Transcript 6.2

6
Differential Equations
Copyright © Cengage Learning. All rights reserved.
6.2
Differential Equations:
Growth and Decay
Copyright © Cengage Learning. All rights reserved.
Objectives
 Use separation of variables to solve a simple
differential equation.
 Use exponential functions to model growth
and decay in applied problems.
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Warm-Up
Rewrite the natural log equation in exponential form:
ln x  5
Simplify the exponential expression:
e 2  e5 
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Differential Equations
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Differential Equations
Analytically, you have learned to solve only two types of
differential equations—those of the forms
y' = f(x) and y'' = f(x).
In this section, you will learn how to solve a more general
type of differential equation.
The strategy is to rewrite the equation so that each variable
occurs on only one side of the equation. This strategy is
called separation of variables.
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Example 1 – Solving a Differential Equation
dy 2 x

dx
y
y dy  2 x dx
Apply separation of variables.
Integrate each side.
So, the general solution is given by: y2 – 2x2 = C.
You can use implicit differentiation with respect to x to check.
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Growth and Decay Models
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Proportionality
2 quantities are proportional if one is a multiple of the other.
2x

y  e , y  2e
2x
5x

y  e , y  5e
5x
y  e , y   Ce
cx
In each example, the rate of
change of y is proportional
to the value of y.
cx
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Growth and Decay Models
In many applications, the rate of change of a variable y is
proportional to the value of y. If y is a function of time t, the
proportion can be written as follows.
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Solve the differential equation by separating the variables:
dy
 ky
dt
dy
 k dt
y

1
dy 
y
 kdt
ye e
C1 kt
y  Ce
kt
ln y  kt  C
y e e
kt C1
So, all solutions of y’ = ky
kt
are
in
the
form
of
y
=
Ce

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Growth and Decay Models
The general solution of this differential equation is given in
the following theorem.
So, if y’ = ky, then y = Cekt
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Example 2 – Using an Exponential Growth Model
The rate of change of y is proportional to y. When t = 0,
y = 2, and when t = 2, y = 4. What is the value of y when t = 3?
Solution:
Because y' = ky, you know that y and t are related by the
equation y = Cekt.
You can find the values of the constants C and k by applying
the initial conditions.
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Example 2 – Solution
cont'd
So, the model is y ≈ 2e0.3466t .
When t = 3, the value of y is 2e0.3466(3) ≈ 5.657.
Figure 6.8
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Growth and Decay Models
Notice that we did not actually have to solve the differential
equation y' = ky, we just used its solution y  Cekt .
The next example demonstrates a problem whose
solution involves the separation of variables technique.
The example concerns Newton's Law of Cooling, which
states that: the rate of change in the temperature of
an object is proportional to the difference between
the object’s temperature and the temperature of
the surrounding medium.
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Example 6 – Newton's Law of Cooling
The rate of change in the temperature of an object is proportional to the difference
between the object’s temperature and the temperature of the surrounding medium.
Let y represent the temperature (in ºF) of an object in a
room whose temperature is kept at a constant 60º. If the
object cools from 100º to 90º in 10 minutes, how much
longer will it take for its temperature to decrease to 80º?
Solution:
From Newton's Law of Cooling, you know that the rate of
change in y is proportional to the difference between
y and 60.
This can be written as
y' = k(y – 60),
80 ≤ y ≤ 100.
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cont'd
Let y represent the temperature (in ºF) of an object in a room whose
temperature is kept at a constant 60º. If the object cools from 100º to 90º in
10 minutes, how much longer will it take for its temperature to decrease to 80º?
To solve this differential equation, use separation of
variables, as follows.
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ln  y  60  kt  C1
The object cools from 100º to 90º
in 10 minutes.
cont'd
Because y > 60, |y – 60| = y – 60, and you can omit the absolute value signs.
Rewriting the solution in exponential form, you have
Using y = 100 when t = 0, you obtain: 100 = 60 + Cek(0)
kt
y

60

40
e
so, 100 = 60 + C, which implies that C = 40 and
Because y = 90 when t = 10,
90 = 60 + 40ek(10)
30 = 40e10k

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cont'd
Let y represent the temperature (in ºF) of an object in a room whose
temperature is kept at a constant 60º. If the object cools from 100º to 90º in 10
minutes, how much longer will it take for its temperature to decrease to 80º?
y  60  40e
So, the model is: y = 60 + 40e–0.02877t
kt
Cooling model
and finally, when y = 80, you obtain
Figure 6.11
So, it will require about 14.09 more minutes for the object
to cool to a temperature of 80º (see Figure 6.11).
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Extra Example– Newton's Law of Cooling
When an object is removed from a furnace and placed in a
room with a constant temperature of 80  F . , its temperature
is 1500  F . One hour after it is removed its temperature is
1120  F . Find its temperature 5 hours after the object it is
removed from the furnace.
Solution:
From Newton's Law of Cooling, you know that the rate of
change in y is proportional to the difference between
y and 80.
This can be written as
y' = k(y – 80),
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Solution con’t.
y  80  Ce kt
y' = k(y – 80)
dy
 k  y  80 
dt
dy
 k dt
 y  80 
dy
  y  80    k dt
ln  y  80  kt  C
y  80  Ce kt
 0, 1500
1500  80  C
C  1420
y  80  1420e kt
1, 1120
1120  80  1420e
5, y 
k 1
k  ln .7324
y  80  1420e5(ln .7324)
When t  5, y  379.236
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6.2 Differential Equations: Growth & Decay
Radioactive material decays exponentially and is measured
in terms of half-life  the number of years required for half
of the atoms in a sample of radioactive material to decay. The
half-life of some common radioactive isotopes are as follows.
Uranium (U238)
Plutonium (Pu239)
Carbon (C14)
Radium (Ra226)
Einsteinium (Es254)
Nobelium (No257)
4,510,000,000 years
24,100 years
5730 years
1620 years
270 days
23 seconds
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6.2 Differential Equations: Growth & Decay
A sample contains 1 gram of radium. How much radium will
remain after 1000 years? (Use a half-life of 1620 years.)
y  Ce
kt
1  Ce
1
k 1, 620
 1e
2
1
1620k
ln  ln e
2
1
ln  1620k
2
k  0
Take the ln of both sides.
k  .0004279
.0004279(1000)
y  1e
y  .652 g
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Suppose that 10 grams of the plutonium isotope Pu-239 was
released in the Chernobyl nuclear accident. How long will it
take for the 10 grams to decay to 1 gram? Pu-239 has
a half life of 24,100 years
y=
Cekt
y  10e
k  0
We know that C = 10 grams at
time t = 0. First, find k.
5 = 10ek(24,100)
So, the model is: y = 10e-.000028761t
1
24,100k
e
2
1
1
k
ln
24,100 2
To find the time it would take for
10 grams to decay to 1 gram, solve
for t in: 1 = 10e-.000028761t
k  .000028761
t = 80,059 years
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Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
 g  x  h  y
dx
Example:
dy
 2 xy 2
dx
dy
 2 x dx
2
y
y 2 dy  2 x dx
h y  0
2
y
 dy   2 x dx
1
 y  C1  x  C2
2
1
  x2  C
y
1
 2
y
x C
Combined
constants of
integration
1
y 2
x C
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
Example:
Separable Differential Equations
dy
2
x2
 2 x 1  y  e
dx
Separable differential equation
1
x2
dy  2 x e dx
2
1 y
1
u  x2
x2
 1  y 2 dy   2 x e dx du  2x dx
1
u
dy

e
 1 y2
 du
tan 1 y  C1  eu  C2
1
tan y  C1  e  C2
x2
1
tan y  e  C
x2
Combined constants of integration
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
Example continued:
dy
2
x2
 2 x 1  y  e
dx
1
tan y  e  C
x2

We now have y as an implicit
function of x.

tan  tan y   tan e  C We can find y as an explicit function
of x by taking the tangent of both
1

y  tan e  C
x2

x2
sides.
Notice that we can not factor out the constant C, because
the distributive property does not work with tangent.
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
Example:
Find the general solution of
dy
x  4dx  xy
2
dy
x
 2
dx
y x  4 
First, separate the variables.
y’s on one side, x’s on the other.

dy
x
 y   x 2  4dx
Second, integrate both sides.

Solve for y.
1
ln y  ln x 2  4  C1
2

Write in exponential form. y  eC1

x
2
 4  or
y  C
x
2
 4
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Extra Exponential Examples:
Money is deposited in an account for which
interest is compounded continuously. If the balance
doubles in 6 years, what is the annual percentage rate?
A = Pert
2P = Pert
2 = e6r
ln 2 = 6r
r  .1155  11.55%
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Homework 6.2:
Day 1: Pg. 418
1-27 odds, 41, 43, 49, 53, 62, 71.
Day 2: MMM 218-220
Day 3: In class Slope Field Match, HW Liberty
High Slope Field WS
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