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1. Euler's method

Warm-up

Three quick questions from earlier lessons.

Discover

$y' = y$ , $y(0) = 1$ . At $(0, 1)$ the slope is 1. Follow the tangent line for a step of $0.5$ . Where do you land?

$(0.5,\ 1.5)$
$(0.5,\ 2)$

Key idea

Walk along short tangent lines.

Euler's method for

y' = f(x, y)

y(x_0) = y_0

, step

h

x_{n+1} = x_n + h

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

. Smaller steps give better estimates.

Worked example

$y' = x + y$ , $y(0) = 1$ , $h = 0.5$ . Estimate $y(1)$ .

y_1 = 1 + 0.5 \cdot 1 = 1.5

Slope at (0, 1) is 1.

y_2 = 1.5 + 0.5 \cdot 2 = 2.5

Slope at (0.5, 1.5) is 2.

y(1) \approx 2.5

Two steps reach x = 1.

Your turn, guided

Your turn: $y' = 2y$ , $y(0) = 3$ , one step with $h = 0.1$ .

y(0.1) \approx 3.6

3 + 0.1 × 6.

Practice ★

$\dfrac{dy}{dx} = 1x + y$ and $y(1) = 3$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 4x + y$ and $y(1) = 8$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 9x + y$ and $y(1) = 4$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .

Watch out!

Practice ★★

$\dfrac{dy}{dx} = y$ and $y(0) = 8$ . Use Euler's method with two steps of size $h = \tfrac{1}{2}$ to estimate $y(1)$ .
$\dfrac{dy}{dx} = y$ and $y(0) = 4$ . Use Euler's method with two steps of size $h = \tfrac{1}{2}$ to estimate $y(1)$ .

The solution of $\dfrac{dy}{dx} = y$ , $y(0) = 1$ , is concave up. Does Euler's method with $h > 0$ over- or underestimate $y(1)$ ?
1. Underestimate: each step follows a tangent line, which lies below a concave-up curve
2. Overestimate
3. It is exact

Exit ticket

Three questions to check you have got it.

$\dfrac{dy}{dx} = 1x + y$ and $y(1) = 3$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 4x + y$ and $y(1) = 8$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .
$\dfrac{dy}{dx} = 9x + y$ and $y(1) = 4$ . Use one step of Euler's method with $h = 0.1$ to estimate $y(1.1)$ .

Summary card

Euler's method

$y_{n+1} = y_n + h\,f(x_n, y_n)$ : slope at the current point × step.

Watch out: update the slope at each new point.

2. Logistic growth: carrying capacity and fastest growth

Warm-up

Three quick questions from earlier lessons.

Discover

A population starts at 10 and follows a logistic model with carrying capacity $k$ . Move $k$ .

Logistic curves starting at 10 with carrying capacity k: slow growth, then fast growth in the middle, then levelling off at k.

Where on the curve is the population growing fastest?

Halfway to $k$
Just before it reaches $k$

Key idea

Growth slows as resources run out.

Logistic model:

\dfrac{dP}{dt} = rP\left(1 - \dfrac{P}{K}\right)

. Equilibria

P = 0

and

P = K

(the carrying capacity). For

0 < P_0

P(t) \to K

. Growth is fastest at

P = \tfrac{K}{2}

Worked example

$\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right)$ .

Carrying capacity

K = 500

Set the rate to 0.

Fastest growth at

P = 250

Half of K.

Maximum rate

50

per unit time

0.4 × 250 × ½.

Your turn, guided

Your turn: $\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{800}\right)$ . Fastest growth at…

P = 400

K/2.

Practice ★

A population follows $\dfrac{dP}{dt} = \dfrac{ 7 }{10}P\left(1 - \dfrac{P}{ 558 }\right)$ . What is its carrying capacity?
A population follows $\dfrac{dP}{dt} = \dfrac{ 9 }{10}P\left(1 - \dfrac{P}{ 428 }\right)$ . What is its carrying capacity?

$\dfrac{dP}{dt} = \dfrac{ 3 }{10}P\left(1 - \dfrac{P}{ 280 }\right)$ . For which population $P$ is it growing fastest?
$\dfrac{dP}{dt} = \dfrac{ 9 }{10}P\left(1 - \dfrac{P}{ 590 }\right)$ . For which population $P$ is it growing fastest?

Watch out!

Practice ★★

$\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{ 750 }\right)$ with $P(0) > 0$ . Tick all the true statements.
1. $P = 0$ and $P = 750$ are equilibrium solutions
2. If $P(0) > 750$ , then $P$ decreases towards 750
3. $\displaystyle\lim_{t \to \infty} P(t) = 750$
4. Starting below 750, $P$ eventually exceeds 750
5. $\displaystyle\lim_{t \to \infty} P(t) = 0.2$

Boss problem. $\dfrac{dP}{dt} = \dfrac{ 2 }{10}P\left(1 - \dfrac{P}{ 388 }\right)$ .

Exit ticket

Three questions to check you have got it.

$\dfrac{dP}{dt} = \dfrac{ 3 }{10}P\left(1 - \dfrac{P}{ 280 }\right)$ . For which population $P$ is it growing fastest?
$\dfrac{dP}{dt} = \dfrac{ 9 }{10}P\left(1 - \dfrac{P}{ 590 }\right)$ . For which population $P$ is it growing fastest?
$\dfrac{dP}{dt} = \dfrac{ 1 }{10}P\left(1 - \dfrac{P}{ 342 }\right)$ . For which population $P$ is it growing fastest?

Summary card

Logistic growth

$P' = rP(1 - P/K)$ ; equilibria $0$ and $K$ .
$P \to K$ ; fastest growth at $K/2$ .

Watch out: $r$ is a rate, $K$ is a population.

3. The logistic solution

Warm-up

Three quick questions from earlier lessons.

Discover

The logistic solution is $P(t) = \dfrac{K}{1 + Ae^{-rt}}$ . What does $P(t)$ approach as $t \to \infty$ ?

Key idea

The explicit solution, fitted to the start.

\dfrac{dP}{dt} = rP\left(1 - \dfrac{P}{K}\right)

P(0) = P_0

has the solution

P(t) = \dfrac{K}{1 + Ae^{-rt}}

with

A = \dfrac{K - P_0}{P_0}

Worked example

$K = 1000$ , $P_0 = 200$ , $r = 0.5$ .

A = 4

(1000 − 200)/200.

P(t) = \dfrac{1000}{1 + 4e^{-0.5t}}

Substitute.

P(0) = 200

Check t = 0: 1000/5.

Your turn, guided

Your turn: $K = 600$ , $P_0 = 100$ . Find $A$ .

A = 5

(600 − 100)/100.

Practice ★

The logistic solution $P(t) = \dfrac{ 203 }{1 + Ae^{-rt} }$ has $P(0) = 58$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 755 }{1 + Ae^{-rt} }$ has $P(0) = 48$ . Find $A$ .

$P(t) = \dfrac{ 277 }{1 + 5e^{-0.3t} }$ . Find the initial population $P(0)$ .
$P(t) = \dfrac{ 742 }{1 + 4e^{-0.3t} }$ . Find the initial population $P(0)$ .

Watch out!

Practice ★★

$\dfrac{dP}{dt} = \dfrac{ 2 }{10}P\left(1 - \dfrac{P}{ 253 }\right)$ . How fast is the population growing when $P = \dfrac{ 253 }{2}$ ?
$\dfrac{dP}{dt} = \dfrac{ 7 }{10}P\left(1 - \dfrac{P}{ 556 }\right)$ . How fast is the population growing when $P = \dfrac{ 556 }{2}$ ?

Exit ticket

Three questions to check you have got it.

The logistic solution $P(t) = \dfrac{ 203 }{1 + Ae^{-rt} }$ has $P(0) = 58$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 755 }{1 + Ae^{-rt} }$ has $P(0) = 48$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 770 }{1 + Ae^{-rt} }$ has $P(0) = 24$ . Find $A$ .

Summary card

The logistic solution

$P(t) = \dfrac{K}{1 + Ae^{-rt}}$ , $A = \dfrac{K - P_0}{P_0}$ , $P \to K$ .

Watch out: check $A$ by computing $P(0)$ .

Answers: Euler's method and logistic models

1. Euler's method

$\frac{17}{5}$
$\frac{46}{5}$
$\frac{53}{10}$

$18$
$9$

Underestimate: each step follows a tangent line, which lies below a concave-up curve

$\frac{17}{5}$
$\frac{46}{5}$
$\frac{53}{10}$

2. Logistic growth: carrying capacity and fastest growth

$558$
$428$

$140$
$295$

$P = 0$ and $P = 750$ are equilibrium solutions · If $P(0) > 750$ , then $P$ decreases towards 750 · $\displaystyle\lim_{t \to \infty} P(t) = 750$

$\frac{97}{5}$

$140$
$295$
$171$

3. The logistic solution

$\frac{5}{2}$
$\frac{707}{48}$

$\frac{277}{6}$
$\frac{742}{5}$

$\frac{253}{20}$
$\frac{973}{10}$

$\frac{5}{2}$
$\frac{707}{48}$
$\frac{373}{12}$