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CHAPTER 5SECTION 5.6
DIFFERENTIAL EQUATIONS:GROWTH AND DECAY
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Separation of Variables This strategy involves
rewriting the eqn. so that each variable occurs on only one side of the eqn.
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1dy
x ydx
1. Solve the differential equation:
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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
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Greg Kelly, Hanford High School, Richland, Washington
Glacier National Park, MontanaPhoto by Vickie Kelly, 2004
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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.
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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
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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.
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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
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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
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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.
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0kty y eExponential Change:
If the constant k is positive then the equation
represents growth. If k is negative then the equation represents decay.
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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?
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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
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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?
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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
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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.
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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
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kty Ce5. Find the equation of the graph shown.
123,
4
2
5
4,5
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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
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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.
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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
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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?
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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