ODE · 1. Rearrange the equation in the general form 2. Show that (Test the homogeneity) 3. Substitute and into the general form of homoge General form: Method of solution: neou,,

CCCChapterhapterhapterhapter 2 2 2 21st ORDER DIFFERENTIAL

EQUATIONS

2

2.1 Types of 12.1 Types of 12.1 Types of 12.1 Types of 1stststst ODE ODE ODE ODE

� Separable Equations

�Homogeneous Equations

� Linear Equations

� Exact Equations

Separable equationsSeparable equationsSeparable equationsSeparable equations

( ) ( ) ( )( )

( )( )

( ) ( )

1. Reform the given differential equation into the general

form of separable equation (above)

2. Rearrange the equation to

3

General form:

Method of solut

. Inte

ion:

grat

, ,

1

g x h ydy dy dyg x h y

dx dx h y dx g x

dy g x dxh y

= = =

=

( ) ( )e both side of equation in step 2

4. Obtain the general solution H y G x CÞ = +

Example 2.1Example 2.1Example 2.1Example 2.1

Solve the differential equation

3

4

3

dy xdx y

=

Separate the equation into the form,

Integrate the both sides,

Rearranging, this equation becom

e s:

3

3

4 2

42 4

:

: 3 4

:

3 4

3 44 2

:

3 82

4 3

1

2

3

Solution

y dy xdx

y dy

Step

Step

S

xdx

y xC

yx c y

tep

=

=

æ ö æ ö÷ ÷ç ç÷ ÷= +ç ç÷ ÷ç ç÷ ÷ç çè ø è ø

= + Þ =

ò ò

General solution is,

where

2

142

43

:

8

4

4,

3 3

x C

S

y x A A

t

C

ep

+ +

æ ö÷ç= + =÷ç ÷çè ø


( )


and given that

24

0 2

y x y

y

¢ =

=


wher

e

3 3

3

2

2

3

4 43 3

43

:

: 4

: 4

4ln

3

:

:

,

1

2

3

4

x xC C

xC

Solution

dyx dx

y

dydy x dx

y

xy C

y e y e e

y De

Step

Step

Step

Step

D e

+

=

=

= +

= Þ = ×

= =

ò ò

( )( )

Given that

Particular soluti

on:

3

3

4 0

3

43

: 0 2

2 2

5

2x

y

De D

e

Ste

y

p =

= Þ\ =

=

HOMOGENEOUS equationsHOMOGENEOUS equationsHOMOGENEOUS equationsHOMOGENEOUS equations

Homogeneous functions are functions where the sum of

the powers of every term are the same.

( ) ��

( ) ��

( ) ��

3 2 3

3

deg3 deg3 deg3

deg3 deg1

deg3 deg0deg3

3 3

1) , 4 3

2) ,

3) , 4 4

f x y x x y y

f x y x y

f x y x y

= + +

= +

= + +

( )

( ) ( )1. Rearrange the equation in the general form

2. Show that (Test the homogeneity)

3. Substitute and into the general

form of homoge

General form:

Method of solution:

neou

,

, ,

dyf x y

dx

f x y f x y

dy dvy xv x v

dx dx

l l

=

=

= = +

s equation.

4. Separate variables and to form separable equation.

5. Integrate both side of equation in step 4

6. Substitute into solution in step 5 and simplify the solution.yv

x v

x=



2 2dy x y

dx xy+

=

( )

( ) ( ) ( )( )( )

( )( )

( )

Test homogeneity,

hence, the equation is homogeneous.

2 2

2 2

2 2 2

2

2 2

:

:

,

,

,

1

Solution

x yf x y

xy

x yf x y

x y

x y

xy

x yf x y

xy

Step

l ll l

l l

l

l

+=

+Þ =

+=

+= =

( )( )

Substitute and into

homegeneous equation,

Rearranging, this equation becomes: (separable

form)

22

2 2 2 2

2

2

:

1

:

1

1

1

2

3

dy dvy xv x v

dx dx

x xvdvx vdx x xv

x x v vvx v

dv vx vdx vdvxdx v

vdv dx

Step

S ep

x

t

= = +

++ =

+ += =

+= -

=

=

( )

( )

( )

( )



by substituting

4

2

2

2

2

2

2 2

:

1

ln2

2 ln

:

2 ln ( )

2 ln

2 ln

5

vdv dxx

vx C

v x C

y yx C v

x

Step

Step

x

yx C

xy x x C

=

= +

= +

æ ö÷ç = + =÷ç ÷çè ø

= +

= +

ò ò


( )Solve the differential equation

2 24y dx xy x dy= -

( )

( )

( ) ( )( )( ) ( )

( )( )

( )

Test homogeneity,

hence, the equation is homogeneous.

2

2

2

2

2

2

2 2

2 2

2

2

:

: ,4

:

,

1

2

4

,4

4

,4

Solution

dy yf x y

dx xy x

yf x y

xy x

yf x y

x y x

y

xy x

yf x y

xy

Step

Step

x

ll l

l l l

l

l

= =-

=-

Þ =-

=-

= =-

( )( )

( )( )

( )

2

2

2 2 2

2

2

2

:4

44

:4

4

4 44

44

2

14

3

xvdvx vdx x xv x

x v vvx v

dv vx vdx v

v vdv vxdx v vdv vxdx vv

dv dxv x

Step

Step

+ =-

= =--

= --

-= -

- -

=-

-=

( )

( )


4

:

4 141 1 14

1ln ln

41

ln4

4 ln 4

vdv dx

v x

dv dxv x

v v x C

v xv C

v xv C

Step

-=

æ ö÷ç - =÷ç ÷çè ø

- = +

= +

= +

ò ò

ò ò


by substituting

wher

e

4ln 4

4 4 4 4

:

4 ln 4 ( )

4 ln 4

5

,

yy Cx

y yC Cx x

y y yx C v

x x x

yy C

x

e e e

e y e y A

Step

e A e-

æ ö÷ç= × + =÷ç ÷çè ø

= +

= ×

= × Þ\ = =


Verify that the differential equation

is homogeneous. Then find the general solution.

2

2

3 4

2 2

dy y xydx xy x

+=

+


( )

Find the solution for the following differential

equation

2 22 4x y dy xydx- =

Example 2.7Example 2.7Example 2.7Example 2.7By using the substitution and ,

show that the given differential equation

can be reduced to a homogeneous equation.

Then, solve it.

1 2

22

x X y Y

dy x ydx y

= + = -

+=

+

linear linear linear linear equationsequationsequationsequations

( ) ( ) ( )

( ) ( )

( ) ( )( ) ( ) ( )

( )

( )

1. Write the equation in the

General form:

Method of solution:

form:

2. Find the integrating factor :

;

p x dx

dya x b x y c x

dx

dyp x y q x

dxb x c x

p x q xa x a x

er r

+ =

+ =

= =

ò=

linear equationslinear equationslinear equationslinear equations

( ) ( )

{ }

( )

3. Multiply both sides by

The left side of the equ. can be simplified as,

4. Integrate both sides of the equ. to obtain

If given an initial condition, substitute into the

:

5.

dyp x y q x

dxd

ydx

y q x dx

r r r

r

r r

r + =

= òsolution.


Solve the following differential equation:

32 5dyx y xdx

+ =

( ) ( )

2

3

32

2

2 ln

ln

2

:

2 5: ,

2 5;

1

:2

5

dxx

x

x

Solution

dy xy

dx x x

xp x q x x

x x

e

e

e

x

Step

Step r

r

æ ö÷ç+ =÷ç ÷çè ø

= = =

ò=

=

=

=

( )

{ }

Multiply both sides by

2

2 2 2 2

2 4

2 4

2 4

52

32

: :

25

2 5

5

:

5

55

3

4

x

dyx x y x xdx x

dyx xy xdx

dx y

Step

Ste

xdx

x y x dx

xx y C

Cy x

x

p

r =

æ ö÷ç+ =÷ç ÷çè ø

+ =

=

=

= +

= +

ò


( ) ( )


1 , 0 1x xdye e y x y

dx+ + = =

( ) ( )

Let

1

:

: ,1 1

;1 1

1

2 : , 1

x

x

x

x

x x

x

x x

edx

xe

e

Solution

dy e xy

dx e e

e

Step

St

xp x q x

e

ep

e

u

e

e er +

æ ö÷ç ÷+ =ç ÷ç ÷ç + +è ø

= =+ +

ò= = +

= x

duu e

ln

ln 1

1

x

xx

u

e

x

du due dx

dx e

e

e

er

+

ò= Þ =

=

=

= +

( ) ( ) ( )

( )

( )( ){ }

( )

( )

( ) ( )

Multiply both sides by

2

2

: 1 :

1 1 1 11 1

1

1

:

1

12

2

3

1 1

4

x

xx x x

x x

x x

x

x

x

x x

e

dy e xe e y e

dx e e

dye e y x

dxd

e y xdx

e y xdx

xe y C

x Cy

Step

S

e e

tep

r = +

æ ö æ ö÷ç ÷ç÷+ + + = + ÷ç ç÷ ÷ç ç÷ç è ø+ +è ø

+ + =

+ =

+ =

+ = +

= ++ +

ò

( )

( )( ) ( )

( ) ( )

( )

When

Particular solution is

2

0 0

2

2

: 0 1 :

01

2 1 1

1 22

2

2 1 1

1 12

21

5

x x

x

y

C

e e

CC

xy

e

St p

x

e

e

ye

=

= ++ +

= Þ\ =

= ++ +

æ ö÷ç= + ÷ç ÷çè ø+

Exercise 2.1Exercise 2.1Exercise 2.1Exercise 2.1

( )


2 cos

cos: sin

x

x

dyx x y e xdx

e x CAnswer y x

x x x

+ - =

æ ö÷ç= + + ÷ç ÷çè ø

exact equationsexact equationsexact equationsexact equations

( ) ( )

( ) ( )

( )

1.

2. Test for exa

Write down your and

Let be the solution,

i) integrate with respect to while holding constant,

to ob

General form

ta

c

:

Method of solution:

tness:

, , 0

, ,

3. ,

M x y dx N x y dy

M x y N x y

M N

F x y

M x y

y x

+ =

¶ ¶=

¶ ¶

( )

( )where is an arbi

in

trary function of

,

( )

.

F i or

y y

Mdx y

f

f= +ò ⋯⋯⋯⋯⋯⋯⋯⋯

( )

( )

ii

where is an arbitrary function

integrate with respect to while holding constant,

to obtain,

Differentiate from with respect to , and equate

the result to and

of

i ( )

)

( )

.

4. )

F Ndy ii

x

N y x

N

x

x

F i y

f

f= +ò ⋯⋯⋯⋯⋯⋯⋯⋯

( )

( )

find

Differentiate from with respect to , and equate

the result to and find

ii

( )

)

y

or

F ii x

M

FN

y

FM

x

x

f

f

¢

¢

¶=

¶

¶=

¶

( ) ( )

( ) ( )

( )

( ) ( )

( ) ( )

( )

Integrate with respect to to find

Integrate with respect to to find

i)

ii)

6. Put back the value of in

or

7. The solution of the equati

equation

on is:

5.

,

F

y y dy

x x dx

F x y

y y y

or

x x x

y x

f f

f f

f

f

f

f

f f

¢=

¢

¢

=

¢

=

ò

ò

C


( ) ( )


2 3 2 2 0dy

x ydx

+ + - =

( ) ( )

( )

( ) ( )

( )

( )

Test exactnes

s:

2

2

:

: 2 3 2 2 0

2 3; 2 2

:

2

2

:

2 3

23

2

2

1

3

3

Solution

x dx y dy

M x N y

MM Ny Exacty xN

xF Mdx y

F x dx y

xF x y

Step

Step

Step

F x x y

f

f

f

f

+ + - =

= + = -

ü¶ ïï= ï¶ ¶ï¶ ï = Þýï ¶ ¶¶ ï= ïïï¶ þ= +

= + +

= + +

= + +

òò

( )( )

( ) ( )( ) ( ) ( )

( )

( )( )

( )

Find

Integrate to get

Put back the value in equation:

where

2

2

2 2

2 2

:

0 0 2 2

2 2

: :

2 2

22

22

:

,

3 2

3 2 ,

4

5

6

FN

y

y y

y y

y y

y y dy y dy

yy y A

y y y A

y F

F x y C

x x y y A C

x x y y B B C A

Step

Step

Step

f

f

f f

f f

f

f

f

¶=

¶¢+ + = -

¢\ = -

¢

¢= = -

= - +

= - +

=

+ + - + =

+ + - = = -

ò ò


( ) ( )( )


Given that,

2 3 23 1 6 0

0 3.

x y dx x y y dy

y

- + + - =

=

( ) ( )

( )

( ) ( )

( )

( )

Test exactness:

2 3 2

2 3 2

2

2

2

3

3

:

: 3 1 6 0

3 1; 6

:

3

3

:

3

2

33

3

1

1

Solution

x y dx x y y dy

M x y N x y y

Mx

M Ny Exacty xN

xx

F Mdx

Step

St

y

F x y dx y

x yF x y

F x y x

ep

Step

y

f

f

f

f

- + + - =

= - = + -

ü¶ ïï= ï¶ ¶ï¶ ï = Þýï ¶ ¶¶ ï= ïïï¶ þ= +

= - +

= - +

= - +

òò

( )( )

( )

( ) ( )( ) ( )

( ) ( )

( )

Find

Integrate to get

3 3 2

3 2 3

2

2

3 2 32

4 :

6

6

6

: :

6

63

3

5

2 3

Step

S

FN

y

x y x y y

y x y y x

y y y

y y

y y dy

y y y dy

y y yy A y A

tep

f

f

f

f f

f f

f

f

¶=

¶

¢+ = + -

¢ = + - -

¢\ = - +

¢

¢=

= - +

= - + + = - + +

òò

( )( )

( )

( ) ( ) ( )

Put back the value in equation:

where

When

33 2

33 2

33 2

33 2

:

,

33

3 ,3

: 0 3 :

30 3 0 3 3

39 27

18

3 183

6

7

y F

F x y C

yx y x y A C

yx y x y B B C A

y

C

C

C

yx

Step

St

y

e

x

p

y

f

=

- - + + =

- - + = = -

=

- - + =

- + =

\ =

\ - - + =


( ) ( )Solve the following differential equation:

cos cos sin 0

: cos sin ,

y y x dx x x y dy

Answer xy x y x B B A C

+ - + - =

+ - = = -

2.2 applications of 12.2 applications of 12.2 applications of 12.2 applications of 1stststst ODE ODE ODE ODE

�Newton’s Law of Cooling

�Modeling of Population Growth

2.2.1 newton2.2.1 newton2.2.1 newton2.2.1 newton’’’’s law of coolings law of coolings law of coolings law of cooling

Definition (by word):Definition (by word):Definition (by word):Definition (by word):

“Newton's Law of Cooling states that the rate the rate the rate the rate

of change of the temperature of an object of change of the temperature of an object of change of the temperature of an object of change of the temperature of an object is

proportional to the difference between its the difference between its the difference between its the difference between its

own temperature and the ambient own temperature and the ambient own temperature and the ambient own temperature and the ambient

temperaturetemperaturetemperaturetemperature (the surroundings temperature)”

Mathematically:

sdT

T Tdt

µ -The rate of change

Its own temp.

Ambient temp.

( ) ( ) 0, 0sdT

k T T T Tdt

= - - =

( ) ktsT t T Ae= +

Separable Equ.

Solution: Represent the temperature of an object at time t.

is decreasing

is increasing

s

s

dTT T Cooling

dtdT

T T Heatingdt

> Þ Þ

< Þ Þ

( )


ln

ln

11 :

:

1

ln

2

,

s

s

s

s

s

s

T T kt C

T T kt C

kts

kt Cs

dTk T T

dt

dT k dtT T

dT k dtT

Step

S

T

T T kt C

e e

e e e

T T Ae

te

T Ae A e

p

T

- - +

- -

-

-

= - -

= --

= --

- = - +

=

= ×

- =

\ = + =

ò ò


constant temperature of At 12 noon

A body of an apparent homicide victim was found in a room that

was kept at a . , the

and at 1temp pm erature it was of the body was

Assume that tem th

e

7

7

.8 50

0

F

F

F

°

° °

Based on the given information, determine the ti

Surrou

perature of the body at the time of death was

98

Initi

me of

al tem

de

Temp. after 1 hour

p. (12

nding temp.: =

noon): =

a.6 th.

0 8

7

0

.

0

:

s

Soluti

F

n

T

o

F

F

T °

°

°

Temp. of normal

(1 pm): =

body: = 1 75

98.6

T F

T F

°

°

( ) ( )

( )

( )

When

Whe

n

0

0

1

:

: , 0

: 0, 80, 70 :

80 70

10

70 1

1

0

: 1, 75

2

:

75 70 10

3

s

kts

s

k

kt

k

Step

Step

S

Solution

dTk T T T T

dt

tep

T T Ae

t T T

Ae

A

T e

t T

e

-

-

-

-

= - - =

Þ = +

= = =

= +

\ =

Þ = +

= =

= +

( )

Determine the time of death, when :

hours

The negative sign indicates t

0.6931

0.6931

0.6931

75 70101

ln2

0.6931

70 10

: 98.6

98.6 70 10

98.6 7010

ln 2.86 0.6931

1.51

4

61

k

t

t

t

e

k

k

T e

T

e

e

t

t

Step

-

-

-

-

-=

æ ö÷ç = -÷ç ÷çè ø

\ =

Þ = +

=

= +

-=

= -

\ = -

hat the time is 1.5161 hrs before 12 noon.

Therefore, the time of death is approximately at 10.29 am.

example 2.15example 2.15example 2.15example 2.15

( ) ( )

Temp. after 1.5 hour (10 pm):

Temp. of

Ini

nor

tial t

mal bo

emp. (8.30 p

dy: =

Wh

Surrounding temp.

=

en

m): =

: =

0

1.5

0

98

:

: , 0

: 0, 78.3

74

1

78.

7

2

.6

3

3

s

t

s

ks

Solution

dTk T T T T

dtT

T F

Step

St

T

T F

T Ae

F

p t Te

T F

-

= - - =

Þ = +

°

= =

°

°

°

( )

0

, 73 :

78.3 73

5.3

70 5.3

s

k

kt

T

Ae

A

T e

-

-

=

= +

\ =

Þ = +

( )

( )

When

Determine the time of death, wh en :

1.5

1.5

1.1118

1.1118

1.1118

: 1.5, 74 :

74 73 5.3

74 735.310

ln 1.553

1.1118

73 5.3

: 98.6

98.6 73 5.3

98.6 735.3

ln 4.8

3

3

4

k

k

t

t

t

t T

e

e

k

Step

S

k

T e

Te

e

p

e

t

-

-

-

-

-

= =

= +

-=

æ ö÷ç = -÷ç ÷çè ø

\ =

Þ = +

=

= +

-=

= -

hours

So, death occurred about 7.05pm.

1.1118

1.42

t

t\ = -


A pie is removed from an oven with temperature and

cooled in a room with temperature . In minutes, the pie

has a temperature of Determine the time required to cool

the pie to tempe

350

75 15

150 .

F

F

F

°

°

°

rature of

minutes

80 .

: 46

F

Answer t

°

=

2.2.1 the population growth and 2.2.1 the population growth and 2.2.1 the population growth and 2.2.1 the population growth and decaydecaydecaydecay

Definition (by word):Definition (by word):Definition (by word):Definition (by word):

“The rate of change of the population is

proportional to the existing population”

Mathematically:

( )dPP t

dtµThe rate

of change

( ) ( )

( )

dPb m P t

dtdP

kP tdt

= -

=

ktP Ae=

Separable Equ.

Solution: Represent the temperature of an object at time t.

0

0

k population growth

k decay of radioactive

> Þ

< Þ

56

Example 2.17:population growthExample 2.17:population growthExample 2.17:population growthExample 2.17:population growth

The population of a country is growing at a rate that is

proportional to the population of the country. The

population in 1990 was 20 million and in 2000, the

population was 22 million. Estimate the population in 2020.

( )

( )

Initial (1990): =

Population after 10years (2000): =

Population after 3

0years (2020):

When

0

10

30

0

0

:

20

22

?

: , 0

: 0, 20 :

20

1

20

2

2

0

kt

k

kt

Solution

P m

P m

P

dPP P P

dtStep

S

P Ae

t P

Ae

te

P e

p

A

-

= =

Þ =

= =

=

\ =

Þ =

( )

( )

When

when :

So, in 2020 the population is about 26.62 million.

10

10

0.009531

0.009531 30

3

4

: 10, 22 :

22 20

2220

22ln 10

20

0.009531

20

: 30

20

26.62

k

k

t

t P

e

e

k

k

P e

t

P e

P million

Step

Step

= =

=

=

æ ö÷ç =÷ç ÷çè ø

\ =

Þ =

=

=

=

radioactive decayradioactive decayradioactive decayradioactive decay

� Radioactive substances randomly emit protons, electors, radiation, and they are transformed in another substance.

� It can be seen that the time rate of change of the amount N of a radioactive substances is proportional to the amount of radioactive substance.

� The half-life is the time, k needed to get

� Using the half-life, we get

1 12 2

0 0 12 1

2

1 1 1 1 1ln ln

2 2 2 2

kt ktP e P e kt k

t

æ ö æ ö÷ ÷ç ç= Þ = Þ = Þ\ =÷ ÷ç ç÷ ÷ç çè ø è ø

( ) 01

02

P P=

12

1 1ln

2

0( )

tt

P t P e

æ ö÷ç ÷ç ÷÷çè ø=

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