Ordinary differential equations · Lecture 21

ODEs I: Taylor, Picard and Euler Methods

Differential equations tell us how a quantity changes, not what it is. Stepping $y' = f(x, y)$ forward: the Taylor series and Picard methods, Euler's method, the modified Euler (Heun) method, and the difference between local and global error.

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 law of change, not a formula

Newton's law of cooling: the temperature $T$ (°C) of tea in a 25 °C room changes at a rate proportional to the difference.

$\dfrac{dT}{dt} = -0.1\,(T - 25)$ per minute, with $T(0) = 90$
We know the rate at every moment, not the temperature itself
This one can be solved by hand; most real models (epidemics, circuits, orbits, neural ODEs) cannot
Idea: start from the known value and take small steps, using the rate to predict each next value

$t$ (min)	$T$ (°C)	rate $dT/dt$
0	90	−6.5
1	?	?
2	?	?
…	…	…
10	?

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

2 / 25

Today's goals

State an initial value problem and read its slope field
Solve by the Taylor series method and by Picard's successive approximations
Apply Euler's method and the modified Euler (Heun) method by hand and in code
Distinguish local and global error; predict how the error changes with $h$
Recognise when a step size is too large (instability)

Last time

Integration rules: rectangle, trapezoid, Simpson.

Today

Integrate an equation whose integrand depends on the unknown: $y(x+h) = y(x) + \int f(t, y(t))\,dt$ .

Ordinary differential equations · you are here

L21 · ODEs I→L22 · ODEs II

Builds on: L3, L20

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

3 / 25

Start here

The idea in one picture

A differential equation gives the slope y′ at every point, not y itself.We know one point: the starting value.Compute the slope there and walk along it for one step h.At the new point compute the slope again and take another step.Repeat: a chain of short straight pieces.Each piece leaves the true curve a little …… and the errors pile up. Smaller steps (or a smarter method) reduce the gap.

In plain words

An ODE tells us the slope of the unknown curve at every point
Start at the known point and walk a short step in that direction
Repeat. Shorter steps follow the true curve better

New words today

ODE: an equation containing y′, the rate of change of y
initial value: the known starting point y(x₀) = y₀
step size h: how far we move each time
Euler's method: new y = old y + h × slope

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

4 / 25

Initial value problems and slope fields

First-order initial value problem

\frac{dy}{dx} = f(x, y), \qquad y(x_0) = y_0

Integrated form: every method approximates this integral

y(x_{n+1}) = y(x_n) + \int_{x_n}^{x_{n+1}} f\big(t, y(t)\big)\,dt

$f(x, y)$ is the slope of the solution through the point $(x, y)$
Drawing short slope segments everywhere gives the slope field; solutions follow it
One initial value picks one solution curve
Unique solution if $f$ is smooth (Lipschitz in $y$ ) near $(x_0, y_0)$

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

5 / 25

The Taylor series method

Expand the solution about x₀ (Lecture 3)

y(x_0 + h) = y_0 + h\,y_0' + \frac{h^2}{2!}y_0'' + \frac{h^3}{3!}y_0''' + \frac{h^4}{4!}y_0^{(4)} + \cdots

Get the derivatives by differentiating the ODE itself. For y′ = x + y:

y' = x + y, \quad y'' = 1 + y', \quad y''' = y'', \quad y^{(4)} = y'''

At $(0, 1)$ : $y' = 1$ , $y'' = 2$ , $y''' = 2$ , $y^{(4)} = 2$
$y(0.1) = 1 + 0.1 + \tfrac{0.01}{2}(2) + \tfrac{0.001}{6}(2) + \tfrac{0.0001}{24}(2) = 1.110342$
Exact: $2e^{0.1} - 1.1 = 1.110342$
Accurate, but each new ODE needs its own derivative formulas, which quickly become messy

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

6 / 25

Picard's method of successive approximations

Iterate on the integral equation (a fixed-point iteration for functions, Lecture 8)

y^{(k+1)}(x) = y_0 + \int_{x_0}^{x} f\big(t, y^{(k)}(t)\big)\,dt, \qquad y^{(0)}(x) = y_0

For y′ = x + y, y(0) = 1

\begin{aligned} y^{(1)} &= 1 + x + \tfrac{x^2}{2} \\ y^{(2)} &= 1 + x + x^2 + \tfrac{x^3}{6} \\ y^{(3)} &= 1 + x + x^2 + \tfrac{x^3}{3} + \tfrac{x^4}{24} \end{aligned}

At $x = 0.1$ : 1.105, 1.110167, 1.110337 → exact 1.110342
Each iteration adds one correct power of $x$
Needs the integrals in closed form: only for simple $f$
Its real value is theoretical: it proves that solutions exist

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

7 / 25

Euler's method

Follow the slope at the current point for one step

y_{n+1} = y_n + h\,f(x_n, y_n), \qquad x_{n+1} = x_n + h

Error per step (local) and after reaching a fixed x (global)

\text{local: } \frac{h^2}{2}y''(\xi) = O(h^2), \qquad \text{global: } O(h)

= first two Taylor terms; = left-rectangle rule for the integral
One evaluation of $f$ per step
$1/h$ steps, each with an $O(h^2)$ error, add up to $O(h)$
Halve $h$ : the error at the end halves (and the work doubles)

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

8 / 25

The method, step by step

Euler's method, step by step

Write the problem as $y' = f(x, y)$ , $y(x_0) = y_0$ ; choose $h$
Evaluate the slope $f(x_n, y_n)$ at the current point
Step: $y_{n+1} = y_n + h\,f(x_n, y_n)$ , $\;x_{n+1} = x_n + h$
Repeat until $x$ reaches the target
Check: repeat with $h/2$ ; the change estimates the error

Pseudocode

input f, x0, y0, h, x_end
x ← x0;  y ← y0
while x < x_end − h/2:
    y ← y + h·f(x, y)
    x ← x + h
    print x, y
return y

The loop condition avoids an extra step caused by rounding in $x + h$ .

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

9 / 25

Explore: stepping along the slope field

Grey dashes: the slope field $f(x, y)$
Red: Euler's polygon; dashed: the exact solution
Press $h \div 2$ and watch the final error halve
Tick Heun and RK4 to preview the better methods

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

10 / 25

Problem 1: Euler's method for $y' = x + y$ , $y(0) = 1$

Use Euler's method with

h = 0.1

to find

y(0.4)

. Compare with the exact solution

y = 2e^x - x - 1

.

$n$	$x_n$	$y_n$	$f(x_n, y_n) = x_n + y_n$	$y_{n+1} = y_n + 0.1f$	exact $y(x_{n+1})$	error
0	0.0	1.0000	1.0000	1.1000	1.110342	0.0103
1	0.1	1.1000	1.2000	1.2200	1.242806	0.0228
2	0.2	1.2200	1.4200	1.3620	1.399718	0.0377
3	0.3	1.3620	1.6620	1.5282	1.583649	0.0554

$y(0.4) \approx 1.5282$ ; the error grows step by step because each step starts from a point that is already wrong.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

11 / 25

Modified Euler (Heun): predict, then correct with the average slope

Predictor (Euler) and corrector (trapezoidal rule)

\begin{aligned} y^{*}_{n+1} &= y_n + h\,f(x_n, y_n) \\ y_{n+1} &= y_n + \frac h2\big[f(x_n, y_n) + f(x_{n+1}, y^{*}_{n+1})\big] \end{aligned}

Error

\text{local } O(h^3), \qquad \text{global } O(h^2)

Slope at the start and an estimated slope at the end, averaged
Two evaluations of $f$ per step; second order: halve $h$ , error $\div 4$
Some textbooks repeat the corrector until it settles (iterated modified Euler)
It is the trapezoidal rule of Lecture 20 with a predicted end value

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

12 / 25

The method, step by step

Modified Euler (Heun), step by step

Slope at the start: $k_1 = f(x_n, y_n)$
Predict with Euler: $y^{*} = y_n + h\,k_1$
Slope at the predicted end point: $k_2 = f(x_n + h, y^{*})$
Correct: $y_{n+1} = y_n + \dfrac h2(k_1 + k_2)$
(Optional) recompute $k_2$ with the corrected value and repeat step 4 until two values agree

Pseudocode

input f, x0, y0, h, x_end
x ← x0;  y ← y0
while x < x_end − h/2:
    k1 ← f(x, y)
    yp ← y + h·k1          # predictor
    k2 ← f(x + h, yp)
    y ← y + h/2·(k1 + k2)  # corrector
    x ← x + h
return y

Twice the work of Euler per step, but far fewer steps for the same accuracy.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

13 / 25

Problem 2: modified Euler for $y' = x + y$ , $y(0) = 1$

Find

y(0.1)

and

y(0.2)

with

h = 0.1

. Then iterate the corrector in the first step.

1

Step 1:

k_1 = 0 + 1 = 1

;

\;y^* = 1 + 0.1(1) = 1.1

;

\;k_2 = 0.1 + 1.1 = 1.2

;

\;y_1 = 1 + 0.05(1 + 1.2) = 1.11

2

Step 2:

k_1 = 0.1 + 1.11 = 1.21

;

\;y^* = 1.231

;

\;k_2 = 0.2 + 1.231 = 1.431

;

\;y_2 = 1.11 + 0.05(2.641) = 1.24205

3

Iterated corrector in step 1:

1.11 \to 1 + 0.05(1 + 0.1 + 1.11) = 1.1105 \to 1.110525 \to 1.110526

y(0.2) \approx 1.24205

(exact 1.242806, error

7.6\times10^{-4}

; Euler's error was

2.3\times10^{-2}

).

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

14 / 25

Local and global error: what halving $h$ does

$h$	Euler $y(1)$	error	Heun $y(1)$	error	RK4 (Lecture 22)	error
0.2	2.976640	0.460	3.405416	$3.1\times10^{-2}$	3.43650227	$6.1\times10^{-5}$
0.1	3.187485	0.249	3.428162	$8.4\times10^{-3}$	3.43655949	$4.2\times10^{-6}$
0.05	3.306595	0.130	3.434382	$2.2\times10^{-3}$	3.43656339	$2.7\times10^{-7}$
0.025	3.370128	0.066	3.436008	$5.6\times10^{-4}$	3.43656364	$1.7\times10^{-8}$

$y' = x + y$ , $y(0) = 1$ , exact $y(1) = 3.436564$ . Halving $h$ divides the global error by 2 (Euler, order 1), 4 (Heun, order 2) and 16 (RK4, order 4).

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

15 / 25

Classical vs numerical

Same kind of problem, two approaches

Calculus

Classical (solve exactly)

$y' = x + y$ : linear, integrating factor $e^{-x}$
$(ye^{-x})' = xe^{-x}$ , so $y = 2e^x - x - 1$
$y(1) = 3.436564$ , exactly

vs

Algorithm

Numerical (Heun, $h = 0.1$ )

10 steps, 20 evaluations of $f$
$y(1) = 3.428162$ (error 0.008)
Same code for any $f(x, y)$

When the classical approach cannot work

$y' = x^2 + y^2$ , $y(0) = 0$ (a Riccati equation)

No elementary formula exists (the solution involves Bessel functions). Numerically: Euler ( $h = 0.1$ ) 0.2925, Heun 0.3518, RK4 0.350234; very accurate value $y(1) = 0.350232$ . Most models in science and engineering are like this.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

16 / 25

Euler and modified Euler in Python and C

Both methods for $y' = x + y$ , $h = 0.1$
Prints each step with the exact value
Change f for any other equation
Lab: the tea-cooling model

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

17 / 25

When Euler goes wrong

Limitations shown failing

$y' = -20y$ : the true solution decays to 0 almost at once
Euler multiplies $y$ by $(1 - 20h)$ each step: $h = 0.12$ gives $-1.4$ , so the numbers explode
Stability needs $|1 + h\lambda| < 1$ , i.e. $h < 0.1$ here, whatever the accuracy
First order is slow: halving the error doubles the work

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

18 / 25

Practice

Practice problem

Problem 3"Given

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

,

y(0) = 2

, find

y(0.2)

by Euler's method with

h = 0.1

, then improve the result with the modified Euler method. Compare with the exact solution

y = e^x + x + 1

."

How to write a complete answer

Euler: $y_1 = 2 + 0.1(2) = 2.2$ ; $\;y_2 = 2.2 + 0.1(2.2 - 0.1) = 2.41$
Modified Euler step 1: $k_1 = 2$ , $y^* = 2.2$ , $k_2 = 2.1$ , $y_1 = 2 + 0.05(4.1) = 2.205$
Step 2: $k_1 = 2.105$ , $y^* = 2.4155$ , $k_2 = 2.2155$ , $y_2 = 2.205 + 0.05(4.3205) = 2.421025$
Exact $e^{0.2} + 1.2 = 2.421403$ : errors 0.0114 (Euler) and 0.00038 (modified Euler); present both as tables

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

19 / 25

When to use it

Euler and modified Euler: method profile

Best case

Teaching, quick sketches, very simple models
Heun: modest accuracy (3–4 digits) with little code
Euler inside game physics and real-time loops, where speed matters more than accuracy

Worst case

Long time spans: errors accumulate (first order)
Stiff equations: unstable unless $h$ is tiny
Oscillations and orbits: Euler adds energy and spirals outwards (Lecture 22)

Best suited for

Prototyping a model
Heun as a building block: predictor–corrector schemes
Taylor and Picard: deriving series solutions by hand; existence proofs

Avoid when

Production simulations (use RK4 or adaptive solvers)
Stiff problems (use implicit methods)
When 6+ digits are needed

Rule of thumb: Euler to understand, Heun for a quick improvement, RK4 (Lecture 22) for real work.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

20 / 25

Comparative analysis: methods for $y' = f(x, y)$

Method	Needs	Evaluations per step	Global order	$y(1)$ error ( $y' = x+y$ , $h = 0.1$ )	Best use
Taylor (4 terms)	derivatives of $f$	several formulas	4	$\approx 4\times10^{-6}$	simple $f$ , hand work
Picard	integrals in closed form	—	series	depends on iterations	theory, series solutions
Euler	$f$ only	1	1	0.249	teaching, real-time loops
Heun (modified Euler)	$f$ only	2	2	$8.4\times10^{-3}$	quick, modest accuracy
RK4 (Lecture 22)	$f$ only	4	4	$4.2\times10^{-6}$	general purpose

RK4 matches the Taylor method's accuracy using only values of $f$ : that is the subject of the next lecture.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

21 / 25

Roadmap

Limitations, and where we fix them

Euler and Heun show the principle; real problems need more accuracy, more stability and systems of equations.

What it does well (keep this)

Step along the slope: $y_{n+1} = y_n + h \cdot$ (slope)
Predictor–corrector idea
Local $O(h^{p+1})$ vs global $O(h^p)$
Check by halving $h$

LimitationHow it is overcomeWhere

Low order: many steps for few digits

→

Runge–Kutta: several slopes per step, order 4

Lecture 22 →

Slopes from earlier steps are thrown away

→

Multistep: Milne, Adams–Bashforth–Moulton

Lecture 22 →

Instability with large steps (stiff problems)

→

Stability regions; implicit methods

Lecture 22 →

Real models have several unknowns

→

Systems: the same methods with vectors

Lecture 22 →

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

22 / 25

Applications

Where this method runs in the real world

Click a card to see an ODE being stepped in a real system.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

23 / 25

Checkpoint

Checkpoint · think, then discuss with your neighbourOne Euler step for

y' = 2y

,

y(0) = 1

,

h = 0.1

. What is

y_1

?

A

1.2

B

1.1

C

1.2214

D

0.2

A.

y_1 = 1 + 0.1 \times 2(1) = 1.2

. The exact value is

e^{0.2} = 1.2214

(option C); Heun gives

1 + 0.05(2 + 2.4) = 1.22

.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

24 / 25

Summary

Key points

IVP $y' = f(x, y)$ , $y(x_0) = y_0$ ; every method approximates $\int f(t, y(t))dt$ over each step
Taylor: expand and differentiate the ODE; Picard: $y^{(k+1)} = y_0 + \int f(t, y^{(k)})dt$
Euler $y_{n+1} = y_n + hf_n$ , global $O(h)$ ; Heun: predictor + trapezoidal corrector, $O(h^2)$
Too large $h$ can make the solution explode: stability, not only accuracy

Opening problem: solved

Tea after 10 minutes: Euler ( $h = 1$ ) gives 47.66 °C, Heun 48.96 °C; the exact value is 48.91 °C. Heun's error is 30 times smaller for twice the work.

Homework

Solve $y' = -2xy^2$ , $y(0) = 1$ to $x = 0.2$ by Euler and by modified Euler ( $h = 0.1$ ). Compare with $y = 1/(1 + x^2)$ .
Find the first three Picard approximations for $y' = 1 + xy$ , $y(0) = 1$ .
Use the Taylor series method (4 terms) to find $y(0.1)$ for $y' = x^2 + y^2$ , $y(0) = 1$ .
For $y' = -50y$ , find the largest $h$ for which Euler's method does not grow. Check it in the lab.

Next lecture

ODEs II: Runge–Kutta and predictor–corrector methods. Fourth-order accuracy, multistep methods, stability, stiff equations and systems.

Md Mahbub E Noor

Lecture 21: ODEs I: Taylor, Picard, Euler, Heun

University of Barishal

25 / 25

ODEs I: Taylor, Picard and Euler Methods

The problem: a law of change, not a formula

Today's goals

The idea in one picture

Initial value problems and slope fields

The Taylor series method

Picard's method of successive approximations

Euler's method

Euler's method, step by step

Explore: stepping along the slope field

Problem 1: Euler's method for $y' = x + y$ , $y(0) = 1$

Modified Euler (Heun): predict, then correct with the average slope

Modified Euler (Heun), step by step

Problem 2: modified Euler for $y' = x + y$ , $y(0) = 1$

Local and global error: what halving $h$ does

Same kind of problem, two approaches

Classical (solve exactly)

Numerical (Heun, $h = 0.1$ )

$y' = x^2 + y^2$ , $y(0) = 0$ (a Riccati equation)

Euler and modified Euler in Python and C

Limitations shown failing

Practice problem

Euler and modified Euler: method profile

Comparative analysis: methods for $y' = f(x, y)$

Limitations, and where we fix them

Where this method runs in the real world

🧠 ResNets are Euler steps

🎮 Game physics

🔌 Charging a capacitor (RC circuit)

🕵️ Forensics: time of death

Checkpoint

Summary

ODEs I: Taylor, Picard and Euler Methods

The problem: a law of change, not a formula

Today's goals

The idea in one picture

Initial value problems and slope fields

The Taylor series method

Picard's method of successive approximations

Euler's method

Euler's method, step by step

Explore: stepping along the slope field

Problem 1: Euler's method for y′=x+yy' = x + yy′=x+y, y(0)=1y(0) = 1y(0)=1

Modified Euler (Heun): predict, then correct with the average slope

Modified Euler (Heun), step by step

Problem 2: modified Euler for y′=x+yy' = x + yy′=x+y, y(0)=1y(0) = 1y(0)=1

Local and global error: what halving hhh does

Same kind of problem, two approaches

Classical (solve exactly)

Numerical (Heun, h=0.1h = 0.1h=0.1)

y′=x2+y2y' = x^2 + y^2y′=x2+y2, y(0)=0y(0) = 0y(0)=0 (a Riccati equation)

Euler and modified Euler in Python and C

Limitations shown failing

Practice problem

Euler and modified Euler: method profile

Comparative analysis: methods for y′=f(x,y)y' = f(x, y)y′=f(x,y)

Limitations, and where we fix them

Where this method runs in the real world

🧠 ResNets are Euler steps

🎮 Game physics

🔌 Charging a capacitor (RC circuit)

🕵️ Forensics: time of death

Checkpoint

Summary

Problem 1: Euler's method for $y' = x + y$ , $y(0) = 1$

Problem 2: modified Euler for $y' = x + y$ , $y(0) = 1$

Local and global error: what halving $h$ does

Numerical (Heun, $h = 0.1$ )

$y' = x^2 + y^2$ , $y(0) = 0$ (a Riccati equation)

Comparative analysis: methods for $y' = f(x, y)$