chapter 2 slides

FIRST ORDER EQUATIONS

F(x, y, y′) = 0

Basic assumption: This equation can be

solved for y′; that is, the equation can be

written in the form

y′ = f(x, y) (1)

1

I. First Order Linear Equations

Equation (1) is a linear equation if

f has the form

f(x, y) = P (x)y + q(x)

where P and q are continuous

functions on some interval I. Thus

y′ = P (x)y + q(x)

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Standard form: The standard form

for a first order linear equation is:

y′ + p(x)y = q(x)

where p and q are continuous

functions on the interval I

3

Solution Method:

Step 1. Establish that the equa-

tion is linear and write it in standard

form

y′ + p(x)y = q(x).

Step 2. Multiply by e∫

p(x) dx :

e∫

p(x) dxy′+p(x)e∫

p(x) dxy = q(x)e∫

p(x) dx

[e∫

p(x) dx y]′

= q(x)e∫

p(x) dx

4

Step 3. Integrate:

e∫

p(x) dx y =∫

q(x)e∫

p(x) dx dx + C.

Step 4. Solve for y :

y = e−∫

p(x) dx∫

q(x)e∫

p(x) dx dx+Ce−∫

p(x) dx.

This is the general solution of the

equation.

5

Examples:

1. Find the general solution:

y′ + 2x y = 4x

Answer: y = 2 + Ce−x2


xy′ + 3y =ln x

x

Answer: y =ln x

2x−

1

4x+

C

x3

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3. Solve the initial-value problem:

y′+(cot x)y = 2cos x, y(π/2) = 3

Answer: y =2

sin x−

cos 2x

sin x

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Important Special Case

y′ + p(x)y = 0.

General solution: y = Ce−∫

p(x) dx.

Since e−∫

p(x) dx 6= 0 for all x, it

follows that:

1. If y(a) = 0 for some a, then

C = 0 and y ≡ 0.

2. If y(a) 6= 0 for some a, then

C 6= 0 and y(x) 6= 0 for all x.

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The term “linear:”

L[y] = y′ + p(x) y

is a linear operator:

L[y1 + y2] = L[y1] + L[y2]

and

L[cy] = cL[y].

9

A related equation: Transforma-

tion into a linear equation:

Bernoulli equations

An equation of the form

y′ + p(x)y = q(x)yk, k 6= 0, 1

is called a Bernoulli equation.

10

The change of variable

v = y1−k

transforms a Bernoulli equation into

the linear equation

v′ + (1 − k)p(x)v = (1 − k)q(x).

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Examples:


y′ − 4y = 2ex √y

Answer: y = (Ce2x − ex)2


xy′ + y = 3x3 y2

Answer: y2 = −32 x3 + Cx

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II. Separable Equations

Equation (1) is a separable equa-

tion if f has the form

f(x, y) = p(x)h(y)

where p and h are continuous

functions.

Thus

y′ = p(x)h(y)

13

Solution Method

Step 1. Establish that the equa-

tion is separable.

Step 2. Divide both sides by h(y)

to “separate” the variables.

1

h(y)y′ = p(x) or q(y)y′ = p(x)

which, in differential form, is:

q(y) dy = p(x) dx.

the variables are “separated.”

14

Step 3. Integrate

∫q(y) dy =

∫p(x) dx + C

Q(y) = P (x) + C

Notes:

1. Q(y) = P (x)+C is the general

solution. Typically, this is an im-

plicit relation; you may or may not

be able to solve it for y.

15

2. h(y) = 0 is a possible source of

singular solutions of the form y = k.

Examples:

1. Find the general solution and

any singular solutions:

y′ − xy2 = x



1

x

dy

dx= ex

√y + 1

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3. Solve the initial-value problem:

y2 ln xdy

dx=

y3 + 1

x, y(e) = 2



dy

dx=

ex−y

1 + ex

17

A related equation: Transforma-

tion into a separable equation:

Homogeneous equations

Equation (1) is a homogeneous equa-

tion if

f(tx, ty) = f(x, y)

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If (1) is homogeneous, then the change

of dependent variable

y = vx, y′ = v + xv′

transforms (1) into a separable equa-

tion:

y′ = f(x, y) → v+xv′ = f(x, vx) = f(1, v)

which can be written

1

f(1, v) − vdv =

1

xdx;

the variables are separated.

19

Examples:


y′ =x2 + y2

2xy


dy

dx=

x2ey/x + y2

xy

20

III. Applications

3.1. Orthogonal Trajectories

Given a one-parameter family of curves

F (x, y, C) = 0.

A curve that intersects each mem-

ber of the family at right angles (or-

thogonally) is called an orthogonal

trajectory of the family.

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If

F (x, y, c) = 0 and G(x, y, K) = 0

are one-parameter families of curves

such that each member of one fam-

ily is an orthogonal trajectory of the

other family, then the two families

are said to be orthogonal trajecto-

ries.

22

A procedure for finding a family of

orthogonal trajectories

G(x, y, K) = 0

for a given family of curves

F (x, y, C) = 0

Step 1. Determine the differential

equation for the given family

F (x, y, C) = 0.

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Step 2. Replace y′ in that equa-

tion by −1/y′; the resulting equa-

tion is the differential equation for

the family of orthogonal trajecto-

ries.

Step 3. Find the general solu-

tion of the new differential equation.

This is the family of orthogonal tra-

jectories.

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Examples

1. Find the family of orthogonal

trajectories of:

y3 = Cx2 + 2

2. Find the orthogonal trajecto-

ries of the family of parabolas with

vertical axis and vertex at the point

(−1,3).

25

3.2. Radioactive Decay/Exponential

Growth

Radioactive Decay

“Experiment”: The rate of decay of

a radioactive material at time t is

proportional to the amount of mate-

rial present at time t. Let A = A(t)

be the amount of radioactive mate-

rial present at time t.

26

Mathematical Model

dA

dt= −r A, r > 0 constant.

r is the decay rate.

Solution: A(t) = A0 e−rt, where

A0 = A(0) the initial amount.

Half-life T =ln 2

r.

27

Example: A certain radioactive ma-

terial is decaying at a rate propor-

tional to the amount present. If a

sample of 50 grams of the mate-

rial was present initially and after

2 hours the sample lost 10% of its

mass,

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find:

1. An expression for the mass of

the material remaining at any time

t.

2. The mass of the material after

4 hours.

3. How long will it take for 75%

of the material to decay?

4. The half-life of the material.

29

Exponential Growth

“Experiment”: Under “ideal” con-

ditions, the rate of increase of a

population at time t is proportional

to the size of the population at time

t. Let P = P (t) be the size of the

population at time t.

30

Mathematical Model

dP

dt= k P, k > 0 constant.

k is the growth rate.

Solution: P (t) = P0 ekt, P0 =

P (0) the initial population.

Doubling time T =ln 2

k.

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Example: In 1980 the world popu-

lation was approximately 4.5 billion

and in the year 2000 it was approx-

imately 6 billion. Assume that the

population increases at a rate pro-

portional to the size of population.

32

1. Find the growth constant

and give the world population at any

time t.

2. How long will it take for the

world population to reach 9 billion

(double the 1980 population)?

3. The world population for 2002

was reported to be about 6.2 billion.

What population does the formula

in (1) predict for the year 2002?

33

Example: It is estimated that the

arable land on earth can support a

maximum of 30 billion people. Ex-

trapolate from the data given in the

previous example to estimate the year

when the food supply becomes in-

sufficient to support the world pop-

ulation.

34

3.3. Newton’s Law of Cooling

“Experiment” The rate of change

of the temperature of an object at

time t is proportional to the dif-

ference between the temperature of

the object u = u(t) and the (con-

stant) temperature σ of the sur-

rounding medium (e.g., air or wa-

ter)

35

Mathematical Model

du

dt= −k(u − σ), k > 0 constant.

Solution:

u(t) = σ + [u0 − σ]e−kt

u0 = u(0) the initial temperature

of the object.

36

Example: Suppose that a corpse is

discovered at 10 p.m. and its tem-

perature is determined to be 85o F .

Two hours later, its temperature is

74o F . If the ambient temperature is

68o F , estimate the time of death.

37

3.6. “Limited” Growth – the

Logistic Equation

“Experiment” Given a population

of size M . The spread of an in-

fectious disease at time t (or infor-

mation, or ...) is proportional to

the product of the number of peo-

ple who have the disease P (t) and

the number of people who do not

M − P (t).

38

Mathematical Model

dP

dt= kP (M − P ), k > 0 constant.

Solution:

P (t) =MR

R + (M − R)e−Mkt

P (0) = R the number of people

who have the disease initially

39

Example: A rumor spreads through

a small town with a population of

5,000 at a rate proportional to the

product of the number of people who

have heard the rumor and the num-

ber who have not heard it. Suppose

that 100 people initiated the rumor

and that 500 people heard it after 3

days.

40

1. How many people will have

heard the rumor at time t?

2. How many people will have

heard the rumor after 8 days?

3. How long will it take for half

the population to hear the rumor?

41

Miscellaneous Examples

1. A 1000-gallon cylindrical tank,

initially full of water, develops a leak

at the bottom. Given that 200 gal-

lons of water leak out in the first 10

minutes, find the amount of water,

A(t), left in the tank t minutes

after the leak develops if the water

drains off a rate proportional to the

product of the time elapsed and the

amount of water present.

42

Answer: A(t) = 1000(4

5

)t2/100.

2. A 1000-gallon conical tank, ini-

tially full of water, develops a leak

at the bottom. Suppose that the

water drains off a rate proportional

to the product of the time elapsed

and the square root of the amount

of water present. Let A(t) be the

amount of water in the tank at time

t.

43

a. Give the mathematical model

(differential equation) which describes

the process.

Answer:dA

dt= kt

√A, k < 0.

b. Find the general solution and

the particular solution that satisfies

A(0) = 1000.

Answer: A(t) =(14kt2 + 10

√10

)2.

44

3. A disease is spreading through

a city of population M (constant).

Let P (t) be the number of peo-

ple who have the disease at time

t. Suppose that R people had

the disease initially and that the rate

at which the disease is spreading at

time t is proportional to product of

the time elapsed and the number of

people who don’t have the disease.

45

a. Give the mathematical model

(differential equation) which describes

the spread of the disease.

Answer:dP

dt= kt(M −P ), k > 0.

b. Find the particular solution that

satisfies P (0) = R.

Answer: P (t) = M −(M −R)e−kt.

46

4. Direction Fields

The direction field for the differential equa-

tion

y′ = x − y

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A sketch of the solution of y′ = x− y that

satisfies the initial condition y(0) = 1

y = x − 1 + 2e−x.

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y′ = y2 − xy + 2x

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Solution curves generated by the initial con-

ditions y(−2) = 1 and y(1) = 2.

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5. Existence and Uniqueness Theo-

rem

Given the initial-value problem

y′ = f(x, y) y(a) = b.

If f and ∂f/∂y are continuous on a

rectangle

R : a−α ≤ x ≤ a+α, b−β ≤ y ≤ b+β, α, β > 0,

then there is an interval

a − h ≤ x ≤ a + h, h ≤ α

on which the initial-value problem has a

unique solution y = y(x).

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