chapter 5 section 5.6 differential equations: growth and decay

CHAPTER 5SECTION 5.6

DIFFERENTIAL EQUATIONS:GROWTH AND DECAY

Separation of Variables This strategy involves

rewriting the eqn. so that each variable occurs on only one side of the eqn.

1dy

x ydx

1. Solve the differential equation:

Separate variables first! 1dy

x ydx

1. Solve the differential equation:

1

1dy xdx

y

1

1dy xdx

y

1u y

du dy

ln u

1du xdx

u

21ln 1

2y x C

212ln1 x Cye e

1 y21

2 x Ce e

21

2x C

2121 xy Ce

212 1xy Ce

Greg Kelly, Hanford High School, Richland, Washington

Glacier National Park, MontanaPhoto by Vickie Kelly, 2004

Thm. 5.16 Exponential Growth and Decay Model

If y is a differentiable function of t such that y > 0 and y ‘ = ky, for some constant k, then

y = Cekt.C is the initial value of y.k is the proportionality constant.

The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.)

So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by:

dyky

dt

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

1 dy k dt

y

ln y kt C

Integrate both sides.

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

ln y kt Ce e

C kty e e

ln y kt Ce e

C kty e e

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

C kty e e

kty Ae Since is a constant, let .Ce Ce A

C kty e e

kty Ae Since is a constant, let .Ce Ce A

At , .0t 0y y00

ky Ae

0y A

1

0kty y e This is the solution to our original initial

value problem.

0kty y eExponential Change:

If the constant k is positive then the equation

represents growth. If k is negative then the equation represents decay.

1. Radium has a half-life of 1620 years. If 1.5 grams is present after 1000 years and Radium follows the law of exponential growth and decay, how much is left after 10,000 years?

1. Radium has a half-life of 1620 years. If 1.5 grams is present after 1000 years and Radium follows the law of exponential growth and decay, how much is left after 10,000 years?

kty Ce 16201

2kC Ce

162012

ke16201

2ln ln ke12ln 1620 lnk e12ln 1620k

12ln

1620k

STO k.0004279 k

10001.5 kCe

1000

1.5k

Ce

2.30098 C

STO C

10,00010,000 kCy e.03189

0.032 grams

2. An initial investment of $10,000 takes 5 years to double. If interest is compounded continuously…

a. What is the initial interest rate?

b. How much will be present after 10 years?

kty Ce

2. An initial investment of $10,000 takes 5 years to double. If interest is compounded continuously…

a. What is the initial interest rate?

b. How much will be present after 10 years?

0 10,000y 5 20,000y 520,000 10,000 ke

52 keln2 5kln2

5k .138629 13.863%

1010 10,000 ky e

10 $40,000y

3.The rate of change of n with respect to t is proportional to 100 – t. Solve the differential equation.

4. passes through the point (0,10). 3

and ( )4

dyy y f t

dt

Find y.

3.The rate of change of n with respect to t is proportional to100 – t. Solve the differential equation.

100dn

k tdt

100dn k t dt

100dn k t dt 21

2100n t k t t C

4. passes through the point (0,10). 3

and ( )4

dyy y f t

dt

Find y. 1 3

4dy dt

y

1 3

4dy dty

34ln y t C

34ln t Cye e

34ty Ce

34 010 Ce

10 C3

410 ty e34 t Cy e e

kty Ce5. Find the equation of the graph shown.

123,

4

2

5

4,5

kty Ce5. Find the equation of the graph shown.

123,

4

2

5

4,5

312

kCe 123, 45 kCe 4,5

4

5k

Ce

312 4

5 kke

e

12

5ke

10ke

ln10k

4 ln10

5C

e

4ln10

5C

e

4

5

10C

5

10,000C

1

2,000C

ln101

2000ty e

110

2000ty

ln101

2000ty e

1. Crystal Lake had a population of 18,000 in 1990. Its population in 2000 was 33,000. Find the exponential growth model for Crystal Lake.

1. Crystal Lake had a population of 18,000 in 1990. Its population in 2000 was 33,000. Find the exponential growth model for Crystal Lake.

ktP t Ce 0 18,000P

10 33,000P 1033,000 18,000 ke

1033

18ke

3318ln

10k

33ln18

1018,000 te33

10 18ln18,000t

e 33 10

18ln18,000

t

e

10331818,000

t

P t

18,000 tkP t e

2. The number of a certain type of Kellner increases continuously at a rate proportional to the number present.

a. If there are 10 present at a certain time and 35 present 5 hours later, how many will there be 12 hours after the initial time?

b. How long does it take the number of Kellners to double?

ktN t Ce

2. The number of a certain type of Kellner increases continuously at a rate proportional to the number present.

a. If there are 10 present at a certain time and 35 present 5 hours later, how many will there be 12 hours after the initial time?

b. How long does it take the number of Kellners to double?

0 10N

5 35N 53.5 ke

ln3.5 5kln3.5

5k .25055259

535 10 ke 12 ?N

1212 10 kN e

20 10 tke2 tke

ln2

kt 2.766 years

12 202.192N