Sequences and Series - University of Technology, Iraq...Infinite Series Infinite series are sequences of a special kind: those in which the nth-term is the sum of the first n terms

Mathematics Second Year Electrical & Electronic Engineering Department Dr. Atheer Alaa Sabri

Sequences and Series

Sequences of Numbers A sequence of numbers is a function whose domain is the set of positive integers.

Example

...,1...,2,1,0 −n for a sequence whose defining rule is 1−= nan

...,1...,31,

21,1

n for a sequence whose defining rule is

nan

1=

The index is the domain of the sequence. While the numbers in the range of the

sequence are called the terms of the sequence, and the number being called the n

n

na th-

term, or the term with index . n

Example n

nan1+

= then the terms are

...,1...,34,

23,2

321

321 nnaaaa

termntermtermterm

n

thrdndst

+====

and we use the notation { as the sequence . }na na

Example

Find the first five terms of the following:

(a) ⎭⎬⎫

⎩⎨⎧

+−

2312

nn

, (b) ⎭⎬⎫

⎩⎨⎧ −−

3

)1(1n

n

, (c) ⎭⎬⎫

⎩⎨⎧

−−

−+

)!12()1(

121

nx n

n

Solution

(a) 179,

147,

115,

83,

51

(b) 125

2,0,272,0,2

(c) !9

,!7

,!5

,!3

,9753 xxxxx −−

1


Example

Find the nth-term of the following:

(a) 43,

32,

21,0 , (b)

44ln,

33ln,

22ln,0 , (c)

163,

92,

41,0 ,

(d) 2

5

2

4

2

3

52,

42,

32,1,2

Solution

(a) n

nan1−

= , (b) nnan

ln= , (c) 2

1n

nan−

= , (d) 2

2n

an

n =

Convergence of Sequences

The fact that { converges to is written as }na L

Lann=

∞→lim or as Lan → ∞→n

and we call the limit of the sequence { }na . If no such limit exists, we say that { }na

diverges.

From that we can say that

1) (Conv.) Lann=

∞→lim

2) (Div.) ∞=∞→ nn

alim

3) (Div.) ⎩⎨⎧

=∞→

2

1limLL

ann

Also, if and both exist and are finite, then nnaA

∞→= lim nn

bB∞→

= lim

i) { } BAba nnn+=+

∞→lim

ii) { } kAkann=

∞→lim

2


iii) { } BAba nnn⋅=⋅

∞→lim

iv) BA

ba

n

n

n=

⎭⎬⎫

⎩⎨⎧

∞→lim , provided 0≠B and is never 0 nb

Example

Test the convergence of the following:

(a) ⎭⎬⎫

⎩⎨⎧

n1

, (b) { }n)1(1 −+ , (c) { }2n , (d) { }nn −+1 ,

(e) ⎭⎬⎫

⎩⎨⎧

++−

62553

2

2

nnnn

, (f) ⎭⎬⎫

⎩⎨⎧

−−

1243 2

nnn

, (g) ⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

−− 4

7332

nn

, (h) ⎭⎬⎫

⎩⎨⎧

−+−

1034227

25

nnnn

,

(i) ⎭⎬⎫

⎩⎨⎧

n

n

52

, (j) ⎭⎬⎫

⎩⎨⎧

nenln

Solution

(a) 01lim =⎟⎠⎞

⎜⎝⎛

∞→ nn (Conv.)

(b) (Div.) ( )⎩⎨⎧

=−+=−+∞→∞→ evenn

oddnn

n

n

n 20

)1(lim1)1(1lim

(c) ( ) ∞=∞→

2lim nn

(Div.)

(d) ( ) ( ) ⎟⎠⎞

⎜⎝⎛

++−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛++++

×−+=−+∞→∞→∞→ nn

nnnnnnnnnn

nnn 11lim

111lim1lim

0111lim =

∞+∞=⎟

⎠⎞

⎜⎝⎛

++=

∞→ nnn (Conv.)

3


(e) 53

625

53

lim625

53lim

222

2

22

2

2

2

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛++

−∞→∞→

nnn

nn

nn

nn

nnnn

nn (Conv.)

(f) ∞==⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

∞→∞→ 03

12

43

lim1243lim

22

22

2

2

nnn

nn

nn

nnn

nn (Div.)

(g) 8116

32

7332lim

44

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

−−

∞→ nn

n (Conv.)

(h) 01013

42

lim103

42lim75

52

27

25

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−+

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

−∞→∞→

nn

nnnn

nnnn

(Conv.)

(i) ∞=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∞→∞→ 5

2ln.2lim52lim

n

n

n

n n (Div.)

(j) 01.1lim/1limlnlim =

∞=⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

∞→∞→∞→ nnnnnn enen

en

(Conv.)

Example

Prove the following limits

(a) 0lnlim =⎟⎠⎞

⎜⎝⎛

∞→ nn

n, (b) ( ) 1lim =

∞→

n

nn , (c) ( ) 1lim /1 =

∞→

n

nx , )0( >x

(d) xn

ne

nx

=⎟⎠⎞

⎜⎝⎛ +

∞→1lim (any x ), (e) 0

!lim =⎟⎟

⎠

⎞⎜⎜⎝

⎛∞→ n

xn

n (any x )

4


Solution

(a) 010

1/1limlnlim ==⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

∞→∞→

nnn

nn

(b) Let , then nn na /1= 0ln1lnln /1 →== n

nna n

n ,

So, 1lim 0ln/1 =→=∞→

een nan

n

(c) Let , then nn xa /1= 0ln1lnln /1 →== x

nxa n

n ,

So, 1lim 0ln/1 =→=∞→

eex nan

n

(d) Let n

n nxa ⎟⎠⎞

⎜⎝⎛ += 1 , then

⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

nxn

nxa

n

n 1ln.1lnln

So, ( )

2

2

/1/1

1

lim/1

/1lnlim1ln.limn

nx

nxn

nxnxn

nnn −

⎟⎠⎞

⎜⎝⎛−⋅⎟

⎠⎞

⎜⎝⎛+=

+=⎟

⎠⎞

⎜⎝⎛ +

∞→∞→∞→

xnx

xn

=+

=∞→ /1

lim ,

Thus, xan

n

eeanx

n →==⎟⎠⎞

⎜⎝⎛ + ln1

(e) 0...321

lim!

lim =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∞→∞→ n

xxxxnx

n

n

n

5


Exercises on Sequences

Find the values of , , and for the following sequences 1a 2a 3a 4a

1) 2

1n

nan−

= 2) !

1n

an = 3) 12

)1( 1

−−

=+

na

n

n

4) nna )1(2 −+= 5)

122+= n

n

na 6) n

n

na2

12 −=

Find a formula for the term of the following sequences thn

1) ,....1,1,1,1,1 −− 2) ,....1,1,1,1,1 −−− 3) ,...25,16,9,4,1 −−

4) ,...251,

161,

91,

41,1 −− 5) ,...24,15,8,3,0 6) ,...1,0,1,2,3 −−−

7) ,...17,13,9,5,1 8) ,...18,14,10,6,2 9) ,...1,0,1,0,1

Which of the following sequences converge and which diverge?

1) nna )1.0(2 += Ans. Converges, 2

2) nnan 21

21+−

= Ans. Converges, − 1

3) 34

4

851

nnnan +

−= Ans. Converges, − 5

4) 1

122

−+−

=n

nnan Ans. Diverges

5) nna )1(1 −+= Ans. Diverges

6


6) ⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ +

=nn

nan11

21

Ans. Converges, 21

7) 12

)1( 1

−−

=+

na

n

n Ans. Converges, 0

8) 1

2+

=n

nan Ans. Converges, 2

9) ⎟⎠⎞

⎜⎝⎛ +=

nan

12

sin π Ans. Converges, 1

10) n

nansin

= Ans. Converges, 0

11) nnna2

= Ans. Converges, 0

12) nnan

)1ln( += Ans. Converges, 0

13) nna /18= Ans. Converges, 1

14) n

n na ⎟

⎠⎞

⎜⎝⎛ +=

71 Ans. Converges, 7e

15) nn na 10= Ans. Converges, 1

16) n

n na

/13⎟⎠⎞

⎜⎝⎛= Ans. Converges, 1

17) nn nna /1

ln= Ans. Diverges

18) n nn na 4= Ans. Converges, 4

7


20) nnna 610!

= Ans. Diverges

21) )/(ln11 n

n na ⎟

⎠⎞

⎜⎝⎛= Ans. Converges, 1−e

22) n

n nna ⎟

⎠⎞

⎜⎝⎛

−+

=1313

Ans. Converges, 3/2e

23) nn

n nxa

/1

12 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

= , 0>x Ans. Converges, x ( )0>x

24) !2

63n

a n

nn

n ××

= − Ans. Converges, 0

25) )tanh(nan = Ans. Converges, 1

26) nn

nan1sin

12

2

−= Ans. Converges,

21

27) )(tan 1 nan−= Ans. Converges,

2π

28) n

n

na21

31

+⎟⎠⎞

⎜⎝⎛= Ans. Converges, 0

29) ( )nnan

200ln= Ans. Converges, 0

30) nnnan −−= 2 Ans. Converges, 21

31) ∫=n

n dxxn

a1

11 Ans. Converges, 0

8


Infinite Series Infinite series are sequences of a special kind: those in which the nth-term is the

sum of the first n terms of a related sequence.

Example

Suppose that we start with the sequence

,...641,

321,

161,

81,

41,

21,1

If we denote the above sequence as , and the resultant sequence of the series as ,

then

na ns

111 == as ,

23

211212 =+=+= aas ,

47

41

2113213 =++=++= aaas ,

as the first three terms of the sequence { }ns .

When the sequence { is formed in this way from a given sequence { by the

rule

}ns }na

∑=

=+++=n

kknn aaaas

121 ...

the result is called an Infinite Series.

The number is called the n∑=

=n

kkn as

1

th partial sum of the series.

Instead of { , we usually write or simply }ns ∑∞

=1nna ∑ na .

The series is said to converge to a number if and only if ∑ na L

9


∑=∞→∞→

==n

kknnn

asL1

limlim

in which case we call the sum of the series and write L

Lan

n =∑∞

=1 or Laaa n =++++ ......21

If no such limit exists, the series is said to diverge.

Geometric Series A series of the form

...... 132 +++++++ −narararara

is called a Geometric Series. The ratio of any term to the one before it is r . If 1<r ,

the geometric series converges to )1/( ra − . If 1≥r , the series diverges unless 0=a .

If , the series converges to 0 . 0=a

Example

Geometric series with 91

=a and 31

=r .

61

)3/1(19/1...

31

311

91...

811

271

91

2 =−

=⎟⎠⎞

⎜⎝⎛ +++=+++

Geometric series with and 4=a21

−=r .

⎟⎠⎞

⎜⎝⎛ −+−+−=−+−+− ...

161

81

41

2114...

41

21124

38

)2/1(14

=+

=

10


Example

Determine whether each series converges or diverges. If it converges, find its sum.

(a) n

n∑∞

=⎟⎠⎞

⎜⎝⎛

0 32

, (b) n

n∑∞

=⎟⎠⎞

⎜⎝⎛

1 23

, (c) ∑∞

=⎟⎠⎞

⎜⎝⎛

1 3cos2

n

nπ, (d) ∑

∞

=⎟⎠⎞

⎜⎝⎛

0 4tan

n

nπ, (e) ∑

∞

=

−1 4

)1(5n

n

n

Solution

(a) Since the series is a geometric series with 132<=r , so the series is convergent with

a sum of 3)3/2(1

1=

−

(b) Since the series is a geometric series with 123>=r , so the series is divergent.

(c) 2/13/cos =π . This is a geometric series with first term and the ratio 11 =a2/1=r ; so the series converges and its sum is 2)

211/(1 =− .

(d) 14/tan =π . This is a geometric series with 1=r , so the series diverges.

(e) This is a geometric series with first term 4/51 −=a and ratio 4/1−=r . So the series

converges and its sum is 1)4/1(1

4/5−=

+− .

Test Convergence of Series with Non-negative Terms

1) The nth- Term Test

If , or if fails to exist, then diverges. 0lim ≠∞→ nn

a nna

∞→lim ∑

∞

=1nna

If converges, then . ∑∞

=1nna 0→na

If , then the test fails. 0lim =∞→ nn

a

11


From the above, it can not be concluded that if then ∑ converges.

The series may diverge even though . Thus is a necessary

but not a sufficient condition for the series to converge.

0→na∞

=1nna

∑∞

=1nna 0→na 0lim =

∞→ nna

∑∞

=1nna

Examples

∑∞

=1

2

nn diverges because ∞→2n ,

∑∞

=

+1

1n n

n diverges because 011

≠→+n

n,

∑∞

=

+−1

1)1(n

n diverges because does not exist, 1)1(lim +

∞→− n

n

∑∞

= +1 52n nn

diverges because 021

52lim ≠=

+∞→ nn

n,

∑∞

=1

1n n

can not be tested by the nth-term test for divergence because 01→

n.

2) The Integral Test Let the function )(xfy = , obtained by introducing the continuous variable x in

place of the discrete variable n in the nth-term of the positive series ∑ , then ∞

=1nna

...

)(1 Conv

DivDiv

cdxxf

⎪⎩

⎪⎨

⎧

∞<<∞−∞−∞+

=∫∞

12


Example

Prove that, for the -series, if is a real constant, the series p p

∑∞

=

+++++=1

...1...31

21

111

nppppp nn

converges if and diverges if 1>p 1≤p .

Solution

To prove this, let

pxxf 1)( =

Then, if 1>p , we have

11

1lim

1 1

1

−=

+−=∫

∞ +−

∞→

−

ppxdxx

bp

b

p

which is finite. Hence, the -series converges if . p 1>p

If 1=p , which is called a harmonic series, we have

...1...31

211 +++++

n,

and the integral test is

+∞==∞→

∞−∫

b

bxdxx 1

1

1 lnlim

which diverges.

Finally, for 1<p , then the terms of the series are greater than the corresponding

terms of the divergent harmonic series. Hence the -series diverges for . p 1<p

Thus, we have a convergence for 1>p , but divergence for 1≤p .

13


Example

Test the convergence of

(a) ∑∞

=1

1n

ne, (b) ∑

∞

=22)(ln

1n nn

Solution

(a) e

eeedxe xx 1)( 1

11

=−−=−= −∞−∞−∞

−∫ (Conv.)

(b) 2ln

12ln

11ln

1)(ln

/1)(ln

122

22

2 =+∞−

=−

==∞∞∞

∫∫ xdx

xxdx

xx (Conv.)

3) The Ratio Test

Let be a series with positive terms, and suppose that ∑ na

ρ=+

∞→n

n

n aa 1lim

Then

The series converges if 1<ρ ,

The series diverges if 1>ρ ,

The series may converge or it may diverge if 1=ρ . (Test fails)

The Ratio Test is often effective when the terms of the series contain factorials of

expressions involving or expressions raised to a power involving . n n

14


Example

Test the following series for convergence or divergence , using the Ratio Test.

(a) ∑∞

=1 )!2(!!

n nnn

, (b) ∑∞

=1 )!2(!!4

n

n

nnn

, (c) ∑∞

=

+0 3

52n

n

n

, (d) ∑∞

=1 3!

nn

n, (e) ∑

∞

=1 !n

n

nn

Solution

(a) If )!2(!!

nnnan = , then

)!22()!1()!1(

1 +++

=+ nnnan and

)12)(22()1)(1(

)!2)(12)(22(!!)!2()!1()!1(1

++++

=++

++=+

nnnn

nnnnnnnn

aa

n

n

141

241

<→++

=nn

(Conv.)

(b) If )!2(

!!4n

nnan

n = , then )!22(

)!1()!1(4 1

1 +++

=+

+ nnna

n

n and

)12)(22()1)(1(4

!!4)!2(

)!2)(12)(22()!1()!1(4 1

1

++++

=×++

++=

++

nnnn

nnn

nnnnn

aa

n

n

n

n

112)1(2→

++

=nn

(Test fails)

(c) If n

n

na3

52 += , then 1

1

1 352

+

+

+

+= n

n

na and

5252

31

3/)52(3/)52( 111

1

++

×=++

=+++

+n

n

nn

nn

n

n

aa

132

12

31

251252

31

<=×→⎟⎟⎠

⎞⎜⎜⎝

⎛×+×+

×= −

−

n

n

(Conv.)

15


(d) If nnna3

!= , then 11 3

)!1(++

+= nn

na and

13

1!

33

)!1(1

1 >∞→+

=×+

= ++ n

nn

aa n

nn

n (Div.)

(e) If !n

nan

n = , then )!1(

)1( 1

1 ++

=+

+ nna

n

n and

n

n

n

n

n

n

nnnnnn

nn

nn

aa

!)1(!)1()1(!

)!1()1( 1

1

+++

=×++

=+

+

17.2111)1( 1 >=→⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +

=+

= enn

nn

n nn

n

n

(Div.)

4) The nth Root Test

Let be a series with for and suppose that ∑ na 0≥na 0nn >

ρ→nna

Then

The series converges if 1<ρ .

The series diverges if 1>ρ .

The test is not conclusive if 1=ρ .

Example

Test the convergence of the following series using the nth Root Test.

(a) ∑∞

=1

1n

nn, (b) ∑

∞

=12

2n

n

n, (c) ∑

∞

=⎟⎠⎞

⎜⎝⎛ −

1

11n

n

n, (d) ∑

∞

=⎟⎠⎞

⎜⎝⎛

+1

2

1n

n

nn

, (e) ∑∞

=⎟⎠⎞

⎜⎝⎛

+1 12

n

n

nn

16


Solution

(a) 1011<→=

nnn

n (Conv.)

(b) ( ) 12122222222 >=→==

nnn

n

nnn (Div.)

(c) 11111 →⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −

nnn

n

(Test fails)

(d) 17.2

11/11

1111

22

<=→⎟⎠⎞

⎜⎝⎛+

=⎟⎠⎞

⎜⎝⎛

+=⎟

⎠⎞

⎜⎝⎛

+=⎟

⎠⎞

⎜⎝⎛

+ ennn

nn

nn nn

nn

n

n

(Conv.)

(e) 121

21

2>→

+=⎟

⎠⎞

⎜⎝⎛

+ nn

nn

n

n

(Div.)

Exercises on Series Find the sum of the following series

1) ∑∞

=

−0 4

)1(n

n

n

Ans.54

2) ∑∞

=1 47

nn Ans.

37

3) ∑∞

=⎟⎠⎞

⎜⎝⎛ +

0 31

25

nnn Ans.

223

4) ∑∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

0 5)1(

21

nn

n

n Ans.6

17

17


5) ∑∞

= +−1 )14)(34(4

n nn Ans. 1

6) ∑∞

= +−122 )12()12(

40n nn

n Ans. 5

7) ∑∞

=⎟⎠⎞

⎜⎝⎛

+−

1 111

n nn Ans. 1

8) ∑∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+1 )1ln(

1)2ln(

1n nn

Ans.2ln

1−

Which of the following series converges and which diverges? Find the sum

of the convergent series.

1) n

n∑∞

=⎟⎠⎞

⎜⎝⎛

0 21

Ans. Converges, 22+

2) ∑∞

=

+−1

1

23)1(

nn

n Ans. Converges, 1

3) ∑∞

=0)cos(

nnπ Ans. Diverges

4) ∑∞

=

−

0

2

n

ne Ans. Converges, 12

2

−ee

5) ∑∞

=1 102

nn Ans. Converges,

92

6) ∑∞

=

−0 3

12n

n

n

Ans. Converges, 23

18


7) ∑∞

=0 1000!

nn

n Ans. Diverges

8) ∑∞

=⎟⎠⎞

⎜⎝⎛

+1 1ln

n nn

Ans. Diverges

9) ∑∞

=⎟⎠⎞

⎜⎝⎛

0n

neπ

Ans. Converges, e−π

π

Which of the following series converges and which diverges?

1) ∑∞

=1 101

nn Ans. Converges (Geometric)

2) ∑∞

= +1 1n nn

Ans. Diverges ( nth-term test)

3) ∑∞

=1

3n n

Ans. Diverges ( p-series)

4) ∑∞

=

−1 8

1n

n Ans. Converges (Geometric)

5) ∑∞

=2

lnn n

n Ans. Diverges (Integral Test)

6) ∑∞

=1 32

nn

n

Ans. Converges (Geometric)

7) ∑∞

= +−

0 12

n n Ans. Diverges (Integral Test)

19


8) ∑∞

= +1 12

n

n

n Ans. Diverges ( nth-term test)

9) ∑∞

=2 lnn nn


10) ( )∑∞

=1 2ln1

nn Ans. Diverges (Geometric)

11) ( )

( )∑∞

= −32 1lnln

/1n nn

n Ans. Converges (Integral Test)

12) ∑∞

=1

1sinn n


13) ∑∞

= +121n

n

n

ee

Ans. Converges (Integral Test)

14) ∑∞

=

−

+12

1

1tan8

n nn

Ans. Converges (Integral Test)

15) ∑∞

= −1 132

n nn


16) ∑∞

=1

2

2nn

n Ans. Converges (Ratio Test)

17) ∑∞

=

−

1!

n

nen Ans. Diverges (Ratio Test)

18) ∑∞

=1

10

10nn

n Ans. Converges (Ratio Test)

19) ∑∞

=⎟⎠⎞

⎜⎝⎛ −

1

31n

n


20


20) ( )( )∑

∞

=

++1 !

21n n

nn Ans. Converges (Ratio Test)

21) ( )∑

∞

=

+1 3!!3

!3n

nnn

Ans. Converges (Ratio Test)

22) ( )∑∞

= +1 !12!

n nn

Ans. Converges (Ratio Test)

23) ( )∑∞

=2 lnnnn

n Ans. Converges (Root Test)

24) ( )( )∑

∞

=12

!n

n

n

nn

Ans. Diverges (Root Test)

25) ( )∑∞

=12

2nn

nn Ans. Converges (Root Test)

21


Alternating Series A series in which the terms are alternately positive and negative.

Example

...)1(...51

41

31

211

1

+−

+−+−+−+

n

n

...2

4)1(...81

41

2112 +

−++−+−+− n

n

...)1(...654321 1 +−++−+−+− + nn

The Convergence Test of Alternating Series The series

...)1( 43211

1 +−+−=−∑∞

=

+ uuuuun

nn

converges if all three of the following conditions are satisfied:

1) The ’s are all positive. nu

2) for all 1+≥ nn uu Nn ≥ , for some integer N .

3) . 0→nu

Example

The alternating harmonic series

∑∞

=

+ +−+−=−1

1 ...41

31

2111)1(

n

n

n

satisfies the three requirements of convergence; it therefore converges.

22


Absolute Convergence A series ∑ converges absolutely (is absolutely convergent) if the

corresponding series of absolute values,

na

∑ na , converges, i.e.,

If ∑∞

=1nna converges, then ∑ converges.

∞

=1nna

Example

The geometric series ...81

41

211 +−+− converges absolutely because the

corresponding series of absolute values ...81

41

211 ++++ converges .

Conditional Convergence

A series that converges but does not converge absolutely converges conditionally.

Example

The alternating harmonic series ...41

31

211 +−+− does not converge absolutely. The

corresponding series of absolute values ...41

31

211 ++++ is the divergent harmonic

series.

Power Series A power series about 0=x is a series of the form

......2210

0+++++=∑

∞

=

nn

n

nn xcxcxccxc

A power series about ax = is a series of the form

23


...)(...)()()( 2210

0+−++−+−+=−∑

∞

=

nn

n

nn axcaxcaxccaxc

in which the center a and the coefficients are constants. ,...,...,,, 210 ncccc

Example

The series ∑ is a geometric series with first term 1 and ratio ∞

=0n

nx x . It converges to

......11

1 2 +++++=−

nxxxx

for 1<x

Convergence of Power Series

If the power series converges for ...2210

0+++=∑

∞

=

xaxaaxan

nn 0≠= cx , then it

converges absolutely for all x with cx < . If the series diverges for , then it

diverges for all

dx =

x with dx > .

The test of power series is done using the Ratio Test.

⎪⎩

⎪⎨

⎧

=><

=+

∞→

FailsDiv

Conv

cc

n

n

n

1.1.1

lim 1 ρ

Notes:

Use the Ratio Test to find the interval where the series converges absolutely.

If the interval of absolute convergence is finite, test the convergence or

divergence at each endpoint. Use the integral test or the Alternating Series Test

for endpoints.

24


If the interval of absolute convergence is Rax <− , the series diverges for

Rax >− (it does not even converge conditionally), because the nth-term does

not approach zero for those values of x .

Example

For what values of x do the following power series converge?

(a) ...32

)1(32

1

1 −+−=−∑∞

=

− xxxnx

n

nn , (b) ...

5312)1(

53

1

121 −+−=

−−∑

∞

=

−− xxx

nx

n

nn

(c) ...!3!2

1!

32

0++++=∑

∞

=

xxxnx

n

n

, (d) ...!3!21! 32

0++++=∑

∞

=

xxxxnn

n

Solution

(a) xxn

nxn

nx

uu

n

n

n

n →+

=×+

=+

+

11

11 .

The series converges absolutely for 1<x . It diverges if 1>x because the nth-term

does not converge to zero. At 1=x , we get the alternating harmonic series

, which converges. At ...4/13/12/11 +−+− 1−=x , we get ...4/13/12/11 −−−−− ,

the negative of the harmonic series; it diverges. So, the series converges for

11 ≤<− x and diverges elsewhere.

(b) 2212

121

121212

12xx

nn

xn

nx

uu

n

n

n

n →+−

=−

×+

= −

++ .

The series converges absolutely for 12 <x . It diverges for because the n12 >x th-

term does not converge to zero. At 1=x , the series becomes ...7/15/13/11 +−+− ,

which converges because it satisfies the three conditions of convergence of

alternating series. It also converges at 1−=x because it is again an alternating series

25


that satisfies the conditions for convergence. The value at 1−=x is the negative of

the value at 1=x . So, the series converges for 11 ≤≤− x and diverges elsewhere.

(c) 01

!)!1(

11 →

+=⋅

+=

++

nx

xn

nx

uu

n

n

n

n for every x .

The series converges absolutely for all x .

(d) ∞→+=+

=+

+ xnxn

xnu

un

n

n

n )1(!)!1( 1

1 unless 0=x .

The series diverges for all values of x except 0=x .

Exercises on Alternating & Power Series

Which of the following series converges and which diverges?

1) ( )∑∞

=

+−1

21 11

n

n

n Ans. Converges

2) ( )∑∞

=

+⎟⎠⎞

⎜⎝⎛−

1

1

101

n

nn n

Ans. Diverges, ∞→an

3) ( )∑∞

=

+−2

1

ln11

n

n

n Ans. Converges

4) ( ) ( )∑∞

=

+−2

21

lnln1

n

n

nn

Ans. Diverges, 21

→na

5) ( )∑∞

=

+

++

−1

1

111

n

n

nn

Ans. Converges

26


Which of the following series converges absolutely, conditionally, and

which diverges?

1) ( ) ( )∑∞

=

+−1

1 1.01n

nn Ans. Converges absolutely

2) ( )∑∞

=

−1

11n

n

n Ans. Converges conditionally

3) ( )∑∞

=

+

++

−1

1

531

n

n

nn

Ans. Diverges, 1→na

4) ( )∑∞

=⎟⎠⎞

⎜⎝⎛−

1

2

321

n

nn n Ans. Converges absolutely

5) ( )∑∞

=

−

+−

12

1

1tan1

n

n

nn

Ans. Converges absolutely

6) ( )∑∞

= +−

1 11

n

n

nn

Ans. Diverges, 1→na

7) ∑∞

=

−1 !

)100(n

n

n Ans. Converges absolutely

8) ∑∞

=1

)cos(n nn

nπ Ans. Converges absolutely

9) ( )∑

∞

=

+−1 )2(

)1(1n

n

nn

nn

Ans. Converges absolutely

10) ( )∑∞

=

−1 !2

)!2(1n

nn

nnn

Ans. Diverges, ∞→na

27


Find the interval of convergence for the following series

1) ∑∞

=0n

nx Ans. 11 <− < x

2) ( ) ( )nn

n x 1410

+−∑∞

=

Ans. 021

<<− x

3) ( )∑

∞

=

−0 10

2n

n

nx Ans. 128 <− < x

4) ∑∞

= +0 2n

n

nnx

Ans. 11 <− < x

5) ∑∞

=1 3nn

n

nnx

Ans. 33 ≤− ≤ x

6) ∑∞

=

−0 !

)1(n

nn

nx

Ans. For all x

7) ∑∞

=

+

0

12

!n

n

nx

Ans. For all x

8) ∑∞

= +02 3n

n

nx

Ans. 11 <− ≤ x

9) ( )∑

∞

=

+0 5

3n

n

nxn Ans. 28 <− < x

10) ∑∞

=0 3nn

nxn Ans. 33 <− < x

11) ∑∞

=⎟⎠⎞

⎜⎝⎛ +

1

11n

nn

xn

Ans. 11 <x − <

28


12) ∑∞

=1n

nn xn Ans. 0=x

13) ( )∑

∞

=

+ +−1

1

22)1(

nn

nn

nx

Ans. 04 ≤x − <

Find the interval of convergence and the sum within this interval for the

following series

1) ( )∑

∞

=

−0

2

41

n

n

nx

Ans. 31 <<− x , 2234

xx −+

2) ∑∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

01

2n

nx

Ans. 160 << x , x−4

2

3) ∑∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛ +0

2

31

n

nx

Ans. 22 <<− x , 223

x−

29


Taylor Series & Maclaurin Series

Let f be a function with derivatives of all orders throughout some interval

containing as an interior point. Then the Taylor Series generated by a f at ax = is

...)(!

)(...)(!2

)())(()()(!

)( )(2

0

)(

+−++−′′

+−′+=−∑∞

=

nn

k

kk

axn

afaxafaxafafaxk

af

The Maclaurin Series generated by f is

...!

)0(...!2

)0()0()0(!

)0( )(2

0

)(

+++′′

+′+=∑∞

=

nn

k

kk

xn

fxfxffxk

f

which is a Taylor series generated by f at 0=x .

Example

Find the Taylor series and the interval of convergence for the following functions

(a) xxf /1)( = at 2=x , (b) )ln()( xxf = at 1=x .

Solution

(a) We need to find ),...2(),2(),2( fff ′′′ . Taking derivatives we get

1)( −= xxf , 21)2( =f ,

2)( −−=′ xxf , 221)2( −=′f ,

3!2)( −=′′ xxf , 321

!2)2(=

′′f,

4!3)( −−=′′′ xxf , 421

!3)2(

−=′′′f

,

30


•••

•••

)1()( !)1()( +−−= nnn xnxf , 1

)(

2)1(

!)2(

+

−= n

nn

nf

.

The Taylor series is

...)2(!

)2(...)2(!2

)2()2)(2()2()()(

2 +−++−′′

+−′+= nn

xn

fxfxffxf

...2

)2()1(...2

)2(2

)2(211

13

2

2 +−

−+−−

+−

−= +n

nn xxx

x

This is a geometric series with first term and ratio 2/1 2/)2( −−= xr . It

converges absolutely for 22 <−x or 40 << x .

(b) )ln()( xxf = , 0)1( =f ,

x

xf 1)( =′ , 1)1( =′f ,

2

1)(x

xf −=′′ , 1)1( −=′′f ,

3

2)(x

xf =′′′ , 2)1( =′′′f ,

4)4( 6)(

xxf −= , , 6)1()4( −=f

31


•••

•••

nnn

xnxf )!1()1()( 1)( −

−= + , , )!1()1()1( 1)( −−= + nf nn

...)1(!

)1(...)1(!2

)1()1)(1()1()()(

2 +−++−′′

+−′+= nn

xn

fxfxffxf

...)1(!

)!1()1(...!4

)1(6!3

)1(2!2

)1()1(0)ln(1432

+−−−

++−

−−

+−

−−+=+

nn

xn

nxxxxx

...)1()1(...4

)1(3

)1(2

)1()1(1432

+−−

++−

−−

+−

−−=+

nn

xn

xxxx

∑∞

=

+

−−

=1

1

)1()1()ln(n

nn

xn

x

111

)1(lim)1(1

)1(lim1

<−=+

−=−

×+−

∞→

+

∞→x

nnx

xn

nx

nn

n

n

So, 111 <−<− x or 20 << x . For 0=x , we get ...4/13/12/11 −−−−−

which diverges because it is the negative of the harmonic series. While, for 2=x , we

get which converges because it is an alternating series that

satisfies the three conditions of convergence of alternating series. So, the region of

convergence will be

...4/13/12/11 +−+−

20 ≤< x .

32


Example

Find the Maclaurin series generated by the following functions

(a) , (b) , (c) xe )cosh(x )sinh(x

Solution

(a) ⇒ xexf =)( 1)0( =f ,

xexf =′ )( ⇒ 1)0( =′f ,

xexf =′′ )( ⇒ 1)0( =′′f ,

xexf =′′′ )( ⇒ 1)0( =′′′f ,

•••

•••

xn exf =)()( ⇒ 1)0()( =nf

∑∞

=

=++++=0

2

!!...

!21

n

nnx

nx

nxxxe .

To find the interval of convergence, we have

101

lim!)!1(

limlim1

1 <=+

=×+

=∞→

+

∞→

+

∞→ nx

xn

nx

aa

nn

n

nn

n

n.

So, the series is convergent for all values of x .

(b) 2

)cosh(xx eex

−+=

⎥⎦

⎤⎢⎣

⎡ −+= ∑∑

∞

=

∞

= 00 !)(

!21

n

n

n

n

nx

nx

⎥⎦

⎤⎢⎣

⎡+−+−+++++= ...)

!3!21(...)

!3!21(

21 3232 xxxxxx

33


⎥⎦

⎤⎢⎣

⎡+++= ...

!42

!222

21 42 xx

)!2(...

!4!21...

!4!212

21 24242

nxxxxx n

++++=⎥⎦

⎤⎢⎣

⎡+++×=

∑∞

=

=0

2

)!2()cosh(

n

n

nxx


10)1)(22(

lim)!2()!22(

limlim2

2

221 <=

++=×

+=

∞→

+

∞→

+

∞→ nnx

xn

nx

aa

nn

n

nn

n

n.


(c) ))(cosh()sinh( xfx ′=

⎟⎟⎠

⎞⎜⎜⎝

⎛++++′=

)!2(...

!4!21

242

nxxxf

n

)!2(

2...!6

6!4

4!2

201253

nnxxxx n−

+++++=

)!12(...

!5!3)sinh(

1253

−++++=

−

nxxxxx

n

∑∞

=

−

−=

1

12

)!12()sinh(

n

n

nxx , or ∑

∞

=

+

+=

0

12

)!12()sinh(

n

n

nxx


10)2)(12(

lim)!12()!12(

limlim2

12

121 <=

+=

−×

+=

∞→−

+

∞→

+

∞→ nnx

xn

nx

aa

nn

n

nn

n

n.


34


Exercises on Taylor Series

Find Maclaurin series for the following functions

1) xe− Ans. ∑∞

=

−0 !

)(n

n

nx

2) x+1

1 Ans. ( )∑

∞

=

−0

1n

nn x

3) )3sin( x Ans. ∑∞

=

++

+−

0

1212

)!12(3)1(

n

nnn

nx

4) )cos(7 x− Ans.( )∑

∞

=

−0

2

)!2(17

n

nn

nx

5) 452 34 +−− xxx Ans. 452 34 +−− xxx

Find the Taylor series generated by f at ax = for the following functions

1) 42)( 3 +−= xxxf , 2=a Ans. ( ) ( ) ( )32 2262108 −+−+−+ xxx

2) 1)( 24 ++= xxxf , 2−=a Ans.( ) ( ) ( )

( )432

2

2822523621

++

+−+++−

x

xxx

3) 2

1)(x

xf = , 1=a Ans. ( ) ( )( )∑∞

=

−+−0

111n

nn xn

4) xexf =)( , 2=a Ans. ( )∑∞

=

−0

2

2!n

nxne

35


Find Maclaurin series for the following functions

1) xe 5− Ans. ∑∞

=

−0 !

)5(n

n

nx

2) )sin(5 x− Ans. ∑∞

=

+

+−−

0

12

)!12()()1(5

n

nn

nx

3) 1cos +x Ans.( ) ( )∑

∞

=

+−0 )!2(

11n

nn

nx

4) xxe Ans. ∑∞

=

+

0

1

!n

n

nx

5) xx cos12

2

+− Ans. ∑∞

=

−2

2

)!2()1(

n

nn

nx

6) )cos( xx ⋅π Ans.( )∑

∞

=

+−0

122

)!2(1

n

nnn

nxπ

7) )(cos2 x Ans.( )∑

∞

= ⋅−

+1

2

)!2(2)2(11

n

nn

nx

8) x

x21

2

− Ans. ∑

∞

=0

2 )2(n

nxx

9) ( )211x−

Ans. ∑∞

=

−

1

1

n

nnx

36


37

Partial Derivatives Functions of Independent Variables

Suppose D is a set of n -tuples of real numbers ),...,,( 21 nxxx . A real-valued

function f on D is a rule that assigns a unique (single) real number

),...,,( 21 nxxxfw =

to each element in D . The set D is the function’s domain. The set of w-values taken

on by f is the function’s range. The symbol w is the dependent variable of f , and f

is said to be a function of the n independent variables 1x to nx . We also call the jx ’s

the function’s input variables and call w the function’s output variable.

Example

The value of 222),,( zyxzyxf ++= at the point )4,0,3( is

525)4()0()3()4,0,3( 222 ==++=f

Domains and Ranges

Example

Function Domain Range

2xyw −= 2xy ≥ [ )∞,0

xyw 1= 0≠xy ( ) ( )∞∪∞− ,00,

xyw sin= Both yx & ),( +∞−∞

or Entire plane [ ]1,1 +−


38

Function Domain Range

222 zyxw ++= Entire space [ )∞,0

222

1zyx

w++

= )0,0,0(),,( ≠zyx ( )∞,0

zxyw ln= Half-space 0>z ( )∞∞− ,

Partial Derivatives

The partial derivative of ),( yxf with respect to x at the point ),( 00 yx is

hyxfyhxffyxf

dxd

xf

hxyx

),(),(lim),( 0000

00),( 00

−+===

∂∂

→

provided the limit exists.

The partial derivative of ),( yxf with respect to y at the point ),( 00 yx is

hyxfhyxffyxf

dyd

yf

hyyx

),(),(lim),( 0000

00),( 00

−+===

∂∂

→


Example

Find the values of xf ∂∂ / and yf ∂∂ / at the point )5,4( − if

13),( 2 −++= yxyxyxf

Solution

To find xf ∂∂ / , we treat y as a constant and differentiate with respect to x

yxyxyxyxxx

f 320032)13( 2 +=−++=−++∂∂

=∂∂

The values of xf ∂∂ / at )5,4( − is 7)5(3)4(2 −=−+


39

To find yf ∂∂ / , we treat x as a constant and differentiate with respect to y

130130)13( 2 +=−++=−++∂∂

=∂∂ xxyxyx

yyf

The values of yf ∂∂ / at )5,4( − is 131)4(3 =+

Example

Find yf ∂∂ / if )sin(),( xyyyxf =

Solution

We treat x as a constant and f as a product of y and )sin(xy

( ) )()sin()sin()sin( yy

xyxyy

yxyyyy

f∂∂

+∂∂

=∂∂

=∂∂

( ) )sin()cos()sin()()cos( xyxyxyxyxyy

xyy +=+∂∂

=

Example

Find xf and yf if xy

yyxfcos

2),(+

=

Solution

We treat f as a quotient

2)cos(

)cos(2)2()cos(

cos2

xy

xyx

yyx

xy

xyy

xfx +

+∂∂

−∂∂

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+∂

∂=

22 )cos(sin2

)cos()sin(2)0)(cos(

xyxy

xyxyxy

+=

+−−+

=


40

2)cos(

)cos(2)2()cos(

cos2

xy

xyy

yyy

xy

xyy

yf y +

+∂∂

−∂∂

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+∂

∂=

22 )cos(cos2

)cos()1(2)2)(cos(

xyx

xyyxy

+=

+−+

=

Example

Find xz ∂∂ / for yxzyz +=− ln

Solution

We differentiate both sides of the equation with respect to x , holding y constant

and treating z as a differentiable function of x

)()()(ln)( yx

xx

zx

yzx ∂

∂+

∂∂

=∂∂

−∂∂

011+=

∂∂

−∂∂

xz

zxzy

11=

∂∂

⎟⎠⎞

⎜⎝⎛ −

xz

zy ⇒

1−=

∂∂

yzz

xz

Example

If x , y and z are independent variables and

)3sin(),,( zyxzyxf +=

then [ ] )3sin()3sin( zyz

xzyxzz

f+

∂∂

=+∂∂

=∂∂

)3cos(3)3()3cos( zyxzyz

zyx +=+∂∂

+=


41

Second Order Partial Derivatives

xxfxf=

∂∂

2

2

, yyfy

f=

∂∂

2

2

, yxfyxf=

∂∂∂2

, xyfxyf=

∂∂∂2

The defining equations are

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=∂∂

xf

xxf2

2

,

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

∂yf

xyxf2

Differentiate first with respect to y , then with respect to x .

( )xyyx ff = Means the same thing

The Mixed Derivative Theorem

If ),( yxf and its partial derivatives xf , yf , xyf , and yxf are defined throughout a

region containing a point ),( ba and are all continuous at ),( ba , then

),(),( bafbaf yxxy =

Example

If xyeyxyxf += cos),( , find

2

2

xf

∂∂

, xyf∂∂

∂2

, 2

2

yf

∂∂

, and yxf∂∂

∂2

,

Solution

xx yeyyeyxxx

f+=+

∂∂

=∂∂ cos)cos( , xey

xf

yxyf

+−=⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=∂∂

∂ sin2

xyexf

xxf

=⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=∂∂

2

2


42

xx eyxyeyxyy

f+−=+

∂∂

=∂∂ sin)cos( , xey

yf

xyxf

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

∂ sin2

yxyf

yyf cos2

2

−=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∂∂

Partial Derivatives of Higher Order

Example

Find yxyzf if yxzxyzyxf 2221),,( +−=

Solution

We first differentiate with respect to the variable y , then x , then y again, and

finally with respect to z 24 xxyzf y +−= , xyzf yx 24 +−= , zf yxy 4−= , 4−=yxyzf

Exercises

Find the Partial Derivatives of the functions with respect to each variable

1) 432),( 2 −−= yxyxf Ans. xfx 4= , 3−=yf

2) ( )( )21),( 2 +−= yxyxf Ans. )2(2 += yxfx , 12 −= xf y

3) ( )21),( −= xyyxf Ans. )1(2 −= xyyfx , )1(2 −= xyxf y

4) 22),( yxyxf +=

Ans.22 yx

xfx+

= ,22 yx

yf y+

=


43

5) yx

yxf+

=1),(

Ans.( )2

1yx

fx +−

= ,( )2

1yx

f y +−

=

6) 1

),(−+

=xy

yxyxf

Ans.2

2

)1(1

−−−

=xy

yfx , 2

2

)1(1

−−−

=xy

xf y

7) ( )1),( ++= yxeyxf

Ans. ( )1++= yxx ef , ( )1++= yx

y ef

8) )ln(),( yxyxf += Ans. yxfx +=

1 , yx

f y +=

1

9) )3(sin),( 2 yxyxf −= Ans. )3cos()3sin(2 yxyxfx −−= ,

)3cos()3sin(6 yxyxf y −−−=

10) yxyxf =),( Ans. 1−= yx yxf , )ln(xxf y

y =

11) ∫=y

x

dttgyxf )(),( ,

( g continuous for all t )

Ans.

)(xgfx −= , )(ygf y =

12) 22 21),,( zxyzyxf −+= Ans. 2yfx = , xyf y 2= , zf z 4−=

13) 22),,( zyxzyxf +−=

Ans. 1=xf , ( ) 2/122 −+−= zyyf y ,

( ) 2/122 −+−= zyzf z

14)

)(sin),,( 1 xyzzyxf −=

Ans. 2221 zyxyzfx

−= ,

2221 zyxxzf y

−= ,

2221 zyxxyf z

−=


44

15)

)32ln(),,( zyxzyxf ++=

Ans.zyx

fx 321++

= ,zyx

f y 322++

= ,

zyxf z 32

3++

=

16) ( )222),,( zyxezyxf ++−=

Ans. ( )2222 zyx

x xef ++−−= ,

( )2222 zyx

y yef ++−−= ,

( )2222 zyx

z zef ++−−=

17) )32tanh(),,( zyxzyxf ++= Ans. )32(2 zyxsechfx ++= ,

)32(2 2 zyxsechf y ++= ,

)32(3 2 zyxsechf z ++=

18) )2cos(),( απα −= ttf Ans. )2sin(2 αππ −−= tft ,

)2sin( απα −= tf

19) )cos()sin(),,( θϕρθϕρ =h Ans.

)cos()sin( θϕρ =h ,

)cos()cos( θϕρϕ =h ,

)sin()sin( θϕρθ −=h

Find the second order Partial Derivatives of the functions with respect to

each variable

1) xyyxyxf ++=),( Ans. 0=xxf , 0=yyf , 1=xyf


45

2) )sin()cos(),( 2 xyyyxyxg ++= Ans. )sin(2 xyygxx −= ,

)cos(yg yy −= ,

)cos(2 xxg xy +=

3) ( )yxyxr += ln),(

Ans. 2)(1yx

rxx +−

= ,

2)(1yx

ryy +−

= ,

2)(1yx

rxy +−

=

Find the mixed Partial Derivatives for the following functions

1) )32ln( yxw += 2) )ln()ln( xyyxew x ++=

3) 43322 yxyxxyw ++= 4) xyxyyxw ++= )sin()sin(

Chain Rule

If ),( yxfw = has continuous partial

derivatives xf and yf and if )(txx = ,

)(tyy = are differentiable functions of t , then

the composite ))(),(( tytxfw = is a

differentiable function of t and

dtdy

yw

dtdx

xw

dtdw

∂∂

+∂∂

=

),( yxfw=

x y

t

yw∂∂

xw∂∂

dtdy

dtdx


46

Example

Use the Chain Rule to find the derivative of xyw =

with respect to t along the path

)cos(tx = & )sin(ty =

What is the derivative’s value at 2/π=t ?

Solution

We apply the Chain Rule to find dtdw / as follows

dtdy

yw

dtdx

xw

dtdw

∂∂

+∂∂

=

( ) ( ))sin()()cos()( tdtdxy

yt

dtdxy

x×

∂∂

+×∂∂

=

( ) ( ))cos()sin( txty ×+−×=

( ) ( ) ( ) ( ))cos()cos()sin()sin( tttt ×+−×=

)(cos)(sin 22 tt +−=

)2cos( t=

We can check the result with a more direct calculation as a function of t

)2sin(21)sin().cos( tttxyw ===

So, )2cos()2cos(221)2sin(

21 ttt

dtd

dtdw

=×=⎟⎠⎞

⎜⎝⎛=

In either case, at a given value of t ,

1cos2

2cos2/

−==⎟⎠⎞

⎜⎝⎛ ×=⎟

⎠⎞

⎜⎝⎛

=

πππtdt

dw


47

Chain Rule for Functions of Three Independent Variables

Example

Find dtdw / if

zxyw += , )cos(tx = , )sin(ty = , tz =

What is the derivative’s value at 0=t ?

Solution

dtdz

zw

dtdy

yw

dtdx

xw

dtdw

∂∂

+∂∂

+∂∂

=

)1)(1()))(cos(())sin()(( ++−= txty

1))))(cos((cos())sin())((sin( ++−= tttt

)2cos(11)(cos)(sin 22 ttt +=++−=

2)0cos(10

=+=⎟⎠⎞

⎜⎝⎛

=tdtdw

),,( zyxfw=

x y

t

yw∂∂

xw∂∂

dtdy

dtdx

z

zw∂∂

dtdz

Here we have three routes from w to t

instead of two, but finding dtdw / is still the

same. Read each route, multiplying derivatives

along the way; then add.

dtdz

zw

dtdy

yw

dtdx

xw

dtdw

∂∂

+∂∂

+∂∂

=


48

Chain Rule for Two Independent Variables and Three Intermediate

Variables

Example

Express rw ∂∂ / and sw ∂∂ / in terms of r and s if

22 zyxw ++= , srx = , sry ln2 += , rz 2=

Solution

)2)(2()2)(2(1)1( zrsr

zzw

ry

yw

rx

xw

rw

++⎟⎠⎞

⎜⎝⎛=

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

rs

rrs

121)2(441+=++=

)0)(2(1)2()1( 2 zss

rsz

zw

sy

yw

sx

xw

sw

+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

2

2sr

s−=

Suppose that ),,( zyxfw = , ),( srgx = ,

),( srhy = , ),( srkz = . If all four functions

are differentiable, then w has partial derivatives

with respect to r and s , given by the formulas

rz

zw

ry

yw

rx

xw

rw

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

sz

zw

sy

yw

sx

xw

sw

∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

),,( zyxfw=

x y

r

yw∂∂

xw∂∂

ry∂∂

rx∂∂

z

zw∂∂

rz∂∂

),,( zyxfw=

x y

s

yw∂∂

xw∂∂

sy∂∂

sx∂∂

z

zw∂∂

sz∂∂


49

Example

Express rw ∂∂ / and sw ∂∂ / in terms of r and s if 22 yxw += , srx −= , sry +=

Solution

ry

yw

rx

xw

rw

∂∂

∂∂

+∂∂

∂∂

=∂∂

)1)(2()1)(2( yx +=

)(2)(2 srsr ++−= r4=

sy

yw

sx

xw

sw

∂∂

∂∂

+∂∂

∂∂

=∂∂

)1)(2()1)(2( yx +−=

)(2)(2 srsr ++−−= s4=

Implicit Differentiation

Suppose that 0),( =yxF is differentiable and that the equation 0),( =yxF

defines y as a differentiable function of x . Then at any point where 0≠yF

y

x

FF

dxdy

−=

Example

Find dxdy / if xyxy sin22 =−

Solution

Take xyxyyxF sin),( 22 −−= . Then

xyxyxyyx

xyxyxyyx

FF

dxdy

y

x

cos2cos2

cos2cos2

−+

=−−−

−=−=


50

Exercises

Find dtdw / at the given value for the following functions

1) 22 yxw += , )cos(tx = , )sin(ty = , at π=t Ans. 0==πtdt

dw

2) zy

zxw += , )(cos2 tx = , )(sin2 ty = ,

tz 1= , at 3=t Ans. 1

3

==tdt

dw

3) )ln(2 zyew x −= , )1ln( 2 += tx , )(tan 1 ty −= , tez = ,

at 1=t Ans. 1

1

+==

πtdt

dw

Answer the following questions:

1) Find uz ∂∂ / and vz ∂∂ / for )ln(4 yez x= , ( ))cos(ln vux = , )sin(vuy =

at the point ( ) ( )4/,2, π=vu .

Ans. )22(ln2 +=uz , )22(ln22 −−=vz

2) Find uw ∂∂ / and vw ∂∂ / for xzyzxyw ++= , vux += , vuy −= ,

uvz = at the point ( ) ( )1,2/1, =vu .

Ans. 3=uz ,

23

−=vz


51

3) Find xu ∂∂ / , yu ∂∂ / and zu ∂∂ / forrqqpu

−−

= , zyxp ++= ,

zyxq +−= , zyxr −+= at the point ( ) ( )1,2,3,, =zyx

Ans. 0=xu , 1=yu , 2−=zu

4) Find rw ∂∂ / if ( )2zyxw ++= , srx −= , )cos( sry += , )sin( srz += at

the point ( ) ( )1,1, −=sr

Ans. 12

5) Find vw ∂∂ / if ( )xyxw /2 += , 12 +−= vux , and 22 −+= vuy , at the

point ( ) ( )0,0, =vu

Ans. 7−

6) Find uz ∂∂ / and vz ∂∂ / if )(tan5 1 xz −= and vex u ln+= at the point

( ) ( )1,2ln, =vu

Ans. 2=uz , 1=vz

Find dxdy / at the given point for the following functions

1) 02 23 =+− xyyx , ( )1,1 Ans. 3/4

2) 0722 =−++ yxyx , ( )2,1 Ans. 5/4−


52

Directional Derivatives

The derivative of f at ),( 000 yxP in the direction of the unit vector juiuu 21 +=

is the number

syxfsuysuxf

dsdf

sPu

),(),(lim 002010

0, 0

−++=⎟

⎠⎞

⎜⎝⎛

→


The directional derivative is also denoted by

( )0Pu fD “The derivative of f in the direction of u at 0P ”

Example

Find the derivative of xyxyxf += 2),( at )2,1(0P in the direction of the unit

vector ( ) ( ) jiu 2/12/1 +=

Solution

syxfsuysuxf

dsdf

sPu

),(),(lim 002010

0, 0

−++=⎟

⎠⎞

⎜⎝⎛

→

s

fssf

s

)2,1(2

12,2

11lim

0

−⎟⎠⎞

⎜⎝⎛ ++

=→

s

sss

s

)211(2

22

12

1lim

22

0

×+−⎟⎠⎞

⎜⎝⎛ +⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ +

=→

s

ssss

s

322

3222

21lim

22

0

−⎟⎟⎠

⎞⎜⎜⎝

⎛+++⎟⎟

⎠

⎞⎜⎜⎝

⎛++

=→


53

250

25

25lim2

5

lim0

2

0=+=⎟

⎠⎞

⎜⎝⎛ +=

+=

→→s

s

ss

ss

So, the rate of change of xyxyxf += 2),( at )2,1(0P in the direction of

( ) ( )ji 2/12/1 +=u is 2/5 .

Gradient Vector

The gradient vector (gradient) of ),( yxf at a point ),( 000 yxP is the vector

jiyf

xff

∂∂

+∂∂

=∇

obtained by evaluating the partial derivatives of f at 0P .

The notation f∇ is read as “gradient of f ” and “del f ”. The symbol ∇ by itself

is read “del”.

Another definition for Directional Derivative is the dot product of the gradient of

f at 0P with the unit vector u

( ) ufdsdf

PPu

•∇=⎟⎠⎞

⎜⎝⎛

0

0,

Example

Find the derivative of )cos(),( xyxeyxf y += at the point )0,2( in the direction

of ji 43 −=v .

Solution

The direction of v is the unit vector obtained by dividing v by its length

525169)4()3( 22 ==+=+=v


54

ji54

53

5−===

vvvu

10))sin(()0,2( 0)0,2( =−=−= exyyef y

x

2022))sin(()0,2( 0)0,2( =×−=−= exyxxef y

y

The gradient of f at )0,2( is

jiji 2)0,2()0,2()0,2(

+=+=∇ yx fff

The derivative of f at )0,2( in the direction of v is therefore

( ) uffDu •∇=)0,2()0,2(

158

53

54

53)2( −=−=⎟

⎠⎞

⎜⎝⎛ −•+= jiji

Properties of the Directional Derivative θcosfuffDu ∇=•∇=

1) The function f increases most rapidly when 1cos =θ or when u is the

direction of f∇ . That is, at each point P in its domain, f increases most rapidly

in the direction of the gradient vector f∇ at P . The derivative in this direction is

( ) fffDu ∇=∇= 0cos

2) Similarly, f decreases most rapidly in the direction of f∇− . The derivative in

this direction is

( ) fffDu ∇−=∇= πcos

3) Any direction u orthogonal to a gradient 0≠∇f is a direction of zero change in

f because θ then equals 2/π and

( ) 002/cos =×∇=∇= fffDu π


55

Example

Find the direction in which ( ) ( )2/2/),( 22 yxyxf +=

a) Increases most rapidly at the point )1,1(

b) Decreases most rapidly at the point )1,1(

c) What are the directions of zero change in f at )1,1(

Solution

(a) The function increases most rapidly in the direction of f∇ at )1,1( . The gradient is

( ) jiji +=+=∇ )1,1()1,1( )( yxf

Its direction is

jiji

jiji

21

21

)1()1( 22+=

++

=++

=u

(b) The function decreases most rapidly in the direction of f∇− at )1,1( , which is

ji2

12

1−−=− u

(c) The direction of zero change at )1,1( are the directions orthogonal to f∇

ji2

12

1+−=n and ji

21

21

−=− n

Gradients and Tangents to Level Curves

At every point ),( 00 yx in the domain of a differentiable function ),( yxf , the

gradient of f is normal to the level curve through ),( 00 yx .

This observation also enables us to find equations for tangent lines to level curves.

They are the lines normal to the gradients. The line through a point ),( 000 yxP normal

to a vector ji BAN += has the equation


56

0)()( 00 =−+− yyBxxA

If N is the gradient ( ) ji ),(),( 0000),( 00yxfyxff yxyx +=∇ , the equation of the

tangent line will be

0))(,())(,( 000000 =−+− yyyxfxxyxf yx

Example

Find an equation for the tangent to the ellipse 24

22

=+ yx at the point )1,2(− .

Solution

The ellipse is a level curve of the function 22

4),( yxyxf +=

The gradient of f at )1,2(− is

jiji 222 )1,2(

)1,2( +−=⎟⎠⎞

⎜⎝⎛ +=∇

−−

yxf

The tangent is the line

0)1)(2()2)(1( =−++− yx ⇒ 42 =+− yx

or 42 −=− yx

Algebra Rules for Gradients

1) Constant Multiple Rule: fkkf ∇=∇ )(

2) Sum Rule: gfgf ∇+∇=+∇ )(

3) Difference Rule: gfgf ∇−∇=−∇ )(

4) Product Rule: fggffg ∇+∇=∇ )(

5) Quotient Rule: 2ggffg

gf ∇−∇

=⎟⎟⎠

⎞⎜⎜⎝

⎛∇


57

Example

yxyxf −=),( yyxg 3),( =

ji −=∇f j3=∇g

1) fyxf ∇=−=−∇=∇ 222)22()2( ji

2) gfyxgf ∇+∇=+=+=+∇ ji 2)2()(

3) gfyxgf ∇−∇=−=−∇=−∇ ji 4)4()(

4) ji )63(3)33()( 2 yxyyxyfg −+=−∇=∇

jjji )63(3)(3 yxyy −++−=

jji )33()(3 yxy −+−=

gffgyxy ∇+∇=−+−= jji 3)()(3

5) ⎟⎟⎠

⎞⎜⎜⎝

⎛−∇=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −∇=⎟⎟

⎠

⎞⎜⎜⎝

⎛∇

31

33 yx

yyx

gf

ji 2331

yx

y−=

22 93)()(3

yyxy

ggffg jji −−−=

∇−∇

jiji 222 331

93

93

yx

yyx

yy

−=−=

Functions of Three Variables

Example

a) Find the derivative of zxyxzyxf −−= 23),,( at )0,1,1(0P in the direction of

kji 632 +−=v .

b) In what directions does f change most rapidly at 0P , and what are the rates of

change in these directions?


58

Solution

(a) The direction of v is obtained by dividing v by its length

749)6()3()2( 222 ==+−+=v

kji76

73

72

+−==vvu

The partial derivatives of f at 0P are

2)3()0,1,1(

22 =−= yxfx , 22)0,1,1(

−=−= xyf y ,

11)0,1,1(

−=−=zf

The gradient of f at 0P is

kji −−=∇ 22)0,1,1(

f

The derivative of f at 0P in the direction of v is therefore

( ) uffDu •∇=)0,1,1()0,1,1(

⎟⎠⎞

⎜⎝⎛ +−•−−= kjikji

76

73

72)22(

74

76

76

74

=−+=

(b) The function increases most rapidly in the direction of kji −−=∇ 22f and

decreases most rapidly in the direction of f∇− . The rate of change in the directions

are, respectively,

39)1()2()2( 222 ==−+−+=∇f and 3−=∇− f


59

Tangent Plane

The tangent plane at the point ),,( 0000 zyxP on the level surface czyxf =),,( of

a differentiable function f is the plane through 0P normal to 0Pf∇ .

( )( ) ( )( ) ( )( ) 0000000 =−+−+− zzPfyyPfxxPf zyx

Normal Line

The normal line of the surface at ),,( 0000 zyxP on the level surface czyxf =),,(

of a differentiable function f is the line through 0P parallel to 0P

f∇ .

tPfxx x )( 00 += , tPfyy y )( 00 += , tPfzz z )( 00 +=

Example

Find the tangent plane and normal line of the surface

09),,( 22 =−++= zyxzyxf

at the point )4,2,1( .

Solution

The tangent plane is the plane through 0P perpendicular to the gradient of f at 0P .

The gradient is

kjikji ++=++=∇ 42)22( )4,2,1(0yxf P

The tangent plane is therefore the plane

0)4()2(4)1(2 =−+−+− zyx , or 1442 =++ zyx

The line normal to the surface at 0P is

tx 21+= , ty 42 += , tz += 4 .


60

Exercises

Find the gradient of the function at the given point

1) xyyxf −=),( , ( )1,2 Ans. ji +−=∇f

2) 2),( xyyxg −= , ( )0,1− Ans. ji +=∇ 2g

3) )ln(2),,( 222 xzzyxzyxf +−+= ,

( )1,1,1

Ans. kji 423 −+=∇f

4) ( ) )ln(),,( 2/1222 xyzzyxzyxf +++=−

,

( )2,2,1 −− Ans.

kji5423

5423

2726

−+−=∇f

Find the derivative of the function at 0P in the direction of A

1) 232),( yxyyxf −= , ( )5,50P , ji 34 +=A Ans. 4−

2) ( ) )2(sec3/),( 12 xyxyxyxg −+−= , ( )1,10P , ji 512 +=A Ans. 13/31

3) xzyzxyzyxf ++=),,( , ( )2,1,10 −P , kji 263 −+=A Ans. 3

4) )cos(3),,( yzezyxg x= , ( )0,0,00P , kji 22 −+=A Ans. 2


61

Find the directions in which the function increase and decrease most

rapidly at 0P . Then find the derivatives of the functions in these directions

1)

22),( yxyxyxf ++= , ( )1,10 −P

Ans. ji2

12

1+−=u ,

( ) 20=Pu fD ,

ji2

12

1−=− u ,

( ) 20

−=− Pu fD

2)

( ) yzyxzyxf −= /),,( , ( )1,1,40P

Ans. k-ji33

133

533

1−=u ,

( ) 330=Pu fD ,

kji33

133

5331

++−

=− u ,

( ) 330

−=− Pu fD

3)

( ) ( ) ( )xzyzxyzyxf lnlnln),,( ++= ,

( )1,1,10P

Ans. ( )kji ++=3

1u ,

( ) 320=Pu fD ,

( )kji ++−

=−31u ,

( ) 320

−=− Pu fD


62

Write an equation for the tangent line at the given point

1) 422 =+ yx , ( )2,2 Ans. 22+−= xy

2) 4−=xy , ( )2,2 − Ans. 4−= xy

Find equations for (a) the tangent plane and

(b) normal line

at the point 0P on the given surface

1) 3222 =++ zyx , ( )1,1,10P Ans. (a) 3=++ zyx

(b) tx 21+= ,

ty 21+= , tz 21+=

2) 02 2 =− xz , ( )2,0,20P Ans. (a) 022 =−− zx (b) tx 42 −= , 0=y ,

tz 22 +=

3) 4)cos( 2 =++− yzeyxx xzπ , ( )2,1,00P Ans. (a) 0422 =−++ zyx

(b) tx 2= , ty 21+= ,

tz += 2

4) 1=++ zyx , ( )0,1,00P Ans. (a) 01=−++ zyx

(b) tx = , ty +=1 , tz =


63

Find an equation for the plane that is tangent to the given surface at the

given point

1) ( )22ln yxz += , ( )0,0,1 Ans. 022 =−− zx

2) xyz −= , ( )1,2,1 Ans. 012 =−+− zyx

Extreme Values and Saddle Points

The extreme values of ),( yxf can occur only at

i. Boundary points of the domain of f and endpoints.

ii. Critical points (interior points where 0== yx ff or points where xf or yf fail

to exist).

If the first- and second-partial derivatives of f are continuous throughout a point

),( ba and 0),(),( == bafbaf yx , the nature of ),( baf can be tested with the

Second Derivative Test:

i. 0<xxf and 02 >− xyyyxx fff at ),( ba ⇒ Local Maximum

ii. 0>xxf and 02 >− xyyyxx fff at ),( ba ⇒ Local Minimum

iii. 02 <− xyyyxx fff at ),( ba ⇒ Saddle Point

iv. 02 =− xyyyxx fff at ),( ba ⇒ Test is inconclusive

The expression 2xyyyxx fff − is called the discriminant of f and written in

determinant form as follows:

yyxy

xyxxxyyyxx ff

fffff =− 2


64

Example

Find the local extreme values of the function

422),( 22 +−−−−= yxyxxyyxf Solution

The function is defined and differentiable for all x and y and its domain has no

boundary points. The function therefore has extreme values only at the points where xf

and yf are simultaneously zero. This leads to

022 =−−= xyfx , 022 =−−= yxf y ,

or 2−== yx .

Therefore, the point )2,2( −− is the only point where f may take on an extreme

value.

2−=xxf , 2−=yyf , 1=xyf .

The discriminant of f at )2,2(),( −−=ba is

314)1()2)(2( 22 =−=−−−=− xyyyxx fff

The combination 0<xxf and 02 >− xyyyxx fff tells us that f has a local

maximum at )2,2( −− . The value of f at this point is 8)2,2( =−−f .

Absolute Maxima and Minima on Closed Bounded Regions

We organize the search for the absolute extreme of a continuous function ),( yxf

on a closed and bounded region R into three steps:

1. List the interior points of R where f may have local maxima and minima and

evaluate f at these points. These are the critical points of f .


65

2. List the boundary points of R where f have local maxima and minima and

evaluate f at these points. These are the critical points of f .

3. Look through the lists for the maximum and minimum values of f . These will be

the absolute maximum and minimum values of f on R .

Solution

(a) Interior points

022 =−= xfx , 022 =−= yf y ,

yielding the single point )1,1(),( =yx . The value of f is

4)1,1( =f

(b) Boundary points

We take the triangle one side at a time

(i) On the segment OA , 0=y . The function

222)0,(),( xxxfyxf −+==

may now be regarded as a function of x defined on the close interval 90 ≤≤ x . Its

extreme values may occur at the endpoints

o )0,9(A

)9,0(B

0=x

0=y

xy −= 9

Example

Find the absolute maximum and minimum

values of 22222),( yxyxyxf −−++=

on the triangular region in the first quadrant bounded

by the lines 0=x , 0=y , and xy −= 9 .


66

0=x where 2)0,0( =f

9=x where 6181182)0,9( −=−+=f

and at the interior points where 022)0,( =−=′ xxf . The only interior point

where 0)0,( =′ xf is 1=x , where

3)0,1()0,( == fxf

(ii) On the segment OB , 0=x and 222),0(),( yyyfyxf −+==

may now be regarded as a function of y defined on the close interval 90 ≤≤ y . Its

extreme values may occur at the endpoints

0=y where 2)0,0( =f

9=y where 6181182)9,0( −=−+=f

and at the interior points where 022),0( =−=′ yyf . The only interior point

where 0),0( =′ yf is 1=y , where

3)1,0(),0( == fyf

(iii) On the segment AB . We have already accounted for the values of f at the

endpoints, so we need only look at the interior points of AB . With xy −= 9

222 21861)9()9(222)9,( xxxxxxxxf −+−=−−−−++=−

Setting 0418)9,( =−=−′ xxxf gives

5.44

18==x

At this value of x

5.45.49 =−=y and 5.20)5.4,5.4(),( −== fyxf


67

Finally, we list all the candidates: 5.20,3,61,2,4 −− . The maximum is 4,

which f assumes at )1,1( . The minimum is 61− , which f assumes at )9,0( and

)0,9( .

Exercises Find all local maxima, local minima, and saddle points for the following

functions

1) 433),( 22 +−+++= yxyxyxyxf Ans. 5)3,3( −=−f local min.

2)

444252),( 22 −++−−= yxyxxyyxf

Ans. 034,

32

=⎟⎠⎞

⎜⎝⎛f local max.

3) 523),( 2 ++++= yxxyxyxf Ans. )1,2(−f saddle point

4) 26375),( 2 +−+−= yxxxyyxf Ans. ⎟⎠⎞

⎜⎝⎛

2569,

56f saddle point

5) 264),( 22 +++−= yyxyxyxf Ans. )1,2(f saddle point

6) yxyxyxyxf 25432),( 22 +−++= Ans. 6)1,2( −=−f local min.

7) 642),( 22 ++−−= yxyxyxf Ans. )2,1(f saddle point

8) xyxyxf 2),( 2 += Ans. )0,0(f saddle point

9) 62),( 33 +−−= xyyxyxf Ans. )0,0(f saddle point,

27170

32,

32

=⎟⎠⎞

⎜⎝⎛−f local max.

10) xyyxxyxf 6326),( 232 ++−= Ans.

0)0,0( =f local min.,

( )1,1f saddle point


68

11) xyyxyxf 43/9),( 33 −−= Ans. )0,0(f saddle point,

8164

34,

94

−=⎟⎠⎞

⎜⎝⎛f local min.

12) 833),( 2233 −−++= yxyxyxf Ans. )0,0(f saddle point,

( ) 212,0 −=f local min.,

( ) 40,2 −=−f local max.,

)2,2(−f saddle point

13) 444),( yxxyyxf −−= Ans. )0,0(f saddle point,

( ) 21,1 =f local max.,

( ) 21,1 =−−f local max.

14) 1

1),( 22 −+=

yxyxf Ans. 1)0,0( −=f local max.

15) )sin(),( xyyxf = Ans. )0,( πnf saddle point

Find the absolute maxima and minima for the following functions on the

given domains

1) 1442),( 22 +−+−= yyxxyxf on the

closed triangular plate bounded by the lines

0=x , 2=y , xy 2= in the first quadrant

Ans. Absolute max.: 1 at ( )0,0 ,

Absolute min.: 5− at ( )2,1

2) 22),( yxyxf += on the closed triangular

plate bounded by the lines 0=x , 0=y ,

22 =+ xy in the first quadrant

Ans. Absolute max.: 4 at ( )2,0 ,

Absolute min.: 0 at ( )0,0


69

3) 26),( 22 +−++= xyxyxyxT on the

rectangular plate 50 ≤≤ x , 03 ≤≤− y

Ans. Absolute max.: 11 at( )3,0 −

Absolute min.: 10− at

( )2,4 −

4) ( ) )cos(4),( 2 yxxyxf −= on the

rectangular plate 31 ≤≤ x ,

4/4/ ππ ≤≤− y

Ans. Absolute max.: 4 at( )0,2

Absolute min.: 2

23 at

⎟⎠⎞

⎜⎝⎛ −

4,3 π

, ⎟⎠⎞

⎜⎝⎛

4,3 π

, ⎟⎠⎞

⎜⎝⎛ −

4,1 π

,

and ⎟⎠⎞

⎜⎝⎛

4,1 π


70

Differential Equations

A Differential Equation is an equation that contains one or more derivatives of a

differentiable function. An equation with partial derivatives is called a Partial

Differential Equation. While, an equation with ordinary derivatives, that is, derivatives

of a function of a single variable, is called an Ordinary Differential Equation.

The order of a differential equation is the order of the equation’s highest order

derivative. A differential equation is linear if it can be put in the form

)()()(...)()( 011

1

1 xFyxadxdyxa

dxydxa

dxydxa n

n

nn

n

n =++++ −

−

−

The degree of a differential equation is the power (exponent) of the equation’s

highest order derivative.

Example

First order, first degree, linear ydxdy 5= , 0sin3 =− x

dxdy

Third order, second degree, nonlinear xedxdy

dxyd

dxyd

=−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛5

2

22

3

3

Solution of First Order Differential Equations

1) Separable Equations A first order differential equations is separable if it can be put in the form

0)()( =+ dyyNdxxM


Steps for Solving a Separable First Order Differential Equation

i. Write the equation in the form 0)()( =+ dyyNdxxM .

ii. Integrate M with respect to x and N with respect to y to obtain an equation

that relates y and x .

Example

Solve the following differential equations

(a) xeydxdy )1( 2+= , (b)

yyyxx

dxdy

cossin)1ln2(

++

= , (c) x

dydxe yx =+

Solution

(a) xeydxdy )1( 2+= ⇒ 0

11

2 =+

− dyy

dxex

Cdyy

dxex =+

− ∫∫ 211

⇒ Cyex =− −1tan

Cey x −=−1tan ⇒ )tan( Cey x −=

(b) yyy

xxdxdy

cossin)1ln2(

++

= ⇒ dxxxdyyyy )1ln2()cos(sin +=+

Cdxxxdyyyy =+−+∫ ∫ )1ln2()cos(sin

Cxdxdxxxdyyydyy =−−+∫ ∫∫∫ )ln(2)cos()sin(

[ ] Cxdxx

xxxdyyyyy =−⎥⎦

⎤⎢⎣

⎡×−×−−+− ∫∫ 2

122

)ln(2)sin(sin)cos(222

Cxxxxyyyy =−+−++−22

lncossin)cos(22

2

Cxxyy =− lnsin 2 .

71


(c) x

dydxe yx =+ ⇒ x

dydxee yx =

yx

edydxxe = ⇒ Cdyedxxe yx =− ∫∫ −

Cedxexe yxx =+− −∫ ⇒ Ceexe yxx =+− −

Notes

1) 0)()()()( 2211 =+ dxygxfdyygxf Separable

2) 0)()(

)()(

2

2

1

1 =+ dxygxfdy

ygxf

Separable

3) [ ] [ dxygxfdyygxf )()()()( 2211 ±+± ] SeparableNot

Example

xxf =)(1 , , )sin()(2 xxf = yyg =)(1 , )tan()(2 yyg =

1) 0)tan()sin( =+ dxyxxydy ⇒ dxyxxydy )tan()sin(−=

dxx

xdyy

y )sin()tan(

−= Separable

2) 0)tan()sin(

=+ dxyxdy

yx

⇒ dxyxdy

yx

)tan()sin(

−=

dxx

xdyy

y )sin()tan(−= Separable

3) ( ) 0)tan()sin()( =+++ dxyxdyyx

( dxyxdyyx )tan()sin()( +−=+ ) SeparableNot

72


Special Type of Separable Equations

If )( cbyaxfdxdy

++= ; then let cbyaxz ++= and the resultant equation may

be reduced to a separable equation.

Example

Solve the differential equation )(tan2 yxdxdy

+=

Solution

yxz += ⇒ dxdy

dxdz

+=1

1−=dxdz

dxdy

⇒ )(tan1 2 zdxdz

=−

1)(tan2 += zdxdz

⇒ )(sec2 zdxdz

=

dxz

dz=

)(sec2 ⇒ dxdzz =)(cos2

Cdxdzz =− ∫∫ )(cos2 ⇒ Cdxdzz=−

+∫∫ 2

)2cos(1

Cxzz =−+ )2sin(41

21

While yxz += , then the solution is

Cxyxyx =−+++ ))(2sin(41)(

21

73


Note

For the differential equation )sec()tan( yxyxdxdy

−−+= , we can not use the

assumption because the difference between the arguments of )tan( yx + and

)sec( yx − . So, the differential equation can not be converted to a separable equation.

Exercises

Find the solution of the following Differential Equations

1) kyy =′ 2) xyy −=′

3) 02 =+−′ ayy 4) 0=+′ byyx

5) ( ) yyxx =′ln 6) 0)2( =−′+ xyyx

7) 12 1 −=′ − yxy 8) )exp(2 2yxyy =′

9) )cot(2 xyy =′ 10) 0)csc( =+′ yy

11) )1)(1( 2yxy ++=′ 12) )(sin5.0 2 xyy ω=′

13) )tanh(xyy =′ 14) )2cos()2sin( xyxy =′

15) )2tan( xyy =′ , 2)0( =y 16) yyx 2)1( =′+ , 1)0( =y

17) yyx 32 =′ , 4)1( =y 18) yxxy =′ )ln( , )4ln()2( =y

19) 32 yey x=′ , 5.0)0( =y 20) 1)1( 2 =′+ yyx , 3)0( −=y

21) θθθ drdr )cos(2)sin( = , 2)2/( =πr 22) Cdxd =/υυ , 00 )( υυ =x

74


Homogeneous Function

If then ),(),( yxfyxf nλλλ = ),( yxf is homogeneous function and

represents the degree of the homogeneous function.

n

Example

For the function then 22),( yxyxf +=

( ) ( )22),( yxyxf λλλλ +=

2222 yx λλ +=

( ) ),(2222 yxfyx λλ =+=

So, the function ),( yxf is homogeneous with degree . 2

Example

For the function then 2),( yxyxf +=

( )2),( yxyxf λλλλ +=

22 yx λλ +=

( )2yx λλ +=

So, the function ),( yxf is not homogeneous.

Example

5),( 22 ++= yxyxf (Non-homogeneous)

xxyxyxf ++= 3),( (Non-homogeneous)

)cos(),( xyyxf = (Non-homogeneous)

)cos(),( 22 yxyxf ±= (Non-homogeneous)

75


⎟⎟⎠

⎞⎜⎜⎝

⎛=

yxyxf cos),( (Homogeneous)

⎟⎟⎠

⎞⎜⎜⎝

⎛=

yxyxf

2

cos),( (Non-homogeneous)

Homogeneous Equations

The differential equation dyyxNdxyxM ),(),( + is homogeneous if M and N

are homogeneous functions of the same degree.

Example

1) 0)( 22 =++ xydydxyx

This is homogeneous because M and N are both homogeneous with degree 2.

2) 0)( 33 =++ xydydxyx

This is not homogeneous because M is homogeneous with degree while 3 N is

homogeneous with degree 2.

3) 0)( 2 =++ dyyxxdx

This is not homogeneous because N is not homogeneous.

Solution of Homogeneous Equations A homogeneous first order differential equation can be put in the form

⎟⎠⎞

⎜⎝⎛=

xyF

dxdy

This equation can be changed into separable equation with the substitutions

76


xyv = ⇒ vxy = ⇒

dxdvxv

dxdy

+=

Then becomes )(vFdxdvxv =+

which can be rearranged algebraically to give

0)(=

−+

vFvdv

xdx

with the variables now separated, the equation can now be solved by integrating with

respect to x and v . We can then return to x and by substituting y xyv /= .

Example

Find the solution of the differential equation

xyyx

dxdy

2

22 +−=

that satisfies the condition . 1)1( =y

Solution

Dividing the numerator and denominator of the right-hand side by 2x gives

)/(2)/(1 2

xyxy

dxdy +

−= ⇒ )(2

1 2

vFvv

dxdy

=+

−=

vvvvFv

21)(

2++=−

vv

vvv

213

212 222 +

=++

=

0)(=

−+

vFvdv

xdx

⇒ 013

22 =+

+vvdv

xdx

The solution of this equation can be written as

77


Cvvdv

xdx

=+

+ ∫∫ 132

2 ⇒ Cvx =++ )31ln(31ln 2

Cvx 3)31ln(ln3 2 =++ ⇒ Cvx 3)31ln(ln 23 =++

Cvx ee 3)31ln(ln 23=++ ⇒ Cvx eee 3)31ln(ln 23

=× +

Cvx ′=+ )31( 23 ⇒ Cxyx ′=⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

23 31

Cxyx ′=+ 23 3

The condition is that when 1=x then 1=y and the constant C′ can be found

C′=+ 23 )1)(1(3)1( ⇒ 4=′C

The final solution is . 43 23 =+ xyx

Reducible to Homogeneous If the differential equation has the form

222

111

cybxacybxa

dxdy

++++

=

Case 1: if 2

1

2

1

bb

aa

= then ybxaz 11 +=

Case 2: if 2

1

2

1

bb

aa

≠ then intersect the two lines 0111 =++ cybxa

), k

and

to find the intersection point ( and let 0222 =++ cybxa h

hXx += ⇒ dXdx = , and kYy += ⇒ dYdy =

78


Example

Solve the differential equation 423564

++++

=xyyx

dydx

Solution

564432

++++

=yxyx

dxdy

21 =a , 42 =a ⇒21

42

2

1 ==aa

31 =b , ⇒ 62 =b21

63

2

1 ==bb

So 2

1

2

1

bb

aa

= ⇒ Case 1

Let yxz 32 += ⇒ dxdy

dxdz 32 +=

⎟⎠⎞

⎜⎝⎛ −= 2

31

dxdz

dxdy

⇒524

32

31

++

=−z

zdxdz

⇒ 252

123+

++

=zz

dxdz

52104123

++++

=z

zzdxdz

⇒ 52227++

=z

zdxdz

⇒ dxdzzz

=++22752

Cdxdzzz

=−++

∫∫ 22752

⇒ Cdxdzz

=−⎟⎠⎞

⎜⎝⎛

+×− ∫∫ 227

179

72

Cdxdzz

dz =−+

××

− ∫∫ ∫ 2277

779

72

Cxzz =−+×− )227ln(499

72

( ) Cxyxyx =−++×−+ 22)32(7ln499)32(

72

79


Example

Solve the differential equation dxyxdyyx )32()32( −+=−+

Solution

3232

−+−+

=yxyx

dxdy

11 =a , 22 =a ⇒21

2

1 =aa

21 =b , ⇒ 12 =b 22

1 =bb

So 2

1

2

1

bb

aa

≠ ⇒ Case 2

x y2+ 3− 0= ….. )1(

x2 y+ 3− 0= ….. )2(x2+ y4+ 6− 0= x2m ym 3± 0=

y3 3− 0=

⇒ 1=ySubstituting into , we get )2(

0312 =−+x ⇒ 1=x The intersection point )1,1(),( =kh .

Let 1+= Xx ⇒ dXdx =

1+= Yy ⇒ dYdy =

3)1()1(23)1(2)1(

−+++−+++

=YXYX

dXdY

31223221

−+++−+++

=YXYX

⇒ YXYX

dXdY

++

=2

2

80


XYXY

dXdY

+

+=

2

21

Let XYv = ⇒ )(

221 vFvv

dXdY

=++

=

0)(=

−+

vFvdv

XdX

vvvvFv

++

−=−2

21)(

vv

vvvv

+−

=+

−−+=

21

2212 22

01

22 =−+

+ dvv

vX

dX ⇒ Cdv

vvX =−+

+ ∫ 12ln 2

1)1()1(

1112

22 −++−

=−

++

=−+

vvBvA

vB

vA

vv

)1()1(2 ++−=+ vBvAv

At ⇒ 1=v23

=B

At ⇒ 1−=v21−

=A

Cdvv

dvv

X =−

++

−+ ∫∫ 1

2/312/1ln

CvvX =−++− )1ln(23)1ln(

21ln

81


82

CXY

XYX =−++− )1ln(

23)1ln(

21ln

Cxy

xyx =−

−−

++−−

−− )1)1()1(ln(

23)1

)1()1(ln(

21)1ln(

Exercises Find the solution of the following Differential Equations

1) yxyx +=′ 2) yxyx 22 +=′

3) 222 xxyyyx ++=′ 4) 222 45 xxyyyx ++=′

5) xyxyy

+−

=′ 6)xyxyy

−+

=′

7) 2)( xyy −=′ 8) 1)tan( −+=′ yxy

9) 51

+−+−

=′xyxyy 10)

xyxyy

21421

++−−

=′

2) Linear First Order Equations A differential equation that can be written in the form

)()( xQyxPdxdy

=+

where P and Q are functions of x , is called a Linear First Order Equation. The

solution is

∫= dxxQxx

y )()()(

1 ρρ

where ∫=dxxPex )()(ρ


Steps for Solving a Linear First Order Equation i. Put it in standard form and identify the functions P and Q .

ii. Find an anti-derivative of )(xP .

iii. Find the integrating factor ∫ . =dxxPex )()(ρ

iv. Find y using the following equation

∫= dxxQxx

y )()()(

1 ρρ

Example

Solve the equation 23 xydxdyx =−

Solution

Step 1: Put the equation in standard form and identify the functions P and Q . To do

so, we divide both sides of the equation by the coefficient of , in this

case

dxdy /

x , obtaining

xyxdx

dy=−

3 ⇒

xxP 3)( −= , xxQ =)( .

Step 2: Find an anti-derivative of )(xP .

)ln(3133)( xdxx

dxx

dxxP −=−=−= ∫∫∫

Step 3: Find the integrating factor )(xρ .

3

1lnlnln3)( 1)( 33

xeeeex xxxdxxP

=====−−∫ρ

Step 4: Find the solution.

∫∫ ⎟⎠⎞

⎜⎝⎛== dxx

xxdxxQx

xy )(1

)/1(1)()(

)(1

33ρρ

83


2332

3 11 xCxCx

xdxx

x −=⎟⎠⎞

⎜⎝⎛ +−== ∫

The solution is the function . 23 xCxy −=

Example

Solve the equation . 0))(tan()1( 12 =−++ − dxxydyx

Solution

Dividing the two sides by dxx )1( 2+

01

)(tan1 2

1

2 =+

−+

+−

xx

xy

dxdy

2

1

2 1)(tan

1 xx

xy

dxdy

+=

++

−

⇒ 211)(x

xP+

= , 2

1

1)(tan

xxQ

+=

−

)(tan1

1)( 12 xdx

xdxxP −=

+= ∫∫

)(tan 1)( xex

−

=ρ

Cdxx

xeye xx ++

= ∫−

−−

2

1)(tan)(tan

1)(tan11

)(tan 1 xz −= ⇒ dxx

dz 211+

=

Czdzeye zx +×= ∫− )(tan 1

Cdzeze zz +−= ∫

Ceze zz +−=

Ceexye xxx +−=−−− − )(tan)(tan1)(tan 111

)(tan

84


Steps for Solving other Form of Linear First Order Equation form There is another form of differential equation that can be written in the

)()( yQxyPdydx

=+

where P and are functions of . The solution is found as follows: Q y

i. Put it in standard form and identify the functions P and Q .

ii. Find an anti-derivative of )(yP .

iii. Find the integrating factor = ∫ dyyPey )()(ρ .

iv. Find x using the following equation

∫= dyyQyy

x )()()(

1 ρρ

Example

the equation .

Solu

Solve 0)(2 22 =−+ dyyxedxe yy

tion

Dividing the differential equation by to get dye y2

022 2 =−+ − ydx yexdy

yyexdydx 222 −=+ ⇒ 2)( =yP ,

,

yyeyQ 22)( −=

ydydyyP 22)( == ∫∫ ydyyP eey 2)()( == ∫ρ

( )( ) Cdyyeee

x yyy += ∫ −22

2 21 ⇒ Cydyxe y += ∫22

Cyxe y +=2

22

2 ⇒ Cyxe y += 22

85


Reducible to Linear The general form

)()()( yfxQyxPdxdy

=+

where the function f is to any power .

Also, it may be in the following form

y n

)()()()( yhxQygxPdxdy

=+

where the function g and are functions of .

Example

h y

Solve the equation )ln( yx 2

xy

dx

Solution

dy=+

Dividing the two sides of the equation by 2y

)ln(112 xydxy

xdy=+

Let ⇒dxdy

ydxdz

2

1−= ⇒

dxdzy

dxdy 2−=

yz 1=

)ln(1 xzxdx

=+− dz

)ln(1 xzxdx

dz−=− ⇒

xP 1−= , )ln(xQ −=

)ln(1)( xdxx

dxxP −= ∫∫ −=

86


xeeeex xxxdxxP 1)(

1ln)ln()ln()( 1

=====⎟⎠⎞

⎜⎝⎛

− −∫ρ

CdxxQxzx += ∫ )()()( ρρ

Cdxxx

zx

+−= ∫ )ln(11

( ) Cxyx

+−=×2

)ln(11 2

⇒ ( ) Cx

xy+−=

2)ln(1 2

Example

Solve the equation )(cos)2sin( 2 yxyxdxdy

=+

Solution

Dividing the two sides of the equation by )(cos2 y

xyyx

dxdy

y=+

)(cos)2sin(

)(cos1

22 ⇒ xy

yyxdxdyy =+

)(cos)cos()sin(2)(sec 2

2

xyyx

dxdyy =+

)cos()sin(2)(sec2 ⇒ xyx

dxdyy =+ )tan(2)(sec2

Let )tan(yz = ⇒dxdyy

dxdz )(sec2= ⇒

dxdz

ydxdy

)(sec12=

xxzdxdz

=+ 2 ⇒ xP 2= , xQ =

22)( xxdxdxxP == ∫∫ ⇒ 2)()( xdxxP eex == ∫ρ

CdxxQxzx += ∫ )()()( ρρ

Cdxxeze xx += ∫ )(22

⇒ Ceyex

x +=2

)tan(2

2

87


Another Form of Reducible to Linear The general form may be as follows

)()()( xfyQxyPdydx

=+

where the function f is x to any power . n

Also, it may be in the following form

)()()()( xhyQxgyPdydx

=+

where the function g and are functions of h x .

Example

Solve the equation dyxyxdxy ))(sin()cos( −=

Solution

Dividing the two sides of the equation by dyy)cos(

)cos()cos()sin( 2

yxx

yy

dydx

−= ⇒ )sec()tan( 2 yxyxdydx

−=−

Dividing by 2x , we get

)sec()tan(112 yy

xdydx

x−=−

Let x

z 1= ⇒

dydx

xdydz

2

1−= ⇒

dydzx

dydx 2−=

)sec()tan( yyzdydz

−=−−

)sec()tan( yyzdydz

=+ ⇒ )tan(yP = , )sec(yQ =

88


( ))cos(ln)cos()sin()tan()( ydy

yydyydyyP −=== ∫∫∫

( ) ( ) )sec()( )cos(1ln

)cos(ln)cos(ln)( 1yeeeey yyydyyP

=====⎟⎟⎠

⎞⎜⎜⎝

⎛− −∫ρ

CdyyQyzy += ∫ )()()( ρρ

Cdyyyx

y +=× ∫ )sec()sec(1)sec(

Cdyyx

y+= ∫ )(sec)sec( 2 ⇒ Cy

xy

+= )tan()sec(

Exercises


1) 3=−′ yy 2) 02 =+′ xyy

3) xeyy 62 =+′ 4) 2424 xxyy −=−′

5) )sin(xyy =+′ 6) )cos(2 xyy =+′

7) kxekyy −=+′ 8) )cot()1( xyy −=′

9) xexyyx 32 =−′ 10) )3sinh(22 xxyyx =+′

11) xeyy =−′ , 0)1( =y 12) 2)1( +=+′ xyy , 0)0( =y

13) 1−=+′ xyxyy 14) 211 yxyxy −− =+′

15) yyxyx +=′ 53102 16) 1−=+′ xyyy

89


Exact Differential Equations Example

If Cyxf =),( and )sin(),( xyyxf = then

0)cos()cos( =+=dxdyxyxxyy

dxdf

, or

0)cos()cos( =+= dyxyxdxxyydf

i.e., 0)cos()cos( =+ dyxyxdxxyy

From the above equation, we see that xfxyyyxM∂∂

== )cos(),( , and

yfxyxyxN∂∂

== )cos(),( . The solution of this differential equation is Cyxf =),( .

Exact Differential Equation Test

A differential equation 0 ),(),( =+ dyyxNdxyxM is said to be exact if for

some function ),( yxf

dfdyyfdx

xfdyyxNdxyxM =

∂∂

+∂∂

=+ ),(),(

is exact if and only if xN

yM

∂∂

=∂∂

Example

The equation 0 is exact because the partial

derivatives

))cos(2()( 22 =+++ dyyxydxyx

yyxyy

M 2)( 22 =+∂∂

=∂∂

, yyxyxx

N 2))cos(2( =+∂∂

=∂∂

are equal.

90


The equation 0 is not exact because the partial

derivatives

))cos(()3( 2 =+++ dyyxdxyx

3)3( =+∂∂

=∂∂ yx

yyM

, xyxxx

N 2))cos(( 2 =+∂∂

=∂∂

are not equal.

Steps for Solving an Exact Differential Equation

i. Match the equation to the form 0),(),( =+ dyyxNdxyxM to identify M and

N .

ii. Integrate M (or N ) with respect to x (or y ), writing the constant of integration

as )(yg (or )(xg ).

iii. Differentiate with respect to y (or x ) and set the result equal to N (or M ) to

find )(yg′ (or )(xg′ ).

iv. Integrate to find )(yg (or )(xg ).

v. Write the solution of the exact equation as Cyxf =),( .

Example

Solve the differential equation

0))cos(2()( 22 =+++ dyyxydxyx .

Solution

Step 1: Match the equation to the form 0),(),( =+ dyyxNdxyxM to identify M .

22),( yxyxM +=

Step 2: Integrate M with respect to x , writing the constant of integration as )(yg .

91


)(3

)(),(),( 23

22 ygxyxdxyxdxyxMyxf ++=+== ∫∫

Step 3: Differentiate with respect to and set the result equal to y N to find )(yg′ .

)(2)(3

23

ygxyygxyxy

′+=⎟⎟⎠

⎞⎜⎜⎝

⎛++

∂∂

)cos(2)(2 yxyygxy +=′+ ⇒ )cos()( yyg =′

Step 4: Integrate to find )(yg .

)sin()cos()( ydyydyyg ==′ ∫∫

Step 5: Write the solution of the exact equation as Cyxf =),( .

Cyxyx=++ )sin(

32

3

Another Solution

Step 1: Match the equation to the form 0),(),( =+ dyyxNdxyxM to identify N .

)cos(2),( yxyyxN +=

Step 2: Integrate N with respect to , writing the constant of integration as y )(xg .

)()sin())cos(2(),(),( 2 xgyxydyyxydyyxNyxf ++=+== ∫∫

Step 3: Differentiate with respect to x and set the result equal to M to find )(xg′ .

( ) )()()sin( 22 xgyxgyxyx

′+=++∂∂

222 )( yxxgy +=′+ ⇒ 2)( xxg =′

92


Step 4: Integrate to find )(xg .

3)(

32 xdxxdxxg ==′ ∫∫

Step 5: Write the solution of the exact equation as Cyxf =),( .

Cyxyx=++ )sin(

32

3

Reducible to Exact

A differential equation 0),(),( =+ dyyxNdxyxM which is not exact can be

made exact by multiplying both sides by a suitable integrating factor ρ . In other words,

the equation

0),(),( =+ dyyxNdxyxM ρρ

is an exact equation for an appropriate choice of ρ .

Method to Find the Integrating Factor

If )(xfN

xN

yM

=∂∂

−∂∂

or Constant then ∫ . =dxxfex )()(ρ

If )(yfM

yM

xN

=∂∂

−∂∂

or Constant then ∫ . =dyyfey )()(ρ

93


Example

Solve the equation 02 =+ xdyydx

Solution

yyxM 2),( = ⇒ 2=∂∂

yM

xyxN =),( ⇒ 1=∂∂

xN

This equation is not exact

)(112 xfxxN

xN

yM

==−

=∂∂

−∂∂

)ln(1)( xdxx

dxxf == ∫∫

xeex xdxxf=== ∫ )ln()()(ρ

Multiplying both sides of the equation by the integrating factor xx =)(ρ , we get

( 02 =+ xdyydxx ) ⇒ 02 2 =+ dyxxydx

which is exact because xyM 2=∂∂

and xxN 2=∂∂

, and the solution is

)(2),( 2 ygyxxydxyxf +== ∫

( ) )()( 22 ygxygyxy

′+=+∂∂

22 )( xygx =′+ ⇒ 0)( =′ yg

Cdyygyg =′= ∫ )()( ⇒ Cyx =2

94


Exercises


1) 0=+ xdyydx 2) 0/)( 2 =− xydxxdy

3) 0)2( =++ dyxedxex yy 4) 0)ln(2 21 =+ − dyxydxyx

5) dyyxdxyx )sin()cosh()cos()sinh( = 6) 03 33 =+ dredre θθ θ

7) 02)1( 2 =++ xydxdyx 8) 04 =− ydxxdy

9) 0)1( =++ dyyxydx 10) 0)2( 2 =+ xedyydx

11) ( ) 0/)3sin()3cos(3 2 =− ydyxdxxy 12) dyydxy )cos()sin( βββ −=

13) 0=− ydxxdy 14) dyydxy )sin()cos(2 πππ =

15) 0)sin(3)cos( =+ dyxdxxy 16) 023 =+ xdyydx

17) 0)/( 2 =+ dyxydx 18) 02 =− − dyedx xy

19) 0)sin(2)cos( =+ dyxdxxy 20) 0)1()1( =+−+ dyxdxy

21) 02 =+ xdyydx 22) 0)cos()sin( =+ dyydxy

23) 0)cos()sec(2 =+ yxdx , 0)0( =y 24) 032 22 =− dyxydxx , 0)1( =y

25) 0)cos()sin(2 =+ dyydxy

2/)0(

,

π=y

26) 02)2( =++ xdydxxyy ,

2)3( =y

95


96

Second Order Linear Homogeneous Equations The linear equation

)()()(...)()( 011

1

1 xFyxadxdyxa

dxydxa

dxydxa n

n

nn

n

n =++++ −

−

−

if 0)( =xF then it is called homogeneous; otherwise it is called non-homogeneous.

Linear Differential Operator It is convenient to introduce the symbol to represent the operation of

differentiation with respect to

Dx . That is, we write )(xDf to mean dxdf / .

Furthermore, we define powers of to mean taking successive derivatives: D

{ } 2

22 )()(

dxfdxDfDxfD == , { } 3

323 )()(

dxfdxfDDxfD ==

)(2)(2)()()()2( 2

222 xf

dxdf

dxfdxfxDfxfDxfDD −+=−+=−+

The Characteristic Equation The linear second order equation with constant real-number coefficients is

022

2

=++ bydxdya

dxyd

or, in operator notation

0)2( 2 =++ ybaDD

0))(( 21 =−− yrDrD


Solution of 022

2

=++ bydxdya

dxyd

Roots & 1r 2r Solution

Real and unequal xrxr eCeCy 21

21 +=

Real and equal xreCxCy 2)( 21 +=

Complex conjugate, βα j± )sincos( 21 xCxCey x ββα +=

Example

Solve the following differential equations:

(a) 022

2

=−+ ydxdy

dxyd

, (b) 0442

2

=++ ydxdy

dxyd

(c) 0642

2

=++ ydxdy

dxyd

, (d) 042

2

=+ ydx

yd

Solution

(a) 022

2

=−+ ydxdy

dxyd

The characteristic equation is

022 =−+ DD

0)2)(1( =+− DD ⇒ 11 =r and 22 −=r

The solution is xx eCeCy 2

21−+=

97


(b) 0442

2

=++ ydxdy

dxyd


0442 =++ DD

0)2( 2 =+D ⇒ 221 −== rr

The solution is xeCxCy 2

21 )( −+=

(c) 0642

2

=++ ydxdy

dxyd


0642 =++ DD

AACBBr

242

2,1−±−

=

224164

)1(2)6)(1(4)4(4 2

2,1−±−

=−±−

=r

2224

284

2,1jr ±−

=−±−

=

222,1 jr ±−= ⇒ 221 jr +−= and 222 jr −−=

⇒ 2−=α and 2=β

The solution is

)2sin2cos( 212 xCxCey x += −

98


(d) 042

2

=+ ydx

yd


042 =+D

0)2)(2( =+− jDjD ⇒ 21 jr = and 22 jr −=

⇒ 0=α and 2=β

The solution is

xCxCy 2sin2cos 21 +=

Exercises


1) 034 =+′−′′ yyy 2) 016 =−′′ yy

3) 016 =+′′ yy 4) 06 =−′−′′ yyy

5) 02 =′+′′ yy 6) 022 =+′−′′ yyy

7) 02 =+′′ yy ω , ( )0≠ω 8) 054 =+′+′′ yyy

9) 0=−′′ yy , , 6)0( =y 4)0( −=′y 10) 09 =−′′ yy , , 2)0( =y 0)0( =′y

11) 034 =+′−′′ yyy , 1)0( −=y ,

5)0( −=′y

12) 023 =+′−′′ yyy , 1)0( −=y ,

0)0( =′y

13) 022 =+′+′′ yyy 14) 044 =+′+′′ yyy

15) 09 =−′′ yy 16) 0126 =+′+′′ yyy

17) 04 =′−′′ yy 18) 01744 =+′+′′ yyy

99


Second Order Non-homogeneous Linear Equations Now, we solve non-homogeneous equations of the form

)(22

2 dyyd xFbydx

adx

=++

The procedure has three basic steps. First, we find the homogeneous solution

s

hy

(h tands for “homogeneous”) of the reduced equation

022

2 dyyd=++ by

dxa

dx

Second, we find a particular solution of the complete equation. Finally, we add

t

py

py o y to form the general solution of the complete equation. So, the final solution is

yyy

h

ph +=

ariation of ParametersV already know the homogeneous solution This method assumes we

)()( 2211 xuCxuCyh +=

The method consists of replacing the constants and by functions and 1C 2C )(1 xv

) and then requiring that the new expression

vuvyh

(2 xv

2211 u+=

and by solving the following two equations

02211 +′ =′vuv u

)(2211 xFuvuv =′′+′′

for the unknown functions and 1v′ 2v′ using the following matrix notation

100


⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡′′

⎥⎦

⎤⎢⎣

⎡′′ )(

0

2

1

21

21

xFvv

uuuu

inally and can be found by integration.

rameters to find the particular solution,

the f

e following equations

1v 2v

In applying the method of variation of pa

F

ollowing steps are taken:

i. Find 1v′ and 2v′ using th

DxFu

uuuu

uxFu0

DxFu

uuuu

xFuu

v )()(0

1

21

21

1

1

2 =

′′

′=′ v )()( 2

21

21

2

2

1−

=

′′

′=′ ,

21

21

uuuu

D′′

= where

ii. Integrate and to find and 1v′ 2v′ 1v 2v .

iii. Write the particular solution as

yp 2211 uvuv +=

xampleE

Solve the equation 6322

2

=−+ ydxdy

dxyd

Solution

omogeneous solution can be found using the reduced equation hyThe h

0322

2 dyy=−+ y

dxdxd

101


0322 =−+ DDThe characteristic equation is and the roots of this equation are

and , so

Then

31 −=r 12 =rxy x

h eCeC 23

1 += −

xeu 31

−= , xeu =2

xxxxx

xx ee 23

−−

eeeee

D 223

433

−−−

=+=−

=

xx

x

x

x

x

ee

ee

ee

v 3221 2

34

6460

−=−

==′ −− , xx

x

x

x

x

eee

ee

e

v −−

−

−

−

−

==−

=′23

46

4630

2

3

2

3

3

2

xx edxev 331 2

123

−=−= ∫ , xx edxev −− −== ∫ 23

23

2

xxxx ee ⎜⎛−+⎟

⎞ −

233

p eeuvuvy ⎟⎠⎞

⎝⎠⎜⎝⎛−=+= −3

21

2211 2−=

223

1 −+=+= − xxph eCeCyyy

Example

Solve the equation

Solu

)ln(2 xeyyy x=+′−′′

tion

The homogeneous solution can be found using the reduced equation hy

02 =+′−′′ yyy


0122 =+− DD

102


0)1( 2 =−D

121 == rr The roots are

The solution is

From that we have , and .

xh eCxCy ( 21 += )

xxh eCxeCy 21 + =

xxexu =)(1

xexu =)(2

xxxxxxxxe

xx

eexexeeeexe

D 2222 )( −=+−=+

=

)ln()ln()ln(0

2

2

21 xe

exe

exee

v x

x

x

xx

x

=−

−=

−=′

)ln()ln()ln(0

2

2

22 xxe

exxe

xeexexe

v x

x

x

xxx

x

−=−

=−

+=′

xxxdxxv −== ∫ )ln()ln(1

dxxxv ∫−= )ln(2

⇒xdxdu = ⇒

2

2xv = )ln(xu = , xdxdv =

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟− ∫dxx ln()ln −=

⎠

⎞⎜⎜⎝

⎛×−= ∫ dxxxx

xxxv

2)

21

2(

2

222

2

)ln(244

)ln(2

2222

xxxxxx−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

103


The particular solution is

2211 uvuvyp += ( ) xx exxxxexxx ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−= )ln(

24)ln(

22

)ln(24

)ln(22

22 x xexexexxex xxx −+−=

xx exxex4

3)ln(2

22

−=

The complete solution is

xxxxph exxexe2= CxeCyyy

43)ln(

2

22

1 −++=+

Undetermined Coefficients

This method gives us the particular solution for selected equations.

Coefficients for Selected Equations of the Form

The Method of Undetermined

)(22

xFbydyayd=++ 2 dxdx

If )(xF has a term of The expression for py

A (Constant) (Another Constant) C rxe rxAe

)sin(kx , )cos(kx )sin()cos( kxCkxB +

cbx ++2 ax FExDx ++2

104


Example xeyy =+′′ 3 Solve the equation

olutionS

omogeneous solution can be found using the reduced equation hyThe h

03 =+′′ yy


032 =+D

31 jr = , and 32 jr −= ⇒ 0=α and 3=β The roots are

So, )3sin()3cos( 21 xCxCyh +=

Since then let

e differential equation we get

xexF =)( xp Aey = ⇒ x

p Aey =′ ⇒ xp Aey =′′

xeyy =+′′ 3Substituting into th

xxx eAeAe =+ 3 ⇒ 13 =+ AA ⇒ 41

=A

xpy =So, e

41

And the complete solution is

xexCxCy41)3sin()3cos( 21 ++=

Important Note

The expression used for should not have any term sim s of the

omogeneous solution. Otherwise, multiply the term that is similar to the homogeneous

ly by

py ilar to the term

h

x until it becosolution repeated mes different.

105


Example

Solve the equation xeyyy 523 =+′−′′

Solution

The homogeneous solution can be found using the reduced equation hy

023 =+′−′′ yyy

The on is characteristic equati

0232 =+− DD

0)2)(1 =−− DD (

11 =r , and 22 =r The roots are

xxh eCeCy 2

21 +=

Since then let xexF 5)( = xp Aey = ⇒ x

p Aey =′ ⇒ xp Aey =′′

xeyyy 523 =+′−′′Substituting into the differential equation we get

xxxx eAeAeAe 523 =+−

(Wrong Answer)

The trouble can be traced to the fact that is already a solution in the

.

The appropriate way is to modify the particular solution to replace

xe50 =xe

homogeneous equation x eCeCy 2+= xh 21

xAe by

Substituting into the differential equation we get

xp Axey =

xxp AeAxey +=′

xxxxx AeAxeAeAeAxey 2+=++=′′ p

xeyyy 523 =+′−′′

( ) ( ) xxxxxx eAxeAeAxeAeAxe 5232 =++−+

106


xx eAe 5=−

5−=A ⇒

So,

The complete solution (general solution) is

Example

xxey 5−=p

xxx xeeCeCy 5221 −+=

(a)

(c)

Solve the equation xeyy 396 =+′−′′ , (b) y )2sin(5 xeyy x −=′−′′

342 xyyy =−′−′′

Solution

(a) T ution can be found using the reduced equation hyhe homogeneous sol

96 +′−′′ yyy 0=


0962 =+− DD

0)3( 2 =−D

21 == rr 3The roots are

Since then let . But,

xh eCxCy 3

21 )( +=

xexF 3)( = xp Aey 3= xAe3

. Again

is similar to the second term of the

hom tion so, let ogeneous solu x3p Axey = xAxe3 is also similar to the first

nally, let

term of the homogeneous solution. Fix

p eAxy 32= ⇒ xxp AxeeAxy 332 23 +=′

107


( ) ( )xxxxp AeAxeAxeeAxy 33332 2669 +++=′′

xAx e xx AeAxe 33 12 32 29 ++

Substituting into the differential equ yy 96 +′ we get

=

ation xey 3=−′′

( ) ( ) xxx eeAxAxe 3323 92 =++ xxxx eAxAeAxeeAx 323332 362129 −++xx eAe 332 =

⇒ 12 =A

21

=A ⇒

xp exy 32

21

So, =

The general solution is xx exeCxCy 32321 2

1)( ++=

(b) The homogeneous solution can be fouhy nd using the reduced equation

0=′−′′ yy


02 =− DD

0)1( =−DD

11 =r , and 02 =r The roots are

21 CeCyh += x

Since then let . But, )2sin(5)( xexF −= x )2sin()2cos( xCxBAey xp ++=

xAe is similar to the first term of the homogeneous solution so, let

108


)2in()2cos( xxBAxey xp += sC+

)2cos(2)2sin(2 xCxBAeAxey xxp +−+=′

)2sin(4)2cos(4 xCxBAeAeAxey xxxp −−++=′′

)2sin(4)2cos(42 xCxBAeAxe xx −−+=

Substituting into the differential equation we get )2sin(5 xeyy x −=′′′ −

( ))2sin(4)2cos(42 xCxBAeAxe xx −−+

( ) )2sin(5)2cos(2)2sin(2 xexCxBAeAxe xxx =+−+− −

( ) ( ) )2sin(5)2sin(42)2cos(24 xexCBxCBAe xx −=−++−

⇒ 5=A , ( ) 024 =+ CB ( ) 142 −=− CB ,

, 101

−=Bor 5=A , 51

= C

So, )2sin(51)2cos(

1015 xxxey x

p +−=

The general solution is

)2sin(51)2cos(

101521 xxxeCeCyyy x== x

ph +−+++

(c) The homogeneous solution can be found using the reduced equation hy

02 =−′−′′ yyy


022 =−− DD

109


110

0)1)(2( =+− DD

The roots are 21 =r , and 12 −=r

xxh eCeCy −+= 2

21

Since then let

34)( xxF =

DCxBxAxyp +++= 23 ⇒ CBxAxyp ++=′ 23 2

BAxyp 26 +=′′

Substituting into the differential equation we get 342 xyyy =−′−′′

( ) ( ) 3232 422326 xDCxBxAxCBxAxBAx =+++−++−+

( ) ( ) ( ) 323 422226232 xDCBxCBAxBAAx =−−+−−++−−

2−=A ⇒

023 =+ BA ⇒ 3=B ⇒ 02)2(3 =+− B

0226 =−− CBA ⇒ 2)2(6 9−=C02)3( =− C−− ⇒

⇒ 2

15=D 022 =−− DCB ⇒ 02)9()3(2 =−−− D

5.7932 23 +−+−= xxxyp So,


5.7932 232

2 +−+−+= − xxxeCeCy xx 1


111

Example

129 2 −+=′′ xxy

02 =D ⇒ 021 == rr ⇒ 21 CxCyh +=

( )CBxAxxyp ++= 22

xyy =′−′′

02 =− DD

0)1( =−DD ⇒ 01 =r and 12 =r ⇒ xh eCCy 21 +=

( )BAxxyp +=

1235 +−=−′′ xeyy x

052 =−D

0)5)(5( =+− DD ⇒ 51 =r and 52 −=r

xxh eCeCy 5

25

1−+=

CBxAey xp ++=

234 3 +=+′−′′ xeyyy

0342 =+− DD

0)1)(3( =−− DD ⇒ 31 =r and 12 =r ⇒ xxh eCeCy 2

31 +=

BAxey xp += 3


112

)cos(66 xeyy x +=+′′

012 =+D ⇒ jr =1 and jr −=2 ⇒ 0=α , 1=β

)sin()cos( 21 xCxCyh +=

( ))sin()cos(3 xCxBxAey xp ++=

xxeyyy =+′−′′ 2

0122 =+− DD

0)1( 2 =−D ⇒ 121 == rr ⇒ ( ) xh eCxCy 21 +=

))(( 2 xp exBAxy +=

)2sin(2 xxyy =+′′

012 =+D ⇒ jr =1 and jr −=2 ⇒ 0=α , 1=β

)sin()cos( 21 xCxCyh +=

( ) ( ))2sin()2cos(2 xxCBxAxyp +++=

Notes:

To find the roots of an equation 0... 12

21

1 =+++++ −−−

nnnnn axaxaxax

r is a root of )(xf if 0)( =rf .

r is a repeated root of )(xf if 0)( =′ rf .

If r is a root then r must be a factor of na .

If r is a root then )(xf is divided by )( rx − .


113

Example

01834 23 =−−+ xxx

Factors of 18 are: ( )18,9,6,3,2,1 ±±±±±±

01618)1(3)1(4)1()1( 23 ≠−=−−+=f

01218)1(3)1(4)1()1( 23 ≠−=−−−−+−=−f

018)2(3)2(4)2()2( 23 =−−+=f ⇒ 21 =r .

383)( 2 −+=′ xxxf

0253)2(8)2(3)2( 2 ≠=−+=′f ⇒ 21 =r is not a repeated root.

2x x6+ 9+

x 2− 3x 24x+ x3− 18− 3xm 22x± 26x x3− 26xm x12± x9 18− x9m 18± 0 0

0962 =++ xx ⇒ ( ) 03 2 =+x ⇒ 332 −== rr .

Higher Order Differential Equation A general differential equation can be put in the form

)(... 01)1(

1)( xFyayayaya n

nn

n =+′+++ −−


114

Homogeneous Higher Order Differential Equation

It is homogeneous if 0)( =xF

Example

06116 =−′+′′−′′′ yyyy

06116 23 =−+− DDD

Factors of 6 are: ( )6,3,2,1 ±±±± .

06)1(11)1(6)1()1( 23 =−+−=f ⇒ 11 =r .

11123)( 2 +−=′ DDDf

0211)1(12)1(3)1( 2 ≠=+−=′f ⇒ 11 =r is not a repeated root.

2D D5− 6+

D 1− 3D 26D− D11+ 6− 3Dm 2D± 25D− D11+ 25D± D5m D6 6− D6m 6± 0 0

0652 =+− DD

0)3)(2( =−− DD ⇒ 22 =r and 33 =r

xrxrxrh eCeCeCy 321

321 ++=

xxxh eCeCeCy 3

32

21 ++=


115

Example

03626 =+′+′′−′′′ yyyy ⇒ 03626 23 =++− DDD

Factors of 36 are: ( )...,3,2,1 ±±±

03336)1(2)1(6)1()1( 23 ≠=++−=f

02736)1(2)1(6)1()1( 23 ≠=+−+−−−=−f

02436)2(2)2(6)2()2( 23 ≠=++−=f

036)2(2)2(6)2()2( 23 =+−+−−−=−f ⇒ 21 −=r

2123)( 2 +−=′ DDDf

0382)2(12)2(3)2( 2 ≠=+−−−=−′f ⇒ 21 −=r is not a repeated root.

2D D8− 18+

D 2+ 3D 26D− D2+ 36+ 3Dm 22Dm 28D− D2+ 28D± D16± D18 36+ D18m 36m 0 0

01882 =+− DD

272648

)1(2)18)(1(4)8(8

24 22

3,2−

=−

=−−

=mmm

AACBBr

243,2 jr m= ⇒ 242 jr += & 243 jr −=

⇒ 4=α & 2=β

( ) ( )( )xCxCeeCy xxrh ββα sincos 321

1 ++=

( ) ( )( )xCxCeeCy xxh 2sin2cos 32

421 ++= −


116

Example

033 =+′+′′+′′′ yyyy

0133 23 =+++ DDD

Factors of 1 are: 1±

081)1(3)1(3)1()1( 23 ≠=+++=f

01)1(3)1(3)1()1( 23 =+−+−+−=−f ⇒ 11 −=r

363)( 2 ++=′ DDDf

03)1(6)1(3)1( 2 =+−+−=−′f ⇒ 12 −=r

66)( +=′′ DDf

06)1(6)1( =+−=−′′f ⇒ 13 −=r

xrxrxrh eCeCeCy 321

321 ++=

xxxh exCxeCeCy −−− ++= 2

321

Example

0168)4( =+′′+ yyy

0168 24 =++ DD

( ) 04 22 =+D

422,1 −=r ⇒ 22,1 jr ±= ⇒ 01 =α & 21=β

424,3 −=r ⇒ 24,3 jr ±= ⇒ 02 =α & 22=β

( ) ( )( ) ( ) ( )( )xCxCexCxCey xxh 24231211 sincossincos 21 ββββ αα +++=

( ) ( )( ) ( ) ( )( )xCxCxxCxCyh 2sin2cos2sin2cos 4321 +++=


117

Example

01632248)4( =+′+′′+′′′+ yyyyy

01632248 234 =++++ DDDD

Factors of 16 are: ( )16,8,4,2,1 ±±±±±

0116)1(32)1(24)1(8)1()1( 234 ≠=+−+−+−+−=−f

016)2(32)2(24)2(8)2()2( 234 =+−+−+−+−=−f ⇒ 21 −=r

3248244)( 23 +++=′ DDDDf

032)2(48)2(24)2(4)2( 23 =+−+−+−=−′f ⇒ 22 −=r

484812)( 2 ++=′′ DDDf

048)2(48)2(12)2( 2 =+−+−=−′′f ⇒ 23 −=r

4824)( +=′′′ DDf

048)2(24)2( =+−=−′′′f ⇒ 24 −=r

xrxrxrxrh eCeCeCeCy 4321

4321 +++=

xxxxh exCexCxeCeCy 23

422

32

22

1−−−− +++=

Non-homogeneous Higher Order Differential Equation A differential equation that has the form

)(... 01)1(

1)( xFyayayaya n

nn

n =+′+++ −−

The solution is

ph yyy +=


118

Example

)sec(xyy =′+′′′

First of all we find the homogeneous solution, i.e.,

0=′+′′′ yy

03 =+ DD

0)1( 2 =+DD ⇒ 01 =r ,

⇒ jr ±=3,2 ⇒ 0=α & 1=β

( ) ( )( )xCxCeeCy xxrh ββα cossin 321

1 ++=

( ) ( )xCxCCyh cossin 321 ++=

Now, we find the particular solution with )sec()( xxF =

332211 uvuvuvyp ++=

11 =u , )sin(2 xu = , )cos(3 xu =

∫= dxDetDv 1

1 , ∫= dxDetDv 2

2 , ∫= dxDetDv 3

3

321

321

321

uuuuuuuuu

Det′′′′′′′′′= ,

32

32

32

1

)(00

uuxFuuuu

D′′′′′′= ,

31

31

31

2

)(00

uxFuuuuu

D′′′′′′= ,

)(00

21

21

21

3

xFuuuuuu

D′′′′′′=


119

)cos()sin()sin()cos(

1)cos()sin(0)sin()cos(0

)cos()sin(1

xxxx

xxxx

xxDet

−−−

+=−−−=

( ) 1)(sin)(cos)(sin)(cos 2222 −=+−=−−= xxxxDet

)sin()cos()cos()sin(

)sec()cos()sin()sec()sin()cos(0

)cos()sin(0

1 xxxx

xxxxxx

xxD

−+=

−−−=

( ) )sec()1()sec()(cos)(sin)sec( 22 xxxxx −=−×=−−=

)cos()sec()sin(0

1)cos()sec(0)sin(00

)cos(01

2 xxx

xxx

xD

−−

+=−−=

)tan()sec()sin( xxx =×=

)sec()sin(0)cos(

1)sec()sin(0

0)cos(00)sin(1

3 xxx

xxxx

D−

+=−

=

1)sec()cos( =×= xx

∫∫∫ ++

×=== dxxxxxxdxxdx

DDv

)tan()sec()tan()sec()sec()sec(1

1

)tan()sec(ln)tan()sec(

)tan()sec()(sec2

xxdxxx

xxx+=

++

= ∫


120

)cos(ln)cos()sin()tan(2

2 xdxxxdxxdx

DDv =−=−== ∫∫∫

xdxdxDDv −=−== ∫∫ 3

3

332211 uvuvuvyp ++=

( ) )cos()sin()cos(ln)tan()sec(ln xxxxxx −++=

The general (complete) solution is

ph yyy +=

( ) ( ) ( ) )cos()sin()cos(ln)tan()sec(lncossin 321 xxxxxxxCxCC −+++++=

Example 23 64 xxyy +=′−′′′

The homogeneous solution is found by solving 0=′−′′′ yy

03 =− DD ⇒ 0)1( 2 =−DD

0)1)(1( =+− DDD ⇒ 01 =r , 12 =r & 13 −=r

xxh eCeCCy −++= 321

To find the particular solution, let

( )DCxBxAxxyp +++= 23 DxCxBxAx +++= 234

DCxBxAxyp +++=′ 234 23 ⇒ CBxAxyp 2612 2 ++=′′

BAxyp 624 +=′′′


121

( ) ( ) 2323 64234624 xxDCxBxAxBAx +=+++−+

44 =− A ⇒ 1−=A

63 =− B ⇒ 2−=B

0224 =− CA ⇒ 02)1(24 =−− C ⇒ 12−=C

06 =− DB ⇒ 0)2(6 =−− D ⇒ 12−=D

⇒ xxxxyp 12122 234 −−−−=


ph yyy += xxxxeCeCC xx 12122 234321 −−−−++= −

Example

)sin(18168)4( xyyy −=+′′−

The homogeneous solution is found using

0168)4( =+′′− yyy ⇒ 0168 24 =+− DD

( ) 04 22 =−D ⇒ 22,1 ±=r & 24,3 ±=r

xxxxh xeCxeCeCeCy 2

42

32

22

1−− +++=

To find the particular solution, let

)sin()cos( xBxAyp += ⇒ )cos()sin( xBxAyp +−=′

)sin()cos( xBxAyp −−=′′ ⇒ )cos()sin( xBxAyp −=′′′

)sin()cos()4( xBxAyp +=


122

( ) ( ))sin()cos(8)sin()cos( xBxAxBxA −−−+

( ) )sin(18)sin()cos(16 xxBxA −=++

( ) ( ) )sin(18)sin(168)cos(168 xxBBBxAAA −=+++++

⇒ 025 =A ⇒ 0=A

⇒ 1825 −=B ⇒ 2518−

=B

⇒ )sin(2518 xyp −=

ph yyy += )sin(25182

42

32

22

1 xxeCxeCeCeC xxxx −+++= −−


1) 23xyy =+′′ 2) 22 xyyy =+′+′′

3) xyyy 2732 =+′+′′ 4) )4sin(30 xyy −=+′′

5) )sin(6 xyy =+′′ 6) )cos(2)sin(34 xxyyy +=+′+′′

7) )cosh(1844 xyyy =+′+′′ 8) )cos(222 xeyyy x=+′−′′

9) )cos(1045)4( xyyy =+′′− 10) xeyyy 32 =−′+′′

11) xxyy +=+′′ 2 12) xeyy =−′′

13) xeyyy =+′−′′ 2 14) 234 124 xxxyyy ++=+′+′′

15) 34122 xyyyy −=−′−′′+′′′ 16) )cos(222 xeyyy x=+′−′′


Unit Step Function

The unit step function can be )(tua 1defined as

a

⎩⎨⎧

><

=atat

tua 10

)(

1

⎩⎨⎧

><

=00

10

)(0 tt

tu

Example

Express the following functions in terms of the unit step function

(a) (b) ⎩⎨⎧

><

=22

68

)(tt

tf⎪⎩

⎪⎨

⎧

><<

<=

btbta

atKtf0

0)(

Solution

(a) ⎩⎨⎧

><

−+=

22

20

8)(tt

tf⎩⎨⎧

><

−=22

10

28tt

)(28 2 tu−=

(b) )]()([)( tutuKtf ba −=

where and ⎩⎨⎧

><

=atat

tua 10

)(⎩⎨⎧

><

=btbt

tub 10

)(

123


Laplace Transform Let )(tf be a function of t . Then the Laplace Transform of )(tf is

{ } ∫∞

−==0

)()()( dtetfsFtf stL

Laplace Transform for some Functions

1)( =tf

{ } { } [ ]s

ees

es

dtetf stst 111).1(1)( 0

00

=−−=−=== ∞−∞

−∞

−∫LL

So, { }skk =L

ate tf =)(

{ }∞

−−∞

−−∞

−

−−=== ∫∫

0

)(

0

)(

0

1. tastasstatat eas

dtedteeeL

{ }asas

eat

−=−

−−=

1)10(1L

)cos(at , )sin(at

)sin()cos( atjate jat +=

{ } { } { )sin()cos( atjate jat LLL += }

{ } 22

1asjas

jasjas

jase jat

++

=++

×−

=L

{ } 2222 asaj

asse jat

++

+=L

By comparison ⇒ { } 22)cos(as

sat+

=L & { } 22)sin(as

aat+

=L

124


( )atat eeattf −−==21)sinh()(

{ } 22222111

21)sinh(

asa

asasas

asasat

−=

−+−+

=⎟⎠⎞

⎜⎝⎛

+−

−=L

( )atat eeattf −+==21)cosh()(

{ } 22222111

21)cosh(

ass

asasas

asasat

−=

−−++

=⎟⎠⎞

⎜⎝⎛

++

−=L

ttf =)(

{ } ∫∞

−=0

. dtett stL

tu = ⇒ dtdu = , dtedv st−= ⇒ stes

v −−=1

{ } ( )02

02

00

11001 ees

es

dtes

estt ststst −

−=−−=+

−= ∞−

∞−

∞−

∞− ∫L

{ } 2

1s

t =L

In general, { } 1

!+= n

n

sntL

)()( tutf =

{ }s

es

dtedtetutu ststst 11)1()()(000

=−

===∞

−∞

−∞

− ∫∫L

)( )( tutf a=

125


{ }s

ees

dtedtetutuas

a

st

a

ststaa

−∞−

∞−

∞− =

−=== ∫∫

1)1()()(0

L

Laplace Transform Properties 1) Linearity

If and { )()( 11 tfsF L= } { })()( 22 tfsF L= then

{ } )()()()( 22112211 sFCsFCtfCtfC +=+L

Example

{ }1

154

3!245)2cos(34 232

+×+

+×−×=+− −

sss

sett tL

15

438

23 ++

+−=

sss

s

2) Shifting Property

If { })( then )( tfsF L=

{ } )()( asFtfeat −=L

If { })( then )( tfsF L=

{ } asesFatf −=− )()(L

Example

{ })2cos( te t−L

Here, )2cos()( ttf = & , then 1−=a4

)( 2 +=

sssF and

126


{ }4)1(

1)1()2cos( 2 +++

=+=−

sssFte tL

3) Derivative Property

If then { )()( tfsF L= }

}

{ } )0()()( fssFtf −=′L

{ } )0()0()()( 2 fsfsFstf ′−−=′′L

4) Integral Property

If then { )()( tfsF L=

ssFduuf

t )()(0

=⎭⎬⎫

⎩⎨⎧∫L

Example

⎭⎬⎫

⎩⎨⎧∫t

duu0

)2sin(L

Here, )2sin()( ttf = then 4

2)( 2 +=

ssF and

)4(2)()2sin( 2

0 +==

⎭⎬⎫

⎩⎨⎧∫ sss

sFduut

L

5) Multiplication by nt


{ } n

nnn

dssFdtft )()1()( −=L

127


Example

{ })sin(2 ttL

Here, )sin()( ttf = then 1

1)( 2 +=

ssF

{ }⎭⎬⎫

⎩⎨⎧

+−=

11)1()sin( 22

222

sdsdttL

222 )1(2

11

+−

=⎭⎬⎫

⎩⎨⎧

+ ss

sdsd

[ ]42

222

42

222

22

2

)1(8)1)(2()1(

)1()2)(1(22)2()1(

11

+++−+

=+

+×+−×+=

⎭⎬⎫

⎩⎨⎧

+ ssss

sssss

sdsd

32

2

)1(26

+−

=ss

6) Division by t


∫∞

=⎭⎬⎫

⎩⎨⎧

s

duuFttf )()(L

Example

⎭⎬⎫

⎩⎨⎧

tt)sin(L

Here, )sin()( ttf = then 1

1)( 2 +=

ssF and

∞−∞

=+

=⎭⎬⎫

⎩⎨⎧

∫ ss

uduut

t )(tan1

1)sin( 12L ⎟

⎠⎞

⎜⎝⎛=−= −−

ss 1tan)(tan

211π

128


7) Initial-Value Property

If then { )()( tfsF L= })(lim)(lim

0ssFtf

st ∞→→=

8) Final-Value Property

If then { )()( tfsF L= })(lim)(lim

0ssFtf

st →∞→=

Gamma Function

The gamma function can be defined as )(nΓ

∫∞

−−=Γ0

1)( dtetn tn

Important Properties of the Gamma Function

)( )1( nnn Γ=+Γ

!)1 for ,....2,1, ( nn =+Γ 0=n

π=⎟⎠⎞

⎜⎝⎛Γ

21

Example

∫∞

−=+Γ0

)1( dtetn tn

ntu = ⇒ dttndu n 1. −= , dtedv t−= ⇒ tev −−=

)()1(0

1

0nndtetnten tnnt Γ=+−=+Γ ∫

∞−−∞−

129


( ) 1)1()1( 0

00

=−−=−==Γ ∞−∞−∞

−∫ eeedte tt

So, 1)1( =Γ

1)1(1)2( =Γ×=Γ

2)2(2)3( =Γ×=Γ

!323)3(3)4( =×=Γ×=Γ

!)1( nn =+Γ

π21

21

21

23

=⎟⎠⎞

⎜⎝⎛Γ=⎟

⎠⎞

⎜⎝⎛Γ

π8

1521

21

23

25

23

23

25

25

25

27

=⎟⎠⎞

⎜⎝⎛Γ××=⎟

⎠⎞

⎜⎝⎛Γ×=⎟

⎠⎞

⎜⎝⎛Γ=⎟

⎠⎞

⎜⎝⎛Γ

{ } 11

)1(!++

+Γ== nn

n

sn

sntL

{ } 2/52/52/52/52/3 4

321

21

23

23

23

25

sssst

π=

⎟⎠⎞

⎜⎝⎛Γ×

=⎟⎠⎞

⎜⎝⎛Γ

=⎟⎠⎞

⎜⎝⎛Γ

=L

Laplace Transform of Periodic Functions

If )(tf is a periodic function with a period of 0>T such that )()( tfTtf =+

then

{ } sT

Tst

e

dtetftf −

−

−=∫

1

)()( 0L

130


Evaluation of Integrals


)()(0

sFdtetf st =∫∞

− ⇒ )0()(0

Fdttf =∫∞

This can be used in finding a lot of integrals. For example

{ }1

1)sin()sin( 20 +

==∫∞

−

stdtte st L

{ }51

1)2(1)sin()sin( 22

0

2 =+

===

∞−∫ s

t tdtte L

⎩⎨⎧

<<<<

−=

πππ

π 20

2)(

tt

tt

tf

Example

Find the Laplace transform for the the

signal shown: π

Solution π π2This is a periodic signal with a period

of

π2=T .

ssT

Tst

esF

e

dtetfsF π2

10

1)(

1

)()( −−

−

−=

−=∫

∫∫ −− −+=π

π

π

π2

01 )2()( dtetdtetsF stst

∫∫∫ −−− −+=π

π

π

π

π

π22

0

2 dtetdtedtet ststst

131


∫∫−−

−−−

−+−+−

=π

π

π

π

π

π

πππ 222

00

2 dts

es

tees

dts

este stst

ststst

and so on.

Solution

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

<<<<<<<<<<

=

:5443322110

:43210

)(

ttttt

tf

( ) ( ) ( ) ....)()(4)()(3)()(2)()()( 54433221 +−+−+−+−= tututututututututf

....)()()()()( 4321 ++++= tututututf

{ } ...)(432

++++=−−−−

se

se

se

setf

ssss

L

( )...1 32 ++++= −−−−

ssss

eees

es

s

ese

−

−

−⋅=1

1

Example

Express the function )(tf in terms of the

unit step function. Then find its Laplace

transform. 1

f )(t

3

2

1 2 3 4

132


Exercises

Find the Laplace Transform of the following functions

1) )cos()( tttf ω= Ans. ( )222

22

)(ωω

+

−=

sssF

2) )sin()( tttf ω= Ans. ( )222

2)(ωω+

=s

ssF

3) )cosh()( tattf = Ans. ( )222

22

)(asassF

−

+=

4) )sinh()( tattf = Ans. ( )222

2)(assasF

−=

5) ttetf 2)( = Ans. ( )212)(−

=s

sF

6) )cos()( 2 tetf t−= Ans. ( ) 122)( 2 ++

+=

sssF

7) ( ))sin()cos()( tBtAetf t ββα += − Ans. 22)()()(

βαβα

++++

=s

BsAsF

8) )()()( tuttf ππ−= Ans.2)(

sesF

sπ−

=

9) )()( ttutf = 2 Ans. sess

sF 22

21)( −⎟⎠⎞

⎜⎝⎛ +=

10) )sin()()( ttutf π= Ans.1

)( 2 +−

=−

sesF

sπ

133


11)

Ans.sesF

s )1(2)(π−−

=

12)

Ans.

π

2

0 t

)1( saes −+)( ksF =

13)

Ans.)1(

1)( sessF −+=

14) ttf =)( , 20 << t Ans.se

se

ssF

ss 2

2

2

2

21)(−−

−−=

15) )sin()( tKtf ω= ωπ /0 << t Ans.)(

)1()( 22

/

ωω ωπ

++

=−

seKsF

s

16) )cos()( tKtf ω= ωπ /20 << t Ans.)(

)1()( 22

/2

ω

ωπ

+−

=−

sesKsF

s

a

k

0 ta2 a3 a4

1

2

3

1 2 3 4

4

t

134


Find the Laplace Transform of the following periodic functions

1) ttf −= π)( , )20( π<< t Ans.)1()1(1)( 22

2

s

s

esesssF π

πππ−

−

−++−

=

2) 224)( ttf −= π , )20( π<< t Ans.)1(

24)24()( 23

222

−++−

= s

s

essessF π

π ππ

3) tetf =)( , )20( π<< t Ans.)1)(1(

1)( 2

)1(2

s

s

esesF π

π

−

−

−−−

=

4) ⎩⎨⎧

<<<<

=πππ2

00

)(ttt

tf Ans.)1(

)1(1

)( 2

2

s

ss

e

es

essF π

ππ π

−

−−

−

−−=

5) ⎩⎨⎧

<<<<

−=

πππ

π 20

)(tt

tt

tf Ans.)1(

)1(1)1()( 2

22

s

sss

e

es

eessF π

ππππ

−

−−−

−

−+−=

Inverse Laplace Transform

)(sF { } )()(1 tfsF =−L )(sF { } )()(1 tfsF =−L

s1

1 22

1as +

aat)sin(

2

1s

t 22 ass+

)cos(at

1

1+ns

!n

t n

22

1as −

a

at)sinh(

as −1

ate 22 ass−

)cosh(at

135


Example

Find )(tf if

(a) , (b) 11)( 2 +

+=

sssF

35)(+

=s

sF , (c) 2)25(1)(+

=s

sF ,

1)2(2)( 2 ++

+=

sssF , (e)

4)1()( 2 −−=

sssF , (f)

)1(1)( 22 +

=ss

sF , (d)

1024)( 2 ++

=ss

sF (g)

Solution

(a)

(b)

tetf 35)( −=

1

111 22 ++

1)( 2 ++=

+=

sss

sssF ⇒ )sin()cos()( tttf +=

{ } )()( asFtfeat −=L then { } )()(1 tfeasF at=−−L . (c) Using the shifting property

Here, we have 2

1)(s

sF = with 25−=a . Since, t=⎭⎬⎫ then

s⎩⎨⎧

2

1−1L

tets

252

1

)25(1 −− ⋅=

⎭⎬⎫

⎩⎨⎧

+L

(d) Here, we have a shifting of with 2−1

)( 21 +=

sssF . So, and )cos()(1 ttf =

tettfs

s 22

1 )cos()(1)2(

2 −− ⋅==⎭⎬⎫

⎩⎨⎧

+++L

136


4)1(1

4)1(1

4)1(11

4)1()(e) ( 2222 −−

+−−

−=

−−+−

=−−

=ss

sss

ss

So,

sF

)2sinh(21)2cosh()( tetetf tt +=

(f) We know that )sin(1

11 t=⎫⎧−2s ⎭

⎬⎩⎨ +

L and using the property of division by s which

means an integration in time domain, we get

)cos(1)sin(1

111 duut

−==⎫⎧ ⋅−L0

2 tss ⎭

⎬⎩⎨ + ∫

Again using the same property we get

( ) )sin()cos(11

110

221 ttduu

ss

t

−=−=⎭⎬⎫

⎩⎨⎧

+⋅ ∫−L

(g9)1(

410112

4)( 22 +=

+−++=sF )

1024

2 +=

++ sssss

This is 9

4)( 21 +=

ssF with a shifting of 1−=a . So, )3sin(

34)(1 ttf = and

)3sin(3

)( tetf = 4 t−

of Inverse Using Partial Fraction Method

Example

Solution

Find if 32

73)( 2 −−+

=ss

ssF )(tf

137


138

Solution

31)3)(1(73

−+

+=

−++

sB

sA

sss

)1()3(73 ++−=+ sBsAs ⇒

First Method

BBsAAss ++−=+ 373 3=+ BA ⇒

⇒ 73 +− B =A

1−=A and 4=B Solving these two equations we get

Second Method

)1(7 ++=+ )3(3 −s A s sB

At 1−=s A473 −=+− 1⇒ −=A we get

3=s we get B479 =+ ⇒ 4=B At

Third Method

14

431

7)1(3)3(73

1

−=−

=−−+−

=−+

=−=ss

sA

44

1613

7)3(3)1(73

3

==++

=++

==ss

sB

34

11

)3)(1(73)(

−+

+−

=−+

+=

ssssssF

o

tt eetf 34)( +−= − S


Example

Find )(tf if )1)(1(

13)( 2 +−+

=ss

ssF

Solution

11)1)(1(13

22 ++

+−

=+−

+s

CBssA

sss

224

1)1(1)1(3

113

21

2 ==++

=++

==ss

sA

So 11

2)1)(1(

1322 ++

+−

=+−

+s

CBssss

s ⇒ )1)(()1(213 2 −+++=+ sCBsss

CCsBsBsss −+−++=+ 22 2213

CsBCsBs −+−++=+ 2)()2(13 2 ⇒ 02 =+ B ⇒ 2−=B

⇒ 3=− BC ⇒ 1=C

112

12

)1)(1(13)( 22 +

+−+

−=

+−+

=s

ssss

ssF

11

12

12)( 22 +

++

−−

=ss

ss

sF

So )sin()cos(22)( ttetf t +−=

Note:

If )(tf has the form of niss

K)( −

then the partial fraction of it will be

ni

nn

i

n

iin

i ssC

ssC

ssC

ssC

ssK

)()(......

)()()( 11

221

−+

−++

−+

−=

− −−

139


( )iss

nin sssFC

=−= )( ( )[ ]

iss

nin sssF

dsdC

=− −= )(

!11

1

( )[ ]iss

nin sssF

dsdC

=− −= )(

!21

2

2

2

or in general ( )[ ]iss

nik

k

kn sssFdsd

kC

=− −= )(

!1

Example

Find )(tf if 3)1(1)(

+−

=sssF

Solution

323 )1()1()1()1(1

++

++

+=

+−

sC

sB

sA

ss

211

−=−=−=s

sC

( ) 11!1

11

=−=−=s

sdsdB

( ) 0)1(211

!21

112

2

==−=−=−= ss ds

dsdsdA

So 32 )1(2

)1(1)(

+−

+=

sssF

⇒ ( )2)( ttetf t −= −

140


Example

22222 )4()4()4(4)(

++

+++

=+−

=s

DCss

BAss

ssF ( )WH .

Another Solution

∫∫ +−=

+− ds

ssds

ss

2222 )4(22

)4(4

)2sin(4

112

2 ts

=⎟⎠⎞

⎜⎝⎛

+−−

=

⇒ )2sin()( tttf −=

Example

)( ⇒ tan)( 1 ssF −= 211)(s

sF+

=′

[ ])()( sFdsdtft −⇔⋅ ⇒

11)( 2 +

⇔⋅−s

tft

)sin()( ttft =⋅− ⇒ t

ttf )sin()( −=

)2 ⇒ ln()( 2 += ssF2

2)( 2 +=′

sssF

( )ttft 2cos2)( =⋅− ⇒ ( )tt

tf 2cos2)( −=

141


2

)(4

−=

−

sesF

s

We know that . Here { } asa esGtuatg −=− )()()(L 4=a

)(2

1 2 tges

t =⇔−

⇒ )4(2)4( −=− tetg

)4()()( )4(24

)4(2 −== −− tuetuetf tt

Exercises

Find the Inverse Laplace Transform of the following functions

1) ss

sF+

= 2

1)( Ans. tetf −−=1)(

2) ss

sF4

1)( 3 += Ans. ( ) 4/)2cos(1)( ttf +=

3) ⎟⎠⎞

⎜⎝⎛

+−

=asas

ssF 1)( Ans. 12)( −= −atetf

4) 24 48)(

sssF

−= Ans. tttf 2)2sinh()( −=

5) 34 21)(

sssF

−= Ans. ( ) 8/221)( 22 ttetf t −−−=

6) ⎟⎠⎞

⎜⎝⎛

++

=111)( 22 s

ss

sF Ans. )sin()cos(1)( ttttf −−+=

142


7) 2/

1)( 2 sssF

+= Ans. 2/22)( tetf −−=

8) 23

1)(kss

sF−

= Ans. ( )kte

ktf kt −−= 11)( 2

9) ss

sF5

5)( 3 −= Ans. ( ) 15cosh)( −= ttf

10) 24 41)(

sssF

−= Ans. tttf

41)2sinh(

81)( −=

11) 222)2()(

ππ

nsnsF++

= Ans. )sin()( 2 tnetf t π−=

12) 1)3(

)( 2 ++=

sssF Ans. ( ))sin(3)cos()( 3 ttetf t −= −

13) s

eesFss )(2)(

42 −− −= Ans.

elsewheret

tf42

02

)(<<

⎩⎨⎧

=

14) 2)(

sesF

as−

= Ans.atatat

tf<>

⎩⎨⎧ −

=0

)(

143


144

15) 2)(

632 )3(s

eeee ssss −−−− +−+sF = Ans.

elsewherettt

ttt

tf633221

06

321

)(<<<<<<

⎪⎪⎩

⎪⎪⎨

⎧

−−−

=

16) 4

)( 2 +=

−

ssesF

sπ

Ans.elsewhere

tttf

π>

⎩⎨⎧

=0

)2cos()(

17) 22

)( 2 ++=

−

ssesF

sπ

Ans.elsewhere

ttetf

t ππ >

⎩⎨⎧−

=−

0)sin(

)(

18) ss

ssF4

12)( 2 ++

= Ans. tetf 423)( −−=

19) 82

3)( 2 −+=

ssssF Ans. tt eetf 242)( += −

20) )1)(3(

123)( 2

2

+−−−

=sssssF Ans. )sin()cos(2)( 3 ttetf t ++=

21) 2)2(410)(

−−

=s

ssF Ans. )42()( 2 −= tetf t

22) 22

23

)52(33)(

++−−+

=ss

ssssF Ans. ( ))2sin(2)2cos()( tttetf t −= −

23) 32

23

)2()1(9147)(

−−−+−

=ss

ssssF Ans. tt ettetf 225.0)( −=


Solution of Differential Equation Using Laplace Transform Here, we use the derivative property as follows:

{ } )()( sYty =L

{ } )0()()( yssYty −=′L

{ } )0()0()()( 2 ysysYsty ′−−=′′L

{ } )0()0()0()()( 23 yysyssYsty ′′−′−−=′′′L

Example

Solve the following differential equation using Laplace transform

(a) tyyy =+′+′′ 2 with , and 0)0( =y 1)0( =′y

(b) 4=+′ yy with 0)0( =y

(c) )sin(tyy =+′ with 1)0( =y

(d) tyyt =+′

(e) 0=+′−′′ yytyt with , and 0)0( =y 1)0( =′y

Solution

(a) Taking the Laplace transform of the two sides, we get

( ) ( ) 22 1)()0()(2)0()0()(

ssYyssYysysYs =+−+′−−

( ) ( ) 22 1)(0)(21)0()(

ssYssYssYs =+−+−−

22 1)()(21)(

ssYssYsYs =++−

( ) 1112)( 22 +=++

ssssY

( ) 2

22 112)(

sssssY +

=++

145


( ) ( ) ( ) ( )2222

2

22

2

1111

121)(

++

+++=

++

=++

+=

sD

sC

sB

sA

sss

sssssY

( )1

11

02

2

=++

==ss

sB

( )( ) ( ) ( )( )

( )2

12

111221

11

04

22

02

2

−=−

=+

++−+=⎥

⎦

⎤⎢⎣

⎡++

=== ss

sssss

ss

dsdA

21

12

2

=+

=−=ss

sD

( ) ( )( ) 21

422121

14

22

12

2

=+−

=+−

=⎥⎦

⎤⎢⎣

⎡ +=

−=−= sss

sssss

sdsdC

( )22 12

1212)(

++

+++

−=

sssssY ⇒ tt etetty −− ⋅+++−= 222)(

(b) Taking the Laplace transform of the two sides, we get

ssYyssY 4)()0()( =+− ⇒

ssYssY 4)(0)( =+−

( )s

ssY 41)( =+ ⇒ ( )14)(+

=ss

sY

( ) 114

++=

+ sB

sA

ss

( ) 41

4

0

=+

==ss

A , and 441

−==−=ss

B

144)(+

−=ss

sY ⇒ tety −−= 44)(

146


(c) Taking the Laplace transform of the two sides, we get

11)()0()( 2 +

=+−s

sYyssY ⇒ ( ) 11

11)( 2 ++

=+s

ssY

( )1

111)( 2

2

+++

=+s

sssY ⇒ ( )( )112)( 2

2

+++

=ss

ssY

( )( ) 11112

22

2

++

++

=++

+s

CBss

Ass

s

( ) 23

1)1(2)1(

12

2

2

12

2

=+−+−

=++

=−=ss

sA

( ) ( )( )112 22 ++++=+ sCBssAs

( ) ( ) CAsCBsBAs +++++=+ 22 2

BA+=1 ⇒21

231 −=−=B , 0=+CB ⇒

21

=C

( ) ( )1

2/12/112/3)( 2 +

+−+

+=

ss

ssY

11

21

121

11

23)( 22 +

⋅++

⋅−+

⋅=ss

ss

sY ⇒ )sin(21)cos(

21

23)( ttety t +−= −

(d) Taking the Laplace transform of the two sides, we get

( ) 2

1)()0()(s

sYyssYdsd

=+−−

⇒ ( ) 2

1)(0)(s

sYssYdsd

=+−−

( ) 2

1)()(s

sYssYdsd

=+−

⇒ ( ) 2

1)()()(s

sYsYsYs =++′−

2

1)(s

sYs =′− ⇒ 3

1)(s

sY −=′ ⇒ 3

1)(sds

sdY −=

147


dss

sdY 3

1)( −= ⇒ ∫

−= ds

ssY 3

1)( ⇒ 221)(s

sY = ⇒ tty21)( =

(e) Taking the Laplace transform of the two sides, we get

( ) ( ) 0)()0()()0()0()(2 =+−−

−′−−− sYyssY

dsdysysYs

dsd

( ) ( ) 0)(0)(10)(2 =+−+−−− sYssY

dsdsYs

dsd

( ) ( ) 0)()(1)(2 =++−− sYssY

dsdsYs

dsd

( ) ( ) 0)()()()2()()(2 =++′+×+′− sYsYsYsssYsYs

( ) ( ) 0)(22)(2 =+−+′+− sYssYss ⇒ ( ) ( ) )(22)(2 sYssYss −=′+−

)(22)()( 2 sYss

sds

sdYsY+−−

==′ ⇒ ( )( )dsss

ssYsdY

112

)()(

−−−

=

∫∫ −= ds

ssYsdY 2)()(

⇒ ( ) )ln(2)(ln ssY −= ⇒ ( ) )ln()(ln 2−= ssY

( ) ⎟⎠⎞

⎜⎝⎛= 2

1ln)(lns

sY ⇒ 2

1)(s

sY = ⇒ tty =)(

Example

Solve the following differential equations

(a) 21 yy −=′ 1)0(1 =y

12 yy =′ 0)0(2 =y

(b) yxdtdx 32 −= 8)0( =x

xydtdy 2−= 3)0( =y

148


Solution

(a) Taking the Laplace transform of the two equations, we get

)()0()( 211 sYyssY −=− ⇒ )(1)( 21 sYssY −=− ⇒ 1)()( 21 =+ sYssY

)()0()( 122 sYyssY =− ⇒ )(0)( 12 sYssY =− ⇒ 0)()( 21 =+− ssYsY

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− 0

1)()(

11

2

1

sYsY

ss

11

10

11

)( 21 +=

−

=s

s

ss

ssY ⇒ )cos()(1 tty =

11

11011

)( 22 +=

−

−=

ss

s

s

sY ⇒ )sin()(2 tty =

(b) Taking the Laplace transform of the two equations, we get

)(3)(2)0()( sYsXxssX −=−

)(3)(28)( sYsXssX −=−

( ) 8)(3)(2 =+− sYsXs )1(...

)(2)()0()( sXsYyssY −=−

)(2)(3)( sXsYssY −=−

( ) 3)(1)(2 =−+ sYssX )2(...

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−38

)()(

1232

sYsX

ss

149


43178

623988

23)1)(2(33)1(8

123213

38

)( 22 −−−

=−+−

−−=

×−−−×−−

=

−−

−=

sss

sss

sss

ss

ssX

41)4)(1(178)(

−+

+=

−+−

=s

Bs

Ass

ssX

5525

4117)1(8

4178

1

=−−

=−−−−

=−−

=−=ss

sA

35

151417)4(8

1178

4

==+−

=+−

==ss

sB

43

15)(

−+

+=

sssX ⇒ tt eetx 435)( += −

43223

6231663

23)1)(2(28)2(3

12323282

)( 22 −−−

=−+−

−−=

×−−−×−−

=

−−

−

=ss

sss

sss

s

ss

s

sY

41)4)(1(223)(

−+

+=

−+−

=sD

sC

ssssY

5525

4122)1(3

4223

1

=−−

=−−−−

=−−

=−=ss

sC

2510

1422)4(3

1223

4

−=−

=+−

=+−

==ss

sD

42

15)(

−−

+=

sssY ⇒ tt eety 425)( −= −

150



1) 04 2 =+′′ yy π , 2)0( =y , 0)0( =′y .

2) 02 =+′′ yy ω , Ay =)0( , By =′ )0( .

3) 082 =−′+′′ yyy , 1)0( =y , 8)0( =′y .

4) 032 =−′−′′ yyy , 1)0( =y , 7)0( =′y .

5) 0=′−′′ yky , 2)0( =y , ky =′ )0( .

6) 02 2 =−′+′′ ykyky , 2)0( =y , ky 2)0( =′ .

7) 04 =+′ yy , 8.2)0( =y

8) )2sin(1721 tyy =+′ , 1)0( −=y .

9) 06 =−′−′′ yyy , 6)0( =y , 13)0( =′y .

10) 041

=−′′ yy , 4)0( =y , 0)0( =′y .

11) 044 =+′−′′ yyy , 1.2)0( =y , 9.3)0( =′y

12) 022 =+′+′′ yyy , 1)0( =y , 3)0( −=′y .

13) teyyy 321127 =+′+′′ , 5.3)0( =y , 10)0( −=′y .

14) teyy −=+′′ 109 , 0)0( =y , 0)0( =′y .

15) 64925.23 3 +=+′+′′ tyyy , 1)0( =y , 5.31)0( =′y

16) )2cos(2956 tyyy =+′−′′ , 2.3)0( =y , 2.6)0( =′y

17) 022 =+′+′′ yyy , 0)0( =y , 1)0( =′y .

18) 0172 =+′+′′ yyy , 0)0( =y , 12)0( =′y .

151


152

19) 054 =+′−′′ yyy , 1)0( =y , 2)0( =′y .

20) 069 =+′−′′ yyy , 3)0( =y , 1)0( =′y .

21) 0102 =+′−′′ yyy , 3)0( =y , 3)0( =′y .

22) 03744 =+′−′′ yyy , 3)0( =y , 5.1)0( =′y

23) 0584 =+′−′′ yyy , 0)0( =y , 1)0( =′y .

24) 025.1 =+′+′′ yyy , 1)0( =y , 5.0)0( −=′y

25) )cos(2 tyy =+′′ , 0)0( =′y, 2)0( =y .

26) 034 =+′−′′ yyy , 3)0( =y , 7)0( =′y .

27) teyyy 22 −=+′+′′ , 0)0( =y , 0)0( =′y .

28) )2nh(1032 si tyyy =−′+′′ , 0)0( =y , 4)0( =′y .

29) ( ))5sin(2)5cos(1025 ttyy −=+′′ , 1)0( =y , 2)0( =′y .

30) 21 yy −=′ , , 12 yy =′ 1)0(1 =y , 0)0(2 =y .

31) )cos(21 tyy =+′ 2 , 012 =+′ yy , 0)0(1 =y , 1)0(2 =y .

, 0)0(3 =y . 32) , , )sinh(221 tyy =′+′

tt eeyy −+=′+′ 231

teyy =′+′ 321)0)0( 21 =(= yy

33) , 02 321 =′+′+′− yyy 2421 +=′+′ tyy ,

2232 +=+′ tyy

0)0()0()0( 321 === yyy

34) , 211 3yyy +=′′ , teyy 44 12 −=′′ 2)0(1 =y , 3)0(1 =′y , 1)0(2 =y , 2)0(2 =′y .

35) )2cos(521 tyy −=+′′ , )2cos(512 tyy =+′′ , 1)0()0( 11 =′= yy , 1)0(2 −=y ,

1)0(2 =′y .


Fourier Series

Periodic Function

A function )(xf is said to have a period T or to be periodic with period T if for

all t , ))( (tfTtf =+ , where T is a positive constant. The least value of 0>T is

called the period of )(tf .

Example

The function )sin(x has period π2 , since )sin()2sin( xx =+ π .

The period of )sin(nx or )cos(nx , where n is a positive integer, is n/2π .

Example

⎩⎨⎧

=<<−<<

−= 10

0550

33

)( Periodx

xxf

⎩⎨⎧

=<<<<

= ππππ

22

00

)sin()( Period

xxx

xf

153


⎪⎩

⎪⎨

⎧=

<<<<<<

= 6644220

010

)( Periodxxx

xf

Exercises

Find the smallest positive period of the following functions

1) )cos(x Ans. π2

2) )sin(x Ans. π2

3) )2cos( x Ans. π

4) )2sin( x Ans. π

154


5) )cos( xπ Ans. 2

6) )sin( xπ Ans. 2

7) )2cos( xπ Ans. 1

8) )2sin( xπ Ans. 1

9) )cos(nx Ans. n/2π

10) )sin(nx Ans. n/2π

11) ⎟⎠⎞

⎜⎝⎛

kxπ2cos Ans. k

12) ⎟⎠⎞

⎜⎝⎛

kxπ2sin Ans. k

13) ⎟⎠⎞

⎜⎝⎛

knxπ2cos Ans.

nk

14) ⎟⎠⎞

⎜⎝⎛

knxπ2sin Ans.

nk

Fourier Series

Let )(tf is a periodic function with a period of T . The Fourier Series or Fourier

Expansion corresponding to )(tf is given by

∑∑∞

=

∞

=

++=1

01

00 )sin()cos()(n

nn

n tnbtnadtf ωω

where the Fourier coefficients and are na nb

∫−

=2/

2/0 )cos()(2 T

Tn dttntf

Ta ω ,...2,1,0=n

155


∫−

=2/

2/0 )sin()(2 T

Tn dttntf

Tb ω ,...2,1,0=n

with Tπω 2

0 =

and ∫−

=2/

2/0 )(1 T

T

dttfT

d

Example

Find the Fourier series corresponding to the function

⎩⎨⎧

=<<<<−

= 105005

30

)( Periodxx

xf

Solution

10=T , 510

220

πππω ===T

23)05(

103

1033

101)(

101)(1 5

0

5

0

5

5

2/

2/0 =−===== ∫∫∫

−−

xdxdxxfdxxfT

dT

T

5

0

5

0

2/

2/0 )

5sin(5

53)

5cos(3

102)cos()(2 nx

ndxnxdxxnxf

Ta

T

Tn

ππ

πω ×=== ∫∫−

156


0)sin(3)0sin()55

sin(3==⎥⎦

⎤⎢⎣⎡ −×= π

ππ

πn

nn

n

∫∫∫ ===−−

5

0

5

5

2/

2/0 )

5sin(3

51)

5sin()(

102)sin()(2 dxnxdxnxxfdxxnxf

Tb

T

Tn

ππω

[ ])cos(13)55

cos(13)5

cos(553 0

5

ππ

ππ

ππ

nn

nn

nxn

−=⎥⎦⎤

⎢⎣⎡ ×−=×=

oddnevenn

n⎩⎨⎧−+

=11

)cos( π ⇒ oddnevenn

nbn

⎩⎨⎧

=π/6

0

The corresponding Fourier series is

∑∑∞

=

∞

=

++=1

01

00 )sin()cos()(n

nn

n xnbxnadxf ωω

( ) ⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+= .......sin

51

53sin

31

5sin6

23)( xxxxf πππ

π

Notes:

0)sin( =πn , 0)2sin( =πn (cos , sin)(0,1)

⎪⎩

⎪⎨

⎧

==

−+=⎟

⎠⎞

⎜⎝⎛

,...11,7,3,...9,5,1

11

0

2sin

nn

evennnπ

⎩⎨⎧−+

=oddnevenn

n11

)cos( π

1)2cos( =πn

⎪⎩

⎪⎨

⎧

==

−+=⎟

⎠⎞

⎜⎝⎛

,...10,6,2,...12,8,4

11

0

2cos

nn

oddnnπ

(-1,0) (1,0)

(0,-1)

157


0 if ,...)cos()sin(2/

2/0

2/

2/0 == ∫∫

−−

T

T

T

T

dttkdttk ωω 3,2,1=k

Proof

2/

2/00

2/

2/0 )cos(1)sin( T

T

T

T

tkk

dttk−

−

−=∫ ωω

ω

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

×−⎟⎠⎞

⎜⎝⎛ ×−=

22cos

22cos1

0

TT

kTT

kk

ππω

( ) 0)cos()cos(1

0

=−−−= ππω

kkk

2/

2/00

2/

2/0 )sin(1)cos( T

T

T

T

tkk

dttk−

−

=∫ ωω

ω

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −

×−⎟⎠⎞

⎜⎝⎛ ×=

22sin

22sin1

0

TT

kTT

kk

ππω

( ) 0)sin()sin(1

0

=−−= ππω

kkk

⎩⎨⎧

=≠

== ∫∫−− nm

nmT

dttntmdttntmT

T

T

T 2/0

)sin()sin()cos()cos(2/

2/00

2/

2/00 ωωωω

where and assume any of the values m n ,...3,2,1Proof

Using the trigonometry )cos(21)cos(

21)cos()cos( BABABA ++−= then

If nm ≠

0))cos((21))cos((

21)cos()cos(

2/

2/0

2/

2/0

2/

2/00 =++−= ∫∫∫

−−−

T

T

T

T

T

T

tnmtnmdttntm ωωωω

158


Also, by using )cos(21)cos(

21)sin()sin( BABABA +−−= then

If nm ≠

0))cos((21))cos((

21)sin()sin(

2/

2/0

2/

2/0

2/

2/00 =+−−= ∫∫∫

−−−

T

T

T

T

T

T

tnmtnmdttntm ωωωω

If , we have nm =

( )∫∫−−

+=2/

2/0

2/

2/00 )2cos(1

21)cos()cos(

T

T

T

T

dttndttntm ωωω

22221)2cos(

21

21 2/

2/0

2/

2/

TTTdttndtT

T

T

T

=⎟⎠⎞

⎜⎝⎛ +=+= ∫∫

−−

ω

( )∫∫−−

−=2/

2/0

2/

2/00 )2cos(1

21)sin()sin(

T

T

T

T

dttndttntm ωωω

22221)2cos(

21

21 2/

2/0

2/

2/

TTTdttndtT

T

T

T

=⎟⎠⎞

⎜⎝⎛ +=−= ∫∫

−−

ω

Note that if 0== nm then TdttntmT

T

=∫−

2/

2/00 )cos()cos( ωω

and 0)sin()sin(2/

2/00 =∫

−

T

T

dttntm ωω

0 )cos()sin(2/

2/00 =∫

−

T

T

dttntm ωω

Proof

Using the trigonometry )sin(21)sin(

21)cos()sin( BABABA ++−=

If nm ≠

159


( ) ( )∫∫∫−−−

=++−=2/

2/0

2/

2/0

2/

2/00 0)(sin

21)(sin

21)cos()sin(

T

T

T

T

T

T

dttnmdttnmdttntm ωωωω

If , we have nm =

( )∫∫−−

==2/

2/0

2/

2/00 02sin

21)cos()sin(

T

T

T

T

dttndttntm ωωω

Example

Find the Fourier series corresponding to the function

⎩⎨⎧

=<<<<−−

= ππ

π2

00

)( Periodxx

kk

xf

Solution

π2=T , 1222

0 ===πππω

T

0)(21)(

21)(1

0

02/

2/0 =⎥

⎦

⎤⎢⎣

⎡+−=== ∫∫∫∫

−−−

π

π

π

π ππkdxdxkdxxfdxxf

Td

T

T

∫−

=2/

2/0 )cos()(2 T

Tn dxxnxf

Ta ω

⎥⎦

⎤⎢⎣

⎡+−= ∫∫

−

π

ππ 0

0

)cos()cos()(22 dxnxkdxnxk

160


0)sin()sin(10

0

=⎥⎦

⎤⎢⎣

⎡+−=

−

π

ππnx

nknx

nk

∫−

=2/

2/0 )sin()(2 T

Tn dxxnxf

Tb ω

⎥⎦

⎤⎢⎣

⎡+−= ∫∫

−

π

ππ 0

0

)sin()sin()(22 dxnxkdxnxk

⎥⎦

⎤⎢⎣

⎡−=

−

π

ππ 0

0

)cos()cos(1 nxnknx

nk

[ ]1)cos()cos(1 +−−= πππ

nnnk

[ ])cos(12 ππ

nn

k−=

oddnevenn

n⎩⎨⎧−+

=11

)cos( π ⇒ oddnevenn

nkbn

⎪⎩

⎪⎨⎧

=π

40

The corresponding Fourier series is

∑∑∞

=

∞

=

++=1

01

00 )sin()cos()(n

nn

n xnbxnadxf ωω

( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +++= .......5sin

513sin

31sin4)( xxxkxf

π

The partial sums are

)sin(41 xkS

π= , ⎥⎦

⎤⎢⎣⎡ += )3sin(

31)sin(4

2 xxkSπ

161


162


Example

Find the Fourier series of the periodic function 2)( xxf = π20 << x

Solution

π2=T ⇒ 1222

0 ===πππω

T

34

321

21)(1 22

0

32

0

2

00

πππ

ππ

==== ∫∫xdxxdxxf

Td

T

∫∫ ==π

πω

2

0

2

00 )cos(

22)cos()(2 dxnxxdxxnxf

Ta

T

n

( ) ( ) 2

2

032

2 4)sin(2)cos(2)sin(1nn

nxn

nxxnnxx =

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛=

π

π

∫∫ ==π

πω

2

0

2

00 )sin(

22)sin()(2 dxnxxdxxnxf

Tb

T

n

( ) ( )nn

nxn

nxxn

nxx ππ

π4)cos(2)sin(2)cos(1

2

032

2 −=

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=

163


So, ∑∑∞

=

∞

=

−+=11

2

2

)sin(4)cos(43

4)(nn

nxn

nxn

xf ππ

⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ +++= ...)2sin(

21)sin(4...)2cos(

41)cos(4

34)(

2

xxxxxf ππ

Exercises

Evaluate the following integrals where ,...2,1,0=n

1) ∫π

0

)sin( dxnx Ans.⎩⎨⎧

oddnevenn

n/20

2) ∫π

dxnxx )sin(−π

0

Ans.⎪⎩

⎪⎨

⎧

===

− ,...4,2,...3,1

0

/2/2

0

nnn

nn

ππ

3) ∫−

2/

2/

)cos(π

π

dxnxx Ans.

4) ∫−

0

)sin(π

dxnxex Ans. ( )( )( )21

11nen n

+−− −π

5) ∫−

π

π

dxnxx )cos(2 Ans. ( )⎩⎨⎧

==

− ,...2,10

/413/2

2

3

nn

nn ππ

164


Find the Fourier Series for the following periodic functions

1)

Ans. ⎟⎠⎞

⎜⎝⎛ −+−+ ...)5cos(

51)3cos(

31)cos(2

21 xxx

π

2)

Ans. ⎟⎠⎞

⎜⎝⎛ ++++ ...)5sin(

51)3sin(

31)sin(2

21 xxx

π

3)

Ans. ⎟⎠⎞

⎜⎝⎛ ++++ ...)5cos(

251)3cos(

91)cos(4

2xxx

ππ

4)

Ans.⎟⎠⎞

⎜⎝⎛ +++− ...)5cos(

251)3cos(

91)cos(2

4xxx

ππ

...)3sin(31)2sin(

21)sin( −+−+ xxx

5)

Ans.⎟⎠⎞

⎜⎝⎛ +++− ...)5cos(

251)3cos(

91)cos(4 xxx

π

⎟⎠⎞

⎜⎝⎛ ++++ ...)5sin(

51)3sin(

31)sin(2 xxx

165


Find the Fourier Series for the following periodic functions

1) ⎩⎨⎧

<<<<−

−=

2/32/2/2/

11

)(ππππ

xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)5cos(

51)3cos(

31)cos(4 xxx

π

2) xxf =)( ππ <<− x Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)3sin(

31)2sin(

21)sin(2 xxx

3) 2)( xxf = ππ <<− x Ans. ⎟⎠⎞

⎜⎝⎛ −+−− ...)3cos(

91)2cos(

41)cos(4

3

2

xxxπ

4) ⎩⎨⎧

<<<<−

−+

=π

πππ

xx

xx

xf0

0)( Ans. ⎟

⎠⎞

⎜⎝⎛ ++++ ...)5cos(

251)3cos(

91)cos(4

2xxx

ππ

5) ⎩⎨⎧

<<<<−

−=

2/32/2/2/

)(ππππ

π xx

xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)5sin(

251)3sin(

91)sin(4 xxx

π

6) ⎩ << 2/32/4/ πππ x⎨⎧ <<−

=2/2/

)(2

2 ππ xxxf Ans.

)3cos(27

4)2cos(21)cos(4

6

2

xxxππ

π+−−

...)4cos(81

−+ x

7) ⎩⎨⎧

<<<<−−

=1001

11

)(xx

xf Ans. ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +++ ...5sin

513sin

31sin4 xxx πππ

π

8) ⎩⎨⎧

<<<<−−

=2002

11

)(xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ ...

25sin

51

23sin

31

2sin4 xxx πππ

π

9) ⎩⎨⎧

<<<<−

=2002

10

)(xx

xf Ans. ⎟⎠

⎞⎜⎝

⎛ +⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+ ...

25sin

51

23sin

31

2sin2

21 xxx πππ

π

10) 2)( xxf = 11 <<− x Ans. ⎟⎠⎞

⎜⎝⎛ −+−− ...)3cos(

91)2cos(

41)cos(4

31

2 xxx ππππ

166


167

⎩ << 10⎨⎧ <<−

=010

)(x

xf11) xx

Ans.⎟⎠⎞

⎜⎝⎛ ++− ...)3cos(

91)cos(2

41

2 xx πππ

⎟⎠⎞

⎜⎝⎛ +−+ ...)2sin(

21)sin(1 xx ππ

π

12) )sin()( xxf π= 10 << x Ans. ⎟⎟⎠

⎞⎜⎜⎝

⎛++− ...)4cos(

)5)(3(1)2cos(

)3)(1(142 xx ππ

ππ

13) xxf =)( 11 <<− x Ans. ⎟⎠⎞

⎜⎝⎛ +++− ...)5cos(

251)3cos(

91)cos(4

21

2 xxx ππππ

14) 21)( xxf −= 11 <<− x Ans. ⎟⎠⎞

⎜⎝⎛ −+−+ ...)3cos(

91)2cos(

41)cos(4

32

2 xxx ππππ

15) ⎩⎨ << 102 xx⎧ <<−−

=011

)(x

xf Ans.⎟⎠⎞

⎜⎝⎛ ++− ...)3cos(

91)cos(4

2 xx πππ

⎟⎠⎞

⎜⎝⎛ +−+ ...)2sin(

21)sin(22 xx ππ

π

16)

311001

1)(

<<<<<<−

⎪⎩

⎪⎨

⎧−=

xxx

xx

xf Ans.

( ) ⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛++⎟

⎠⎞

⎜⎝⎛−

23cos

91cos

21

2cos4

43

2

xxx ππππ

( ) ...3cos181

25cos

251

++⎟⎠⎞

⎜⎝⎛+ xx ππ


Odd and Even Functions

A function )(xf is odd if )()( xfxf −=− . Thus 3x , xxx 23 35 +− , )sin(x ,

)3tan( x are odd functions. The figure below is an example of an odd function.

A function )(xf is even if )()( xfxf =− . Thus 4x , 542 26 +− xx , )cos(x ,

are even functions. The figure below is an example of an even function. xx ee −+

while the figure below is neither odd nor even function.

168


Example

Classify each of the following functions according as they are even, odd, or neither

even nor odd.

(a) ⎩⎨⎧

<<−<<

−=

0330

22

)(x

xxf 6=Period

(b) ⎩⎨⎧

<<<<

=πππ2

00

)cos()(

xxx

xf π2=Period

(c) )10()( xxxf −= 100 << x , 10=Period

Solution

(a)

From the figure above it is seen that )()( xfxf −=− , so that the function is odd.

(b)

From the above figure it is seen that the function is neither even nor odd

169


(c)

From the figure above it is seen that )()( xfxf =− , so that the function is even.

Note:

In the Fourier series corresponding to an odd function, only sine terms can be

present. In the Fourier series corresponding to an even function, only cosine terms (and

possibly a constant) can be present.

Exercises

Are the following functions even, odd, or neither even nor odd?

1) xe Ans. Neither even nor odd

2) 2xe Ans. Even

3) )sin(nx Ans. Odd

4) )sin(xx Ans. Even

5) xx /)cos( Ans. Odd

6) )ln(x Ans. Neither even nor odd

7) ( )2sin x Ans. Even

170


8) )(sin2 x Ans. Even

9) x Ans. Even

10) )sin(2 nxx Ans. Odd

11) 2xx + Ans. Neither even nor odd

12) xe− Ans. Even

13) )cosh(xx Ans. Odd

Are the following functions, which are assumed to be periodic, even, odd,

or neither even nor odd?

1) xxf =)( ππ <<− x Ans. Odd

2) xxxf =)( ππ <<− x Ans. Odd

3) ⎩⎨⎧

<<<<

=πππ2

00

)(xxx

xf Ans. Neither even nor odd

4) ⎩⎨⎧

<<<<−

=2/32/2/2/

0)(

ππππ

xxx

xf Ans. Odd

5) 3)( xxf = ππ <<− x Ans. Odd

6) xexf 4)( −= ππ <<− x Ans. Neither even nor odd

7) 3)( xxxxf −= ππ <<− x Ans. Odd

8) π

π<<<<−

⎩⎨⎧

+−+

=x

xx

xxf

00

)1/(1)1/(1

)( 2

2

Ans. Odd

171


Find the Fourier Series for the following periodic functions (Even & Odd)

1) ⎩⎨⎧

<<<<−

=2/32/2/2/

01

)(ππππ

xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ −+−+ ...)5cos(

51)3cos(

31)cos(2

21 xxx

π

2) ⎩⎨⎧

<<<<−

−=

2/32/2/2/

)(ππππ

π xx

xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)5sin(

251)3sin(

91)sin(4 xxx

π

3) ⎩⎨⎧

<<<<−−

=π

πxx

xx

xf0

0)( Ans. ⎟

⎠⎞

⎜⎝⎛ +++− ...)5cos(

251)3cos(

91)cos(4

2xxx

ππ

4) 4

)(2xxf = ππ <<− x Ans. ...)3cos(

91)2cos(

41)cos(

12

2

+−+− xxxπ

5) xxf −=π)( ππ <<− x Ans. ⎟⎠⎞

⎜⎝⎛ ++++ ...)5cos(

251)3cos(

91)cos(4

2xxx

ππ

6) ⎩⎨⎧

<<<<−

=2002

02

)(xx

xf Ans. ⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛− ...

25sin

51

23sin

31

2sin41 xxx πππ

π

Half Range Fourier Sine or Cosine Series A half range Fourier sine or cosine series is a series in which only sine terms or

only cosine terms are present respectively. When a half range series corresponding to a

given function is desired, the function is generally defined in the interval )2/,0( T

(which is half of the interval ),0( T , thus accounting for the name half range) and then

the function is specified as odd or even, so that it is clearly defined in the other half of

the interval. In such case, we have for odd functions (Sine Series)

172


00 =d , , 0=na dttntfT

bT

n ∫=2/

00 )sin()(4 ω

while for even functions (Cosine Series)

∫=2/

00 )(2 T

dttfT

d , dttntfT

aT

n ∫=2/

00 )cos()(4 ω , 0=nb

Example

Find the Fourier series for the periodic function

)sin()( xxf = π<< x0

Solution

π=T ⇒ 2220 ===

πππω

T

Since the function is even then 0=nb

( )ππππ

ππ 2012)cos(2)sin(2)(2 2/

0

2/

0

2/

00 =−=

−=== ∫∫ xdxxdxxf

Td

T

dxnxxdxxnxfT

aT

n ∫∫ ==2/

0

2/

00 )2cos()sin(4)cos()(4 π

πω

( ) ( )dxxndxxn ∫∫ ++−=2/

0

2/

0

)21(sin2)21(sin2 ππ

ππ

173


( ) ( ) 0

2/21)21(cos

21)21(cos2

ππ ⎥⎦⎤

⎢⎣⎡

++

+−−

=n

xnn

xn

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

+

⎟⎠⎞

⎜⎝⎛ +

−−

⎟⎠⎞

⎜⎝⎛ −

−+

+−

=n

n

n

n

nn 212

)21(cos

212

)21(cos

211

2112

ππ

π

14/4

41/4

4121212

222 −−

=−

=−

−++×=

nnnnn ππ

π

⎟⎠⎞

⎜⎝⎛ ++−= ...)4cos(

151)2cos(

3142)( xxxf

ππ

Example

Expand xxf =)( , 20 << x in a half range (a) sine series, (b) cosine series

Solution

(a) To get a sine series the function must be an odd function. So, we extend the given

function to have an odd function. This is called the odd extension of )(xf .

174


4=T ⇒24

220

πππω ===T

Since now the function is odd then 00 =d , and 0=na

dxxnxdxxnxfT

bT

n ∫∫ ⎟⎠⎞

⎜⎝⎛==

2

0

2/

00 2

.sin44)sin()(4 πω

( )2

022 2

.sin4)1(2.cos2

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

−⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

=xn

nxn

nx π

ππ

π

oddn

evenn

n

nn

n⎪⎪⎩

⎪⎪⎨

⎧−

=−

=

π

ππ

π 4

4

)cos(4

Then ⎟⎠⎞

⎜⎝⎛−

= ∑∞

= 2.sin)cos(4)(

1

xnnn

xfn

πππ

⎥⎦⎤

⎢⎣⎡ −⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛= ...

2.3sin

31

2.2sin

21

2.sin4 xxx πππ

π

(b) To get a cosine series the function must be an even function. So, we extend the given

function to have an even function. This is called the even extension of )(xf .

175


4=T ⇒24

220

πππω ===T

Since now the function is even then 0=nb

122

142)(2

2

0

22

0

2/

00 ==== ∫∫

xxdxdxxfT

dT

dxxnxdxxnxfT

aT

n ∫∫ ⎟⎠⎞

⎜⎝⎛==

2

0

2/

00 2

.cos44)cos()(4 πω

( ) ( )2

022 2

.cos412.sin2

⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−

−⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

xnn

xnn

x ππ

ππ

( )oddn

evenn

n

nn

⎪⎪⎩

⎪⎪⎨

⎧

−=−=

22

228

01)cos(4

π

ππ

Then ( )∑∞

=⎟⎠⎞

⎜⎝⎛−+=

1220 2

.sin1)cos(4)(n

xnnn

dxf πππ

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−= ...

2.5sin

251

2.3sin

91

2.sin81)( 2

xxxxf ππππ

Complex Notation for Fourier Series

The Fourier series for )(tf can be written in complex notation as

∑∞

−∞=

=n

tjnneCtf 0)( ω

where ∫−

−=2/

2/

0)(1 T

T

tjnn dtetf

TC ω

176


Example

Write an expression for the function )(xf in terms of the complex exponential

Fourier series.

1 2

1

-1

-1-2

Solution

2=T ⇒ πππω ===2

220 T

∫∫ −− ==2

00

)(21)(1

0 dxexfdxexfT

C xjnT

xjnn

πω ( ) ( )∫∫ −− −=2

1

1

0

1211

21 dxedxe xjnxjn ππ

( ) ( )ππππ

ππjnjnnjjn e

jneee

jn−−−− −=−++−= 111

21 2

evenn

oddnjnCn

⎪⎪

⎩

⎪⎪

⎨

⎧

=0

2π

∑∞

−∞=

=n

xjnneCxf 0)( ω

⎟⎠⎞

⎜⎝⎛ −−−−+++= −−− .....

51

31.....

51

312 5353 xjxjxjxjxjxj eeeeee

jππππππ

π

177


Exercises

Find the Fourier Sine Series for the following periodic functions

1) xxf =)( π<< x0 Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)3sin(

31)2sin(

21)sin(2 xxx

2) xxf =)( 10 << x Ans. ⎟⎠⎞

⎜⎝⎛ −+− ...)3sin(

31)2sin(

21)sin(2 xxx πππ

π

3) ⎩ <<⎨⎧ <<

=ππ

ππ x2/2/

xxxf

2/0)( Ans.

)2sin(21)sin(21 xx −⎟

⎠⎞

⎜⎝⎛ +

π

...)4sin(41)3sin(

92

31

+−⎟⎠⎞

⎜⎝⎛ −+ xx

π

4) ⎩ <<− 4/8/4/ ππ⎨⎧ <<

=8/0

)(π

π xxxx

xf Ans.⎜⎝⎛ +− )20sin(

251)12sin(

91)4sin(1 xxx

π

⎟⎠⎞+− ...)28sin(

491 x

Find (a) the Fourier Cosine Series, (b) the Fourier Sine Series

f xx −=2)( 2<< 01) x Ans.

(a) ⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+ ...

25cos

251

23cos

91

2cos81 2

xxx ππππ

(b) ( ) ( ) ⎟⎠

⎞⎜⎝

⎛ +⎟⎠⎞

⎜⎝⎛++⎟

⎠⎞

⎜⎝⎛ ...2sin

41

23sin

31sin

21

2sin4 xxxx ππππ

π

2) ⎩ << 212 x⎨⎧ <<

=101

)(x

xf Ans.

(a) ⎟⎠

⎞⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛− ...

25cos

51

23cos

31

2cos2

23 xxx πππ

π

(b) ( ) ⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+−⎟

⎠⎞

⎜⎝⎛ ...

25sin

51

23sin

31sin

31

2sin6 xxxx ππππ

π

178


179

3) xx =)( L xf <<0 Ans.

(a) ⎟⎠

⎞⎜⎝

⎛ +⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛− ...5cos

2513cos

91cos4

2 2 Lx

Lx

LxLL πππ

π

(b) ⎟⎠

⎞⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ ...3sin

312sin

21sin2

Lx

Lx

LxL πππ

π

4) xxf −=π)( << x π0 Ans.

(a) ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ ++++ ...5cos

2513cos

91cos4

2xxx

ππ

(b) ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +++ ...3sin

312sin

21sin2 xxx

Find the Complex Form of the Fourier Series for the following periodic

functions

f x x π <<− π=)( 1) x Ans. ∑∞

≠−∞=

−

0

)1(

nn

jnxn

en

j

2) xexf =)( ππ <<− x Ans. ∑∞

−∞= ++

−n

jnxn enjn

211)1()sinh(

ππ

3) 2)( xxf = π <<− πx Ans. ∑∞

≠−∞=

−+

0

2

2 )1(23

nn

jnxn

en

π

4) xxf π20 << x Ans. ∑∞

≠−∞=

+

0

1

nn

jnxen

jπ =)(


Matrices

Linearly Dependent & Linearly Independent Vectors

The vectors are linearly dependent if and only if mvvvv ,...,,, 321

0=m...321 vvvv . If 0...321 ≠mvvvv then are

linearly independent.

mvvvv ,...,,, 321

Example

, , ( )1,6,31 −=v ( )4,2,82 −=v ( )1,1,13 −=v

Since 41

261

1116

81412

3141126

183

−−+

−−

−−

−=

−−−

( ) ( ) ( ) 06822406224168423 ≠−=−−−=+−+−−−=

Then , , are linearly independent 1v 2v 3v

( )6,4,2 , ( )3,3,1 , 1 =v 2 =v ( )3,2,13 =v

Since 3634

13624

13323

2336234112

+−=

( ) ( ) ( ) 060618121212692 =−−=−+−−−=

Then , , are linearly dependent 1v 2v 3v

180


Linearly Dependent & Linearly Independent Functions

If are functions of myyyy ,...,,, 321 x then are linearly dependent

if where

myyyy ,...,,, 321

( ),...,,, 321 =myyy 0yw

( ))1()1()1(

21

21

321

...::::

...

...

,...,,,

21

−−−

′′′=

mmm

m

m

m

myyy

yyyyyy

yyyyw

if then are linearly independent. ( ) 0,...,,, 321 ≠myyyyw myyyy ,...,,, 321

Example

xey =1 , 22 xy =

( ) 0222

),( 22

21 ≠−=−== xxeexxexe

xeyyw xxx

x

x

So, and are linearly independent. 1y 2y

xe , xe y 41 = y 22 =

0882424

),( 2221 =−== xx

xx

xx

eeeeee

yyw

So, and are linearly dependent. 1y 2y

181


Exercises Are the Following Vectors Linearly Dependent or Independent?

1) ( )4,3 , ( ) 3,4 Ans. Independent

2) ( )5,4,1 , ( ), ( ) 8,4,4 0,3,3 − Ans. Dependent

3) ( )0,1,0 , ( ), ( ) 0,1,2 7,4,3 Ans. Independent

4) ( )4,0,2,3 − , ( ), ( ), 1,0,0,5 1,0,1,6− ( )3,0,0,2 Ans. Independent

Eigen Values & Eigen Vectors Let A be an matrix, a real or complex number nn× λ is called an Eigen value of

A if det( − I ) 0=A λ and with [ ] 0=− XIA λ then is called an Eigen vector with

respect to

X

λ where I is the identity matrix.

Example

Find the Eigen values and Eigen vectors of A if

(a) , (b) ⎥⎦

⎤⎢⎣

⎡−−

=21

21A ⎥

⎦

⎤⎢⎣

⎡=

1231

A , (c)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−=

100110011

A

Solution

(a) To find Eigen values, we have ( ) 0det =− IA λ then

02121

=−−−

−λ

λ ⇒ ( )( ) 0221 =+−−− λλ ⇒ 0222 2 =+++−− λλλ

⇒ 02 =+ λλ ⇒ ( ) 01 =+λλ ⇒ 01 =λ , 12 −=λ

182


To find the Eigen vectors, we have [ ] 0=− XIA λ

For 01 =λ ⇒ 021

21

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−− x

x

02 21 =+ xx ⇒ 21 2xx −=

Let The Eigen vector for 12 =x ⇒ 21 −=x ⇒ 01 =λ is ⎥⎦

⎤⎢⎣

⎡−12

For 12 −=λ ⇒ 011

22

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−− x

x

022 21 =+ xx ⇒ 21 22 xx −= ⇒ 21 xx −=

Let ⇒ 12 =x 11 −=x The Eigen vector for ⇒ 12 −=λ is ⎥⎦

⎤⎢⎣

⎡−11

(b) To find Eigen values, we have ( ) 0det =− IA λ then

012

31=

−−

λλ

⇒ ⇒ ( ) 061 2 =−− λ 0621 2 =−+− λλ

⇒ 0522 =−− λλ

⇒ 611 +=λ , 612 −=λ


For 611 +=λ ⇒ 062

36

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−

xx

036 21 =+− xx ⇒ 21 36 xx −=− ⇒ 21 63 xx =

183


Let ⇒ 12 =x6

31 =x The Eigen vector for ⇒ 611 +=λ is

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

16

3

For 612 −=λ ⇒ 062

36

2

1 =⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡xx

036 21 =+ xx ⇒ 21 36 xx −= ⇒ 21 63 xx −

=

Let ⇒ 12 =x63

1−

=x The Eigen vector for ⇒ 612 −=λ is ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−

163

(c) To find Eigen values, we have ( ) 0det =− IA λ then

0100110011

=−−

−−−

λλ

λ ⇒ ( ) 0

1010

11011

1 =−−

+−−

−−

λλλ

λ

⇒ ( ) ( ) 011 2 =+−− λλ

⇒ 11 =λ , 12 =λ , and 13 −=λ


For 12,1 =λ ⇒ 0200

100010

3

2

1

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

xxx

02 =− x , 03 =x , 02 3 =− x

184


The Eigen vector for 12,1 =λ is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

00α

Each choice of α gives us an Eigen vector associated with 1=λ

For 13 −=λ ⇒ 0000120012

3

2

1

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

xxx

02 21 =− xx ⇒ 212 xx =

02 32 =+ xx ⇒ 322 xx −=

Let ⇒ 11 =x 22 =x ⇒ 43 −=x

The Eigen vector for 13 −=λ is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 421

Exercises Find the Eigen values and Eigen vectors of the following matrices

1) ⎥⎦

⎤⎢⎣

⎡− 3001

Ans. 1, ⎥⎦

⎤⎢⎣

⎡01

; 3− , ⎥⎦

⎤⎢⎣

⎡10

2) ⎥⎦

⎤⎢⎣

⎡0000

Ans. 0 , any non-zero vector

3) ⎥⎦

⎤⎢⎣

⎡− 3443

Ans. 5 , ⎥⎦

⎤⎢⎣

⎡12

; 5− , ⎥⎦

⎤⎢⎣

⎡− 21

185


4) ⎥⎦

⎤⎢⎣

⎡−1201

Ans. 1, ⎥⎦

⎤⎢⎣

⎡11

; 1− , ⎥⎦

⎤⎢⎣

⎡10

5)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

600080004

Ans. 4 , ; 8, ; 6 ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

001

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

010

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

100

6)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

000011011

Ans. 0 , , ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

100

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

011

2− ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−01

1

7) ⎥⎦

⎤⎢⎣

⎡−4.00

02 Ans. 2− , ⎥

⎦

⎤⎢⎣

⎡01

; , 4.0 ⎥⎦

⎤⎢⎣

⎡10

8) ⎥⎦

⎤⎢⎣

⎡− 4204

Ans. 4 , ⎥⎦

⎤⎢⎣

⎡14

; 4− , ⎥⎦

⎤⎢⎣

⎡10

9) ⎥⎦

⎤⎢⎣

⎡−−

6925

Ans. 4− , ⎥⎦

⎤⎢⎣

⎡92

; 3, ⎥⎦

⎤⎢⎣

⎡11

10) ⎥⎦

⎤⎢⎣

⎡ −8.06.06.08.0

Ans. 6.08.0 j+ , ⎥⎦

⎤⎢⎣

⎡− j1

; , 6.08.0 j− ⎥⎦

⎤⎢⎣

⎡j1

11) ⎥⎦

⎤⎢⎣

⎡4221

Ans. 5 , ⎥⎦

⎤⎢⎣

⎡21

; 0 , ⎥⎦

⎤⎢⎣

⎡−12

12)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−100000004

Ans. 4 , ; 0 , ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

001

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

010

1− ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

100

186


13)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

70252

226 0 Ans. 3, ; , ; ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−12

26

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

221

9⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 212

14)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

246042001

Ans. 1,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

1023

; ; 4 , 1 2 ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

100

15)

⎥⎦

⎤

⎢⎢⎢

⎣

⎡−

−

221612324

⎥⎥ Ans. 3− , ; , ,

17) Ans. ,

⎥⎦

⎤

⎢⎢⎢

⎣

⎡

−121

⎥⎥ 5

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

103

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

012

16)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−745872

2513 Ans. 9 ,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−12

2

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−−

46204220

42240220

4

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−

1131

; , ; , 0

⎥⎥

⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

1111

8⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

313

1

,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

11

11

; 4−

18) Ans. ,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

− 6241030200110002

2

⎥⎥

⎢⎢

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

11688

−⎢⎢

; , ; , 3

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

2900

6−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

1000

; , 1

⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

4070

, ;

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡−−

1133

3,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

113

3

,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

0001

,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

0010

19) ⎥⎥

⎢⎢

−− 4100 ⎥⎤

⎢⎡

−−

1201001201

; 5−Ans. 1−

⎥⎦

⎢⎣ −− 1400

187


Gauss limination E Method The system of linear equations is denoted by

BAX =

Awhere is a matrix

and are vectors

solved by using the Gauss elimination method as shown bellow.

Example

X BThis system is

The system

56432

21

21

=+=−

xxxx

can be written as

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡ −53

6421

2

1

xx

.

o, by using the Gauss elimination method as follows: S

⎥⎦

⎤⎢⎣

⎡ −53

6421

21 RR −4 → ⎥⎦

⎤⎢⎣

⎡ 3 −−−

714021

⇒ ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− 7

314021

2

1

xx

⇒ 71432

2

21

=−=−

xxx

⇒ 7−14 2 =x ⇒21

2 −=x ⇒ 232121 =+⎟⎠⎞

⎜⎝⎛−=x

⎥⎦

⎤⎢⎣− 2/1

⎡ 2

The solution is

188


Example

Solve the system

23 =−+ zyx

12 =+− zyx

02 =−+ zyx

Solution

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣ −−

−

012

121112311

zyx

⇒⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

012

121112311

⎡

31

212RRRR

+−+−

→→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

23

2

210730311

⇒

32 3RR + → ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

93

2

1300730311

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⎥

⎦

⎤

⎢⎢⎢

⎣

−−

93

2

1300730311

zyx

⇒91337323

−=−=+−=−+

zzyzyx

⎡

⎥⎥

913 −=z ⇒139

−=z

⇒13

813

211313633

31 −

=−

=⎟⎠⎞

⎜⎝⎛ −=y ⎟

⎠⎞

⎜⎝⎛−+=

139733y

137

1382726

138

13932 − +32 ==+⎟⎠⎞

⎜⎝⎛−+

=−+= yzx

189


Matrix Inverse We can use the Gauss elimination method to find the inverse o e know

that a matrix

f a matrix. W

A and its inverse 1−A must satisfy the equation IAA =× −1 (identity

atrix). The following example illustrates the procedure.

Exa

m

mple

Find 1−A if

⎥⎦

⎤⎢⎣

⎡=

4112

A

Solution

⎥⎦

⎤⎢⎣

⎡1001

4112

⇒ 121 R →

⎥⎦

⎤⎢⎣

⎡1002/1

412/11

→ ⎥⎦

⎤⎢⎣

⎡− 12/1

02/12/702/11

⇒ 27

2 R → ⎥⎦

⎤⎢⎣

⎡− 7/27/1

02/1102/11

12 R−R

21 21 RR − →

⎥⎦

⎤⎢⎣

⎡−

−7/27/17/17/4

1001

⇒ ⎥⎦

⎤⎢⎣

⎡−

−=−

7/27/17/17/41A

Note:

The matrix A has an inverse if 0≠A

190


Example

Find 1−A if

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−=

204201

312A

Solution

⎥⎥⎥

⎦

⎤

⎢⎣⎢⎢⎡

−−

100010001

204201

312 ⇒

121 R →

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

100010002/1

2042012/32/11

13

12

4RRRR

−−

→→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−

−

102012/1002/1

4202/72/10

2/32/11

22R →⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−

−

102021002/1

4207102/32/11

23

21

2

21

RR

RR

−

+

→

→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

140021010

1000710201

191


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

10/15/20021010

100710201

3101 R →

32

31

72

RRRR

++

→→

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

10/15/2010/75/415/15/10

100010001

Exercises

Solve the Following Systems of Linear Equations using Gauss Elimination 1)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=−

10/15/2010/75/415/15/10

1A ⇒

53 −=+ yx

632 =+ yx

2) 82 −=− yx

135 −=+ yx

3) 125 =− yx

2286 =+ yx

4) 432 =+ yx

423 −=+ yx

5) 1723 −=+ yx

010 =+ yx

6) 42 =+− yx

3843 =+ yx

192


027 =−− zyx

039 =−− zyx

0742 =−+ zyx

8) 23 −=+− zyx

625 =++ zyx

032 =++ zyx

7)

9) 082 =−+ zyx

0532 =+− zyx

01223 =−+ zyx

10) 1335 −=−+ zyx

1223 −=−+ zyx

822 =+− zyx

11) 1349 −=++ zyx

125 =++ zyx

1437 =++ zyx

12) 12 −=− zy

113 =+ zx

6242 =+− zyx

13) 224 =− zy

2926 =+− zyx

24484 −+ zy =x

14) 0214 4 =−− zyx

06218 =−− zyx

01484 =−+ zyx

15) 2−=+ zy

1264 −=+ zy

2=++ zyx

16) 832 =−+ zyx

325 =+ zx

078 =+− zyx

17) 2444 =+ zy

62113 =−− zy −x

18176 =+− zyx

18) 723 =−+− zyx

333 −=+ zx

zyx 22 ++

19) 14 =− yx

152 =+− yx

20) 923 =− yx

36 −=+− yx

135 =++ zyx 22) 33 =−+ zyx 21) 2

22 =++− zyx 1322 =−+ zyx

0=++ zyx 22 −=−+− zyx

193


Rank

r of a matrix A is the highest order of the matrix with 0≠A The rank

Example

Find the rank of the following matrices

, (b) , (c)

tion

(a) ⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

=220432321

X⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=

221213432

Y⎥

⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

963642321

Z ⎡

Solu

(a) ( ) ( ) 242642862232

22243

1 =+−=−−−=−=X ⇒ 3=r .

(b) ( ) ( ) ( ) 01642634222113

42123

32221

2 =+++−−=−

+−

−=Y

0713

≠−= , so the rank 232

=r . Since

(c) 063

423

9362

296

641 =

−−+

−−−

−−=Z

Since all the minor matrices of degree inants of zero, then the rank 2 have determ

1=r .

194


Exercises Determine the Rank of the Following Matrices

2=1) ⎥⎦

⎤⎢⎣

⎡0574

Ans. r

2) ⎥⎦

⎤⎢⎣

⎡−971304

Ans. 2=r

3) ⎥⎦

⎤⎢⎣

⎡ −1002

314 Ans. 2=r

4) Ans. 2=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

624019

r

3=5)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−− 431822306318

Ans. r

6) ⎥⎦

⎤⎢⎣

⎡−9331

Ans. 2=r

7) ⎥⎦

⎤⎢⎣

⎡3121

6280

9 Ans. 2=r

8) Ans. 2=⎥⎦⎥⎥⎤

⎢⎢⎢

⎣

⎡ −

844541033

r

195


196

9)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

630021

Ans. 1=r

10)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

555570413120

Ans. 3=r

11)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−054503

430 Ans. r 2=

12) Ans. 2=⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

040204020408

r

13) Ans. 3=⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

3703700783378500301

r

14) Ans. 4=⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

481622168424816

16842

r


Multiple Integrals

Double Integral over Rectangular Region

If ),( yxf is continuous throughout the rectangular region :R bxa ≤≤ ,

, then dyc ≤≤

∫ ∫∫ ∫∫∫ ==b

a

d

c

d

c

b

aR

dxdyyxfdydxyxfdAyxf ),(),(),(

Double Integral over Nonrectangular Region

Let ),( yxf be continuous on a region . R

1. If R is defined by bxa ≤ , )(≤ )( 21 xgyxg ≤≤ , with 1g and 2g are continuous

on [ ]ba, , then

dxdyyxfdAyxfb

a

xg

xgR∫ ∫∫∫ =

)(

)(

2

1

),(),(

2. If R is defined by dyc ≤ , )(≤ )( 21 yhxyh ≤≤ , with 1h and 2h are continuous on

[ ]dc, , then

dydxyxfdAyxfd

c

yh

yhR∫ ∫∫∫ =

)(

)(

2

1

),(),(

Finding Limits of Integration

To evaluate ∫∫R

dAyxf ),( and if we integrate first with respect to and then with

respect to

y

x , do the following:

197


o 1

1

x

y

1=+ y

x

R

122 =+ yx

o 1

1

x

y

xy −=1

R 21 xy −=

L

Enters

Leaves

0 1

1

x

y

xy −=1

R 21 xy −=

L

Enters

Leaves

Smallest x is 0=x

Largest x is 1=x

1) Sketch. Sketch the region of integration and label

the bounding curves.

2) Find the y-limits of integration. Imagine a vertical

line cutting through in the direction of

. Mark the -values where enters

. These are the -limits of integration

and are usually functi

Lincreasing

and leaves

R

y

y y

ons of

L

x .

3) Find the x-limits of integration.

Choose x -limits that include all the

vertical lines through . The

integral becomes

R

198


dxdyyxfdAyxfx

xR∫ ∫∫∫

−

−

=1

0

1

1

2

),(),(

To evaluate the same double integral as an iterated integral with the order of

integration reversed, use horizontal lines instead of vertical lines in steps 2 and 3. the

integral is

dydxyxfdAyxfy

yR∫ ∫∫∫

−

−

=1

0

1

1

2

),(),(

Example

Sketch the region of integration for the integral

( ) dxdyxx

x∫ ∫ +2

0

2

2

24

and write the equivalent integral with the order of integration reversed.

0 1

1

x

y

y

Largest yis 1y =

x −=1

Leaves

R 21 yx −=

L

Smallest y is 0= Enters y

199


Solution

0 2

4

x

y

2xy = xy 2=

)4,2(

0 2

4

x

y

yx =2yx =

)4,2(

To find limits for integrating in the reverse order, we imagine a horizontal line

passing from left to right through the region. It enters at 2/yx = and leaves at

yx = . To include all such lines must be from to . y 0 4

So,

( ) ( ) dydxxdxdyxy

y

x

x∫ ∫∫ ∫ +=+4

0 2/

2

0

2

24242

Example

Find the volume of the prism whose base is the triangle in the xy -plane bounded

by the x -axis and the lines xy = and 1=x and whose top lies in the plane

yxyxfz −−== 3),(

Solution

( ) ∫∫ ∫=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−−=

1

0 0

21

0 0 233 dxyxyydxdyyxV

xy

y

x

122

32

331

0

321

0

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

=

=

∫x

x

xxdxxx

200


When the order of the integration is reversed, the integral of the volume is

( ) ∫∫ ∫=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−−=

1

0

121

0

1

233 dyxyxxdydxyxV

x

yxy

∫∫ ⎟⎠⎞

⎜⎝⎛ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛++−−−=

1

0

21

0

22

234

25

23

213 dyyydyyyyy

1212

25 1

0

32 =⎟⎠⎞

⎜⎝⎛ +−=

=

=

y

y

yyv

Example

Calculate ∫∫R

dAx

x)sin( where is the triangle in the R xy -plane bounded by the x -

axis, the line xy = , and the line 1=x .

Solution

We integrate first with respect to and then with respect to y x .

1

0

1

0

1

0 0

1

0 0

)cos()sin()sin()sin( xdxxdxx

xydxdyx

x xy

y

x

−==⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛∫∫∫ ∫

=

=

46.01)1cos( =+−=

Area The area of a closed, bounded plane region is R

∫∫=R

dAA

201


202

Example

Find the area of the region bounded by R xy = and in the first quadrant. 2xy =

Solution

( )∫∫∫ ∫ −===1

0

21

0

1

02

2

dxxxdxydydxA x

x

x

x

61

32

1

0

32

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

xx

Example

Find the area of the region enclosed by the parabola and the line R 2xy =

2+= xy .

Solution

( )∫∫∫ ∫−−

+

−

+

−+===2

1

22

1

22

1

2

222

dxxxdxydydxA x

x

x

x

29

32

2

2

1

32

=⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−

xxx

On the other hand, reversing the order of integration results in dividing the region

into two parts as follows:

∫∫∫∫ +=21

21RR

dxdyAdxdyAA

∫ ∫∫ ∫−−

+=4

1 2

1

0

y

y

y

y

dxdydxdy

Documents

Sequences and Series - University of Technology, Iraq...Infinite Series Infinite series are sequences of a special kind: those in which the nth-term is the sum of the first n terms