ODE Lecture 5

7/25/2019 ODE Lecture 5

1/39


2/39


3/39

By the term mathematical modeling, we

mean the process whereby the behavior of areal-life system or phenomenon, whetherphysical, sociological, or even economics isdescribed by a set of Mathematical relations,after approximation and idealizations.Construction of Mathematical model startswith


4/39

i) Identification of the system we are

modeling. This requires selection of somevariables as important to understanding ordescribing the behavior of the system andignoring other as marginal or irrelevant to

this understanding.


5/39

ii)Making some realistic assumption, orhypothesis, about the system we aredescribing. These assumptions will alsoinclude any empirical laws that may beapplicable to the system.

We translate these assumptions intomathematical language which usuallyresults in a set of equations and /or

inequalities. This gives us a mathematicalmodel which will be an approximation of theactual real-life problem.


6/39

Once we have constructed the mathematicalmodel, we are faced with the problem ofsolving it so as to obtain the unknown variableor variables in terms of the known quantities.

So Modeling steps are the steps thatlead from the physical situation to amathematical formulation and solution, thenphysical interpretation of the result.

In this section we solve some model thatare described by linear first order differentialequations.


7/39

Let x(t) denote the population at any time t.We construct a mathematical model ofpopulation x(t) which grows or decays at a

rate proportional tox.

Setting up a mathematical model

kxdt

dx

Growth and Decay

Step 1


8/39

Step 2

Solving the differential equation obtained.

Step 3

Interpretation of the result.

kxdt

dx

. (1)


9/39

Integrating (1)

kt

cktxkt

(2)

for decay

for growth


10/39

c is any arbitrary constant and can be

determined by using initial condition andvalue of kchanges from problem to problem.


11/39

Example

Bacterial Growth

A culture initially has N0 number of bacteria.At t= 1hr the number of bacteria is measuredto be (3/2)N

0

. If the rate of growth is

proportional to the number of bacteriapresent, determine the time necessary for thenumber of bacteria to triple.


12/39

.

Solution

We first solve the differential equation

kdt

dN. (1)

Subject to N(0)=N0. Then we use theempirical condition to determine theconstant of proportionality k . Now eq (1) is

both separable and linear. When it is put intothe form

0dN

kNdt

...(2)


13/39

We can see by inspection that the integratingfactor is . Multiplying (2) with we getkte

ktd e Ndt

Integrating (3)

.(3)

kt .. .(4)

k te

it follows that, at

so equation (4) takes the form

tcceN

0

0


14/39

kteNtN0

.(5)

so equation (4) takes the form

at t= 1 , we have

kteNN 00

2

3

kte2

3

4055.0)2

3ln(k

Thus0.4055

0( ) t

N t N e


15/39

To find the time at which the bacteria havetripled we solve

t4055.03ln

hrt 71.24055.0

3ln

teNN

4055.0

00

te 4055.0


16/39

Example

Bacteria in a certain culture increases at a rateproportional to the number present. If thenumber of bacteria is 2000 after one hour and4000 after two hours. What was their initial

numbers?

Solution

Let x be the number of bacteria at time t, then

dxkx

dt


17/39

ktcetx )(

Solution of this equation is same as in theabove example i.e.

. (1)

Here we have two conditions

(1) 2000(2) 4000

xx

Using these two conditions in eq (1), we get

values of c & k.

ln 2tx t e


18/39

Now we have to find the initial number when t

= 0. x . (2)

Equation (2) is the required initial number

Radioactive Decay

Radioactive decay is the process in whichan unstable atomic nucleus loses energy by

emitting radiation in the form of particles orelectromagnetic waves


19/39

i) Radium is a radioactive chemical elementwhich has symbol Ra.

ii) Radium is an alkaline earth metal. It isextremely radioactive.

iii) Radium is used in medicines to produceradon gas which in turn is use as a cancertreatment.

iv) Radium is over one million times moreradioactive than the same mass of Uranium.


20/39

Example

Radium decomposes at a rate proportional tothe quantity of radium present. If 100 mgreduces to 90 mg in 20 years. How many milligrams will remain at the end of 1000 years?

How long will it take for one-half of the originalamount to decompose?

Solution

Letx mg denote the amount of Radium presentat time t.


21/39

Here we the differential equation

kxdtdx

Solution of this equation is

ktcetx )( (1)

Here we have two conditions

(0) 100

(20) 90

x

x

.....(2a)

.(2b)


22/39

Appling (2a) in (1) we get

cUsing value of c in equation (1) we get

ktx t e ...(3)

Appling (2b) in (1) we get2090 100 ke

k


23/39

So equation (3) becomes

t

etx0005.0

100)(

Now at t = 1000 gives

txHence after 1000 years nearly 61 mg ofradium will remain.

t = ? Whenx = 50


24/39

Example

Initially there were 100mg of a radioactivesubstance present. After 6hrs the mass decreased

by 3%. If the rate of decay is proportional to the

amount of the substance present at any time, find

the amount remaining after 24 hrs?

Solution

Let be the initial amount, and A(t) be

amount at any time.

0A


25/39

( ) 1 0 0A t m g(1)

( ) 97A t mg (2)

at t = 0

at t = 6

24 , ( ) ?t hr A t

Governing equation is

dAkA

dt


26/39

Its solution isk tA t c e

Using condition (1) in eq (3) we get

(3)

c (4)

Using the value of c in eq (3), we getktA t e

Using condition (2) in eq (4) we get

697 100 ke

0.00507k


27/39

Using the value of k in eq (4) for t = 24 hrs,

we get)24(00507.0100)( etA

mgtA 5.88)(


28/39

Half Life

The half life of a radioactive substance is

the time in which half of the given amountwill disappear. Hence it measures theradioactive decay process .

OrThe time taken for one half of a

radioactive substance to decay is calledhalf life of the substance.


29/39

Example

Determine the half-life of the radioactivesubstance described in last example.

Solution0( )

2

AA t

0.00507( )0 1002

tAe

Eq (1) 0.00507( )100 1002

te


30/39

1ln 0 .0 0 5 0 7

2t

Example

The breeder reactor converts stable uranium238 into the isotope platinum 239. After 15years it is determined that 0.043% of the initial

amount of the platinum has disintegrated. Findthe Half Life of this isotope


31/39

Let A denote the amount of platinum remainingat any time t. Governing equation is

d Ak A

d twith

0A A .. ..(1)

Its solution is

ktA t ce ..(2)

Solution


32/39

Using above condition (1) in eq (2) we get

0( ) kt

A t A e

0AIf 0.043% of have disintegrated, thenremaining amount is 99.957% Or 0.99957

00.99957A A

Now at t = 15

Using this condition in eq (3) we get

(3)

15

0 00.99957 kA A e


33/39

ln 0.99957

15k

0 . 0 0 0 0 2 8 6 7k

0.00002867

0( ) tA t A e

150.99957 ke


34/39

24180t yrs

1ln

2

0.0002867t

1ln 0.00002867

2t

0.0000286700

2

tAA e

0( )2

AA t

Now the half life in the corresponding value oftime for which


35/39

Carbon Dating

About 1950, the chemist Willard deviseda method of using radio active carbon as ameans of determining the approximate ageof fossils, such as remains of woods, bones,

etc.

The half life of a radio active carbon isabout 5600 years. The ratio of Carbon 14 to

ordinary Carbon in the atmosphere isconstant, and this ratio also applies toanimals and plants tissue as long as they arealive.


36/39

However, when the animals or plants die,

the Carbon-14 starts to decay throughradioactivity, while the ordinary carbon doesnot disappear.

Thus knowing the ratio of C-14 to ordinarycarbon we can determine an approximation tothe year of death.

For his work Libby won the Nobel Prize forchemistry in 1960.


37/39

A wooden archeological specimen contains15% of the original Carbon-14. Using a half lifeof 5730 years. Determine the approximate ageof the specimen.

Example

Solution

Governing equation is

dA

kAdtwith

0)0( AA ..(1)


38/39

Its solution is

( ) k tt ce . ..(2)

Using condition (1) in eq (2) we get.

0( ) kt

t A e

5730, ( )2

oA

t A tat .....(3)

Substituting condition (3) in eq (2) we get.

57300

02

kA

A e

1ln 5730

2k


39/39

k

0( ) 0.15A t AWhen

0.000120968

0 0(0.15) tA A e

ln(0.15) 0.000120968t

15682t years

ODE Lecture 5

Documents

Transcript of ODE Lecture 5