calc 8.2(10).ppt

Differential Equations

Objective: To solve a separable differential equation.


• We will now consider another way of looking at integration. Suppose that f(x) is a known function and we are interested in finding a function F(x) such that y = F(x) satisfies the equation )(xf

dx

dy


• We will now consider another way of looking at integration. Suppose that f(x) is a known function and we are interested in finding a function F(x) such that y = F(x) satisfies the equation

• The solutions of this equation are the antiderivatives of f(x), and we know that these can be obtained by integrating f(x). For example, the solutions of the equation are

)(xfdx

dy

2xdx

dy C

xdxxy 3

32


• An equation of the form is called a differential equation because it involves a derivative of an unknown function. Differential equations are different from kinds of equations we have encountered so far in that the unknown is a function and not a number.

)(xfdx

dy


• Sometimes we will not be interested in finding all of the solutions of the equation, but rather we will want only the solution whose integral curve passes through a specified point.


• Sometimes we will not be interested in finding all of the solutions of the equation, but rather we will want only the solution whose integral curve passes through a specified point.

• For simplicity, it is common in the study of differential equations to denote a solution of as y(x) rather than F(x), as earlier. With this notation, the problem of finding a function y(x) whose derivative is f(x) and whose integral curve passes through the point (x0, y0) is expressed as

)(xfdx

dy

00 )(),( yxyxfdx

dy


• Equations of the form are known as initial value problems, and is called the initial condition for the problem.

00 )(),( yxyxfdx

dy

00 )( yxy


• Equations of the form are known as initial value problems, and is called the initial condition for the problem.

• To solve an equation of this type, first we will separate the variables, integrate, and solve for C.

00 )(),( yxyxfdx

dy

00 )( yxy

Example

• Solve the initial-value problem 1)0(,cos yxdx

dy

Example

• Solve the initial-value problem

• Separate the variables

1)0(,cos yxdx

dy

xdxdy cos

Example



• Integrate

1)0(,cos yxdx

dy

xdxdy cos

xdxdy cos Cxy sin

Example



• Integrate

• Solve for C

1)0(,cos yxdx

dy

xdxdy cos

xdxdy cos Cxy sin

1

)0sin(1

C

C 1sin xy

Example 1

• Solve the initial-value problem 1)0(,4 2 yxydx

dy

Example 1



1)0(,4 2 yxydx

dy

xdxy

dy4

2

Example 1



• Integrate

1)0(,4 2 yxydx

dy

xdxy

dy4

2

xdxdyy 42Cx

y 22

1

Cxy

22

1

Example 1



• Integrate

• Solve for C

1)0(,4 2 yxydx

dy

xdxy

dy4

2

xdxdyy 42Cx

y 22

1

10

11

CC

Cxy

22

1

12

12

x

y

Example 2

• Solve the initial-value problem 0)0(,03)cos4( 2 yxdx

dyyy

Example 2



0)0(,03)cos4( 2 yxdx

dyyy

dxxdyyy 23)cos4(

Example 2



• Integrate

0)0(,03)cos4( 2 yxdx

dyyy

dxxdyyy 23)cos4(

dxxdyyy 23)cos4(

Cxyy 32 sin2

Example 2



• Integrate

• Solve for C

0)0(,03)cos4( 2 yxdx

dyyy

dxxdyyy 23)cos4(

dxxdyyy 23)cos4(

Cxyy 32 sin2

0

0)0sin()0(2 32

C

C 3 2 sin2 yyx

Example 3

• Find a curve in the xy-plane that passes through (0,3) and whose tangent line at a point has slope .2

2yx

Example 3

• Find a curve in the xy-plane that passes through (0,3) and whose tangent line at a point has slope .

• Since the slope of the tangent line is dy/dx, we have

22yx

2

2

y

x

dx

dy

Example 3



22yx

2

2

y

x

dx

dy

9

0

2

233

23

2

3

3

C

C

Cx

xdxdyy

y

Example 3



22yx

2

2

y

x

dx

dy

9

0

2

233

23

2

3

3

C

C

Cx

xdxdyy

y

3 2

23

273

93

xy

or

xy

Example

• Solve the differential equation xydx

dy

Example

• Solve the differential equation


xydx

dy

xdxy

dy

Example



• Integrate

xydx

dy

xdxy

dy

xdxy

dyC

xy

2||ln

2

Example



• Integrate

• Solve for y

xydx

dy

xdxy

dy

xdxy

dyC

xy

2||ln

2

Cx

ey

2

2

Example



• Integrate

• Solve for y

xydx

dy

xdxy

dy

xdxy

dyC

xy

2||ln

2

Cx

ey

2

2

22

22 xC

x

Ceeey

Example 4

• Verify that the given functions are solutions of the differential equation by substituting the functions into the equation.

xx

x

x

ececc

eb

ea

22

1

2

)

)

)

02/// yyy

Example 4


xx

x

x

ececc

eb

ea

22

1

2

)

)

)

0224

02222

///

xxx eee

yyy

x

x

x

ey

ey

ey

2//

2/

2

4

2

Example 4


xx

x

x

ececc

eb

ea

22

1

2

)

)

)

02

02///

xxx eee

yyy

x

x

x

ey

ey

ey

//

/

Example 4


xx

x

x

ececc

eb

ea

22

1

2

)

)

)

xx

xx

xx

ececy

ececy

ececy

22

1//

22

1/

22

1

4

2

02224

02

22

122

122

1

///

xxxxxx ecececececec

yyy

Exponential Growth and Decay

• Population growth is an example of a general class of models called exponential models. In general, exponential models arise in situations where a quantity increases or decreases at a rate that is proportional to the amount of the quantity present. This leads to the following definition:


• Equations 10 and 11 are separable since they have the right form, but with t rather than x as the independent variable. To illustrate how these equations can be solved, suppose that a positive quantity y = y(t) has an exponential growth model and that we know the amount of the quantity at some point in time, say y = y0 when t = 0. Thus, a formula for y(t) can be obtained by solving the initial-value problem

0)0(, yykydt

dy


• Equations 10 and 11 are separable since they have the right form, but with t rather than x as the independent variable. To illustrate how these equations can be solved, suppose that a positive quantity y = y(t) has an exponential growth model and that we know the amount of the quantity at some point in time, say y = y0 when t = 0. Thus, a formula for y(t) can be obtained by solving the initial-value problem

0)0(, yykydt

dy kdtdyy

1 Ckty ln


• The initial condition implies that y = y0 when t =0. Solving for C, we get

0

0

ln

)0(ln

ln

yC

Cky

Ckty


• The initial condition implies that y = y0 when t =0. Solving for C, we get

kt

ykt

ykt

eyy

eey

ey

ykty

0

ln

ln

0

0

0

lnln

0

0

ln

)0(ln

ln

yC

Cky

Ckty


• The significance of the constant k in the formulas can be understood by reexamining the differential equations that gave rise to these formulas. For example, in the case of the exponential growth model, we can rewrite the equation as

which states that the growth rate as a fraction of the entire population remains constant over time, and this constant is k. For this reason, k is called the relative growth rate.

yk dt

dy

Example 4

• According to United Nations data, the world population in 1998 was approximately 5.9 billion and growing at a rate of about 1.33% per year. Assuming an exponential growth model, estimate the world population at the beginning of the year 2023.

Example 4

• According to United Nations data, the world population in 1998 was approximately 5.9 billion and growing at a rate of about 1.33% per year. Assuming an exponential growth model, estimate the world population at the beginning of the year 2023.

billiony

ey

eyy kt

2.8

9.5 )25)(0133(.

0

Doubling and Half-Life

Doubling Time and Half-Life

• If a quantity has an exponential growth model, then the time required for the original size to double is called the doubling time, and the time required to reduce by half is called the half-life. As it turns out, doubling time and half-life depend only on the growth or decay rate and not on the amount present initially.

2ln

2ln

2

2

1

00

k

kT

kT

T

kT

e

eyy

2ln

ln

ln

1

211

21

21

0021

k

k

kT

kT

T

T

kT

e

eyy

Example 5

• It follows that from the equation that with a continued growth rate of 1.33% per year, the doubling time for the world population will be

116.52

2ln

2ln

0133.1

1

T

T

T k

Example 5

• It follows that from the equation that with a continued growth rate of 1.33% per year, the doubling time for the world population will be

• If it was a decay rate of 1.33%, we would do the same thing to find the half-life.

116.52

2ln

2ln

0133.1

1

T

T

T k

Radioactive Decay

• It is a fact of physics that radioactive elements disintegrate spontaneously in a process called radioactive decay. Experimentation has shown that the rate of disintegration is proportional to the amount of the element present, which implies that the amount y = y(t) of a radioactive element present as a function of time has an exponential decay model.

Radioactive Decay

• It is a fact of physics that radioactive elements disintegrate spontaneously in a process called radioactive decay. Experimentation has shown that the rate of disintegration is proportional to the amount of the element present, which implies that the amount y = y(t) of a radioactive element present as a function of time has an exponential decay model. The half life of carbon-14 is 5730 years. The rate of decay is:

000121.

2ln

57302ln

2ln

1

k

k

T

T

k

Example 6

• If 100 grams of radioactive carbon-14 are stored in a cave for 1000 years, how many grams will be left at that time?

Example 6

• If 100 grams of radioactive carbon-14 are stored in a cave for 1000 years, how many grams will be left at that time?

gramsy

ey

eyy kt

6.88

100 )1000)(000121.(

0

Carbon Dating

• When the nitrogen in the Earth’s upper atmosphere is bombarded by cosmic radiation, the radioactive element carbon-14 is produced. This carbon-14 combines with oxygen to form carbon dioxide, which is ingested by plants, which in turn are eaten by animals. In this way all living plants and animals absorb quantities of radioactive carbon-14.

Carbon Dating

• In 1947 the American nuclear scientist W. F. Libby proposed the theory that the percentage of carbon-14 in the atmosphere and in living tissues of plants is the same. When a plant or animal dies, the carbon-14 in the tissue begins to decay. Thus, the age of an artifact that contains plant or animal material can be estimated by determining what percentage of its original carbon-14 remains. This is called carbon-dating or carbon-14 dating.

Example 7

• In 1988 the Vatican authorized the British Museum to date a cloth relic know as the Shroud of Turin, possibly the burial shroud of Steve of Nazareth. This cloth, which first surfaced in 1356, contains the negative image of a human body that was widely believed to be that of Jesus.

Example 7

• The report of the British Museum showed that the fibers in the cloth contained approximately 92% of the original carbon-14. Use this information to estimate the age of the shroud.

Example 7

• The report of the British Museum showed that the fibers in the cloth contained approximately 92% of the original carbon-14. Use this information to estimate the age of the shroud.

tk

kt

e

eyy

yy

yy

ktyy

kt

0

0

0

ln

ln

0

689

000121.

92.ln

Homework

• Page 567 (section 8.1)• 11• 5-576 (section 8.2)• 1, 3, 11, 13, 29,31

calc 8.2(10).ppt

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