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1dy y

kydt A

The simplest model of population growth is dy/dt = ky, according to which populations grow exponentially. This may be true over short periods of time, but it is clear that no population can increase without limit. Therefore, population biologists use a variety of other differential equations that take into account environmental limitations to growth such as food scarcity and competition between species. One widely used model is based on the logistic differential equation:

Here k > 0 is the growth constant, and A > 0 is a constant called the carrying capacity. Our next slide shows a typical S-shaped solution of a logistic differential equation. As in the previous section, we also denote dy/dt by

ln

kt

dykdt y

Ce

kt Cy

y

.y

,dy

y F t ydt

Separation

of Variables

Solutions of the logistic equation with y0 < 0 are not relevant to populations because a population cannot be negative (see Exercise 18).

, where

1

y t

dy yky

dt A

Next Slide

The slope field shows clearly that there are three families of solutions, depending on the initial value y0 = y (0).

Slope field for 1dy y

kydt A

• If y0 > A, then y (t) is decreasing and approaches A as t → ∞.

• If 0 < y0 < A, then y (t) is increasing and approaches A as t → ∞.

• If y0 < 0, then y (t) is decreasing and lim .t b

y t

converges to a stable eqilibrium and

diverges from an unstable equilibrium.

y t

has constant solutions: y = 0 and y = A. They correspond to the roots of ky(1 − y/A) = 0, and they satisfy because when y is a constant. Constant solutions are called equilibrium or steady-state solutions. The equilibrium solution y = A is a stable equilibrium because every solution with initial value y0 close to A approaches the equilibrium y = A as t → ∞. By contrast, y = 0 is an unstable equilibrium because every nonequilibrium solution with initial value y0 near y = 0 either increases to A or decreases to −∞.

1dy y

kydt A

0y

1dy y

kydt A

1 ???dy y

ky y tdt A

GraphWe'll show this on the next slide!

1 /

1 1

dykdt

y y A

dyy y A

1 2

1 2

1 2

2 1

1

1 / 1 /

1 /1 /

1 / 1 /

1 1 /

1; 0 1

B B

y y A y y A

y y A B By y A

y y A y y A

B y A B y

y A B y BA

Having described the solutions qualitatively, let us now find the nonequilibrium solutions explicitly using separation of variables. Assuming that y 0 and y A, we have ≠ ≠

1dy y

kydt A

1 /

dykdt

y y A

Let a # which

drops a parameter

y 1/ 1 1 1

1 / 1 /

A

y A A y A A y

Partial Fractions

1 1ln ln

ln kt

dy kdt y y A kt Cy y A

y ykt C Ce

y A y A

kdt

For t = 0, we have a useful relation between C and the initial value y0 = y (0):

0

0

ktye

y AC

A

yC

y

So is easy to find!CSolve for ...y

1 1 1 /

kt

kt kt

kt kt

kt kt

kt kt kt

y y A Ce

y yCe ACe

y yCe ACe

ACe ACe Ay

Ce Ce e C

As C 0, we may divide by Cekt to obtain the general nonequilibrium solution:

≠

11 /kt

dy y Aky y t

dt A e C

1 kt

kt

e

Ce C

So is easy to find!y t

l

T

o

hi

gi

s ge

stic

ts us th

s differ

e solution to ou

ential equat

r

ion.

Solve 0.3 4 with initial condition 0 1.y y y y

We'll rewrite 0.3 4 as 1.2 14

dy yy y y y y y

dt

0

0

11 /kt

ydy y Aky y t C

dt A e C y A

1.2

thus, 1.2 & 4

4so,

1 /t

k A

ye C

0

0

1

3

yC

y AC

1.2

4

1 3 ty

e

0.40.4

1000 500 1

500 or1 9 500 1000 9

tt

ee

0.4 ln 99

0.5.493 s

4 yearte t

0.4

0.4 & 1000

0.4 11000

1000

1 /t

k A

dP Py

dt

ye C

0

00.4

1000

1 9

1

9 ty

yC

y A eC

Deer Population A deer population grows logistically with growth constant k = 0.4 year−1 in a forest with a carrying capacity of 1000 deer.(a) Find the deer population P(t) if the initial population is P0 = 100.(b) How long does it take for the deer population to reach 500?

0

0

1 & 1 /kt

ydy y Aky y C

dt A e C y A

ktyCe

y A

Solve for t

`

Same formula for any

point in the plane.

0

0

2 22

0.8959 0.89590.8959

1

1 /0.1 1

0 1/10 and 0.9 9

1

1 92 2 1 5 1

2 1 95 5 1 9 2

' (1 )

0.89

61

2 ln6

3 3 1 4 11 9

4 4 1 9 3 27

59

3.679 days

kt

kt

k kk

t tt

y te C

yy C C

y A

y te

y e ee

k

y t

y t ky

e e

y

k

te