SPECIALIST MATHSDifferential Equations

Week 3

Problem Solving with Separable Differential Equations

Example 1 (Ex 8E1)• Fluid is flowing through thick walled tube.

• It maintains the temperature on the inner wall of the pipe at 600oC.

• Heat is lost through the tube per unit length according to the formula:

• Calculate the external temperature of the tube.

2.0 and 680 where2 kqdr

dTkrq

0.02m0.04m

r

Solution to Example 1

dr

dTkrq 2

kr

q

dr

dT

2

rk

q

dr

dT 1

2

rdr

dT 1

2.02

680

rdr

dT 11700

drr

T11700

crT ln1700

2.0 ,680 kq

Solution to Example 1 continued

crT ln1700

02.0 when 600 rCT o

c 02.0ln1700

600

02.0ln1700

600

c

02.0ln1700

600ln1700

rT

Solution to Example 1 continued again

02.0ln1700

600ln1700

rT

04.0 when ? rT

02.0ln1700

60004.0ln1700

T

CT o9.224

Example 2 (Ex 8E2)• The rate of decay of a radioisotope is

proportional to the number of particles present.

• If the number of particles decreases by 10% after 50 years, find the percentage left after 200 years.

Solution to example 2present particles ofnumber toalproportion isdecay of Rate

kNdt

dN therefore

kdt

dN

N

1

kdtdtdt

dN

N

1

kdtdN N

1

cktN lnckteN

ckteeN ktceeN

c

kt

eA

AeN

where

,

Solution to example 2 continued0 be particles ofnumber original let the N

0 ,0 when ie NNt ktAeN

00 AeN

AN 0

kteNN 0 therefore

09.0 ie 10%,by

decreases ,50when

NN

Nt

50000.9 therefore keNN

ke500.9

500.9 ke

501

(0.9)ke

50)(0.9 ke

Solution to example 2 continued again

500 )9.0( therefore

tNN

kteNN 0

200when t

40 9.0NN

656.00NN

0656.0 NN

remaining 65.6% ie

501

(0.9)ke

50200

0 )9.0(NN

Example 3 (Ex 8E2)• A man falls out of a plane and accelerates

toward the earth with an acceleration given by

• Terminal velocity is a constant velocity he will eventually reach.

• Calculate his terminal velocity.

1-ms 5

8.9v

a

Solution to Example 3

5

8.9v

dt

dv

495 vdt

dv

149

5

dt

dv

v

149

5

dtdt

dt

dv

v

149

5

dtdv

v

ct49ln1

51

v

1 ct49ln5 v

c5

49ln t

v

49 5c

t

ev

Solution to Example 3 continued

49 5c

t

ev

v49 5

t

Ae

49 5

tceev

49 5

t

Aev

0 when 0 tv

490 0Ae

490 A

49A

4949 5

t

ev

0 , tas

occurs velocity terminal

5

t

e

049 v

49v

-1ms 49 velocity terminal

Example 4 (Ex 8E2)• The rate at which temperature changes is

proportional to the differences between its temperatures.

• A hot metal bar is 700oC at 12:00pm when taken out of the furnace. Its temperature falls to 200oC at 2:00pm.

• If the air temperature is 20oC, calculate its temperature at 3:00pm.

Solution to Example 4

rTTdt

dT

rTTdt

dTk

k1

dt

dT

TT r

k1

dt

dT

TT r

kdt1

dtdt

dT

TT r

kdt1

dTTT r

cktTT r ln

cktr eTT

Solution to Example 4ckt

r eeTT ktc

r eeTT kt

r AeTT

700 ,0when

and 20

Tt

Tr

020700 Ae

680A

kteT 68020

200 ,2when Tt

268020200 ke

ke2680180

Solution to Example 4 continued

ke2680180

kteT 68020

ke2

680

180

234

9 ke

21

34

9

ke

tkeT

68020 2

34

968020

t

T

? ,3when Tt

23

34

968020

T

CT 061.112

Growth and Decay Problemsky

dx

dy form theof equations aldifferenti are These

kdx

dy

y

1

kdxdxdx

dy

y

1

kdxdyy

1

ckxy ln

ckxey

ckxeey

kxAey

problem.decay 0

problem,growth 0,For

k

k

Growth & Decay GraphskxAey kxAey

growth. 0,For k decay. 0,For ky

x

y

x0y

0y

Logistic Differential Equations• The function starts off growing exponentially,

then flattens out approaching a limiting value “A”.

• This happens with populations as the recourses available to support it are limited.

A

0P

P

t

)(P t

P)(' P kt

A

P1)(' P kt

Logistic Differential Equations• Logistic differential equations are of the form:

• When P is small, an exponential growth.

• When a horizontal line.

A

PkP

dt

dP1

kPdt

dP

A

P ,0

0 ,0P

A-1 ,

dt

dPAP

A Handy Algebraic Manipulation

PAP

11 Proof

PAP

PPA

PAP

A

PAP

11

PAP

A

Example 5 (Ex 8F)

The relative growth rate of snakes on Groute Island is given by:

(a) What is the environments carrying capacity?

(b) Solve the differential equation if there are 2280 snakes originally.

(c) When would you expect the population to reach 4000 snakes.

51601

100

1 PP

dt

dP

A

PkP

dt

dP1

Solution to Example 5

. is population limiting 1For AA

PkP

dt

dP

snakes 5160 is population limiting the

51601

100

1For (a)

PP

dt

dP

51601

100

1 (b)

PP

dt

dP

5160

5160

100

1 PP

dt

dP

5160

5160

100

1 PP

dt

dP

5160

5160

100

11 P

dt

dP

P

Pdt

dP

P 5160

100

15160

100

1

5160

5160

dt

dP

PP

100

1

5160

5160

dt

dP

PP

PP

5160

11

PP

PP

5160

5160

PP

5160

5160

100

1

5160

11 therefore

dt

dP

PP

dtdtdt

dP

PP 100

1

5160

11

dtdPPP 100

1

5160

11

ctP

P

100

1

1

5160lnln

ctPP 100

15160lnln

ctP

P

100

1

5160ln

cte

P

P

100

1

5160

ctee

P

P 100

1

5160

ctee

P

P

100

1

15160

ctee

P

P

100

1

15160

ctee

P

P

P

100

15160

tbe

P100

1

15160

tbe

P100

1

15160

tbe

P100

1

15160

tbe

P

100

1

1

1

5160

tbe

P100

1

1

5160

tbe

P100

1

1

5160

2280 ,0when Pt

01

51602280

be

b

1

51602280

516012280 b

516012280 b

516022802280 b

228051602280 b

28802280 b

2280

2880b

263.1b

tbe

P100

1

1

5160

te

P100

1

263.11

5160

? ,4000 when (c) PP

te 100

1

263.11

51604000

te 100

1

263.11

51604000

5160263.114000 100

1

te

4000

5160263.11 100

1

t

e

29.1263.11 100

1

t

e

29.0263.1 100

1

t

e

263.1

29.0100

1

t

e

23.0100

1

t

e

23.0ln100

1 t

23.0ln100

1 t

23.0ln100t

years 147t

Example 6 (Ex 8F)

Returning from a visit to a remote planet, Captain Boss brings with him a rare disease that infects the crew of the ship, spreading according to a logistic equation.

(1) If there are 126 crew members write a differential equation governing the spread.

(2) After 3 days 15 members have the disease, write a formula for the spread in terms of t.

(3) When is the rate of infection greatest.

(1)

1261

NkN

dt

dN

cktNN 126lnln

126

126 NkN

dt

dN

kdt

dN

NN

126

126

kdt

dN

NN

126

11

kdtdtdt

dN

NN 126

11

kdtdNNN 126

11

cktN

N

126ln

ckteN

N 126

ckteN

N 126

ckteN

N

1126

ckteN

N

N

126

ckteeN

1126

ktbeN

1126

ktbe

N

1

1

126

ktbeN

1

126

(2) When t = 0, N = 1

01

1261

be

1261 b125b

kteN

1251

126

When t = 3, N = 15

ke 31251

12615

126125115 3 ke

4.81251 3 ke

4.7125 3 ke

0529.03 ke

0529.03 ke

0529.0ln3 k

0529.0ln3

1k

942.0k

teN

942.01251

126

(3) When is the rate of infection greatest.

126

2kNkN

dt

dN

dt

dNkN

dt

dNk

dt

Nd

126

22

2

dt

dNkNk

dt

Nd

632

2

1261

NkN

dt

dN

126

2kNkN

dt

dN

0 when occurs this2

2

dt

Nd

dt

dNkNk

630

630

kNk

kkN

63

k

kN

63

63N

63hen greatest w is rateinfection N

te 942.01251

12663

126125163 942.0 te

21251 942.0 te

1125 942.0 te

008.0942.0 te

008.0ln942.0 t

008.0ln942.0 t

008.0ln942.0

1t

days 1.5t

This Week

• Text book pages 294 – 307.

• Exercise 8E2 Q1 – 12

• Exercise 8F Q1 - 4.

• Review Sets 8A -8D