)) = f(0.2,\, 0.8308) = 0.8308 - 0.04 + 1 = 1.7908$$ $$y_1 = 0.5 + \frac{0.2}{6}\big(1.5 + 2(1.64) + 2(1.654) + 1.7908\big) = 0.5 + \frac{0.2}{6}(9.8788) = 0.82929333$$

Step 3: Second step, $x_1 = 0.2 \to x_2 = 0.4$

$$k_1 = f(0.2,\, 0.82929333) = 0.82929333 - 0.04 + 1 = 1.78929333$$ $$k_2 = f(0.3,\; 0.82929333 + 0.1(1.78929333)) = f(0.3,\, 1.00822267) = 1.91822267$$ $$k_3 = f(0.3,\; 0.82929333 + 0.1(1.91822267)) = f(0.3,\, 1.02111560) = 1.93111560$$ $$k_4 = f(0.4,\; 0.82929333 + 0.2(1.93111560)) = f(0.4,\, 1.21551645) = 2.05551645$$ $$y_2 = 0.82929333 + \frac{0.2}{6}\big(1.78929333 + 2(1.91822267) + 2(1.93111560) + 2.05551645\big) = 1.21407621$$

Step 4: Errors

$n$	$x_n$	$y_n$ (RK4)	$y(x_n)$ exact	Error
0	0.0	0.50000000	0.50000000	0
1	0.2	0.82929333	0.82929862	$5.3\times10^{-6}$
2	0.4	1.21407621	1.21408765	$1.1\times10^{-5}$

Halving to $h = 0.1$ would shrink these by a factor of about $2^4 = 16$, not $2$ — that is what “fourth order” means.

$$\boxed{y(0.4) \approx 1.21407621,\qquad \text{error} \approx 1.1\times10^{-5}}$$

Practice 3 (system): Convert $y'' + 2y' + 2y = 0$, $y(0) = 1$, $y'(0) = 0$ to a first-order system and take one RK4 step with $h = 0.1$ ▼

Solution

Step 1: Convert to a system

Let $u = y$ and $v = y'$. Then

$$u' = v = f(t,u,v),\qquad v' = -2v - 2u = g(t,u,v),\qquad u(0) = 1,\; v(0) = 0$$

Step 2: Vector RK4 — every stage advances both components together

$$k_i \text{ for } u,\qquad \ell_i \text{ for } v,\qquad \begin{aligned} k_1 &= f(t_0,u_0,v_0), & \ell_1 &= g(t_0,u_0,v_0)\\ k_2 &= f\!\left(t_0+\tfrac h2, u_0+\tfrac h2 k_1, v_0+\tfrac h2 \ell_1\right), & \ell_2 &= g\!\left(t_0+\tfrac h2, u_0+\tfrac h2 k_1, v_0+\tfrac h2 \ell_1\right) \end{aligned}$$

and similarly for the third and fourth stages. Never update $u$ using $k$'s while leaving $v$ frozen — both must move at every stage.

Step 3: Evaluate the stages ($h/2 = 0.05$)

Stage	$(u,v)$ used	$k_i = v$	$\ell_i = -2v - 2u$
1	$(1,\,0)$	$0$	$-2$
2	$(1 + 0.05(0),\; 0 + 0.05(-2)) = (1,\,-0.1)$	$-0.1$	$-1.8$
3	$(1 + 0.05(-0.1),\; 0 + 0.05(-1.8)) = (0.995,\,-0.09)$	$-0.09$	$-1.81$
4	$(1 + 0.1(-0.09),\; 0 + 0.1(-1.81)) = (0.991,\,-0.181)$	$-0.181$	$-1.62$

Step 4: Update both components

$$u_1 = 1 + \frac{0.1}{6}\big(0 + 2(-0.1) + 2(-0.09) + (-0.181)\big) = 1 + \frac{0.1}{6}(-0.561) = 0.99065$$ $$v_1 = 0 + \frac{0.1}{6}\big(-2 + 2(-1.8) + 2(-1.81) + (-1.62)\big) = \frac{0.1}{6}(-10.84) = -0.18066667$$

Step 5: Compare with the exact solution

The characteristic roots are $-1 \pm i$, so $y = e^{-t}(\cos t + \sin t)$ satisfies both initial conditions, and $y' = -2e^{-t}\sin t$.

$$y(0.1) = 0.99065001,\qquad y'(0.1) = -0.18066602$$ $$\boxed{u_1 = 0.99065\ \ (\text{error } 1.1\times10^{-8}),\qquad v_1 = -0.18066667\ \ (\text{error } 6.4\times10^{-7})}$$

Exam tip: for an $m$-component system RK4 needs $4m$ scalar evaluations per step, but the weights $\tfrac16(1,2,2,1)$ and the $\tfrac h2$ offsets are identical to the scalar case. Only the bookkeeping changes.

Practice Problems

Test your understanding with these multiple-choice questions. Click a choice to check your answer.

Progress:

0 / 6 correct

Q1: Computing $k_1$ — for $y' = x^2 + y$ with $y(0) = 1$ and $h = 0.2$, what is $k_1$?

Q2: Understanding $k_2$ — at which point is the second RK4 slope evaluated?

Q3: In $y_{n+1} = y_n + \tfrac{h}{6}\left(k_1 + 2k_2 + 2k_3 + k_4\right)$, what fraction of the total weight does $k_2$ carry?

Q4: What is the local truncation error of RK4 on a single step?

Q5: Take one RK4 step for $y' = -2y$, $y(0) = 1$ with $h = 0.5$. What is $y_1$?

Q6: To reduce the global error of RK4 by a factor of about $256$, by what factor should you decrease $h$?

Quick Reference Card

RK4 Formula at a Glance

\begin{align} k_1 &= f(x_n, y_n)\\ k_2 &= f\!\left(x_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_1\right)\\ k_3 &= f\!\left(x_n + \tfrac{h}{2}, y_n + \tfrac{h}{2}k_2\right)\\ k_4 &= f(x_n + h, y_n + hk_3)\\ y_{n+1} &= y_n + \tfrac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{align}

Key Facts

Order: Fourth-order method, $O (h^{4})$ global error
Function calls: 4 evaluations of $f$ per step
Local truncation error: $O (h^{5})$
Halve $h$ : Error is divided by $2^4 = 16$
Stability: Good stability for most problems; region depends on the problem
Best used when: You need high accuracy without adaptive step control