Ordinary differential equations · Lecture 22

ODEs II: Runge–Kutta and Predictor–Corrector Methods

Fourth-order accuracy from values of $f$ alone (Runge–Kutta), reusing past slopes (Milne and Adams–Bashforth–Moulton), systems of equations, and the stability limits that decide which method survives a stiff problem.

Md Mahbub E Noor · Assistant Professor
CSE-3107 Numerical Methods · B.Sc. 3rd Year 1st Semester
Department of CSE, University of Barishal

Why this lecture

The problem: a spring that must not gain energy

A mass on a spring: position $x$ , velocity $y$ . Without friction it should oscillate for ever with the same energy $x^2 + y^2 = 1$ .

Equations: $x' = y$ , $\;y' = -x$ , start $(1, 0)$ : two unknowns, a system
Euler, $h = 0.1$ , ten periods: energy grows from 1 to 517; the mass flies off
Heun: 1.016 (better, still drifting)
Orbits, circuits, molecules and epidemics need methods that stay accurate over thousands of steps

method ( $h = 0.1$ )	energy after 10 periods
exact	1
Euler	517.4
Heun	1.0158
RK4	?

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

2 / 25

Today's goals

Derive the idea of Runge–Kutta methods and apply RK4 by hand and in code
Solve systems and second-order equations by the same methods
Apply Milne's and Adams–Bashforth–Moulton predictor–corrector methods
Explain stability, stiff equations and why implicit methods exist
Choose a solver for a given problem

Last time

Euler (order 1) and Heun (order 2); instability when $h$ is too large.

Today

Higher order with more slopes per step, or with slopes from earlier steps.

Ordinary differential equations · you are here

L21 · ODEs I→L22 · ODEs II

Builds on: L16, L20, L21

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

3 / 25

Start here

The idea in one picture

Euler used one slope, measured only at the start of the step.RK4 samples four slopes. k₁: the slope at the start.k₂: the slope at the middle of the step, reached using k₁.k₃: the middle again, reached using the better slope k₂.k₄: the slope at the end, reached using k₃.Take one step with a weighted average: the middle slopes count double.Result 2.708 against the true 2.718. Euler with the same step: 2.00.

In plain words

Euler uses the slope at the start only, so it drifts
Runge–Kutta samples the slope at several places inside the step
A weighted average of those slopes is far more accurate

New words today

k₁ … k₄: the four sampled slopes
RK4: the fourth-order Runge–Kutta method
predictor–corrector: guess the next value, then improve it
stability: errors stay bounded instead of exploding

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

4 / 25

The Runge–Kutta idea: average several slopes inside the step

Second order: two slopes, weights chosen to match Taylor up to h²

\begin{aligned} k_1 &= hf(x_n, y_n), \qquad k_2 = hf(x_n + ah,\, y_n + ak_1) \\ y_{n+1} &= y_n + \Big(1 - \tfrac{1}{2a}\Big)k_1 + \tfrac{1}{2a}k_2 \end{aligned}

Two members of the family

\begin{aligned} a = 1:&\;\; y_{n+1} = y_n + \tfrac12(k_1 + k_2) \;\;(\text{Heun}) \\ a = \tfrac12:&\;\; y_{n+1} = y_n + k_2 \;\;(\text{midpoint}) \end{aligned}

Same accuracy as the Taylor method of order 2, but no derivatives of $f$
Each $k$ is a trial slope × $h$ ; later slopes use earlier ones
Matching more Taylor terms needs more slopes: four slopes give order 4
Heun from Lecture 21 is RK2

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

5 / 25

The classical fourth-order Runge–Kutta method (RK4)

Four slopes per step

\begin{aligned} k_1 &= hf(x_n, y_n) & k_2 &= hf\big(x_n + \tfrac h2,\, y_n + \tfrac{k_1}{2}\big) \\ k_3 &= hf\big(x_n + \tfrac h2,\, y_n + \tfrac{k_2}{2}\big) & k_4 &= hf(x_n + h,\, y_n + k_3) \end{aligned}

Weighted average: weights 1, 2, 2, 1

y_{n+1} = y_n + \tfrac16\big(k_1 + 2k_2 + 2k_3 + k_4\big), \qquad \text{global error } O(h^4)

Slopes at the start, twice at the middle, at the end
Weights 1, 2, 2, 1 = Simpson's 1, 4, 1 with the middle split in two (Lecture 20)
Halve $h$ : error $\div 16$
Four evaluations per step, but huge steps are allowed: the workhorse of scientific computing

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

6 / 25

The method, step by step

RK4, step by step

$k_1 = h\,f(x_n, y_n)$ : slope at the start
$k_2 = h\,f(x_n + h/2,\; y_n + k_1/2)$ : midpoint, using $k_1$
$k_3 = h\,f(x_n + h/2,\; y_n + k_2/2)$ : midpoint again, using $k_2$
$k_4 = h\,f(x_n + h,\; y_n + k_3)$ : end point, using $k_3$
$y_{n+1} = y_n + (k_1 + 2k_2 + 2k_3 + k_4)/6$ ; $\;x_{n+1} = x_n + h$ ; repeat

Pseudocode

input f, x0, y0, h, x_end
x ← x0;  y ← y0
while x < x_end − h/2:
    k1 ← h·f(x, y)
    k2 ← h·f(x + h/2, y + k1/2)
    k3 ← h·f(x + h/2, y + k2/2)
    k4 ← h·f(x + h, y + k3)
    y ← y + (k1 + 2k2 + 2k3 + k4)/6
    x ← x + h
return y

Works unchanged for systems if $y$ and the $k$ 's are vectors.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

7 / 25

Problem 1: one RK4 step for $y' = x + y$ , $y(0) = 1$ , $h = 0.1$

Compute

y(0.1)

by RK4 and compare with the exact value

2e^{0.1} - 1.1

.

1

k_1 = 0.1(0 + 1) = 0.1

2

k_2 = 0.1\big(0.05 + 1 + 0.05\big) = 0.11

3

k_3 = 0.1\big(0.05 + 1 + 0.055\big) = 0.1105

4

k_4 = 0.1\big(0.1 + 1 + 0.1105\big) = 0.12105

5

y_1 = 1 + \tfrac16(0.1 + 0.22 + 0.221 + 0.12105) = 1.1103417

Exact

1.1103418

: RK4 is correct to 7 digits after one step (Euler: 1.1, Heun: 1.11).

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

8 / 25

Explore: Euler, Heun and RK4 side by side

With $h = 0.2$ RK4 sits on the exact curve; Euler falls far behind
Halve $h$ and compare how fast each error falls (2, 4, 16)
Periodic example $y' = y\cos x$ : RK4 stays on track with $h = 0.5$
Logistic growth: the S-curve of populations and of sigmoid units

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

9 / 25

Systems and higher-order equations

Several unknowns: the same formulas with vectors

\begin{aligned} \mathbf{y}' &= \mathbf{F}(x, \mathbf{y}), \qquad \mathbf{k}_1 = h\mathbf{F}(x_n, \mathbf{y}_n), \;\dots \\ \mathbf{y}_{n+1} &= \mathbf{y}_n + \tfrac16(\mathbf{k}_1 + 2\mathbf{k}_2 + 2\mathbf{k}_3 + \mathbf{k}_4) \end{aligned}

A second-order equation becomes a first-order system

y'' = g(x, y, y') \;\Longrightarrow\; u' = v,\;\; v' = g(x, u, v)

Spring $x'' = -x$ : $x' = y$ , $y' = -x$
Pendulum $\theta'' = -\tfrac gL\sin\theta$ : no closed-form solution for large swings
Compute all components of $\mathbf{k}_1$ before any of $\mathbf{k}_2$
Every ODE solver library works with systems in this form

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

10 / 25

Explore: systems in the phase plane

Left: the path in the $(x, y)$ plane; right: $x(t)$ and $y(t)$
Euler on the spring spirals outwards; RK4 stays on the circle
Predator–prey: the conserved quantity tests the solver
Switch methods and step sizes and watch the "conserved quantity" line

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

11 / 25

Predictor–corrector methods: reuse earlier slopes

Milne: predictor (open Newton–Cotes) and corrector (Simpson's rule)

\begin{aligned} y^{P}_{n+1} &= y_{n-3} + \frac{4h}{3}\big(2f_n - f_{n-1} + 2f_{n-2}\big) \\ y^{C}_{n+1} &= y_{n-1} + \frac h3\big(f_{n-1} + 4f_n + f_{n+1}\big) \end{aligned}

Adams–Bashforth (predictor) and Adams–Moulton (corrector), fourth order

\begin{aligned} y^{P}_{n+1} &= y_n + \frac{h}{24}\big(55f_n - 59f_{n-1} + 37f_{n-2} - 9f_{n-3}\big) \\ y^{C}_{n+1} &= y_n + \frac{h}{24}\big(9f_{n+1} + 19f_n - 5f_{n-1} + f_{n-2}\big) \end{aligned}

$f_k = f(x_k, y_k)$ are slopes already computed: about 2 new evaluations per step (RK4 needs 4)
Not self-starting: the first 3 steps come from RK4
Predictor − corrector difference estimates the error for free
Milne can be unstable for decaying solutions; Adams is the standard choice

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

12 / 25

The method, step by step

A predictor–corrector method, step by step

Get $y_0, y_1, y_2, y_3$ (starting values) by RK4; compute $f_0, \dots, f_3$
Predict $y^{P}_{n+1}$ with the predictor formula
Evaluate $f^{P}_{n+1} = f(x_{n+1}, y^{P}_{n+1})$
Correct: $y^{C}_{n+1}$ with the corrector formula (optionally repeat steps 3–4 once)
Store $f_{n+1} = f(x_{n+1}, y^{C}_{n+1})$ , shift the window of four slopes, repeat

Pseudocode (Adams–Bashforth–Moulton)

y[0..3] ← RK4 starting values
F[k] ← f(x[k], y[k]),  k = 0 … 3
for n = 3, 4, …:
    yp ← y[n] + h/24·(55F[n] − 59F[n−1]
                     + 37F[n−2] − 9F[n−3])
    fp ← f(x[n+1], yp)
    y[n+1] ← y[n] + h/24·(9fp + 19F[n]
                          − 5F[n−1] + F[n−2])
    F[n+1] ← f(x[n+1], y[n+1])
    err ≈ (19/270)·|y[n+1] − yp|

Two new $f$ values per step; the error estimate costs nothing.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

13 / 25

Problem 2: Milne and Adams for $y' = x + y$ at $x = 0.4$

Starting values from RK4 (

h = 0.1

):

y_0 = 1

,

y_1 = 1.110342

,

y_2 = 1.242805

,

y_3 = 1.399717

. Find

y(0.4)

by Milne's and by Adams' method.

quantity	Milne	Adams–Bashforth–Moulton
slopes $f_0, f_1, f_2, f_3$	1, 1.210342, 1.442805, 1.699717	same
predictor	$1 + \tfrac{0.4}{3}(2.420684 - 1.442805 + 3.399434) = 1.583642$	$1.399717 + \tfrac{0.1}{24}(\ldots) = 1.583640$
$f^{P} = 0.4 + y^{P}$	1.983642	1.983640
corrector	$1.242805 + \tfrac{0.1}{3}(1.442805 + 6.798868 + 1.983642) = 1.583649$	$1.399717 + \tfrac{0.1}{24}(\ldots) = 1.583649$
exact $2e^{0.4} - 1.4$	1.583649	1.583649

Both correctors agree with the exact value to 6 decimals, using only 2 new evaluations of $f$ (RK4 would use 4).

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

14 / 25

Stability and stiff equations

Test equation y′ = λy; amplification factor per step, z = hλ

\begin{aligned} \text{Euler: } y_{n+1} &= (1 + z)\,y_n \\ \text{RK4: } y_{n+1} &= \Big(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24}\Big)y_n \end{aligned}

Implicit (backward) Euler: stable for every h > 0 when λ < 0

y_{n+1} = y_n + h\,f(x_{n+1}, y_{n+1}) \;\Rightarrow\; y_{n+1} = \frac{y_n}{1 - z}

Stable when |factor| ≤ 1: Euler needs $-2 \le h\lambda \le 0$ ; RK4 about $-2.79 \le h\lambda$
Stiff: a fast decaying part (large $|\lambda|$ ) forces a tiny $h$ even after it has died out
Implicit methods solve an equation each step (Newton, Lecture 9–10) but take large steps safely
Milne's corrector has a parasitic root of size $> 1$ : errors grow for decaying solutions

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

15 / 25

Classical vs numerical

Same kind of system, two approaches

Calculus

Classical (solve exactly)

Spring $x'' = -x$ : characteristic roots $\pm i$
$x = \cos t$ , $y = -\sin t$
Energy exactly 1 for ever

vs

Algorithm

Numerical (RK4, $h = 0.1$ )

628 steps over ten periods
Largest error in $x$ : $5\times10^{-5}$
Energy 0.999991 (Euler: 517)

When the classical approach cannot work

An epidemic (SIR model): $S' = -0.3SI$ , $\;I' = 0.3SI - 0.1I$

No formula gives $I(t)$ . RK4 with a step of one day: the infected fraction peaks at 30.4% on day 27, and about 94% of people are eventually infected. This is how epidemic forecasts were computed in 2020.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

16 / 25

RK4 for a single equation and for a system

A reusable rk4_step for $y' = f(x, y)$
The same steps for the spring system $(x, y)$
Compares the energy after 10 periods with Euler
Lab: Adams–Bashforth–Moulton and the epidemic model

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

17 / 25

When explicit methods break

Limitations shown failing

Stiff: $y' = -50(y - \cos x)$ follows $\cos x$ smoothly, yet RK4 explodes for $h = 0.06$ ( $h\lambda = -3 < -2.79$ )
Euler already fails at $h = 0.05$ ; backward Euler is fine even with $h = 0.1$
Singularities: $y' = y^2$ blows up at $x = 1$ ; a fixed-step method steps right past it
Fixed $h$ wastes work where the solution is smooth: real solvers adapt $h$

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

18 / 25

Practice

Practice problem

Problem 3"Use the fourth-order Runge–Kutta method to find

y(0.1)

and

y(0.2)

for

\dfrac{dy}{dx} = y - x

,

y(0) = 2

, with

h = 0.1

."

How to write a complete answer

Write the four formulas and say whether the $k$ 's include $h$
Step 1: $k_1 = 0.2$ , $k_2 = 0.205$ , $k_3 = 0.20525$ , $k_4 = 0.210525$ ; $\;y_1 = 2.205171$
Step 2 from $(0.1, 2.205171)$ : $y_2 = 2.421403$
Compare with the exact $e^{0.2} + 1.2 = 2.421403$ : correct to 6 decimals; give a one-line comparison with Euler (2.41) and modified Euler (2.421025)

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

19 / 25

When to use it

Runge–Kutta and predictor–corrector methods: method profile

Best case

RK4: smooth non-stiff problems; 6–8 digits with moderate $h$ ; self-starting, easy step changes
Adams–Bashforth–Moulton: expensive $f$ (2 evaluations per step) with a free error estimate
Systems of any size, written once in vector form

Worst case

Stiff problems: explicit methods need tiny steps
Milne for decaying solutions: unstable
Solutions with singularities or sudden jumps (fixed $h$ )

Best suited for

Orbits, mechanics, circuits, chemical kinetics (non-stiff), epidemics
Adaptive pairs (RK45 / Dormand–Prince) in SciPy and MATLAB
Multistep methods in long simulations where $f$ is costly

Avoid when

Stiff systems (use implicit methods: backward Euler, BDF, Radau)
Very long Hamiltonian simulations (use symplectic integrators)
Fixed $h$ when the solution changes scale (use adaptive step size)

Rule of thumb: start with an adaptive RK (RK45); if it takes absurdly small steps, the problem is stiff: switch to an implicit solver.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

20 / 25

Comparative analysis: ODE methods

Method	Order	$f$ per step	Self-starting?	Stability ( $h\lambda$ real)	Error at $x = 1$ ( $y' = x+y$ , $h = 0.1$ )
Euler	1	1	yes	$[-2, 0]$	0.249
Heun / RK2	2	2	yes	$[-2, 0]$	$8.4\times10^{-3}$
RK4	4	4	yes	$[-2.79, 0]$	$4.2\times10^{-6}$
Milne P–C	4	2	no (needs RK4 start)	weakly unstable	$6\times10^{-7}$
Adams–Bashforth–Moulton	4	2	no (needs RK4 start)	about $[-1.25, 0]$	$3.6\times10^{-6}$
Backward Euler (implicit)	1	1 + equation solve	yes	all $h\lambda < 0$	0.30

No method wins everywhere: RK4 for general use, Adams when $f$ is expensive, implicit methods when the problem is stiff.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

21 / 25

Roadmap

End of Module 6: from solving to fitting

We can now simulate models. The last module asks the reverse question: given data, which model and which parameters?

What it does well (keep this)

RK4: order 4 from function values alone
Systems and higher-order equations
Predictor–corrector with a free error estimate
Stability limits and stiffness

LimitationHow it is overcomeWhere

Model parameters (e.g.

\beta

,

\gamma

of an epidemic) are unknown

→

Fit them to data: least squares

Lecture 23 →

Choosing a model that fits without overfitting

→

Polynomial, exponential fits; validation

Lecture 24 →

Training is itself an ODE:

w' = -\nabla L(w)

→

Gradient descent = Euler on the gradient flow

Lecture 25 →

Step size too large makes training explode

→

Learning rate vs stability (today's

|1 + h\lambda| < 1

)

Lecture 25 →

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

22 / 25

Applications

Where this method runs in the real world

Click a card to see an RK or multistep solver in action.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

23 / 25

Checkpoint

Checkpoint · think, then discuss with your neighbourRK4 for

y' = y

,

y(0) = 1

,

h = 0.1

:

k_1 = 0.1

. What is

k_2

?

A

0.105

B

0.11

C

0.1

D

0.1105

A.

k_2 = hf(y_0 + k_1/2) = 0.1(1 + 0.05) = 0.105

. Then

k_3 = 0.10525

,

k_4 = 0.110525

,

y_1 = 1.1051708

(exact

e^{0.1} = 1.1051709

).

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

24 / 25

Summary

Key points

RK4: $k_1, \dots, k_4$ at start, middle, middle, end; $y_{n+1} = y_n + \frac16(k_1 + 2k_2 + 2k_3 + k_4)$ ; order 4
Systems: same formulas with vectors; $y'' = g$ becomes $u' = v$ , $v' = g$
Milne and Adams–Bashforth–Moulton: predictor + corrector from past slopes, 2 evaluations per step, need RK4 starting values
Stability: $|$ amplification $| \le 1$ ; stiff problems need implicit methods

Opening problem: solved

RK4 with $h = 0.1$ keeps the spring's energy at 0.999991 after ten periods (Euler: 517, Heun: 1.016). The largest position error is $5\times10^{-5}$ .

Homework

Use RK4 with $h = 0.1$ to find $y(0.2)$ for $y' = x^2 + y^2$ , $y(0) = 1$ .
Write $y'' + 0.5y' + 4y = 0$ , $y(0) = 1$ , $y'(0) = 0$ as a system and take one RK4 step with $h = 0.1$ .
Using the RK4 values of Problem 3 at $x = 0, 0.1, 0.2, 0.3$ , find $y(0.4)$ by the Adams–Bashforth–Moulton method.
For $y' = -100y$ , find the largest stable $h$ for Euler and for RK4. Check in the lab.

Next lecture

Module 7: least squares and linear regression. From solving models to fitting them to data, the core of machine learning.

Md Mahbub E Noor

Lecture 22: ODEs II: Runge–Kutta, Milne, Adams

University of Barishal

25 / 25

ODEs II: Runge–Kutta and Predictor–Corrector Methods

The problem: a spring that must not gain energy

Today's goals

The idea in one picture

The Runge–Kutta idea: average several slopes inside the step

The classical fourth-order Runge–Kutta method (RK4)

RK4, step by step

Problem 1: one RK4 step for $y' = x + y$ , $y(0) = 1$ , $h = 0.1$

Explore: Euler, Heun and RK4 side by side

Systems and higher-order equations

Explore: systems in the phase plane

Predictor–corrector methods: reuse earlier slopes

A predictor–corrector method, step by step

Problem 2: Milne and Adams for $y' = x + y$ at $x = 0.4$

Stability and stiff equations

Same kind of system, two approaches

Classical (solve exactly)

Numerical (RK4, $h = 0.1$ )

An epidemic (SIR model): $S' = -0.3SI$ , $\;I' = 0.3SI - 0.1I$

RK4 for a single equation and for a system

Limitations shown failing

Practice problem

Runge–Kutta and predictor–corrector methods: method profile

Comparative analysis: ODE methods

End of Module 6: from solving to fitting

Where this method runs in the real world

🦠 Epidemic forecasting (SIR)

🛰️ Satellite and planet orbits

🐍 SciPy solve_ivp and MATLAB ode45

🦊 Predator and prey (Lotka–Volterra)

Checkpoint

Summary

ODEs II: Runge–Kutta and Predictor–Corrector Methods

The problem: a spring that must not gain energy

Today's goals

The idea in one picture

The Runge–Kutta idea: average several slopes inside the step

The classical fourth-order Runge–Kutta method (RK4)

RK4, step by step

Problem 1: one RK4 step for y′=x+yy' = x + yy′=x+y, y(0)=1y(0) = 1y(0)=1, h=0.1h = 0.1h=0.1

Explore: Euler, Heun and RK4 side by side

Systems and higher-order equations

Explore: systems in the phase plane

Predictor–corrector methods: reuse earlier slopes

A predictor–corrector method, step by step

Problem 2: Milne and Adams for y′=x+yy' = x + yy′=x+y at x=0.4x = 0.4x=0.4

Stability and stiff equations

Same kind of system, two approaches

Classical (solve exactly)

Numerical (RK4, h=0.1h = 0.1h=0.1)

An epidemic (SIR model): S′=−0.3SIS' = -0.3SIS′=−0.3SI, I′=0.3SI−0.1I\;I' = 0.3SI - 0.1II′=0.3SI−0.1I

RK4 for a single equation and for a system

Limitations shown failing

Practice problem

Runge–Kutta and predictor–corrector methods: method profile

Comparative analysis: ODE methods

End of Module 6: from solving to fitting

Where this method runs in the real world

🦠 Epidemic forecasting (SIR)

🛰️ Satellite and planet orbits

🐍 SciPy solve_ivp and MATLAB ode45

🦊 Predator and prey (Lotka–Volterra)

Checkpoint

Summary

Problem 1: one RK4 step for $y' = x + y$ , $y(0) = 1$ , $h = 0.1$

Problem 2: Milne and Adams for $y' = x + y$ at $x = 0.4$

Numerical (RK4, $h = 0.1$ )

An epidemic (SIR model): $S' = -0.3SI$ , $\;I' = 0.3SI - 0.1I$