ADVANCED ENGINEERING MATHEMATICS...These functions are useful to solve many mathematical problems in...

ADVANCED ENGINEERING MATHEMATICS 2130002 – 5th Edition

Darshan Institute of Engineering and Technology

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Roll No. :

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DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002

I N D E X

UNIT WISE ANALYSIS FROM GTU QUESTION PAPERS ..................................... 5

LIST OF ASSIGNMENT ...................................................................................... 6

UNIT 1 – INTRODUCTION TO SOME SPECIAL FUNCTIONS ............................... 8

1). METHOD – 1: EXAMPLE ON BETA FUNCTION AND GAMMA FUNCTION............................ 9

2). METHOD – 2: EXAMPLE ON BESSEL’S FUNCTION .....................................................................15

UNIT-2 » FOURIER SERIES AND FOURIER INTEGRAL .................................... 16

3). METHOD – 1: EXAMPLE ON FOURIER SERIES IN THE INTERVAL (𝐂, 𝐂 + 𝟐𝐋) ...............18

4). METHOD – 2: EXAMPLE ON FOURIER SERIES IN THE INTERVAL (−𝐋, 𝐋) .......................21

5). METHOD – 3: EXAMPLE ON HALF COSINE SERIES IN THE INTERVAL (𝟎, 𝐋) .................26

6). METHOD – 4: EXAMPLE ON HALF SINE SERIES IN THE INTERVAL (𝟎, 𝐋) .......................27

7). METHOD – 5: EXAMPLE ON FOURIER INTERGRAL ...................................................................29

UNIT-3A » DIFFERENTIAL EQUATION OF FIRST ORDER ................................ 32

8). METHOD – 1: EXAMPLE ON ORDER AND DEGREE OF DIFFERENTIAL EQUATION .....33

9). METHOD – 2: EXAMPLE ON VARIABLE SEPARABLE METHOD ............................................35

10). METHOD – 3: EXAMPLE ON LEIBNITZ’S DIFFERENTIAL EQUATION ................................37

11). METHOD – 4: EXAMPLE ON BERNOULLI’S DIFFERENTIAL EQUATION............................39

12). METHOD – 5: EXAMPLE ON EXACT DIFFERENTIAL EQUATION ..........................................40

13). METHOD – 6: EXAMPLE ON NON-EXACT DIFFERENTIAL EQUATION...............................42

14). METHOD – 7: EXAMPLE ON ORTHOGONAL TREJECTORY......................................................44

UNIT-3B » DIFFERENTIAL EQUATION OF HIGHER ORDER............................. 46

15). METHOD – 1: EXAMPLE ON HOMOGENEOUS DIFFERENTIAL EQUATION ......................48

16). METHOD – 2: EXAMPLE ON NON-HOMOGENEOUS DIFFERENTIAL EQUATION ...........52

17). METHOD – 3: EXAMPLE ON UNDETERMINED CO-EFFICIENT..............................................55

18). METHOD – 4: EXAMPLE ON WRONSKIAN .....................................................................................56

19). METHOD – 5: EXAMPLE ON VARIATION OF PARAMETERS ...................................................57


I N D E X

20). METHOD – 6: EXAMPLE ON CAUCHY EULER EQUATION ....................................................... 60

21). METHOD – 7: EXAMPLE ON FINDING SECOND SOLUTION .................................................... 61

UNIT-4 » SERIES SOLUTION OF DIFFERENTIAL EQUATION ............................ 62

22). METHOD – 1: EXAMPLE ON SINGULARITY OF DIFFERENTIAL EQUATION .................... 63

23). METHOD – 2: EXAMPLE ON POWER SERIES METHOD............................................................ 64

24). METHOD – 3: EXAMPLE ON FROBENIUS METHOD ................................................................... 67

UNIT-5 » LAPLACE TRANSFORM AND IT’S APPLICATION ................................ 70

25). METHOD – 1: EXAMPLE ON DEFINITION OF LAPLACE TRANSFORM ............................... 74

26). METHOD – 2: EXAMPLE ON LAPLACE TRANSFORM OF SIMPLE FUNCTIONS ............... 75

27). METHOD – 3: EXAMPLE ON FIRST SHIFTING THEOREM ....................................................... 77

28). METHOD – 4: EXAMPLE ON DIFFERENTIATION OF LAPLACE TRANSFORM ................. 79

29). METHOD – 5: EXAMPLE ON INTEGRATION OF LAPLACE TRANSFORM........................... 81

30). METHOD – 6: EXAMPLE ON INTEGRATION OF A FUNCTION ............................................... 83

31). METHOD – 7: EXAMPLE ON L. T. OF PERIODIC FUNCTIONS ................................................. 85

32). METHOD – 8: EXAMPLE ON SECOND SHIFTING THEOREM .................................................. 88

33). METHOD – 9: EXAMPLE ON LAPLACE INVERSE TRANSFORM ............................................. 89

34). METHOD – 10: EXAMPLE ON FIRST SHIFTING THEOREM..................................................... 91

35). METHOD – 11: EXAMPLE ON PARTIAL FRACTION METHOD ............................................... 92

36). METHOD – 12: EXAMPLE ON SECOND SHIFTING THEOREM................................................ 95

37). METHOD – 13: EXAMPLE ON INVERSE LAPLACE TRANSFORM OF DERIVATIVES ...... 96

38). METHOD – 14: EXAMPLE ON CONVOLUTION PRODUCT ........................................................ 97

39). METHOD – 15: EXAMPLE ON CONVOLUTION THEROREM .................................................... 99

40). METHOD – 16: EXAMPLE ON APPLICATION OF LAPLACE TRANSFORM ...................... 101

UNIT-6 » PARTIAL DIFFERENTIAL EQUATION AND IT’S APPLICATION ......... 104

41). METHOD – 1: EXAMPLE ON FORMATION OF PARTIAL DIFFERENTIAL EQUATION 105

42). METHOD – 2: EXAMPLE ON DIRECT INTEGRATION ............................................................. 107


I N D E X

43). METHOD – 3: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE ............................... 110

44). METHOD – 4: EXAMPLE ON LAGRANGE’S DIFFERENTIAL EQUATION .......................... 112

45). METHOD – 5: EXAMPLE ON NON-LINEAR PDE ........................................................................ 114

46). METHOD – 6: EXAMPLE ON SEPARATION OF VARIABLES .................................................. 116

47). METHOD – 7: EXAMPLE ON CLASSIFICATION OF 2ND ORDER PDE.................................. 117

8 GTU QUESTION PAPERS OF AEM – 2130002……..……………...………………***

SYLLABUS OF AEM – 2130002……..……………..…………..………..……………….***


U ni t wi se a na ly sis fr om GT U q ues ti on pap er s

UNIT WISE ANALYSIS FROM GTU QUESTION PAPERS

Unit Number ⟼ 1 2 3 4 5 6

W – 14 4 28 28 7 28 24

S – 15 - 35 14 14 28 28

W – 15 3 30 25 14 31 16

S – 16 9 28 26 8 31 17

W – 16 3 30 21 14 31 16

S – 17 2 15 38 11 26 27

W – 17 3 13 35 14 26 28

S – 18 - 22 31 14 24 28

Average ⟼ 3 25 27 12 28 23

*GTU Weightage ⟼ 4 10 20 6 15 15

*Unit weightage out of 70 marks.

Unit No. Unit Name Level GTU Hour

1 Introduction to Some Special Function Easy 2

2 Fourier Series and Fourier Integral Medium 5

3 Differential equation and It’s Application Medium 11

4 Series Solution of Differential Equation Easy 3

5 Laplace Transform and It’s Application Hard 9

6 Partial Differential Equation Hard 12


Li s t of As si gnm en t

LIST OF ASSIGNMENT

Assignment No. Unit No. Method No.

1 4 2

6 6

2 2 1, 2

3 2 3, 4, 5

4 3B ALL METHODS

5 3A ALL METHODS

6 5 Proof of Formulae

7 5 GTU asked examples (Method No. 1 to 8)

8 5 GTU asked examples (Method No. 9 to 16)


[ 7 ]


U N I T - 1 » I n t r o d u c t i o n T o S o m e S p e c i a l F u n c t i o n [ 8 ]

UNIT 1 – INTRODUCTION TO SOME SPECIAL FUNCTIONS

INTRODUCTION:

Special functions are particular mathematical functions which have some fixed notations due

to their importance in mathematics. In this Unit we will study various type of special

functions such as Gamma function, Beta function, Error function, Dirac Delta function etc.

These functions are useful to solve many mathematical problems in advanced engineering

mathematics.

BETA FUNCTION:

If m > 0, n > 0, then Beta function is defined by the integral ∫ xm−1(1 − x)n−1dx1

0 and is

denoted by β(m, n) OR B(m, n).

𝐁(𝐦, 𝐧) = ∫ 𝐱𝐦−𝟏(𝟏 − 𝐱)𝐧−𝟏𝐝𝐱

𝟏

𝟎

Properties:

(1) Beta function is a symmetric function. i.e. B(m, n) = B(n, m), where m > 0, n > 0.

(2) B(m, n) = 2∫ sin2m−1 θ cos2n−1 θdθπ

20

(3) ∫ sinp θ cosq θ dθ =1

2⋅ B (

p+1

2,q+1

2)

π

20

(4) B(m, n) = ∫xm−1

(1+x)m+n

∞

0 dx

GAMMA FUNCTION:

If n > 0, then Gamma function is defined by the integral ∫ e−xxn−1dx∞

0 and is denoted by ⌈n.

⌈𝐧 = ∫ 𝐞−𝐱 𝐱𝐧−𝟏 𝐝𝐱

∞

𝟎

Properties:

(1) Reduction formula for Gamma Function ⌈(n + 1) = n⌈n ; where n > 0.


U N I T - 1 » I n t r o d u c t i o n T o S o m e S p e c i a l F u n c t i o n [ 9 ]

(2) If n is a positive integer, then ⌈(n + 1) = n!

(3) Second Form of Gamma Function ∫ e−x2x2m−1dx =

1

2

∞

0⌈m

(4) Relation Between Beta and Gamma Function, B(m, n) =⌈m⌈n

⌈(m+n). W – 15 ; W – 16

(5) ∫ sinp θ cosq θdθ =1

2

⌈(p+1

2)⌈(

q+1

2)

⌈(p+q+2

2)

π

20

(6) ⌈(n +1

2) =

(2n)!√π

n!4n for n = 0,1,2,3,…

Examples: For n = 0, ⌈1

2= √π W – 16

For n = 1, ⌈3

2=

√π

2 For n = 2, ⌈

5

2=

3√π

4

(7) Legendre’s duplication formula. S – 16

⌈n ⌈(n +1

2) =

√π

22n−1 ⌈(2n) OR ⌈(n + 1) ⌈n +

1

2=

√π

22n ⌈(2n + 1)

(8) Euler’s formula : ⌈n ⌈(1 − n) =π

sinnπ ; 0 < n < 1

METHOD – 1: EXAMPLE ON BETA FUNCTION AND GAMMA FUNCTION

C 1 Find B(4,3).

𝐀𝐧𝐬𝐰𝐞𝐫:1

60

T 2 Find B (

9

2,7

2) .

𝐀𝐧𝐬𝐰𝐞𝐫:5π

2048

S – 16

H 3 State the relation between Beta and Gamma function. W – 15 W – 16

H 4 State Duplication (Legendre) formula. S – 16

C 5 Find ⌈

7

2 .

𝐀𝐧𝐬𝐰𝐞𝐫:15 √π

8

S – 16


U N I T - 1 » I n t r o d u c t i o n T o S o m e S p e c i a l F u n c t i o n [ 1 0 ]

H 6 Find ⌈

13

2 .

𝐀𝐧𝐬𝐰𝐞𝐫:10395 √π

64

W – 15

T 7 Find ⌈

5

4 ⌈3

4 .

𝐀𝐧𝐬𝐰𝐞𝐫: π

2√2

ERROR FUNCTION AND COMPLEMENTARY ERROR FUNCTION:

The error function of x is defined as below and is denoted by erf(x).

𝐞𝐫𝐟(𝐱) =𝟏

√𝛑∫𝐞−𝐭

𝟐𝐝𝐭

𝐱

−𝐱

=𝟐

√𝛑∫𝐞−𝐭

𝟐𝐝𝐭

𝐱

𝟎

The complementary error function is denoted by erfc(x) and defined as

𝐞𝐫𝐟𝐜(𝐱) =𝟐

√𝛑∫ 𝐞−𝐭

𝟐𝐝𝐭

∞

𝐱

Properties:

(1) erf(0) = 0

(2) erfc(0) = 1

(3) erf(∞) = 1

(4) erf(−x) = −erf(x)

(5) erf(x) + erfc(x) = 1

UNIT STEP FUNCTION (HEAVISIDE’S FUNCTION): W – 14 ; W – 16

The Unit Step Function is defined by

u(x − a) = {1 ; x ≥ a

0 ; x < a

.

It is also denoted by H(x − a) or ua(x).



PULSE OF UNIT HEIGHT:

The pulse of unit height of duration T is

defined by

f(x) = {1 ; 0 ≤ x ≤ T

0 ; x > T

.

SINUSOIDAL PULSE FUNCTION:

The sinusoidal pulse function is defined by

f(x) = {

sin ax ; 0 ≤ x ≤π

a

0 ; x >π

a

RECTANGLE FUNCTION: (W – 17)

A Rectangular function f(x) on ℝ is defined by

f(x) = {1 ; a ≤ x ≤ b

0 ; otherwise

GATE FUNCTION:

A Gate function fa(x) on ℝ is defined by

fa(x) = {1 ; |x| ≤ a

0 ; |x| > a

.

Note that gate function is symmetric about axis

of co-domain.

Gate function is also a rectangle function.

𝐓

1

f(x)

x

𝐟(𝐱)

𝟎 𝛑

𝐚

x

𝟏

𝐚

1

𝐟(𝐱)

x 𝐛

−𝐚

1

𝐟(𝐱)

x 𝐚



SIGNUM FUNCTION:

The Signum function is defined by

f(x) =

{

−1 ; x < 0

0 ; x = 0

1 ; x > 0

.

IMPULSE FUNCTION:

An impulse function is defined as below,

f(x) =

{

0 ; x < a

1

ε ; a ≤ x ≤ a + ε

0 ; x > a + ε

DIRAC DELTA FUNCTION(UNIT IMPULSE FUNCTION): W – 14

A Dirac delta Function δ(x − a) is defined by δ(x − a) = limε→0

f(x) .

Where, f(x) is an impulse function, which is defined as

f(x) =

{

0 ; x < a

1

ε ; a ≤ x ≤ a + ε

0 ; x > a + ε

.

PERIODIC FUNCTION:

A function f is said to be periodic, if f(x + p) = f(x) for all x.

If smallest positive number of set of all such p exists, then that number is called the

Fundamental period of f(x).

−𝟏

1

f(x)

x

𝟏𝛆

𝐟(𝐱)

x 𝐚 + 𝛆 𝐚 0



Note:

(1) Constant function is periodic without Fundamental period.

(2) Sine and Cosine are Periodic functions with Fundamental period 2π.

SQUARE WAVE FUNCTION:

A square wave function f(x) of period "2a" is defined by

f(x) = { 1 ; 0 < x < a−1 ; a < x < 2a

.

SAW TOOTH WAVE FUNCTION: (W – 17)

A saw tooth wave function f(x) with period a

is defined as f(x) = x ; 0 ≤ x < a.

TRIANGULAR WAVE FUNCTION:

A Triangular wave function f(x) having period "2a" is defined by

f(x) = { x ; 0 ≤ x < a

2a − x ; a ≤ x < 2a

.

−𝟏

1

f(x)

a

-a 2a

3a x

a

f(x)

a 2a 3a x



FULL RECTIFIED SINE WAVE FUNCTION:

A full rectified sine wave function with period "π" is defined as

f(x) = sin x ; 0 ≤ x < π.

HALF RECTIFIED SINE WAVE FUNCTION:

A half wave rectified sinusoidal function with period "2π" is defined as

f(x) = {sin x ; 0 ≤ x < π

0 ; π ≤ x < 2π

.

𝐟(𝐱)

𝛑 𝟐𝛑 𝟑𝛑 −𝛑

`

𝟎 𝟒𝛑 −𝟐𝛑

x

1 `

`

a

f(x)

a 2a 3a -a -2a 4a x

𝐟(𝐱)

𝛑

`

`

𝟐𝛑 −𝛑 𝟎 x

𝟏



BESSEL’S FUNCTION:

A Bessel’s function of 1st kind of order n is defined by

Jn(x) =xn

2n⌈(n + 1)[1 −

x2

2(2n + 2)+

x4

2 ∙ 4(2n + 2)(2n + 4)− ⋯ ] = ∑

(−1)k

k! ⌈(n + k + 1)(x

2)n+2k

∞

k=0

METHOD – 2: EXAMPLE ON BESSEL’S FUNCTION

C 1 Determine the value J12

(x).

𝐀𝐧𝐬𝐰𝐞𝐫:√2

πx sin x

S – 16

H 2 Determine the value J(−

12)(x).


πx cos x

C 3 Determine the value J32

(x).


πx (sin x

x− cos x)

S – 16

H 4 Determine the value J(−

32)(x).


πx (cos x

x+ sin x)

T 5 Using Bessel’s function of the first kind, Prove that J0(0) = 1.

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆


U N I T - 2 » F o u r i e r S e r i e s a n d F o u r i e r I n t e g r a l [ 1 6 ]

UNIT-2 » FOURIER SERIES AND FOURIER INTEGRAL

BASIC FORMULAE:

Leibnitz’s Formula (Take, Given polynomial function as “u”)

∫𝐮 ∙ 𝐯 𝐝𝐱 = 𝐮 𝐯𝟏 − 𝐮′ 𝐯𝟐 + 𝐮′′ 𝐯𝟑 − 𝐮′′′ 𝐯𝟒 +⋯

Where, u′, u′′, … are successive derivatives of u and v1, v2, … are successive integrals of

v.

Choice of u and v is as per LIATE order.

Where,

L means Logarithmic Function I means Invertible Function

A means Algebraic Function T means Trigonometric Function

E means Exponential Function

When Function is Exponential Function:

∫𝐞𝐚𝐱 𝐬𝐢𝐧 𝐛𝐱 𝐝𝐱 =𝐞𝐚𝐱

𝐚𝟐 + 𝐛𝟐[𝐚 𝐬𝐢𝐧 𝐛𝐱 − 𝐛 𝐜𝐨𝐬 𝐛𝐱] + 𝐜

∫𝐞𝐚𝐱 𝐜𝐨𝐬 𝐛𝐱 𝐝𝐱 =𝐞𝐚𝐱

𝐚𝟐 + 𝐛𝟐[𝐚 𝐜𝐨𝐬 𝐛𝐱 + 𝐛 𝐬𝐢𝐧 𝐛𝐱] + 𝐜

When Function is Trigonometric Function:

𝟐𝐬𝐢𝐧 𝐚 𝐜𝐨𝐬 𝐛 = 𝐬𝐢𝐧(𝐚 + 𝐛) + 𝐬𝐢𝐧(𝐚 − 𝐛)

𝟐𝐜𝐨𝐬 𝐚 𝐬𝐢𝐧 𝐛 = 𝐬𝐢𝐧(𝐚 + 𝐛) − 𝐬𝐢𝐧(𝐚 − 𝐛)

𝟐𝐜𝐨𝐬 𝐚 𝐜𝐨𝐬 𝐛 = 𝐜𝐨𝐬(𝐚 + 𝐛) + 𝐜𝐨𝐬(𝐚 − 𝐛)

𝟐𝐬𝐢𝐧 𝐚 𝐬𝐢𝐧 𝐛 = − 𝐜𝐨𝐬(𝐚 + 𝐛) + 𝐜𝐨𝐬(𝐚 − 𝐛)

NOTE (FOR EVERY, 𝐧 ∈ ℤ)

cosnπ = (−1)n sin nπ = 0 cos(2n + 1)π

2= 0

cos2nπ = (−1)2n = 1 sin 2nπ = 0 sin(2n + 1)π

2= (−1)n



INTRODUCTION:

We know that Taylor’s series representation of functions are valid only for those functions

which are continuous and differentiable. But there are many discontinuous periodic

functions of practical interest which requires to express in terms of infinite series containing

“sine” and “cosine” terms.

Fourier series, which is an infinite series representation in term of “sine” and “cosine” terms,

is a useful tool here. Thus, Fourier series is, in certain sense, more universal than Taylor’s

series as it applies to all continuous, periodic functions and discontinuous functions.

Fourier series is a very powerful method to solve ordinary and partial differential equations,

particularly with periodic functions.

Fourier series has many applications in various fields like Approximation Theory, Digital

Signal Processing, Heat conduction problems, Wave forms of electrical field, Vibration

analysis, etc.

Fourier series was developed by Jean Baptiste Joseph Fourier in 1822.

DIRICHLET CONDITION FOR EXISTENCE OF FOURIER SERIES OF 𝐟(𝐱):

(1) f(x) is bounded.

(2) f(x) is single valued.

(3) f(x) has finite number of maxima and minima in the interval.

(4) f(x) has finite number of discontinuity in the interval.

NOTE:

At a point of discontinuity the sum of the series is equal to average of left and right hand

limits of f(x) at the point of discontinuity, say x0.

i. e. f(x0) =f(x0 − 0) + f(x0 + 0)

2



FOURIER SERIES IN THE INTERVAL (𝐜, 𝐜 + 𝟐𝐋):

The Fourier series for the function f(x) in the interval (c, c + 2L) is defined by

𝐟(𝐱) =𝐚𝟎

𝟐+∑ [𝐚𝐧 𝐜𝐨𝐬 (

𝐧𝛑𝐱

𝐋) + 𝐛𝐧 𝐬𝐢𝐧 (

𝐧𝛑𝐱

𝐋)]

∞

𝐧=𝟏

Where the constants a0, an and bn are given by

a0 =1

L∫ f(x) dx

c+2L

c

an =1

L∫ f(x) cos (

nπx

L) dx

c+2L

c

bn =1

L∫ f(x) sin (

nπx

L) dx

c+2L

c

METHOD – 1: EXAMPLE ON FOURIER SERIES IN THE INTERVAL (𝐂, 𝐂 + 𝟐𝐋)

C 1 Find the Fourier series for f(x) = x2 in (0,2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟒

𝟑+∑ [

𝟒

𝐧𝟐𝛑𝟐𝐜𝐨𝐬(𝐧𝛑𝐱) −

𝟒

𝐧𝛑𝐬𝐢𝐧(𝐧𝛑𝐱) ]

∞

𝐧=𝟏

H 2 Find the Fourier series to represent f(x) = 2x − x2 in (0,3).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑[ −𝟗

𝐧𝟐𝛑𝟐𝐜𝐨𝐬 (

𝟐𝐧𝛑𝐱

𝟑) +

𝟑

𝐧𝛑𝐬𝐢𝐧 (

𝟐𝐧𝛑𝐱

𝟑) ]

∞

𝐧=𝟏

S – 16

C 3 Obtain the Fourier series for f(x) = e−x in the interval 0 < x < 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = (𝟏 − 𝐞−𝟐)

𝟐+∑

(𝟏 − 𝐞−𝟐)

𝐧𝟐𝛑𝟐 + 𝟏[ 𝐜𝐨𝐬 𝐧𝛑𝐱 + (𝐧𝛑) 𝐬𝐢𝐧 𝐧𝛑𝐱 ]

∞

𝐧=𝟏

T 4 Find the Fourier series of the periodic function f(x) = π sin πx. Where 0 <

x < 1 , p = 2l = 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = 𝟐 +∑𝟒

𝟏 − 𝟒𝐧𝟐𝐜𝐨𝐬(𝟐𝐧𝛑𝐱)

∞

𝐧=𝟏

T 5 Find Fourier Series for f(x) = x2 ; where 0 ≤ x ≤ 2π

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟒𝛑𝟐

𝟑+ ∑[

𝟒

𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 −

𝟒𝛑

𝐧𝐬𝐢𝐧 𝐧𝐱 ]

∞

𝐧=𝟏



H 6 Show that, π − x =

π

2+∑

sin 2nx

n

∞

n=1

, when 0 < x < π .

C 7 Obtain the Fourier series for f(x) = (

π−x

2)2

in interval 0 < x < 2π.

Hence prove that π2

12=

1

12−

1

22+

1

32− ⋯.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑𝟐

𝟏𝟐+∑

𝟏

𝐧𝟐𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

W – 14

H 8 Find Fourier Series for f(x) = e−x where 0 < x < 2π.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟏 − 𝐞−𝟐𝛑

𝟐𝛑+∑

𝟏 − 𝐞−𝟐𝛑

𝛑(𝐧𝟐 + 𝟏)[ 𝐜𝐨𝐬 𝐧𝐱 + 𝐧 𝐬𝐢𝐧𝐧𝐱 ]

∞

𝐧=𝟏

H 9 Find Fourier Series for f(x) = eax in (0,2π); a > 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝐞𝟐𝐚𝛑 − 𝟏

𝟐𝐚𝛑+∑

𝐞𝟐𝐚𝛑 − 𝟏

𝛑(𝐧𝟐 + 𝐚𝟐)[𝐚 𝐜𝐨𝐬 𝐧𝐱 − 𝐧 𝐬𝐢𝐧𝐧𝐱 ]

∞

𝐧=𝟏

S – 18

H 10 Develop f(x) in a Fourier series in the interval (0,2) if f(x) = {

x, 0 < x < 1

0, 1 < x < 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟏

𝟒+∑ [

(−𝟏)𝐧 − 𝟏

𝐧𝟐𝛑𝟐𝐜𝐨𝐬(𝐧𝛑𝐱) +

(−𝟏)𝐧+𝟏

𝐧𝛑𝐬𝐢𝐧(𝐧𝛑𝐱) ]

∞

𝐧=𝟏

C 11 For the function f(x) = {

x; 0 ≤ x ≤ 2

4 − x: 2 ≤ x ≤ 4, find its Fourier series. Hence

show that 1

12+

1

32+

1

52+⋯ =

π2

8.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = 𝟏 +∑𝟒 [(−𝟏)𝐧 − 𝟏]

𝛑𝟐𝐧𝟐

∞

𝐧=𝟏

𝐜𝐨𝐬 (𝐧𝛑𝐱

𝟐)

W – 15

T 12 Find the Fourier series for periodic function with period 2 of

f(x) = {πx, 0 ≤ x ≤ 1

π (2 − x), 1 ≤ x ≤ 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑

𝟐+∑

𝟐[(−𝟏)𝐧− 𝟏]

𝛑𝐧𝟐𝐜𝐨𝐬(𝐧𝛑𝐱)

∞

𝐧=𝟏



H 13 Find the Fourier series of f(x) = {

x2 ; 0 < x < π

0 ; π < x < 2π.

𝐀𝐧𝐬𝐰𝐞𝐫:

𝐟(𝐱) =𝛑𝟐

𝟔+∑[

𝟐(−𝟏)𝐧 𝐜𝐨𝐬𝐧𝐱

𝐧𝟐+𝟏

𝛑{−

𝛑𝟐(−𝟏)𝐧

𝐧+𝟐(−𝟏)𝐧

𝐧𝟑−

𝟐

𝐧𝟑} 𝐬𝐢𝐧𝐧𝐱 ]

∞

𝐧=𝟏

C 14 Find the Fourier Series for the function f(x) given by

f(x) = {−π , 0 < x < π

x − π , π < x < 2π

. Hence show that ∑1

(2n + 1)2=π2

8

∞

n=0

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝛑

𝟒+ ∑[

(𝟏 − (−𝟏)𝐧)

𝐧𝟐𝛑𝐜𝐨𝐬(𝐧𝐱) +

(−𝟏)𝐧 − 𝟐

𝐧𝐬𝐢𝐧(𝐧𝐱) ]

∞

𝐧=𝟏

W – 16 S – 18

C 15 Determine the Fourier series to represent

the periodic function as shown in the figure.


𝟐−∑

𝐬𝐢𝐧𝐧𝐱

𝐧

∞

𝐧=𝟏

DEFINITION:

Odd Function: A function is said to be Odd Function if 𝐟(−𝐱) = −𝐟(𝐱).

Even Function: A function is said to be Even Function if 𝐟(−𝐱) = 𝐟(𝐱).

NOTE:

If f(x) is an even function defined in (– l, l), then ∫ f(x)l

–l dx = 2∫ f(x)

l

0 dx.

If f(x) is an odd function defined in (– l, l), then ∫ f(x)l

–l dx = 0.

FOURIER SERIES FOR ODD & EVEN FUNCTION:

Let, f(x) be a periodic function defined in (– L, L)

f(x) is even, bn = 0; n = 1,2,3,… f(x) is odd, a0 = 0 = an; n = 1,2,3,…

𝐟(𝐱) =𝐚𝟎

𝟐+ ∑ 𝐚𝐧 𝐜𝐨𝐬 (

𝐧𝛑𝐱

𝐋)∞

𝐧=𝟏 𝐟(𝐱) = ∑ 𝐛𝐧 𝐬𝐢𝐧 (𝐧𝛑𝐱

𝐋)∞

𝐧=𝟏

𝝅

f(x)

𝟐𝝅

\

pi

x 𝟒𝝅

\

pi



Where,

a0 =2

L∫ f(x) dx

L

0

an =2

L∫ f(x) cos (

nπx

L) dx

L

0

Where,

bn =2

L∫ f(x) sin (

nπx

L) dx

L

0

Sr. No. Type of Function Example

1. Even Function

x2, x4, x6, … i. e. xn, where n is even.

Any constant. e.g. 1,2, π…

cos ax

Graph is symmetric about Y − axis.

|x|, |x3|, |cos 𝑥|, …

f(−x) = f(x)

2. Odd Function

x, x3, x5 , … i. e. xm, where m is odd.

sin ax

Graph is symmetric about Origin.

f(−x) = −f(x)

3. Neither Even nor Odd eax

axm + bxn ; n is even & m is odd number.

METHOD – 2: EXAMPLE ON FOURIER SERIES IN THE INTERVAL (−𝐋, 𝐋)

C 1 Find the Fourier series of the periodic function f(x) = 2x.

Where−1 < x < 1 , p = 2l = 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟒(−𝟏)𝐧+𝟏

𝐧𝛑𝐬𝐢𝐧(𝐧𝛑𝐱)

∞

𝐧=𝟏

W – 16



H 2 Find the Fourier expansion for function f(x) = x − x3 in −1 < x < 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟏𝟐(−𝟏)𝐧+𝟏

𝐧𝟑𝛑𝟑𝐬𝐢𝐧 𝐧𝛑𝐱

∞

𝐧=𝟏

C 3 Find the Fourier series for f(x) = x2 in −l < x < l.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝐥𝟐

𝟑+ ∑

𝟒 𝐥𝟐(−𝟏)𝐧


𝐧𝛑𝐱

𝐥)

∞

𝐧=𝟏

H 4 Expand f(x) = x2 − 2 in – 2 < x < 2 the Fourier series.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝟐

𝟑+ ∑

𝟏𝟔(−𝟏)𝐧


𝐧𝛑𝐱

𝟐)

∞

𝐧=𝟏

C 5 Find the Fourier series of f(x) = x2 + x Where −2 < x < 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟒

𝟑+ ∑[

𝟏𝟔(−𝟏)𝐧

𝐧𝟐𝛑𝟐 𝐜𝐨𝐬 (𝐧𝛑𝐱

𝟐)+

𝟒(−𝟏)𝐧+𝟏


𝐧𝛑𝐱

𝟐)]

∞

𝐧=𝟏

T 6 Find the Fourier expansion for function f(x) = x − x2 in −1 < x < 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝟏

𝟑+ ∑[

𝟒(−𝟏)𝐧+𝟏

𝐧𝟐𝛑𝟐𝐜𝐨𝐬(𝐧𝛑𝐱) +

𝟐(−𝟏)𝐧+𝟏

𝐧𝛑𝐬𝐢𝐧(𝐧𝛑𝐱)]

∞

𝐧=𝟏

C 7 Obtain the Fourier series for f(x) = x2 in the interval – π < x < π and hence

deduce that

(i)∑1

n2=π2

6

∞

n=1

. (ii)∑(−1)n+1

n2

∞

n=1

=π2

12.


𝟑+ ∑

𝟒(−𝟏)𝐧


∞

𝐧=𝟏

S – 16 S – 18

T 8 Find the Fourier series expansion of f(x) = |x|; – π < x < π .


𝟐+∑

𝟐 [(−𝟏)𝐧 − 𝟏]

𝛑𝐧𝟐𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

W – 16

T 9 Find the Fourier series of f(x) = x3 ; x ∈ (−π, π).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐(−𝟏)𝐧+𝟏

𝐧[𝛑𝟐 −

𝟔

𝐧𝟐] 𝐬𝐢𝐧 𝐧𝐱

∞

𝐧=𝟏



H 10 Find the Fourier series of f(x) = x − x2; – π < x < π.

Deduce that: 1

12−

1

22+

1

32−

1

42+⋯ =

π2

12 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝛑𝟐

𝟑+∑ [

𝟒(−𝟏)𝐧+𝟏

𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +

𝟐(−𝟏)𝐧+𝟏

𝐧𝐬𝐢𝐧 𝐧𝐱]

∞

𝐧=𝟏

S – 17

H 11 Find the Fourier series of f(x) = x + x2; – π < x < π.

Deduce that: 1 +1

22+

1

32+

1

42+ ⋯ =

π2

6 .


𝟑+ ∑[

𝟒(−𝟏)𝐧

𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +

𝟐(−𝟏)𝐧+𝟏


∞

𝐧=𝟏

W – 17

C 12 Find the Fourier series of f(x) = x + |x|; – π < x < π.


𝟐+ ∑[

𝟐[(−𝟏)𝐧 − 𝟏]

𝛑𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +

𝟐(−𝟏)𝐧+𝟏


∞

𝐧=𝟏

W – 14 W – 15

H 13 Find the Fourier series to representation ex in the the interval(−π, π).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝐞𝛑 − 𝐞−𝛑

𝟐𝛑+∑

(𝐞𝛑 − 𝐞−𝛑) (−𝟏)𝐧

𝛑(𝐧𝟐 + 𝟏)[𝐜𝐨𝐬𝐧𝐱 + 𝐧 𝐬𝐢𝐧 𝐧𝐱]

∞

𝐧=𝟏

C 14

Determine the Fourier expansion of f(x) =

{

0, − 2 < x < −1

1 + x, − 1 < x < 0

1 − x, 0 < x < 1

0, 1 < x < 2

.


𝟒+

𝟒

𝛑𝟐∑

𝟏

𝐧𝟐(𝟏 − 𝐜𝐨𝐬

𝐧𝛑

𝟐) (𝐜𝐨𝐬

𝐧𝛑𝐱

𝟐)

∞

𝐧=𝟏

T 15 Find the Fourier series for periodic function f(x) with period 2

Where f(x) = {−1,−1 < x < 0

1, 0 < x < 1

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐− 𝟐(−𝟏)𝐧

𝛑𝐧𝐬𝐢𝐧 𝐧𝛑𝐱

∞

𝐧=𝟏



C 16 Find Fourier series expansion of the function given by

f(x) = { 0, −2 < x < 0

1, 0 < x < 2

.


𝟐+∑ [

𝟏 − (−𝟏)𝐧


𝐧𝛑𝐱

𝟐)]

∞

𝐧=𝟏

T 17 Find the Fourier series for periodic function with period 2, which is given

below f(x) = {0,−1 < x < 0

x, 0 < x < 1

.


𝟒+∑ [

(−𝟏)𝐧− 𝟏

𝐧𝟐𝛑𝟐𝐜𝐨𝐬 𝐧𝛑𝐱 +

(−𝟏)𝐧+𝟏

𝐧𝛑𝐬𝐢𝐧𝐧𝛑𝐱]

∞

𝐧=𝟏

C 18 Find the Fourier series expansion of the function

f(x) = {−π,−π ≤ x ≤ 0

x , 0 ≤ x ≤ π

. Deduce that ∑1

(2n − 1)2

∞

n=1

=π2

8.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝛑

𝟒+ ∑[

(−𝟏)𝐧 − 𝟏

𝛑𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +

𝟏 − 𝟐(−𝟏)𝐧


∞

𝐧=𝟏

S – 16

T 19 Obtain the Fourier Series for the function f(x) given by

f(x) = {0 , −π ≤ x ≤ 0

x2 , 0 ≤ x ≤ π

. Hence prove 1 −1

4+1

9−

1

16+⋯ =

π2

12.


𝐟(𝐱) =𝛑𝟐

𝟔+ ∑[

𝟐(−𝟏)𝐧 𝐜𝐨𝐬 𝐧𝐱

𝐧𝟐+𝟏

𝛑{−

𝛑𝟐(−𝟏)𝐧

𝐧+𝟐(−𝟏)𝐧

𝐧𝟑−

𝟐

𝐧𝟑} 𝐬𝐢𝐧 𝐧𝐱]

∞

𝐧=𝟏

H 20 Find the Fourier Series for the function f(x) given by

f(x) = {−π , −π < x < 0

x − π , 0 < x < π

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = −𝟑𝛑

𝟒+ ∑[

(−𝟏)𝐧 − 𝟏

𝛑𝐧𝟐𝐜𝐨𝐬𝐧𝐱 +

(−𝟏)𝐧+𝟏


∞

𝐧=𝟏



T 21 Find Fourier series for 2π periodic function f(x) = {

−k , −π < x < 0

k , 0 < x < π .

Hence deduce that 1 −1

3+1

5−1

7+⋯ =

π

4.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐𝐤[𝟏 − (−𝟏)𝐧]

𝐧𝛑𝐬𝐢𝐧 𝐧𝐱

∞

𝐧=𝟏

C 22 If f(x) = {

π + x , −π < x < 0

π − x , 0 < x < π , f(x) = f(x + 2π),for all x then expand f(x)

in a Fourier series.


𝟐+∑

𝟐

𝛑𝐧𝟐[𝟏 − (−𝟏)𝐧] 𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

H 23 Find the Fourier Series for the function f(x) given by

f(x) =

{

1 +

2x

π ; −π ≤ x ≤ 0

1 −2x

π ; 0 ≤ x ≤ π

. Hence prove 1

12+

1

32+

1

52+⋯ =

π2

8.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟒

𝛑𝟐𝐧𝟐[𝟏 − (−𝟏)𝐧] 𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

HALF RANGE SERIES:

If a function f(x) is defined only on a half interval (0, L) instead of (c, c + 2L), then it is

possible to obtain a Fourier cosine or Fourier sine series.

HALF RANGE COSINE SERIES IN THE INTERVAL (𝟎, 𝐋):

𝐟(𝐱) =𝐚𝟎

𝟐+∑𝐚𝐧 𝐜𝐨𝐬 (

𝐧𝛑𝐱

𝐋)

∞

𝐧=𝟏

Where the constants a0 and an are given by

a0 =2

L∫ f(x) dx

L

0

an =2

L∫ f(x) cos (

nπx

L) dx

L

0



METHOD – 3: EXAMPLE ON HALF COSINE SERIES IN THE INTERVAL (𝟎, 𝐋)

H 1 Find Fourier cosine series for f(x) = x2; 0 < x ≤ π .


𝟑+ ∑

𝟒(−𝟏)𝐧


∞

𝐧=𝟏

S – 15

H 2 Find Half-range cosine series for f(x) = (x − 1)2 in 0 < x < 1.


𝟑+∑

𝟒

𝐧𝟐𝛑𝟐𝐜𝐨𝐬(𝐧𝛑𝐱)

∞

𝐧=𝟏

S – 15

C 3 Find a cosine series for f(x) = π − x in the interval 0 < x < π


𝟐+ ∑

𝟐[(−𝟏)𝐧+𝟏+ 𝟏]

𝛑𝐧𝟐𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

W – 17

H 4 Find a cosine series for f(x) = ex in 0 < x < L.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝐞𝐋 − 𝟏

𝐋+∑

𝟐𝐋[𝐞𝐋(−𝟏) 𝐧 − 𝟏]

𝐧𝟐𝛑𝟐 + 𝐋𝟐𝐜𝐨𝐬 (

𝐧𝛑𝐱

𝐋)

∞

𝐧=𝟏

C 5 Find Half-range cosine series for f(x) = e−x in 0 < x < π.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟏 − 𝐞𝛑

𝛑+∑

𝟐[𝐞−𝛑(−𝟏)𝐧+𝟏 + 𝟏]

𝛑(𝐧𝟐 + 𝟏)𝐜𝐨𝐬 𝐧𝐱

∞

𝐧=𝟏

W – 15

C 6 Find Half range cosine series for sin x in (0, π)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟐

𝛑+ ∑[

−(−𝟏)𝟏+𝐧 + 𝟏

𝟏 + 𝐧+−(−𝟏)𝟏−𝐧 + 𝟏

𝟏 − 𝐧] 𝐜𝐨𝐬𝐧𝐱

∞

𝐧=𝟏

, 𝐚𝟏 = 𝟎 W – 14 S – 18

T 7

Find Half-Range cosine series for f(x) = {kx ; 0 ≤ x ≤

l

2

k(l − x);l

2≤ x ≤ l

.

And hence deduce that ∑1

(2n − 1)2=π2

8

∞

n=1

.

Answer: f(x) =kl

4+2kl

π2∑

1

n2[2 cos (

nπ

2) − 1 − (−1)n]

∞

n=1

cos (nπx

l)

S – 16



HALF RANGE SINE SERIES IN THE INTERVAL (𝟎, 𝐋):

𝐟(𝐱) =𝐚𝟎

𝟐+∑𝐚𝐧 𝐜𝐨𝐬 (

𝐧𝛑𝐱

𝐋)

∞

𝐧=𝟏

Where the constants a0 and an are given by

a0 =2

L∫ f(x) dx

L

0

an =2

L∫ f(x) cos (

nπx

L) dx

L

0

METHOD – 4: EXAMPLE ON HALF SINE SERIES IN THE INTERVAL (𝟎, 𝐋)

C 1 Find the Half range sine series for f(x) = 2x, 0 < x < 1.


𝐧𝛑(−𝟏)𝐧+𝟏 𝐬𝐢𝐧 𝐧𝛑𝐱

∞

𝐧=𝟏

S – 15

H 2 Expand πx − x2 in a half-range sine series in the interval (0, π) up to first

three terms.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟒[(−𝟏)𝐧+𝟏 + 𝟏]

𝐧𝟑𝛑𝐬𝐢𝐧𝐧𝐱

∞

𝐧=𝟏

T 3 Find half range sine series of f(x) = x3 , 0 < x < π.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐

𝐧(−𝟏)𝐧 [

𝟔

𝐧𝟐− 𝛑𝟐] 𝐬𝐢𝐧 𝐧𝐱

∞

𝐧=𝟏

S – 17

H 4 Find Half-range sine series for f(x) = ex in 0 < x < π.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐𝐧[𝐞𝛑(−𝟏)𝐧+𝟏 + 𝟏]

𝛑(𝟏 + 𝐧𝟐)𝐬𝐢𝐧 𝐧𝐱

∞

𝐧=𝟏

S – 15

C 5 Find the sine series f(x) = {

x ; 0 < x <π

2

π − x ;π

2< x < π

.


𝐧𝟐𝛑𝐬𝐢𝐧 (

𝐧𝛑

𝟐) 𝐬𝐢𝐧𝐧𝐱

∞

𝐧=𝟏

W – 14



H 6 Find Half range sine series for cos2x in (0, π).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟐𝐧[(−𝟏)𝐧+𝟏 + 𝟏]

𝛑(𝐧𝟐 − 𝟒)𝐬𝐢𝐧 𝐧𝐱

∞

𝐧=𝟏𝐧≠𝟐

, 𝐛𝟐 = 𝟎 W – 15

FOURIER INTEGRALS

Fourier Integral of f(x) is given by

f(x) = ∫[A(ω) cosωx + B(ω) sinωx]

∞

0

dω

Where, A(ω) =1

π∫ f(x) cosωx

∞

−∞

dx & B(ω) =1

π∫ f(x) sinωx

∞

−∞

dx

FOURIER COSINE INTEGRAL

Fourier Cosine Integral of f(x) is given by

f(x) = ∫ A(ω) cosωx

∞

0

dω

Where, A(ω) =2

π∫ f(x) cosωx

∞

0

dx

FOURIER SINE INTEGRAL

Fourier Sine Integral of f(x) is given by

f(x) = ∫ B(ω) sin ωx

∞

0

dω

Where, B(ω) =2

π∫ f(x) sin ωx

∞

0

dx



METHOD – 5: EXAMPLE ON FOURIER INTERGRAL

C 1 Using Fourier integral, Prove that

∫cosωx + ωsin ωx

1 + ω2dω

∞

0

=

{

0 ; x < 0

π2 ; x = 0

πe−x ; x > 0.

S – 15

C 2 Find the Fourier integral representation of f(x) = {

1 ; |x| < 1

0 ; |x| > 1.

Hence calculate the followings.

a) ∫sin λ cos λx

λdλ

∞

0

b) ∫sin ω

ωdω

∞

0

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∫𝟐𝐬𝐢𝐧𝛚

𝛑𝛚𝐜𝐨𝐬𝛚𝐱

∞

𝟎

𝐝𝛚 𝐚) {

𝛑

𝟐 ; |𝐱| < 𝟏

𝟎 ; |𝐱| > 𝟏

𝐛)𝛑

𝟐

W – 14 W – 16

H 3 Find the Fourier integral representation of f(x) = {

2 ; |x| < 2

0 ; |x| > 2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∫𝟒𝐬𝐢𝐧 𝟐𝛚𝐜𝐨𝐬𝛚𝐱

𝛑𝛚

∞

𝟎

𝐝𝛚

S – 16 S – 17

C 4 Find the Fourier cosine integral of f(x) = e−kx (x > 0, k > 0).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∫𝟐𝐤

𝛑(𝐤𝟐 +𝛚𝟐)𝐜𝐨𝐬𝛚𝐱

∞

𝟎

𝐝𝛚 W – 16

H 5 Find the Fourier cosine integral of f(x) =π

2e−x, x ≥ 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∫𝟏

(𝟏 + 𝛚𝟐)𝐜𝐨𝐬𝛚𝐱

∞

𝟎

𝐝𝛚 W – 15

C 6 Using Fourier integral prove that

∫1 − cosωπ

ω

∞

0

sin ωx dω = {

π

2 ; 0 < x < π

0 ; x > π.

H 7 Express f(x) = {

1 ; 0 ≤ x ≤ π

0 ; x > π.

as a Fourier Sine integral and hence evaluate∫1−cosλπ

λ

∞

0sin λx dλ .

W – 17



T 8 Find Fourier cosine and sine integral of f(x) = {

sin x ; 0 ≤ x ≤ π

0 ; x > π.


𝐚) 𝐟(𝐱) = ∫𝟐(𝟏 + 𝐜𝐨𝐬𝛚𝛑)

𝛑 (𝟏 − 𝛚𝟐)𝐜𝐨𝐬𝛚𝐱

∞

𝟎

𝐝𝛚 ; 𝐀(𝟏) = 𝟎

𝐛) 𝐟(𝐱) = ∫𝟐𝐬𝐢𝐧𝛚𝛑

𝛑 (𝟏 − 𝛚𝟐)𝐬𝐢𝐧𝛚𝐱 𝐝𝛚

∞

𝟎

; 𝐁(𝟏) = 𝟏

T 9 Show that ∫

λ3 sin λx

λ4 + 4

∞

0

dλ =π

2e−x cos x, x > 0. W – 15

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆


[ 3 1 ]


U N I T - 3 A » D i f f e r e n t i a l E q u a t i o n o f F i r s t O r d e r [ 3 2 ]

UNIT-3A » DIFFERENTIAL EQUATION OF FIRST ORDER

INTRODUCTION:

A differential equation is a mathematical equation which involves differentials or differential

coefficients. Differential equations are very important in engineering problem. Most

common differential equations are Newton’s Second law of motion, Series RL, RC, and RLC

circuits, etc.

Mathematical modeling reduces many Natural phenomenon (real world problem) to

differential equation(s).

In this chapter, we will study, the method of obtaining the solution of ordinary differential

equation of first order.

DEFINITION: DIFFERENTIAL EQUATION:

An eqn. which involves differential co-efficient is called a Differential Equation.

e.g. d2y

dx2+ x2

dy

dx+ y = 0

DEFINITION: ORDINARY DIFFERENTIAL EQUATION:

An eqn. which involves function of single variable and ordinary derivatives of that function

then it is called an Ordinary Differential Equation.

e.g. dy

dx+ y = 0

DEFINITION: PARTIAL DIFFERENTIAL EQUATION:

An eqn. which involves function of two or more variables and partial derivatives of that

function then it is called a Partial Differential Equation.

e.g. ∂y

∂x+

∂y

∂t= 0

DEFINITION: ORDER OF DIFFERENTIAL EQUATION:

The order of highest derivative which appeared in a differential equation is “Order of D.E”.

e.g. (dy

dx)2+

dy

dx+ 5y = 0 has order 1.

DEFINITION: DEGREE OF DIFFERENTIAL EQUATION:



When a D.E. is in a polynomial form of derivatives, the highest power of highest order

derivative occurring in D.E. is called a “Degree of D.E.”.

e.g. (dy

dx)2+

dy

dx+ 5y = 0 has degree 2.

NOTE:

To determine the degree, the D.E has to be expressed in a polynomial form in the derivatives.

If the D.E. cannot be expressed in a polynomial form in the derivatives, the degree of D.E. is

not defined.

METHOD – 1: EXAMPLE ON ORDER AND DEGREE OF DIFFERENTIAL EQUATION

C 1

Find order and degree of d2y

dx2= [y + (

dy

dx)2

]

14

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐, 𝟒

H 2 Find order and degree of y = x

dy

dx+

x

dydx

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏, 𝟐

C 3 Find order and degree of (

d2y

dx2)

3

= [x + sin (dy

dx)]

2

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐, 𝐔𝐧𝐝𝐞𝐟𝐢𝐧𝐞𝐝

S – 15

H 4 Define order and degree of the differential equation. Find order and degree

of differential equation √x2d2y

dx2+ 2y =

d3y

dx3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑, 𝟐

S – 17

H 5 Find order and degree of differential equation dy = (y + sinx)dx .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏, 𝟏 S – 16



SOLUTION OF A DIFFERENTIAL EQUATION:

A solution or integral or primitive of a differential equation is a relation between the

variables which does not involve any derivative(s) and satisfies the given differential

equation.

GENERAL SOLUTION (G.S.):

A solution of a differential equation in which the number of arbitrary constants is equal to

the order of the differential equation, is called the General solution or complete integral or

complete primitive.

PARTICULAR SOLUTION:

The solution obtained from the general solution by giving a particular value to the arbitrary

constants is called a particular solution.

SINGULAR SOLUTION:

A solution which cannot be obtained from a general solution is called a singular solution.

LINEAR DIFFERENTIAL EQUATION:

A differential equation is called “LINEAR DIFFERENTIAL EQUATION” if the dependent

variable and every derivatives in the equation occurs in the first degree only and they should

not be multiplied together.

Examples:

(1) d2y

dx2+ x2

dy

dx+ y = 0 is linear.

(2) d2y

dx2+ y

dy

dx+ y = 0 is non-linear.

(3) d2y

dx2+ x2 (

dy

dx)2+ y = 0 is non-linear.

A Linear Differential Equation of first order is known as Leibnitz’s linear Differential

Equation

i.e. dy

dx+ P(x)y = Q(x) + c OR

dx

dy+ P(y)x = Q(y) + c



TYPE OF FIRST ORDER AND FIRST DEGREE DIFFERENTIAL EQUATION:

Variable Separable Equation

Homogeneous Differential Equation

Linear(Leibnitz’s) Differential Equation

Bernoulli’s Equation

Exact Differential Equation

VARIABLE SEPARABLE EQUATION:

If a differential equation of type dy

dx= f(x, y) can be converted into M(x)dx = N(y)dy, then it

is known as a Variable Separable Equation.

The general solution of a Variable Separable Equation is

∫𝐌(𝐱)𝐝𝐱 = ∫𝐍(𝐲)𝐝𝐲 + 𝐜

Where, c is an arbitrary constant.

NOTE:

For convenience, the arbitrary constant can be chosen in any suitable form for the answers.

e. g. in the form logc, tan−1 c , ec, sin c, etc.

REDUCIBLE TO VARIABLE SEPARABLE EQUATION:

If a differential equation of type dy

dx= f(x, y) can be converted into

dy

dx= φ(

y

x) then it can be

converted into variable separable equation by taking y = vx &dy

dx= x

dv

dx+ v.

METHOD – 2: EXAMPLE ON VARIABLE SEPARABLE METHOD

C 1 Solve: 9 y y′ + 4 x = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟗𝐲𝟐 + 𝟒𝐱𝟐 = 𝐜 S – 16

H 2 Solve dy

dx= ex−y + x2e−y by variable separable method.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐲 = 𝐞𝐱 +𝐱𝟑

𝟑+ 𝐜

S – 18



C 3 Solve: xy′ + y = 0 ; y(2) = −2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 ⋅ 𝐲 = −𝟒

H 4 Solve: L

dI

dt+ RI = 0, I(0) = I0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐈 = 𝐈𝟎 ⋅ 𝐞−𝐑𝐋𝐭

T 5 Solve: (1 + x)ydx + (1 − y)xdy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠(𝐱𝐲) + 𝐱 − 𝐲 = 𝐜

C 6 Solve the following differential equation using variable separable method.

3 ex tany dx + (1 + ex) sec2 y dy = 0

𝐀𝐧𝐬𝐰𝐞𝐫: (𝟏 + 𝐞𝐱)−𝟑 = 𝐜 ⋅ 𝐭𝐚𝐧𝒚

W – 17

H 7 Solve: ex tany dx + (1 − ex) sec2y dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: (𝟏 − 𝐞𝐱)−𝟏 𝐭𝐚𝐧 𝐲 = 𝐜

T 8 Solve: xy′ = y2 + y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲

𝐲 + 𝟏= 𝐱𝐜

C 9 Solve: xy

dy

dx= 1 + x + y + xy .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 − 𝐥𝐨𝐠(𝟏 + 𝐲) = 𝐥𝐨𝐠 𝐱 + 𝐱 + 𝐜

H 10 Solve: tany

dy

dx= sin(x + y) + sin(x − y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝐞𝐜 𝐲 = −𝟐𝐜𝐨𝐬 𝐱 + 𝐜

H 11 Solve: 1 +

dy

dx= ex+y .

𝐀𝐧𝐬𝐰𝐞𝐫:−(𝐞−𝐱−𝐲) = 𝐱 + 𝐜

T 12 Solve:

dy

dx= cosx cosy − sinx siny .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭𝐚𝐧 (𝐱 + 𝐲

𝟐) = 𝐱 + 𝐜

H 13 Solve: x

dy

dx= y + x e

y x .

𝐀𝐧𝐬𝐰𝐞𝐫:− (𝐞−𝐲𝐱) = 𝐥𝐨𝐠 𝐱 + 𝐜

C 14 Solve:

dy

dx=y

x+ tan (

y

x) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝐢𝐧(y/x) = 𝐱 ⋅ 𝐜



LEIBNITZ’S ( LINEAR ) DIFFERENTIAL EQUATION:

Form - 1 Form -2

Form of differential

equation

dy

dx+ P(x)y = Q(x)

dx

dy+ P(y)x = Q(y)

Integrating factor I. F. = e∫P(x)dx I. F. = e∫P(y)dy

Solution y (I. F. ) = ∫Q(x) (I. F. )dx + c x (I. F. ) = ∫Q(y) (I. F. )dy + c

METHOD – 3: EXAMPLE ON LEIBNITZ’S DIFFERENTIAL EQUATION

C 1 Solve: y′ + y sin x = ecosx .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 𝐞−𝐜𝐨𝐬𝐱 = 𝐱 + 𝐜

H 2 Solve:

dy

dx+

1

x2y = 6 e

1x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 𝐞−𝟏𝐱 = 𝟔 𝐱 + 𝐜

H 3 Solve: y′ + 6x2y =

e−2x3

x2, y(1) = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 𝐞𝟐𝐱𝟑= (𝟏 −

𝟏

𝐱)

C 4 Solve: (x + 1)

dy

dx− y = (x + 1)2e3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲

𝐱 + 𝟏=𝐞𝟑𝐱

𝟑+ 𝐜

W – 16

H 5 Solve:

dy

dx+ y = x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 𝐞𝐱 = 𝐞𝐱 𝐱 − 𝐞𝐱 + 𝐜

H 6 Solve: x

dy

dx+ (1 + x)y = x3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 𝐲 𝐞𝐱 = 𝐱𝟑𝐞𝐱 − 𝟑 𝐱𝟐 𝐞𝐱 + 𝟔 𝐱 𝐞𝐱 − 𝟔 𝐞𝐱 + 𝐜

C 7 Solve:

dy

dx+ (cot x)y = 2cos x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 ⋅ 𝐬𝐢𝐧𝐱 = −𝐜𝐨𝐬𝟐𝐱

𝟐+ 𝐜

S – 16



H 8 Solve:

dy

dx+ y tanx = sin2x , y(0) = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 ⋅ 𝐬𝐞𝐜𝐱 = −𝟐𝐜𝐨𝐬 𝐱 + 𝟐 S – 17

T 9 Solve:

dy

dx+

4x

x2 + 1y =

1

(x2 + 1)3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 ⋅ (𝐱𝟐 + 𝟏)𝟐= 𝐭𝐚𝐧−𝟏 𝐱 + 𝐜

BERNOULLI’S DIFFERENTIAL EQUATION:

A differential equation of the form dy

dx+ P(x)y = Q(x)yn OR

dx

dy+ P(y)x = Q(y) xn is known

as Bernoulli’s Differential Equation. Where, n ∈ ℝ − {0,1} such differential equation can be

converted into linear differential equation and accordingly can be solved.

EQUATION REDUCIBLE TO LINEAR DIFFERENTIAL EQUATION FORM:

CASE 1 : A differential equation of the form dy

dx+ P(x)y = Q(x) yn…… (1)

Dividing both sides of equation (1) by yn,

We get, y−ndy

dx+ P(x)y1−n = Q(x)____(2)

Let, y1−n = v

⟹ (1− n)y−ndy

dx=dv

dx

⟹ y−ndy

dx=

1

(1 − n) dv

dx

By Eqn (2), 1

(1−n) dv

dx+ P(x)v = Q(x)

⟹dv

dx+ P(x)(1 − n)v = Q(x)(1 − n)

Which is Linear Differential equation and accordingly can be solved.

CASE 2 : A differential of form dy

dx+ P(x)f(y) = Q(x) g(y) ……(3)

Dividing both sides of equation (3) by "g(y)" ,

We get, 1

g(y)

dy

dx+ P(x)

f(y)

g(y)= Q(x) ……(4)

Let, f(y)

g(y)= v



Differentiate with respect to x both the sides,

Eqn (4) becomes Linear Differential equation and accordingly can be solved.

METHOD – 4: EXAMPLE ON BERNOULLI’S DIFFERENTIAL EQUATION

H 1 Solve:

dy

dx+ y = −

x

y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 𝐞𝟐𝐱 = −𝐱 𝐞𝟐𝐱 +𝐞𝟐𝐱

𝟐+ 𝐜

T 2 Solve the following Bernoulli’s equation dy

dx+

y

x=

y2

x2

𝐀𝐧𝐬𝐰𝐞𝐫:𝟏

𝐱𝐲= −

𝟏

𝟐𝐱𝟐+ 𝐜

W – 17

C 3 Solve:

dy

dx+1

xy = x3y3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏

𝐱𝟐 𝐲𝟐= −𝐱𝟐 + 𝐜

W – 15

H 4 Solve:

dy

dx+y

x= x2y6 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲−𝟓 𝐱−𝟓 =𝟓

𝟐𝐱−𝟐 + 𝐜

H 5 Solve:

dy

dx+1

x=ey

x2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝐲

𝐱=

𝟏

𝟐 𝐱𝟐+ 𝐜

S – 15

C 6 Solve:

dy

dx−tan y

1 + x= (1 + x)ex sec y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝐢𝐧 𝐲

𝟏 + 𝐱= 𝐞𝐱 + 𝐜

H 7 Solve:

dy

dx+ x sin 2y = x3 cos2 y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐱𝟐𝐭𝐚𝐧 𝐲 =

(𝐱𝟐 − 𝟏)𝐞𝐱𝟐

𝟐+ 𝐜

H 8 Solve:

dy

dx− 2 y tan x = y2 tan2 x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝐞𝐜𝟐𝐱

𝐲= −

𝐭𝐚𝐧𝟑𝐱

𝟑+ 𝐜



C 9 Solve: (x3 y2 + x y)dx = dy .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐱𝟐

𝟐

𝐲= (𝟐 − 𝐱𝟐)𝐞

𝐱𝟐

𝟐 + 𝐜

T 10 Solve: x

dy

dx+ y = y2 log x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐥𝐨𝐠 𝐱 + 𝟏) + 𝐜 𝐱 𝐲 = 𝟏

H 11 Solve: ey

dy

dx+ ey = ex .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐱+𝐲 =𝐞𝟐𝐱

𝟐+ 𝐜

EXACT DIFFERENTIAL EQUATION:

A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is said to be Exact Differential

Equation if it can be derived from its primitive by direct differential without any further

transformation such as elimination etc.

The necessary and sufficient condition for differential equation to be exact is 𝛛𝐌

𝛛𝐲=

𝛛𝐍

𝛛𝐱.

Where first order continuous partial derivative of M and N must be exist at all points

of f(x, y).

The general solution of Exact Differential Equation is

∫ M dxy=constant

+∫(terms of N free from x)dy = c


METHOD – 5: EXAMPLE ON EXACT DIFFERENTIAL EQUATION

H 1 Solve: (x3 + 3xy2)dx + (y3 + 3x2y)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟒

𝟒+𝟑𝐱𝟐𝐲𝟐

𝟐+𝐲𝟒

𝟒= 𝐜

W – 14



C 2 Check whether the given differential equation is exact or not (x4 − 2xy2 +

y4)dx − (2x2y − 4xy3 + siny)dy

𝐀𝐧𝐬𝐰𝐞𝐫:𝐱𝟓

𝟓− 𝐱𝟐𝐲𝟐 + 𝐱𝐲𝟒 + 𝐜𝐨𝐬𝐲 = 𝐜

W – 17

H 3 Solve: (x2 + y2)dx + 2xydy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟑

𝟑+ 𝐲𝟐𝐱 = 𝐜

H 4 Solve: 2 x y dx + x2 dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟐𝐲 = 𝐜

C 5 Solve:

dy

dx=x2 − x − y2

2xy .

𝐀𝐧𝐬𝐰𝐞𝐫:𝐱𝟑

𝟑−𝐱𝟐

𝟐− 𝐱𝐲𝟐 = 𝐜

W – 15

H 6 Solve: yexdx + (2y + ex)dy = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝐞𝐱 + 𝐲𝟐 = 𝐜 S – 15

H 7 Solve: (ey + 1) cos x dx + ey sin x dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐞𝐲 + 𝟏) 𝐬𝐢𝐧𝐱 = 𝐜

H 8 Test for exactness and solve :[(x + 1)ex − ey]dx − xeydy = 0, y(1) = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱(𝐞𝐱 − 𝐞𝐲) = 𝐞 − 𝟏

C 9 Solve:

dy

dx+

ycosx + siny + y

sinx + xcosy + x= 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 𝐬𝐢𝐧 𝐱 + 𝐱 𝐬𝐢𝐧 𝐲 + 𝐱𝐲 = 𝐜 W – 16

H 10 Solve:

y2

xdx + (1 + 2ylogx)dy = 0, x > 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 𝐥𝐨𝐠 𝐱 + 𝐲 = 𝐜

H 11 Solve: (y2exy2+ 4x3)dx + (2xyexy

2− 3y2)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐱𝐲𝟐+ 𝐱𝟒 − 𝐲𝟑 = 𝐜

NON-EXACT DIFFERENTIAL EQUATION:

A differential equation which is not exact differential equation is known as Non-Exact

Differential Equation. i.e. if ∂M

∂y≠

∂N

∂x then given equation is Non-Exact Differential Equation.



INTEGRATING FACTOR:

A differential equation which is not exact can be made exact by multiplying it by a suitable

function of x and y. Such a function is known as Integrating Factor.

SOME STANDARD RULES FOR FINDING I.F.:

(1) If Mx + Ny ≠ 0 and the given equation is Homogeneous, then I. F. =1

Mx+Ny .

(2) If Mx − Ny ≠ 0 and the given equation is of the form f(x, y) y dx + g(x, y) x dy = 0

(OR Non-Homogeneous), then I. F. =1

Mx−Ny .

(3) If 1

N(∂M

∂y−

∂N

∂x) = f(x) (i. e. function of only x), then I. F. = e∫ f(x) dx

(4) If 1

M(∂N

∂x−

∂M

∂y) = g(y) (i. e. function of only y), then I. F. = e∫g(y)dy

Further, Multiply I.F. to given differential equation. So, It will converted to Exact Differential

Equation.

Now, we will get new M′(x, y) and N′(x, y).

Where, M′(x, y) = M(x, y) ⋅ (I. F. ) & N′(x, y) = N(x, y) ⋅ (I. F. )

Later on it can be solved same method as used for exact differential equation.

i.e.

∫ M′ dxy=constant

+∫(terms of N′ free from x)dy = c


METHOD – 6: EXAMPLE ON NON-EXACT DIFFERENTIAL EQUATION

H 1 Solve x2y dx − (x3 + xy2)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: −𝐱𝟐

𝟐𝐲𝟐+ 𝐥𝐨𝐠𝐲 = 𝐜

W – 14

C 2 Solve: (x2y − 2xy2)dx − (x3 − 3x2y)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱

𝐲− 𝟐 𝐥𝐨𝐠 𝐱 + 𝟑𝐥𝐨𝐠𝐲 = 𝐜



T 3 Solve: (x3+y3) dx − x y2dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 𝐱 −𝐲𝟑

𝟑𝐱𝟑= 𝐜

C 4 Solve (x4 + y4)dx − xy3dy = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠𝐱 −𝐲𝟒

𝟒𝐱𝟒= 𝐜

S – 18

H 5 Solve: (x + y)dx + (y − x)dy = 0 .


𝟐𝐥𝐨𝐠(𝐱𝟐 + 𝐲𝟐) + 𝐭𝐚𝐧−𝟏 (

𝐱

𝐲) = 𝐜

C 6 Solve: (x2y2 + 2)ydx + (2 − x2y2)xdy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠𝐱

𝐲−

𝟏

𝐱𝟐𝐲𝟐= 𝟐𝐜

W – 14

T 7 Solve: y(1 + xy)dx + x(1 + xy + x2y2)dy = 0 .


𝟐𝐱𝟐𝐲𝟐+

𝟏

𝐱𝐲− 𝐥𝐨𝐠𝐲 = 𝐜

H 8 Solve: y(xy + 2x2y2)dx + x(xy − x2y2)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫:−𝟏

𝐱𝐲+ 𝐥𝐨𝐠

𝐱𝟐

𝐲= 𝟑𝐜

T 9 Solve: (x2+y2 + x)dx + xydy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝐱𝟒 + 𝟔𝐱𝟐𝐲𝟐 + 𝟒𝐱𝟑 = 𝟏𝟐𝐜

C 10 Solve: (x2+y2 + 3)dx − 2xydy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟐 − 𝐲𝟐 − 𝟑 = 𝐜𝐱 S – 17

C 11 Solve: (3x2y4 + 2xy)dx + (2x3y3 − x2)dy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟑𝐲𝟐 +𝐱𝟐

𝐲= 𝐜

DEFINITION: ORTHOGONAL TRAJECTORY:

A curve which cuts every member of a given family at right angles is a called an Orthogonal

Trajectory.



METHODS OF FINDING ORTHOGONAL TRAJECTORY OF 𝐟(𝐱, 𝐲, 𝐜) = 𝟎

(1) Differentiate f(x, y, c) = 0… (1) w.r.t. x.

(2) Eliminate c by using eqn …(1) and its derivative.

(3) Replace dy

dx by −

dx

dy. This will give you differential equation of the orthogonal

trajectories.

(4) Solve the differential equation to get the equation of the orthogonal trajectories.

METHODS OF FINDING ORTHOGONAL TRAJECTORY OF 𝐟(𝐫, 𝛉, 𝐜) = 𝟎

(1) Differentiate f(r, θ, c) = 0… (1) w.r.t. θ.

(2) Eliminate c by using eqn …(1) and its derivative

(3) Replace dr

dθ by −r2

dθ

dr. This will give you differential eqn of the orthogonal

trajectories.

(4) Solve the differential equation to get the equation of the orthogonal trajectories.

METHOD – 7: EXAMPLE ON ORTHOGONAL TREJECTORY

C 1 Find orthogonal trajectories of y = x2 + c .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 𝐱 + 𝟐𝐲 = 𝐜

H 2 Find the orthogonal trajectories of the family of circles x2 + y2 = c2 .


𝐲= 𝐜 S – 16

T 3 Find the orthogonal trajectories of the family of circles x2 + y2 = 2cx .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱𝟐 + 𝐲𝟐 = 𝟐𝐚𝐲

C 4 Find Orthogonal trajectories of 𝑟 = 𝐚 (𝟏 − 𝐜𝐨𝐬 𝛉) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐫 =𝐜

𝟐(𝟏 + 𝐜𝐨𝐬 𝛉)

W – 17

H 5 Find Orthogonal trajectories of rn = an cosnθ .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐫𝐧 = 𝐜𝐧 𝐬𝐢𝐧 𝐧𝛉

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆




U N I T - 3 B » D i f f e r e n t i a l E q u a t i o n o f H i g h e r O r d e r [ 4 6 ]

UNIT-3B » DIFFERENTIAL EQUATION OF HIGHER ORDER

INTRODUCTION:

Many engineering problems such as Oscillatory phenomena, Bending of beams, etc. leads to

the formulation and solution of Linear Ordinary Differential equations of second and higher

order.

In this chapter we will study, the method of obtaining the solution of Linear Ordinary

Differential equations (homogeneous and nonhomogeneous) of second and higher order.

HIGHER ORDER LINEAR DIFFERENTIAL EQUATION:

A linear differential equation with more than one order is known as Higher Order Linear

Differential Equation.

A general linear differential equation of the nth order is of the form

P0dny

dxn+ P1

dn−1y

dxn−1+ P2

dn−2y

dxn−2+ ⋯+ Pny = R(x)……… (A)

Where, P0 , P1 , P2 , … are functions of x.

HIGHER ORDER LINEAR DIFFERENTIAL EQUATION WITH CONSTANT CO-EFFICIENT:

The nth order linear differential equation with constant co-efficient is

a0dny

dxn+ a1

dn−1y

dxn−1+ a2

dn−2y

dxn−2+ ⋯+ any = R(x)………(B)

Where, a0, a1 , a2, … are constants.

NOTATIONS:

Eq. (B) can be written in operator form by taking D ≡d

dx as below,

a0Dny + a1D

n−1y + a2Dn−2y + ⋯+ any = R(x) ………(C)

OR

[f(D)]y = R(x)……… (D)

NOTE:

An nth order linear differential equation has n linear independent solution.



AUXILIARY EQUATION:

The auxiliary equation for nth order linear differential equation

a0Dny + a1D

n−1y + a2Dn−2y + ⋯+ any = R(x)

is derived by replacing D by m and equating with 0.

i. e. a0mn + a1m

n−1 + a2mn−2 + ⋯+ an = 0

COMPLIMENTARY FUNCTION (C.F. -- 𝐲𝐜 ):

A general solution of [f(D)]y = 0 is called complimentary function of [f(D)]y = R(x).

PARTICULAR INTEGRAL (P.I. -- 𝐲𝐩):

A particular integral of [f(D)]y = R(x) is y =1

f(D)R(x).

GENERAL SOLUTION [𝐲 (𝐱)] OF HIGHER ORDER LINEAR DIFFERENTIAL EQUATION:

G. S. = C. F + P. I. i.e. y(x) = yc + 𝑦𝑝

NOTE:

In case of higher order homogeneous differential equation, complimentary function is same

as general solution.

FORMULA:

(1) a3 − b3 = (a − b)(a2 + ab + b2) (2) a3 + b3 = (a + b)(a2 − ab + b2)

(3) (a + b)3 = a3 + b3 + 3ab(a + b) (4) (a − b)3 = a3 − b3 − 3ab(a − b)

METHOD FOR FINDING C.F. OF HIGHER ORDER DIFFERENTIAL EQUATION:

Consider, a0Dny + a1D

n−1y + a2Dn−2y + ⋯+ any = R(x)

The Auxiliary equation is a0mny + a1m

n−1y + a2mn−2y + ⋯+ any = 0

Let, m1 , m2 , m3 , … be the roots of auxiliary equation.

Case Nature of the “n”

roots L.I. solutions General Solutions

1) m1 ≠ m2 ≠ m3 ≠ ⋯ em1x, em2x, em3x, … y = c1em1x + c2e

m2x + c3em3x + ⋯



2) m1 = m2 = m3 = m em1x, xem2x, x2em3x y = (c1 + c2x + c3x2)emx

3) m1 = m2 = m3 = m

m4 ≠ m5 , …

emx, x emx, x2emx,

em4x, em5x, …

y = (c1 + c2x + c3x2)emx

+c4em4x + c5e

m5x + ⋯

4) m = p ± iq epx cos qx , epx sin qx, y = epx(c1 cosqx + c2 sin qx)

5) m1 = m2 = p ± iq

epx cosqx , xepx cos qx,

epx sin qx , xepx sin qx,

y = epx[(c1+ c2x) cosqx

+(c3 + c4x) sin qx]

METHOD – 1: EXAMPLE ON HOMOGENEOUS DIFFERENTIAL EQUATION

C 1 Solve: y′′ + y′ − 2y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞−𝟐𝐱 + 𝐜𝟐𝐞

𝐱

T 2 Solve: y′′ − 9y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝟑𝐱 + 𝐜𝟐𝐞

−𝟑𝐱

H 3 Solve 𝑦′′′ − 6𝑦′′ + 11𝑦′ − 6𝑦 = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞

𝟐𝐱 + 𝐜𝟑𝐞𝟑𝐱

S – 18

H 4 Solve: y′′ − 6y′ + 9y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱) 𝐞3𝐱

C 5 Solve y′′′ − 3y′′ + 3y′ − y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱 + 𝐜𝟑𝐱𝟐) 𝐞𝐱

C 6 Solve:

d4y

dx4− 18

d2y

dx2+ 81y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟑𝐱 + (𝐜𝟑 + 𝐜𝟒𝐱)𝐞

−𝟑𝐱

C 7 Solve: 16y′′ − 8y′ + 5y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞𝐱𝟒 (𝐜𝟏 𝐜𝐨𝐬

𝐱

𝟐+ 𝐜𝟐 𝐬𝐢𝐧

𝐱

𝟐)

H 8 Solve: (D4 − 1)y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞−𝐱 + 𝐜𝟐𝐞

𝐱 + (𝐜𝟑𝐜𝐨𝐬𝐱 + 𝐜𝟒𝐬𝐢𝐧𝐱)

C 9 Solve: y′′′ − y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐞−

𝟏𝟐𝐱 (𝐜𝟐 𝐜𝐨𝐬

√𝟑

𝟐𝐱 + 𝐜𝟑 𝐬𝐢𝐧

√𝟑

𝟐𝐱)



C 10 Solve: y′′ − 5y′ + 6y = 0; y(1) = e2 , y′(1) = 3e2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞𝟑𝐱−𝟏

H 11 Solve: y′′ + 4y′ + 4y = 0; y(0) = 1, y′(0) = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝟏 + 𝟑𝐱)𝐞−𝟐𝐱 S – 16

T 12 Solve: y′′ − 4y′ + 4y = 0; y(0) = 3, y′(0) = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝟑 − 𝟓𝐱)𝐞𝟐𝐱 W – 14

H 13 Solve: y′′′ − y′′ + 100y′ − 100y = 0; y(0) = 4, y′(0) = 11, y′′(0) = −299 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞𝐱 + 𝐬𝐢𝐧𝟏𝟎𝐱 + 𝟑𝐜𝐨𝐬 𝟏𝟎𝐱

T 14 Solve: (D4 + K4)y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝒚 = 𝐞𝐤

√𝟐𝐱{𝐜𝟏𝐜𝐨𝐬

𝐤

√𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧

𝐤

√𝟐𝐱} + 𝐞

−𝐤

√𝟐𝐱{𝐜𝟑𝐜𝐨𝐬

𝐤

√𝟐𝐱 + 𝐜𝟒𝐬𝐢𝐧

𝐤

√𝟐𝐱}

METHOD OF FINDING THE PARTICULAR INTEGRAL:

There are many methods of finding the particular integral 1

f(D) R(x), we shall discuss

following four main methods,

(1) General Methods

(2) Short-cut Methods involving operators

(3) Method of Undetermined Co-efficient

(4) Method of Variation of parameters

GENERAL METHODS:

Consider the differential equation

a0Dny + a1D

n−1y + a2Dn−2y + ⋯+ any = R(x)

⟹ f(D)y = R(x)

∴ Particular Integral = yp =1

f(D) R(x)

Particular Integral may be obtained by following two ways:

(1). Method of Factors:

The operator 1

f(D) may be factorized into n linear factors; then the P.I. will be



P. I. =1

f(D) R(x) =

1

(D −m1)(D − m2) ………… . (D − mn)R(x)

Now, we know that,

1

D −mnR(x) = emnx∫R(x) e−mnxdx

On operating with the first symbolic factor, beginning at the right, the particular integral will

have form

P. I. = 1

(D − m1)(D − m2)………… . (D − mn−1)emnx∫R(x) e−mnxdx

Then, on operating with the second and remaining factors in succession, taking them from

right to left, one can find the desired particular integral.

(2). Method of Partial Fractions:

The operator 1

f(D) may be factorized into n linear factors; then the P.I. will be

P. I. =1

f(D) R(x) = (

A1

D −m1

+A2

D − m2

+⋯+An

D −mn

) R(x)

= A11

D − m1R(x) + A2

1

D −m2R(x) + ⋯+ An

1

D −mnR(x)

Using 1

D−mnR(x) = emnx ∫R(x) e−mnxdx ,we get

P. I. = A1em1x∫R(x) e−m1xdx + A2e

m2x∫R(x) e−m2xdx +⋯+ Anemnx∫R(x) e−mnxdx

Out of these two methods, this method is generally preferred.

SHORTCUT METHOD:

R(x) = eax

P. I. =1

f(D)eax =

1

f(a)eax, if f(a) ≠ 0

If f(a) = 0 , P. I. =1

f(D)eax =

x

f′(a)eax, if f ′(a) ≠ 0

In general, If fn−1(a) = 0 , P. I. =1

f(D)eax =

xn

fn(a)eax, if fn(a) ≠ 0

R(x) = sin(ax + b)



P. I. =1

f(D2)sin(ax + b) =

1

f(−a2)sin(ax + b) , if f(−a2) ≠ 0

If f(−a2) = 0 , P. I. =1

f(D2)sin(ax + b) =

x

f′ (−a2)sin(ax + b) , if f′(−a2) ≠ 0

If f ′(−a2) = 0 , P. I. =1

f(D2)sin(ax + b) =

x2

f"(−a2)sin(ax + b) , if f"(−a2) ≠ 0 and so on…

R(x) = cos(ax + b)

P. I. =1

f(D2)cos(ax + b) =

1

f(−a2)cos(ax + b) , if f(−a2) ≠ 0

If f(−a2) = 0 , P. I. =1

f(D2)cos(ax + b) =

x

f′(−a2)cos(ax + b) , if f′(−a2) ≠ 0

If f ′(−a2) = 0 , P. I. =1

f(D2)cos(ax + b) =

x2

f"(−a2)cos(ax + b) , if f"(−a2) ≠ 0 and so on…

R(x) = xm;m > 0

In this case convert f(D) in the form of 1 + ϕ(D) or 1 − ϕ(D) form.

P. I. =1

f(D)xm =

1

1+ϕ(D)xm = [{1 − ϕ(D) + [ϕ(D)]2− . . . } ]xm

(Using Binomial Theorem)

NOTE:

(1) 1

1+𝑥= (1 + 𝑥)−1 = 1− 𝑥 + 𝑥2 − ⋯

(2) 1

1−x= (1 − x)−1 = 1 + x + x2 +⋯

(3) (1 + h)n = 1 + n ⋅h1

1 !+ n ⋅ (n − 1) ⋅

h2

2 !+ n ⋅ (n − 1) ⋅ (n − 2) ⋅

h3

3 !+⋯

R(x) = eax V(X) , Where V(X) is a function of x.

P. I. =1

f(D)eax V(x) = eax

1

f(D + a)V(x)



METHOD – 2: EXAMPLE ON NON-HOMOGENEOUS DIFFERENTIAL EQUATION

C 1 Solve:

d2y

dx2+dy

dx− 12y = e6x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏𝐞−𝟒𝐱 + 𝐜𝟐𝐞

𝟑𝐱) +𝟏

𝟑𝟎𝐞𝟔𝐱

H 2 Solve y′′ − 3y′ + 2y = e3x


𝟐𝐱 +𝐞𝟑𝐱

𝟐

S – 18

H 3 Solve: (D2 − 2D + 1)y = 10ex .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝐱 + 𝟓𝐱𝟐𝐞𝐱

S – 15

H 4 Solve:

d2y

dx2+ 2

dy

dx− 35y = 12e5x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞−𝟕𝐱 + 𝐜𝟐𝐞

𝟓𝐱 + 𝐱𝐞𝟓𝐱

T 5 Solve: (D3 − 7D + 6)y = e2x .


𝟐𝐱 + 𝐜𝟑𝐞−𝟑𝐱 +

𝐱

𝟓𝐞𝟐𝐱

C 6 Solve: y′′′ − 3y′′ + 3y′ − y = 4et .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐭 + 𝐜𝟑𝐭𝟐)𝐞𝐭 +

𝟐

𝟑𝐭𝟑𝐞𝐭

W – 14

C 7 Solve: y′′ − 6y′ + 9y = 6e3x − 5 log2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱) 𝐞𝟑𝐱 + 𝟑𝐱𝟐𝐞𝟑𝐱 −

𝟓

𝟗 𝐥𝐨𝐠 𝟐

C 8 Solve: (D2 − 49)y = sinh 3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞−𝟕𝐱 + 𝐜𝟐𝐞

𝟕𝐱 −𝟏

𝟒𝟎𝐬𝐢𝐧𝐡 𝟑𝐱

H 9 Find the complete solution of

d3y

dx3+ 8y = cosh(2x) .


𝐲 = 𝐜𝟏𝐞−𝟐𝐱 + 𝐞𝐱[𝐜𝟐 𝐜𝐨𝐬(√𝟑𝐱) + 𝐜𝟑 𝐬𝐢𝐧(√𝟑𝐱)] +

𝟏

𝟑𝟐𝐞𝟐𝐱 +

𝐱

𝟐𝟒𝐞−𝟐𝐱

W – 15

C 10 Solve: (D3 − 3D2 + 9D − 27)y = cos3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝟑𝐱 + 𝐜𝟐 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟑 𝐬𝐢𝐧 𝟑𝐱 −

𝐱

𝟑𝟔(𝐜𝐨𝐬𝟑𝐱 + 𝐬𝐢𝐧𝟑𝐱)

W – 16

H 11 Solve (D2 + 9)y = cos4x

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬𝟑𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟑𝐱 −𝟏

𝟕𝐜𝐨𝐬𝟒𝐱

S – 18



T 12 Find the steady state oscillation of the mass-spring system governed by the

equation y′′ + 3y′ + 2y = 20 cos2t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞−𝟐𝐭 + 𝐜𝟐𝐞

−𝐭 + 𝟑𝐬𝐢𝐧𝟐𝐭 − 𝐜𝐨𝐬𝟐𝐭

T 13 Solve: (D2 + 4)y = sin 2x , given that y = 0 and dy

dx= 2 when x = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 =𝟗

𝟖𝐬𝐢𝐧 𝟐𝐱 −

𝐱

𝟒𝐜𝐨𝐬 𝟐𝐱

C 14 Solve: (D2 − 4D + 3)y = sin 3x cos2x .


𝟑𝐱 +𝐬𝐢𝐧 𝐱 + 𝟐 𝐜𝐨𝐬 𝐱

𝟐𝟎+𝟏𝟎𝐜𝐨𝐬 𝟓𝐱 − 𝟏𝟏 𝐬𝐢𝐧 𝟓𝐱

𝟖𝟖𝟒

H 15 Solve complementary differential equation d2y

dx2−

6dy

dx+ 9y = sin x cos2x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟑𝐱 +

𝐜𝐨𝐬 𝟑𝐱

𝟑𝟔−𝟑 𝐜𝐨𝐬 𝐱 + 𝟒 𝐬𝐢𝐧𝐱

𝟏𝟎𝟎

W – 15

T 16 Solve: (D2 + 9)y = cos3x + 2 sin 3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 −𝐱

𝟑𝐜𝐨𝐬𝟑𝐱 +

𝐱

𝟔𝐬𝐢𝐧𝟑𝐱

S – 16

C 17 Solve: y′′ + 2y′ + 3y = 2x2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞−𝐱(𝐜𝟏𝐜𝐨𝐬√𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧√𝟐𝐱) + ( 𝟐

𝟑𝐱𝟐 −

𝟖

𝟗𝐱 +

𝟒

𝟐𝟕)

W – 14

H 18 Solve: (D3 − D)y = x3 .


−𝐱 + 𝐜𝟑 −𝐱𝟒

𝟒− 𝟑𝐱𝟐

W – 16

H 19 Solve: (D3 − D2 − 6D)y = x2 + 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 + 𝐜𝟐𝐞𝟑𝐱 + 𝐜𝟑𝐞

−𝟐𝐱 −𝐱𝟑

𝟏𝟖+𝐱𝟐

𝟑𝟔−𝟐𝟓𝐱

𝟏𝟎𝟖

T 20 Find the general solution of the following differential equation

d3y

dx3− 2

dy

dx+ 4y = excosx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐞−𝟐𝐱 + 𝐞𝐱(𝐜𝟐𝐜𝐨𝐬𝐱 + 𝐜𝟑𝐬𝐢𝐧𝐱) +

𝐱 𝐞𝐱

𝟐𝟎(𝟑 𝐬𝐢𝐧 𝐱 − 𝐜𝐨𝐬 𝐱)

W – 17

C 21 Solve: (D3 − D2 + 3D + 5)y = ex cos3x .


𝐲 = 𝐜𝟏𝐞−𝐱 + 𝐞𝐱(𝐜𝟐 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟑 𝐬𝐢𝐧 𝟐𝐱) −

𝐞𝐱

𝟔𝟓(𝟑 𝐬𝐢𝐧 𝟑𝐱 + 𝟐𝐜𝐨𝐬 𝟑𝐱)

H 22 Solve: (D2 − 5D + 6)y = e2x sin 2x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝟑𝐱 + 𝐜𝟐𝐞

𝟐𝐱 + 𝐞𝟐𝐱 (−𝟏

𝟓𝐬𝐢𝐧 𝟐𝐱 +

𝟏

𝟏𝟎𝐜𝐨𝐬 𝟐𝐱)



C 23 Solve: (D2 − 2D + 1)y = x2e3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝐱 +

𝐞𝟑𝐱

𝟒(𝐱𝟐 − 𝟐𝐱 +

𝟑

𝟐)

H 24 Solve: (D4 − 16)y = e2x + x4 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝟐𝐱 + 𝐜𝟐𝐞

−𝐱 + 𝐜𝟑 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟒 𝐬𝐢𝐧 𝟐𝐱 +𝐱

𝟑𝟐𝐞𝟐𝐱 −

𝐱𝟒

𝟏𝟔−

𝟑

𝟑𝟐

S – 17

T 25 Solve:

d4y

dt4− 2

d2y

dt2+ y = cos t + e2t + et .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐭)𝐞−𝐭 + (𝐜𝟑 + 𝐜𝟒𝐭)𝐞

𝐭 + (𝐜𝐨𝐬𝐭

𝟒+𝐞𝟐𝐭

𝟗+𝐭𝟐𝐞𝐭

𝟖)

C 26 Solve: (D2 + 16)y = x4 + e3x + cos3x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟒𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟒𝐱 +𝟏

𝟏𝟔(𝐱𝟒 −

𝟑𝐱𝟐

𝟒+

𝟑

𝟑𝟐) +

𝐞𝟑𝐱

𝟐𝟓+𝐜𝐨𝐬 𝟑𝐱

𝟕

H 27 Solve: y′′ + 4y = 8e−2x + 4x2 + 2; y(0) = 2, y′(0) = 2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝐨𝐬𝟐𝐱 + 𝟐𝐬𝐢𝐧𝟐𝐱 + 𝐞−𝟐𝐱 + 𝐱𝟐

METHOD OF UNDETERMINED CO-EFFICIENT:

This method determines P.I. of f(D)y = R(x). In this method we will assume a trial solution

containing unknown constants, which will be obtained by substitution in f(D)y = R(x). The

trial solution depends upon R(x) (the RHS of the given equation f(D)y = R(x) .

Let the given equation be f(D)y = R(x) …... (A)

∴ The general solution of (A) is Y = YC + YP

Here we guess the form of YP depending on X as per the following table.

RHS of 𝐟(𝐃)𝐲 = R(x) Form of Trial Solution

1. R(x) = eax YP = Aeax

e.g. R(x) = e2x

R(x) = e2x − 3e−x

YP = Ae2x

YP = Ae2x + Be−x

2. R(x) = sin ax

YP = A sin ax + B cos ax R(x) = cos ax

e.g. R(x) = cos3x YP = A sin 3x + B cos3x



R(x) = 2 sin(4x − 5) YP = A sin(4x − 5) + B cos(4x − 5)

3.

R(x) = a + bx + cx2 + dx3

R(x) = ax2 + bx

R(x) = ax + b

R(x) = c

YP = A + Bx + Cx2 + Dx3

YP = A + Bx + Cx2

YP = A + Bx

YP = A

4. R(x) = eax sin bx

YP = eax(A sin bx + B cos bx) R(x) = eax cosbx

5. R(x) = xeax

R(x) = x2eax

YP = eax(A + Bx)

YP = eax(A + Bx + Cx2)

6. R(x) = x sin ax

R(x) = x2 cos ax

YP = sin ax (A + Bx) + cos ax (C + Dx)

YP = sin ax (A + Bx + Cx2) + cos ax (D + Ex + Fx2)

METHOD – 3: EXAMPLE ON UNDETERMINED CO-EFFICIENT

C 1 Solve: y′′ + 4y = 4 e2𝑥

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟐𝐱 +𝟏

𝟐𝐞𝟐𝐱

C 2 Solve: y′′ + 4y = 2sin3x.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟐𝐱 −𝟐

𝟓𝐬𝐢𝐧𝟑𝐱

C 3 Solve: y′′ + 9𝑦 = 2x2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 +𝟐

𝟗𝐱𝟐 −

𝟒

𝟖𝟏

S – 17

H 4 Solve y′′ − 2y′ + 5y = 5x3 − 6x2 + 6x by method of undetermined

coefficients.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞𝐱(𝐜𝟏𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟐𝐱) + 𝐱𝟑

S – 18

T 5 Solve: y′′ + 4y′ = 8x2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏+𝐜𝟐𝐞−𝟒𝐱 +

𝐱

𝟒−

𝐱𝟐

2+

𝟐

𝟑𝐱𝟑

S – 16

C 6 Solve: y′′ − 2y′ + y = ex + x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐱 𝐜𝟐)𝐞𝐱 +

𝐱𝟐𝐞𝐱

𝟐+ 𝐱 + 𝟐



H 7 Solve the following differential equation using the method of

undetermined coefficient : d2y

dx2+ 2

dy

dx+ 4y = 2x2 + 3e−x

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞−𝐱(𝐜𝟏𝐜𝐨𝐬√𝟑𝐱 + 𝐜𝟐𝐬𝐢𝐧√𝟑𝐱) −𝟏

𝟐𝐱 +

𝟏

𝟐𝐱𝟐 + 𝐞−𝐱

W – 17

DEFINITION: WRONSKIAN:

Wronskian of the n function y1,y2, …,yn is defined and denoted by the determinant

W(y1, y2, … yn) = |

y1 y2y1′ y2

′ ⋯yny′n

⋮ ⋮ ⋱ ⋮

y1(𝑛−1)

y2(𝑛−1) ⋯ yn

(𝑛−1)

|

THEOREM:

Let y1, y2, … ynbe differentiable functions defined on some interval I. Then

(1) y1, y2, … yn are linearly independent on I if and only if W(y1, y2, … yn) ≠ 0 at least

one value of x ∈ I.

(2) y1, y2, … yn are linearly dependent on I then W(y1, y2, … yn) = 0 for all x ∈ I.

METHOD – 4: EXAMPLE ON WRONSKIAN

H 1 Find wronskian of x, log x , x(log x)2 ; x > 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐋𝐢𝐧𝐞𝐚𝐫 𝐈𝐧𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭

C 2 Find the wronskian ex, e−x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐋𝐢𝐧𝐞𝐚𝐫 𝐈𝐧𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭

METHOD OF VARIATION OF PARAMETERS:

The process of replacing the parameters of an analytic expression by functions is called

variation of parameters.

Consider, y′′ + p(x) y′ + q(x) y = R(x). Where p, q and R(x) are the functions of x.

The general solution of SECOND order differential equation by the method of variation of

parameters is



y(x) = yc + yp

Where, yc = c1y1 + c2y2

yp(x) = −y1∫y2R(x)

Wdx +y2∫y1

R(x)

Wdx

Note that, y1and y2 are the solutions of y" + py′ + qy = 0, W = |y1 y2y′1 y′2

| ≠ 0.

Consider, y′′′ + p(x) y′′ + q(x) y′ + s(x) y = R(x). Where p, q, s and R(x) are the functions

of x.

The general solution of THIRD order differential equation by the method of variation of

parameters is

y(x) = yc + yp

Where, yc = c1y1 + c2y2 + 𝑐3𝑦3 and yp(x) = A(x)y1 + B(x)y2 + C(x)y3

Note that,

A(x) = ∫(y2y3′ − y3y2

′ ) R(x)

W dx

B(x) = ∫(y3y1′ − y1y3

′ ) R(x)

W dx

C(x) = ∫(y1y2′ − y2y1

′ ) R(x)

W dx

Also, W = |

y1 y2 y3y1′ y2

′ y3′

y1′′ y2

′′ y3′′| ≠ 0 .

METHOD – 5: EXAMPLE ON VARIATION OF PARAMETERS

C 1 Solve: (D2 − 4D + 4)y =

e2x

x5

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟐𝐱 +

𝟏

𝟏𝟐

𝐞𝟐𝐱

𝐱𝟑

H 2 Solve: y′′ − 3y′ + 2y = ex

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏𝐞𝟐𝐱 + 𝐜𝟐𝐞

𝐱) − 𝐞𝐱 − 𝐱𝐞𝐱 W – 16



H 3 Solve: (D2 − 1)y = x ex

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞

−𝐱 ) +𝐞𝐱

𝟒𝐱𝟐 −

𝐞𝐱

𝟖(𝟏 − 𝟐𝐱)

S – 17

T 4 Use variation of parameter to find general soln of y′′ − 4y′ + 4y =

e2x

x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐞𝟐𝐱(𝐜𝟏 + 𝐜𝟐𝐱 − 𝐱 + 𝐱 𝐥𝐨𝐠 𝐱) S – 17

C 5 Solve: y′′ + 2y′ + y = e−x cos x

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱 − 𝐜𝐨𝐬 𝐱)𝐞−𝐱

T 6 Solve: (D2 − 4D + 4)y =

e2x

1 + x2

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟐𝐱 − 𝐞𝟐𝐱

𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝐱𝟐) + 𝐱𝐞𝟐𝐱(𝐭𝐚𝐧−𝟏 𝐱)

C 7 Find the solution of y′′ + a2y = tan ax by variation of parameter.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬𝐚𝐱 + 𝐜𝟐𝐬𝐢𝐧𝐚𝐱 + (− 𝐜𝐨𝐬 𝐚𝐱 𝐥𝐨𝐠(𝐬𝐞𝐜𝐚𝐱 + 𝐭𝐚𝐧𝐚𝐱)

𝐚𝟐)

S – 16

H 8 Find solution of

d2y

dx2+ 9y = tan 3x using variation of parameter .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧𝟑𝐱 −𝐜𝐨𝐬𝟑𝐱

𝟗𝐥𝐨𝐠(𝐬𝐞𝐜 𝟑𝐱 + 𝐭𝐚𝐧𝟑𝐱)

W – 15

T 9 Find the solution of y′′ + 4y = 4 tan 2x by variation of parameter.


𝐲 = 𝐜𝟏𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟐𝐱 + −𝐜𝐨𝐬 𝟐𝐱 𝐥𝐨𝐠(𝐬𝐞𝐜 𝟐𝐱 + 𝐭𝐚𝐧 𝟐𝐱)

S – 18

H 10 Solve: y′′ + y = sec x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱 + 𝐱 𝐬𝐢𝐧 𝐱 + 𝐜𝐨𝐬 𝐱 𝐥𝐨𝐠(𝐜𝐨𝐬 𝐱)

C 11 Solve differential equation using variation of parameter y′′ + 9y = sec 3x.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 +𝟏

𝟑𝐱 𝐬𝐢𝐧 𝟑𝐱 +

𝟏

𝟗𝐜𝐨𝐬 𝟑𝐱 𝐥𝐨𝐠 𝐜𝐨𝐬 𝟑𝐱

S – 15

H 12 Solve: (D2 + a2)y = cosec ax

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬𝐚𝐱 + 𝐜𝟐𝐬𝐢𝐧𝐚𝐱 + (−𝐱 𝐜𝐨𝐬 𝐚𝐱

𝐚+

𝟏

𝐚𝟐𝐬𝐢𝐧𝐚𝐱 𝐥𝐨𝐠|𝐬𝐢𝐧 𝐱|)

H 13 Solve the following differential equation d2y

dx2+ y = sinx using the method

of variation of parameters.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱 −𝐱 𝐜𝐨𝐬 𝐱

𝟐+𝐜𝐨𝐬 𝐱 𝐬𝐢𝐧 𝟐𝐱

𝟒−𝐜𝐨𝐬 𝐱 𝐜𝐨𝐬 𝟐𝐱

𝟒

W – 17



C 14 Solve:

d3y

dx3+dy

dx= cosecx

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 + 𝐜𝟐𝐜𝐨𝐬𝐚𝐱 + 𝐜𝟑𝐬𝐢𝐧𝐚𝐱 + 𝐥𝐨𝐠(𝐜𝐨𝐬𝐞𝐜 𝐱 − 𝐜𝐨𝐭 𝐱)

−𝐜𝐨𝐬 𝐱 (𝐥𝐨𝐠 𝐬𝐢𝐧𝐱) − 𝐱 𝐬𝐢𝐧𝐱

CAUCHY – EULER EQUATION

An equation of the form

xndny

dxn+ a1x

n−1dn−1y

dxn−2+ a2x

n−2dn−2y

dxn−2+⋯+ an−1x

dy

dx+ any = R(x)

Where a1, a2 , … , an are constants and R(x) is a function of x, is called Cauchy’s homogeneous

linear equation.

STEPS TO CONVERT CAUCHY-EULER EQ. TO LINEAR DIFFERENTIAL EQ.:

To reduce the above Cauchy – Euler Equation into a linear equation with constant

coefficients, we use the transformation x = ez so that z = logx.

Now, z = log x ⟹dz

dx=

1

x

Now,dy

dx=

dy

dz

dz

dx

⟹dy

dx=1

x

dy

dz

⟹ xdy

dx=dy

dz= D y,where D =

d

dz

Similarly, x2d2y

dx2= D(D − 1)y & x3

d3y

dx3= D(D − 1)(D − 2)y

Using this transformation, the given equation reduces to

[D(D − 1)(D − 2) … (D − n + 1) + a1D(D − 1) …(D − n + 2) +⋯+ an−1D + an]y

= f(ez)

This is a linear equation with constant coefficients, which can be solved by the methods

discussed earlier.



METHOD – 6: EXAMPLE ON CAUCHY EULER EQUATION

C 1 Solve: (x2D2 − 3xD + 4)y = 0; y(1) = 0, y′(1) = 3

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝟑 𝐱𝟐 𝐥𝐨𝐠 𝐱

H 2 Solve: x2y′′ − 2.5xy′ − 2y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐱𝟒 + 𝐜𝟐

𝟏

√𝐱

T 3 Solve: x2y′′ − 4xy′ + 6y = 21x−4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐱𝟐 + 𝐜𝟐𝐱

𝟑 +𝟏

𝟐𝐱−𝟒

H 4 Solve: (x2D2 − 3xD + 4)y = x2; y(1) = 1, y′(1) = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝟏 − 𝟐 𝐥𝐨𝐠 𝐱)𝐱𝟐 +𝟏

𝟐𝐱𝟐(𝐥𝐨𝐠 𝐱)𝟐

C 5 Solve: x3y′′′ + 2x2y′′ + 2y = 10 (x +

1

x)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏 𝐱−𝟏 + 𝐱{𝐜𝟐 𝐜𝐨𝐬(𝐥𝐨𝐠 𝐱) + 𝐜𝟑 𝐬𝐢𝐧(𝐥𝐨𝐠 𝐱)} + 𝟓𝐱 + 𝟐𝐱−𝟏 𝐥𝐨𝐠 𝐱

H 6 Solve: (x2D2 − 3xD + 3)y = 3 ln x − 4

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐱 + 𝐜𝟐𝐱𝟑 + 𝐥𝐧 𝐱

C 7 Solve: x2D2y − xDy + y = sin(logx)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏 + 𝐜𝟐𝐥𝐨𝐠𝐱) 𝐞𝐥𝐨𝐠𝐱 +

𝟏

𝟐𝐜𝐨𝐬(𝐥𝐨𝐠𝐱)

S – 17

T 8 Solve the following Cauchy-Euler equation

x2d2y

dx2+ x

dy

dx+ y = logx ∙ sin(logx)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝐨𝐬(𝐥𝐨𝐠𝐱) + 𝐜𝟐𝐬𝐢𝐧(𝐥𝐨𝐠𝐱) −(𝐥𝐨𝐠𝐱)𝟐

𝟒𝐜𝐨𝐬(𝐥𝐨𝐠𝐱)+

𝐥𝐨𝐠𝒙

𝟒𝐬𝐢𝐧𝐥𝐨𝐠 𝒙

W – 17

H 9 Solve: x2

d2y

dx2+ 4x

dy

dx+ 2y = x2sin (log x)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = (𝐜𝟏𝐞−𝟐𝐥𝐨𝐠𝐱 + 𝐜𝟐𝐞

−𝐥𝐨𝐠𝐱) −𝐱𝟐

𝟏𝟕𝟎(𝟕𝐜𝐨𝐬(𝐥𝐨𝐠𝐱) − 𝟏𝟏𝐬𝐢𝐧(𝐥𝐨𝐠𝐱))

W – 16

T 10 Solve completely the differential equation x2

d2y

dx2− 6x

dy

dx+ 6y = x−3logx.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐱𝟔 + 𝐜𝟐𝐱 +

𝟏

𝟑𝟔𝐱𝟑(𝐥𝐨𝐠 𝐱 +

𝟏𝟑

𝟑𝟔 )

W – 15

C 11 Solve: x2

d2y

dx2− x

dy

dx+ 2y = x log x

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐱[𝐜𝟏 𝐜𝐨𝐬(𝐥𝐨𝐠𝐱) + 𝐜𝟐 𝐬𝐢𝐧(𝐥𝐨𝐠 𝐱)] + 𝐱 𝐥𝐨𝐠 𝐱



HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION:

Reduction of order method for Linear second order O.D.E.

Step-1 Convert given D.E. into d2y

dx2+ P(x)

dy

dx + Q(x)y = 0 and find P(x) & Q(x).

Step-2 Find U.

U =1

y12 e−∫P dx

Step-3 Find V.

V = ∫U dx

Step-4 Second solution y2 = V ∙ y1

Finally, General solution is y = c1 y1 + c2 y2 .

METHOD – 7: EXAMPLE ON FINDING SECOND SOLUTION

C 1 Find second solution of x2𝑦′′ − x y′ + y = 0, y1 = x.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 = 𝐱 𝐥𝐨𝐠 𝐱

H 2 Find second solution of x2y′′ − 4 x y′ + 6 y = 0 , y1 = x2 ; x > 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 = 𝐱𝟑

T 3 Find second solution of x y′′ + 2 y′ + x y = 0, y1 =

sin x

x.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝟐 = −𝐜𝐨𝐬 𝐱

𝐱

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆


U N I T - 4 » S e r i e s S o l u t i o n o f D i f f e r e n t i a l E q u a t i o n [ 6 2 ]

UNIT-4 » SERIES SOLUTION OF DIFFERENTIAL EQUATION

INTRODUCTION:

If homogeneous linear differential equation has constant coefficients, it can be solved by

algebraic methods, and its solutions are elementary functions known from

calculus (ex, cos x , etc … ), as we know from unit-3. However, if such an equation has

variable coefficients it must usually be solved by other methods (for example Euler-Cauchy

equation).

There are some linear differential equations which do not come in this category. In such

cases we have to find a convergent power series arranged according to powers of the

independent variable, which will approximately express the value of the dependent

variables.

We devote an entire Unit-4 to two standard methods of solution and their applications:

Power Series Method and Frobenius Method (an extension of Power Series method).

Before actually proceeding to solve linear ordinary differential equations with polynomial

coefficient, we will look at some of the basic concepts which require for their study.

DEFINITION: POWER SERIES:

An infinite series of the form

∑ak(x − x0)k

∞

k=0

= a0 + a1(x − x0) + a2(x − x0)2 + ⋯

is called a power series in (x − x0).

DEFINITION: ANALYTIC FUNCTION:

A function is said to be analytic at a point x0 if it can be expressed in a power series near x0.

DEFINITION: ORDINARY AND SINGULAR POINT:

Let P0(x)d2y

dx2+ P1(x)

dy

dx+ P2(x)y = 0 be the given differential equation with variable co-

efficient.

Dividing by P0(x),

d2y

dx2+P1(x)

P0(x)

dy

dx+P2(x)

P0(x)y = 0 .



Let, P(x) =P1(x)

P0(x) & Q(x) =

P2(x)

P0(x)

∴d2y

dx2+ P(x)

dy

dx+ Q(x)y = 0 .

DEFINITION: ORDINARY POINT AND SINGULAR POINT:

A point x0 is called an ordinary point of the differential equation if the functions P(x) and

Q(x) both are analytic at x0.

If at least one of the functions P(x) or Q(x) is not analytic at x0 then x0 is called a singular

point.

DEFINITION: REGULAR SINGULAR POINT AND IRREGULAR SINGULAR POINT:

A singular point x0 is called regular singular point if both (x − x0)P(x) and (x − x0)2Q(x) are

analytic at x0 otherwise it is called an irregular singular point.

METHOD – 1: EXAMPLE ON SINGULARITY OF DIFFERENTIAL EQUATION

C 1 Find singularity of y′′ + (x2 + 1)y′ + (x3 + 2x2 + 3x)y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐍𝐨 𝐬𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐨𝐢𝐧𝐭𝐬

H 2 Find singularity of y′′ + exy′ + sin(x2)y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐍𝐨 𝐬𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐩𝐨𝐢𝐧𝐭𝐬

C 3 Find singularity of x3y′′ + 5xy′ + 3y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟎 𝐢𝐬 𝐚𝐧 𝐈𝐫𝐫𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭.

H 4 Find singularity of (1 − x2)y′′ − 2xy′ + n(n + 1)y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝟏 & − 𝟏 𝐚𝐫𝐞 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭. S – 17

H 5 Find singularity of x3(x − 1)y′′ + 3(x − 1)y′ + 7xy = 0


𝐱 = 𝟏 𝐢𝐬 𝐚 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭 & 𝐱 = 𝟎 𝐢𝐬 𝐚𝐧 𝐈𝐫𝐫𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭.

S – 17

T 6 Find singularity of (x2 + 1)y′′ + xy′ − xy = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱 = 𝐢 , −𝐢 𝐚𝐫𝐞 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭.

C 7 Find singularity of 2x(x − 2)2y′′ + 3xy′ + (x − 2)y = 0.


𝐱 = 𝟎 𝐢𝐬 𝐚 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭 & 𝐱 = 𝟐 𝐢𝐬 𝐚𝐧 𝐈𝐫𝐫𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭.

W – 15



H 8 Find singularity of x(x + 1)2y′′ + (2x − 1)y′ + x2y = 0.


𝐱 = 𝟎 𝐢𝐬 𝐚 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭 & 𝐱 = −𝟏 𝐢𝐬 𝐚𝐧 𝐈𝐫𝐫𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭

T 9 x = 0 is a regular singular point of 2x2y" + 3xy′ + (x2 − 4)y = 0 say true

or false.

Answer: True

S – 16

POWER SERIES SOLUTION AT AN ORDINARY POINT:

A power series solution of a differential equation P0(x)d2y

dx2+ P1(x)

dy

dx+ P2(x)y = 0 at an

ordinary point x0 can be obtained using the following steps.

Step-1: Assume that

y = ∑ ak(x − x0)k

∞

k=0= a0 + a1x + a2x

2 + a3x3 + a4x

4 + a5x5 + ⋯

is the solution of the given differential equation.

Differentiating with respect to 𝑥 we get,

⟹ 𝑦′ =𝑑𝑦

𝑑𝑥= a1 + 2a2x + 3a3x

2 + 4a4x3 + 5a5x

4 +⋯

⟹ 𝑦′′ =𝑑2𝑦

𝑑𝑥2= 2a2 + 6a3x + 12a4x

2 + 20a5x3 +⋯

Step-2: Substitute the expressions of y,dy

dx, and

d2y

dx2 in the given differential equation.

Step-3: Equate to zero the co-efficient of various powers of x and find a2, a3, a4… etc. in terms

of a0 and a1.

Step-4: Substitute the expressions of a2, a3, a4, … in

y = a0 + a1x + a2x2 + a3x

3 + a4x4 + a5x

5 +⋯ which will be the required solution.

METHOD – 2: EXAMPLE ON POWER SERIES METHOD

H 1 y′ + 2xy = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 − 𝐚𝟎 𝐱𝟐 +

𝟏

𝟐𝐚𝟎𝐱

𝟒 −𝟏

𝟔𝐚𝟎 𝐱

𝟔 +⋯



H 2 y′′ + y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 −𝟏

𝟐 𝐚𝟎 𝐱

𝟐 −𝟏

𝟔 𝐚𝟏 𝐱

𝟑 +𝟏

𝟐𝟒 𝐚𝟎 𝐱

𝟒 +𝟏

𝟏𝟐𝟎 𝐚𝟏 𝐱

𝟓 + ⋯

C 3 y′′ + xy = 0 in powers of x.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 −𝟏

𝟔𝐚𝟎 𝐱

𝟑 −𝟏

𝟏𝟐𝐚𝟏 𝐱

𝟒 +𝟏

𝟏𝟖𝟎𝐚𝟎 𝐱

𝟔 + ⋯ W – 16 S – 16

T 4 y′′ + x2y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫

𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 −𝟏

𝟏𝟐𝐚𝟎 𝐱

𝟒 −𝟏

𝟐𝟎𝐚𝟏 𝐱

𝟓 +𝟏

𝟔𝟕𝟐𝐚𝟎 𝐱

𝟖 +𝟏

𝟏𝟒𝟒𝟎𝐚𝟏𝐱

𝟗 +⋯

H 5 y′′ = y′ .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 +𝟏

𝟐 𝐚𝟏 𝐱

𝟐 +𝟏

𝟔 𝐚𝟏 𝐱

𝟑 +𝟏

𝟐𝟒 𝐚𝟏 𝐱

𝟒 +𝟏

𝟏𝟐𝟎 𝐚𝟏 𝐱

𝟓 + ⋯

C 6 Find the power series solution about x = 0 of y” + xy′ + x2y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 {𝟏 −𝟏

𝟏𝟐𝐱𝟒 +

𝟏

𝟗𝟎𝐱𝟔 + ⋯} + 𝐚𝟏 {𝐱 −

𝟏

𝟔𝐱𝟑 −

𝟏

𝟒𝟎𝐱𝟓 +⋯}

H 7 y′′ − 2xy′ + 2py = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 − 𝐩 𝐚𝟎 𝐱𝟐 +

(𝟏 − 𝐩)

𝟑𝐚𝟏𝐱

𝟑 − 𝐩 (𝟐 − 𝐩)

𝟔𝐚𝟎 𝐱

𝟒

+ (𝟏 − 𝐩) (𝟑 − 𝐩)

𝟑𝟎𝐚𝟏𝐱

𝟓 +⋯

T 8 (1 − x2)y′′ − 2xy′ + 2y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 − 𝐚𝟎 𝐱𝟐 −

𝟏

𝟑𝐚𝟎 𝐱

𝟒 −𝟏

𝟓𝐚𝟎𝐱

𝟔 +⋯ W – 14

C 9 d2y

dx2(1 − x2) − x

dy

dx+ py = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏𝐱 −𝐩

𝟐 𝐚𝟎𝐱

𝟐 +(𝟏 − 𝐩)

𝟔𝐚𝟏 𝐱

𝟑 −𝐩(𝟒 − 𝐩)

𝟐𝟒𝐚𝟎 𝐱

𝟒

+(𝟗 − 𝐩)(𝟏 − 𝐩)

𝟏𝟐𝟎𝐚𝟏 𝐱

𝟓 + ⋯

C 10 (1 + x2)y′′ + xy′ − 9y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 +𝟗

𝟐𝐚𝟎 𝐱

𝟐 +𝟒

𝟑𝐚𝟏 𝐱

𝟑 +𝟏𝟓

𝟖𝐚𝟎 𝐱

𝟒 −𝟕

𝟏𝟔𝐚𝟎𝐱

𝟔 +⋯ S – 15

H 11 (x2 + 1)y′′ + xy′ − xy = 0 near x = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏 𝐱 + 𝐚𝟎 𝐱𝟑

𝟔− 𝐚𝟏

𝐱𝟑

𝟔+ (

𝐚𝟏

𝟏𝟐) 𝐱𝟒 − (

𝟑

𝟒𝟎) 𝐚𝟎 𝐱

𝟓+.. S – 17



H 12 (x − 2)y′′ − x2y′ + 9y = 0 .


𝐲 = 𝐚𝟎 (𝟏 +𝟗 𝐱𝟐

𝟒+𝟗 𝐱𝟑

𝟐𝟒+𝟗𝟎 𝐱𝟒

𝟒+⋯ ) + 𝐚𝟏 (𝐱 +

𝟏𝟖 𝐱𝟑

𝟐𝟒+𝟏𝟒 𝐱𝟒

𝟒+ ⋯)

W – 15

H 13 (1 − x2)y" − 2xy′ + 2y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐚𝟎 + 𝐚𝟏𝐱 − 𝐚𝟎𝐱𝟐 −

𝐚𝟎

𝟑𝐱𝟒 + ⋯

W – 17

FROBENIUS METHOD:

Frobenius Method is used to find a series solution of a differential equation near regular

singular point.

Step-1: If x0 is a regular singular point, we assume that the solution is

y = ∑ ak(x − x0)m+k

∞

k=0

Differentiating with respect to x ,we get

⟹dy

dx= ∑ (m + k)ak(x − x0)

m+k−1∞

k=0

⟹d2y

dx2= ∑ (m + k)(m + k − 1)ak(x − x0)

m+k−2∞

k=0

Step-2: Substitute the expressions of y ,dy

dx, and

d2y

dx2 in the given differential equation.

Step-3: Equate to zero the co-efficient of least power term in (x − x0) , which gives a

quadratic equation in m, called Indicial equation.

The format of the series solution depends on the type of roots of the indicial equation.

Here we have the following three cases:

Case-I Distinct roots not differing by an integer.(m1 ≠ m2 & m1 −m2 ∉ Z)

o When m1 −m2 ∉ Z, i.e. difference of m1and m2 is not a positive or negative

integer.In this case, the series solution is obtained corresponding to both values of

m. Let the solutions be y = y1 and y = y2, then the general solution is y = c1y1 +

c2y2.



Case-II Equal roots. (m1 = m2)

o In this case, we will have only one series solution. i.e. y = y1 = ∑ ak(x − x0)m+k∞

k=0

o In terms of a0and the variable m. The general solution is y = c1(y1)m + c2 (dy1

dm)m

Case-III Distinct roots differing by an integer. (m1 ≠ m2 & m1 − m2 ∈ Z)

o When m1 −m2 ∈ Z, i.e. difference of m1 & m2is a positive or negative integer.Let

the roots of the indicial equation be m1 & m2 with m1 < m2. In this case, the

solutions corresponding to the values m1& m2 may or may not be linearly

independent.Smaller root must be taken as m1 .

Here we have the following two possibilities.

o One of the co-efficient of the series becomes indeterminate for the smaller root m1

and hence the solution for m1 contains two arbitrary constants. In this case, we will

not find solution corresponding to m2 .

o Some of the co-efficient of the series becomes infinite for the smaller root m1 ,then

it is required to modify the series by replacing a0 by a0(m+ m1). The two linearly

independent solutions are obtained by substituting m = m1 in the modified form of

the series for y and in dy

dm obtained from this modified form.

METHOD – 3: EXAMPLE ON FROBENIUS METHOD

C 1 4x

d2y

dx2+ 2

dy

dx+ y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐚𝟎 (𝟏 −𝐱

𝟐+𝐱𝟐

𝟐𝟒− ⋯) + 𝐜𝟐𝐚𝟎√𝐱(𝟏 −

𝐱

𝟔+

𝐱𝟐

𝟏𝟐𝟎−⋯)

T 2 xy′′ + y′ − y = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝒚 = 𝐜𝟏𝐚𝟎 (𝟏 + 𝐱 +𝐱𝟐

𝟒+ ⋯) + 𝐜𝟐𝐚𝟎 𝐥𝐨𝐠 𝐱 (𝟏 + 𝐱 +

𝐱𝟐

𝟒+⋯ )

−𝟐𝐜𝟐𝐚𝟎 (𝐱 +𝟑

𝟖𝐱𝟐 +⋯ )

H 3 (x2 − x)y′′ − xy′ + y = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐜𝟎𝐱 + 𝐜𝟐𝐜𝟎(𝐱 𝐥𝐨𝐠 𝐱 + 𝟏 − 𝟑𝐱)



H 4 2x2y′′ + xy′ + (x2 − 1)y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐱 (𝟏 −𝐱𝟐

𝟏𝟒+

𝐱𝟒

𝟔𝟏𝟔−⋯ )+ 𝐜𝟐

𝟏

√𝐱(𝟏 +

𝐱𝟐

𝟐+

𝐱𝟒

𝟒𝟎+⋯)

W – 16

C 5 3x

d2y

dx2+ 2

dy

dx+ y = 0

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐚𝟎 (𝟏 + 𝐱 +𝟏

𝟖𝐱𝟐 + ⋯)+ 𝐜𝟐𝐚𝟎𝐱

𝟏

𝟑 (𝟏 −𝟑

𝟖𝐱 +

𝟑

𝟏𝟏𝟐𝐱𝟐 −⋯ )

W – 17

T 6 8x2y′′ + 10xy′ − (1 + x)y = 0


𝐲 = 𝐜𝟏𝐚𝟎𝐱𝟏𝟒 (𝟏 +

𝟏

𝟏𝟒𝐱 +

𝟏

𝟔𝟏𝟔𝐱𝟐 +⋯ )+

𝐜𝟐𝐚𝟎

𝐱𝟐(𝟏 +

𝟏

𝟐𝐱 +

𝟏

𝟒𝟎𝐱𝟐 + ⋯)

S – 18

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆


[ 6 9 ]


U N I T - 5 » L a p l a c e T r a n s f o r m a n d I t ’ s A p p l i c a t i o n [ 7 0 ]

UNIT-5 » LAPLACE TRANSFORM AND IT’S APPLICATION

INTRODUCTION:

Pierre Simon Marquis De Laplace (1749-1827) was French mathematician.

The word Transform itself indicates about the conversion of one form to another form. The

Laplace transforms is a versatile tool for solving differential equations as it transforms

differential equations to algebraic equations and algebraic equations are comparatively

easier to solve.

By using Laplace transform we can find particular solution of a differential equation without

determining the general solution. Laplace useful to finding solution of the problems related

with mechanical or electrical driving force which has discontinuities, impulsive or

complicated periodic functions. Also we can find solutions of a system of ODE, PDE and

integral equations.

In this unit, we will first develop the concept of Laplace transform through definition,

properties and theorems. Then we will study about the solution of some differential

equation(s).

DEFINITION: LAPLACE TRANSFORM

Let f(t) be a given function defined for all t ≥ 0, then the Laplace transform of f(t) is denoted

by ℒ{f(t)} or f(̅s) or F(S), and is defined as

𝓛{𝐟(𝐭)} = ∫ 𝐞−𝐬𝐭𝐟(𝐭)𝐝𝐭∞

𝟎

, provided the integral exist.

PROPERTIES OF LAPLACE TRANSFORMS:

Linearity property of Laplace transform: 𝓛{𝛂 ∙ 𝐟(𝐭) + 𝛃 ∙ 𝐠(𝐭)} = 𝛂 ∙ 𝓛{𝐟(𝐭)} + 𝛃 ∙ 𝓛{𝐠(𝐭)}

Proof: By definition, ℒ{f(t)} = ∫ e−st∞

0f(t) dt

ℒ{α ∙ f(t) + β ∙ g(t)} = ∫ e−st∞

0

[α ∙ f(t) + β ∙ g(t)] dt

= ∫ e−st∞

0

∙ α ∙ f(t) dt + ∫ e−st∞

0

∙ β ∙ g(t) dt

= α ∙ ℒ{f(t)} + β ∙ ℒ{g(t)}



Change of Scale of Laplace transform: If 𝓛{𝐟(𝐭)} = 𝐅(𝐬), then 𝓛{𝐟(𝐛𝐭)} =𝟏

𝐛𝐅 (

𝐬

𝐛).

Proof: By definition, ℒ{f(bt)} = ∫ e−st∞

0f(bt) dt

Take bt = u then dt =du

b then by above, we have

ℒ{f(bt)} = ∫ e−sub

∞

0

f(u)du

b

=1

b∫ e

−(sb)u

∞

0

f(u) du

=1

bF (

s

b)

Thus, ℒ{f(bt)} =1

bF (

s

b) .

Example: 𝐈𝐟 𝓛{𝐟(𝐭)} =𝐬

𝐬𝟐+𝐤𝟐, 𝐟𝐢𝐧𝐝 𝓛{𝐟(𝟐𝐭)}.

Solution: ℒ{f(2t)} =1

2

(s

2)

(s

2)2+k2

=s

s2+4k2

LAPLACE TRANSFORM OF SOME STANDARD FUNCTION:

(1). 𝓛{𝐭𝐧} =𝟏

𝐬𝐧+𝟏1n ; 𝐧 > −𝟏 OR 𝓛{𝐭𝐧} =

𝐧 !

𝐬𝐧+𝟏 ; 𝐧 𝐢𝐬 𝐢𝐧𝐭𝐞𝐠𝐞𝐫

Proof: By definition, L{f(t)} = ∫ e−st∞

0f(t) dt

ℒ{tn} = ∫ e−st∞

0

tn dt

Let, st = x ⟹ sdt = dx & hence, When t ⟶ 0 ⇒ x → 0 and t ⟶ ∞ ⇒ x → ∞

⟹ ℒ{tn} = ∫ e−x∞

0

xn

sndx

s

=1

sn+1∫ e−x

∞

0

xn dx

⟹ ℒ{tn} =1

sn+11n ( ∵ By definition of Gamma function n = ∫ e−x

∞

0xn−1 dx )

If 𝐧 is a positive integer, then 𝐧! = 1n



⟹ ℒ{tn} =n!

sn+1

(2). 𝓛{𝟏} =𝟏

𝐬.


0f(t) dt

ℒ{1} = ∫ e−st∞

0

dt = [e−st

−s]0

∞

=0 − 1

−s=1

s

(3). 𝓛{𝐞𝐚𝐭} =𝟏

𝐬−𝐚 , 𝐬 > 𝐚 (S – 15)


0f(t) dt

ℒ{eat} = ∫ e−st∞

0

eat dt

= ∫ e−(s−a)t∞

0

dt = [e−(s−a)t

−(s − a)]0

∞

= [0 − 1

−(s − a)] [When t ⟶ ∞⟹ e−(s−a)t ⟶ 0 (∵ s > a ⇒ s − a > 0)]

⟹ ℒ{eat} =1

s − a

(4). 𝓛{𝐞−𝐚𝐭} =𝟏

𝐬+𝐚 , 𝐬 > −𝐚 (W – 14)


0f(t) dt

ℒ{e−at} = ∫ e−st∞

0

e−at dt

= ∫ e−(s+a)t∞

0

dt

= [e−(s+a)t

−(s + a)]0

∞

= [0 − 1

−(s + a)] [When t ⟶ ∞⟹ e−(s+a)t ⟶ 0(∵ s > −a ⟹ s+ a > 0)]

=1

s + a



⟹ ℒ{e−at} =1

s + a

(5). 𝓛{𝐬𝐢𝐧 𝐚𝐭} =𝐚

𝐬𝟐+𝐚𝟐 , 𝐬 > 𝟎 and 𝐚 is a constant.


0f(t) dt

ℒ{sin at} = ∫ e−st∞

0

sin at dt

= [e−st

s2 + a2(−s sin at − a cos at)]

0

∞

= 0 −1

s2 + a2(−a) [When t ⟶ ∞⟹ e−st ⟶ 0(∵ s > 0)]

⟹ ℒ{sin at} =a

s2 + a2

(6). 𝓛{𝐜𝐨𝐬 𝐚𝐭} =𝐬

𝐬𝟐+𝐚𝟐 , 𝐬 > 𝟎 and 𝐚 is a constant.


0f(t) dt

ℒ{cos at} = ∫ e−st∞

0

cos at dt

= [e−st

s2 + a2(−s cos at + a sin at)]

0

∞

= 0 −1

s2 + a2(−s) [When t ⟶ ∞ ⟹ e−st ⟶ 0 (∵ s > 0)]

⟹ ℒ{cos at} =s

s2 + a2

(7). 𝓛{𝐬𝐢𝐧𝐡 𝐚𝐭} =𝐚

𝐬𝟐−𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐𝐨𝐫 (𝐬 > |𝐚|) (S – 15)

Proof: ℒ{sinh at} = ℒ {eat−e−at

2}

=1

2[ℒ{eat} − ℒ{e−at}]

=1

2[

1

s − a−

1

s + a] =

1

2[s + a − s + a

s2 − a2]

⟹ ℒ{sinh at} =a

s2 − a2



(8). 𝓛{𝐜𝐨𝐬𝐡 𝐚𝐭} =𝐬

𝐬𝟐−𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐 𝐨𝐫 (𝐬 > |𝐚|)

Proof: L{cosh at} = ℒ {eat+e−at

2}

=1

2[ℒ{eat} + ℒ{e−at}]

=1

2[

1

s − a+

1

s + a] =

1

2[s + a + s − a

s2 − a2]

⟹ ℒ{cosh at} =s

s2 − a2

METHOD – 1: EXAMPLE ON DEFINITION OF LAPLACE TRANSFORM

H 1 Find the Laplace transform of f(t) = {

0 ; 0 ≤ t < 3

4 ; t ≥ 3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝐞−𝟑𝐬

𝐬

C 2 Given that f(t) = {

t + 1 ; 0 ≤ t ≤ 2

3 ; t ≥ 2. Find ℒ{f(t)}.

𝐀𝐧𝐬𝐰𝐞𝐫: −𝐞−𝟐𝐬

𝐬𝟐+𝟏

𝐬+

𝟏

𝐬𝟐

H 3 Find the Laplace transformation of f(x) = {

et ; 0 < t < 1

0 ; t > 1 .

𝐀𝐧𝐬𝐰𝐞𝐫:𝐞𝟏−𝐬

𝟏 − 𝐬−

𝟏

𝟏 − 𝐬

C 4 Find the Laplace transform of f(t) = {

0 ; 0 < t < π

sin t ; t > π .

𝐀𝐧𝐬𝐰𝐞𝐫: −𝐞−𝛑𝐬

𝐬𝟐 + 𝟏

S – 15

H 5 Find the Laplace transform of f(t) = {

cost 0 < t < π

sint t > π .


𝐬𝟐 + 𝟏(𝐞−𝛑𝐬 (𝐬 − 𝟏) + 𝐬)



SOME IMPORTANT FORMULAE:

2 sinA cosB = sin(A + B) + sin(A − B) sin(α + β) = sin α cosβ + cosα sin β

2 cosA sinB = sin(A + B) − sin(A − B) sin(α − β) = sin α cosβ − cosα sin β

2 cosA cosB = cos(A + B) + cos(A − B) cos(α + β) = cosα cosβ − sin α sin β

2 sinA sinB = cos(A − B) − cos(A + B) cos(α − β) = cosα cosβ + sin α sin β

cos3A =cos3A + 3 cosA

4 cos2A =

1 + cos2A

2

sin3A =3 sin A − sin 3A

4 sin2 A =

1 − cos 2A

2

cosh at =eat + e−at

2 cos at =

eiat + e−iat

2

sinh at =eat − e−at

2 sin at =

eiat − e−iat

2i

METHOD – 2: EXAMPLE ON LAPLACE TRANSFORM OF SIMPLE FUNCTIONS

C 1 Find the Laplace transform of t3 + e−3t + t1

2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑!

𝐬𝟒+

𝟏

𝐬 + 𝟑+

√𝛑

𝟐𝐬𝟑𝟐

H 2 Find the Laplace transform of sint

2+ 2t + t

4

3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐

(𝟒𝐬𝟐 + 𝟏)+

𝟏

𝐬 − 𝐥𝐨𝐠 𝟐+𝟒 3/1

𝟗𝐬𝟕𝟑

H 3 Find the Laplace transform of t5 + e−100t + cos 5t.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓!

𝐬𝟔+

𝟏

𝐬 + 𝟏𝟎𝟎+

𝐬

𝐬𝟐 + 𝟐𝟓

C 4 Find the Laplace transform of 100t + 2t10 + sin 10t.


𝐬 − 𝐥𝐨𝐠𝐞𝟏𝟎𝟎+𝟐 ∙ 𝟏𝟎 !

𝐬𝟏𝟏+

𝟏𝟎

𝐬𝟐 + 𝟏𝟎𝟎

C 5 Find the Laplace transformation of f(t) = cosh2 3t .


𝟐𝐬+

𝐬

𝟐(𝐬𝟐 − 𝟑𝟔)



H 6 Find ℒ[(2t − 1)2].

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟖

𝐬𝟑−

𝟒

𝐬𝟐+𝟏

𝐬

C 7 Find the Laplace transformation (i) sin(ωt + α) (ii)cos (ωt + b)


(𝐢) 𝐜𝐨𝐬𝛂 𝛚

𝐬𝟐 + 𝛚𝟐+ 𝐬𝐢𝐧𝛂

𝐬

𝐬𝟐 +𝛚𝟐

(𝐢𝐢)𝐜𝐨𝐬𝐛 𝐬

𝐬𝟐 +𝛚𝟐− 𝐬𝐢𝐧𝐛

𝛚

𝐬𝟐 + 𝛚𝟐

C 8 Find ℒ{sin 2t cos 2t}.


𝐬𝟐 + 𝟏𝟔

H 9 Find ℒ{sin 2t sin 3t}.


𝟐[

𝐬

𝐬𝟐 + 𝟏−

𝐬

𝐬𝟐 + 𝟐𝟓]

C 10 Find Laplace transform of cos2(at)

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝟐 + 𝟐𝐚𝟐

𝐬(𝐬𝟐 + 𝟒𝐚𝟐)

C 11 Find the Laplace transform of sin23t.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟖

𝐬(𝐬𝟐 + 𝟑𝟔)

W – 14

H 12 Find the Laplace transform of cos22t.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝟐 + 𝟖

𝐬(𝐬𝟐 + 𝟏𝟔)

H 13 Find ℒ{cos2 t}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝟐 + 𝟐

𝐬(𝐬𝟐 + 𝟒)

S – 18

H 14 Find Laplace transform of sin3(at).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔𝐚𝟑

(𝐬𝟐 + 𝐚𝟐)(𝐬𝟐 + 𝟗𝐚𝟐)

C 15 Find the Laplace transform of (i)sin32t (ii) cos32t.

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐢)𝟒𝟖

(𝐬𝟐 + 𝟒)(𝐬𝟐 + 𝟑𝟔) (𝐢𝐢)

𝐬𝟑 + 𝟐𝟖𝐬

(𝐬𝟐 + 𝟒)(𝐬𝟐 + 𝟑𝟔)

H 16 Compute ℒ{cos t cos2t cos 3t}.


𝟒𝐬+

𝐬

𝟒(𝐬𝟐 + 𝟑𝟔)+

𝐬

𝟒(𝐬𝟐 + 𝟏𝟔)+

𝐬

𝟒(𝐬𝟐 + 𝟒)



THEOREM. FIRST SHIFTING THEOREM:

Statement: If 𝐋{𝐟(𝐭)} = 𝐅(𝐬), then show that 𝓛{𝐞𝐚𝐭𝐟(𝐭)} = 𝐅(𝐬 − 𝐚).


0f(t) dt

Now, ℒ{eatf(t)} = ∫ e−steat∞

0f(t) dt

= ∫ e−(s−a)t∞

0

f(t) dt

Since, s and a are constants. s − a is also a constant.

Thus, ℒ{eatf(t)} = F(s − a).

Corollary: If 𝓛{𝐟(𝐭)} = 𝐅(𝐬), then show that 𝐋{𝐞−𝐚𝐭𝐟(𝐭)} = 𝐅(𝐬 + 𝐚).

METHOD – 3: EXAMPLE ON FIRST SHIFTING THEOREM

T 1 Find ℒ(e−3tf(t)), if ℒ(f(t)) =s

(s − 3)2 .

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐬 + 𝟑)

𝐬𝟐

H 2 Find ℒ(e−3tt3 2⁄ ) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑√𝛑

𝟒(𝐬 + 𝟑)𝟓𝟐

C 3 By using first shifting theorem, obtain the value of ℒ{(t + 1)2et}.


(𝐬 − 𝟏)𝟑+

𝟐

(𝐬 − 𝟏)𝟐+

𝟏

𝐬 − 𝟏

C 4 Find ℒ(e2t sin 3𝑡).

𝐀𝐧𝐬𝐰𝐞𝐫:𝟑

(𝒔 − 𝟐)𝟐 + 𝟗

S – 18

H 5 Obtain Laplace transform of e2tsin2t


𝟐[

𝟏

𝐬 − 𝟐−

𝐬 − 𝟐

𝐬𝟐 − 𝟒𝐬 + 𝟖]

S – 17

C 6 Find Laplace transform of e−2t(sin4t + t2).

𝐀𝐧𝐬𝐰𝐞𝐫: (𝟒

𝐬𝟐 + 𝟒𝐬 + 𝟐𝟎+

𝟐

(𝐬 + 𝟐)𝟑)

W – 14



H 7 Find Laplace transform of e−3t(2 cos5t − 3 sin 5t).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐬 − 𝟗

(𝐬 + 𝟑)𝟐 + 𝟐𝟓

T 8 Find Laplace transform of e4t(sin 2t cos t).


𝟐 [

𝟑

(𝐬𝟐 − 𝟖𝒔 + 𝟐𝟓)+

𝟏

(𝒔𝟐 − 𝟖𝒔 + 𝟏𝟕)]

T 9 Find the Laplace transformation of f(t) =

cos2t sint

e2t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑

𝟐

𝟏

(𝐬𝟐 + 𝟒𝐬 + 𝟏𝟑)−𝟏

𝟐

𝟏

(𝐬𝟐 + 𝟒𝐬 + 𝟓)

H 10 Find ℒ{cosh 2t cos2t}.


𝟐 [

𝐬 − 𝟐

(𝐬 − 𝟐)𝟐 + 𝟒+

𝐬 + 𝟐

(𝐬 + 𝟐)𝟐 + 𝟒]

T 11 Find the Laplace transformation of f(t) = e−3tcosh4t sin3t


𝟐{

𝟏

(𝐬𝟐 + 𝟏𝟒𝐬 + 𝟓𝟖)+

𝟏

(𝐬𝟐 − 𝟐𝐬 + 𝟏𝟎)}

C 12 Find the Laplace transform of f(t) = t2cos h πt.


(𝐬 − 𝛑)𝟑+

𝟏

(𝐬 + 𝛑)𝟑

W – 14

H 13 Find the Laplace transform of t3 cosh 2t.


(𝐬 − 𝟐)𝟒+

𝟑

(𝐬 + 𝟐)𝟒

THEOREM. DIFFERENTIATION OF LAPLACE TRANSFORM:

Statement: Ifℒ{f(t)} = F(s), then show that ℒ{tnf(t)} = (−1)ndn

dsn F(s) ; n = 1,2,3, …

Proof: By definition, F(s) = ∫ e−st∞

0f(t) dt

⟹ dn

dsn F(s) =

dn

dsn ∫ e−st

∞

0

f(t) dt

= ∫ [ ∂n

∂sn e−st]

∞

0

f(t) dt

= ∫ (−1)(t) ∂n−1

∂sn−1 e−stf(t) dt

∞

0



= ∫ (−1)2(t)2 ∂n−2

∂sn−2 e−stf(t) dt

∞

0

Continuing in this way, we have

= (−1)n∫ e−st(t)nf(t) dt∞

0

= (−1)nℒ{tnf(t)}

Thus, ℒ{tnf(t)} = (−1)n dn

dsn F(s), n = 1,2,3, …

METHOD – 4: EXAMPLE ON DIFFERENTIATION OF LAPLACE TRANSFORM

C 1 Find the value of ℒ{t sin at}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐚𝐬

(𝐬𝟐 + 𝐚𝟐)𝟐

S – 16

H 2 Find the value of ℒ{t sin 2t}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝐬

(𝐬𝟐 + 𝟒)𝟐

S – 15

H 3 Find the value of ℒ{t cosh t}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 + 𝐬𝟐

(𝐬𝟐 − 𝟏)𝟐

H 4 Find the Laplace transform of (i)t2 sin πt (ii)t2 sin 2t.

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐢)𝟐𝛑(𝟑𝐬𝟐 − 𝛑𝟐)

(𝛑𝟐 + 𝐬𝟐)𝟑 (𝐢𝐢)

𝟒(𝟑𝐬𝟐 − 𝟒)

(𝐬𝟐 + 𝟒)𝟑

C 5 Find ℒ{t2 sin 4t}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟖(𝟑𝐬𝟐 − 𝟏𝟔)

(𝐬𝟐 + 𝟏𝟔)𝟑

C 6 Find ℒ{t sin 3t cos2t}

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓𝐬

(𝐬𝟐 + 𝟐𝟓)𝟐+

𝐬

(𝐬𝟐 + 𝟏)𝟐

W – 16

C 7 Find ℒ(t2 cos2 2t).


𝐬𝟑+𝐬(𝐬𝟐 − 𝟒𝟖)

(𝐬𝟐 + 𝟏𝟔)𝟑



H 8 Find the Laplace transform of f(t) = t2 cosh 3t.


(𝐬 − 𝟑)𝟑+

𝟏

(𝐬 + 𝟑)𝟑

S – 16

H 9 Find ℒ{t(sin t − t cos t)}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟖𝐬

(𝐬𝟐 + 𝟏)𝟑

W – 15

H 10 Obtained ℒ{eat t sin at}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐚(𝐬 − 𝐚)

[(𝐬 − 𝐚)𝟐 + 𝐚𝟐]𝟐

C 11 Find ℒ(t e−t cos ht).


𝟐[𝟏

𝐬𝟐+

𝟏

(𝐬 + 𝟐)𝟐]

H 12 Find the Laplace transform of t e4t cos 2t.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝟐 − 𝟖𝐬 + 𝟏𝟐

(𝐬𝟐 − 𝟖𝐬 + 𝟐𝟎)𝟐

S – 17

THEOREM. INTEGRATION OF LAPLACE TRANSFORM:

Statement: If 𝐋{𝐟(𝐭)} = 𝐅(𝐬) and if Laplace transform of 𝐟(𝐭)

𝐭 exists, then 𝓛 {𝐟(𝐭)

𝐭} = ∫ 𝐅(𝐬)𝐝𝐬

∞

𝐬 .

Proof: By definition, F(s) = ∫ e−stf(t) dt∞

0

Integrating both sides with respect to "s" in the range s to ∞.

∫ F(s)ds∞

s

= ∫ (∫ e−st f(t) dt∞

0

) ds∞

s

= ∫ ∫ e−st f(t) ds dt∞

s

∞

0

= ∫ (∫ e−stds∞

s

)∞

0

f(t) dt

= ∫ (e−st

−t)s

∞

f(t) dt∞

0

= ∫ (0− e−st

−t) f(t) dt

∞

0



= ∫ e−st [f(t)

t] dt

∞

0

= ℒ {f(t)

t}

Thus, ℒ {f(t)

t} = ∫ F(s)ds

∞

s

.

METHOD – 5: EXAMPLE ON INTEGRATION OF LAPLACE TRANSFORM

C 1 Find ℒ {

sin ωt

t} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝛑

𝟐− 𝐭𝐚𝐧−𝟏 (

𝐬

𝛚)

H 2 Find ℒ {

sin 2t

t} .


𝟐− 𝐭𝐚𝐧−𝟏 (

𝐬

𝟐)

C 3 Find ℒ {et

sin t

t} .

𝐀𝐧𝐬𝐰𝐞𝐫: (𝛑/𝟐) − 𝐭𝐚𝐧−𝟏(𝐬 − 𝟏)

H 4 Find ℒ {

cos3t

t} .

𝐀𝐧𝐬𝐰𝐞𝐫: ∞ W – 14

C 5 Find ℒ {

1 − et

t} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 (𝐬 − 𝟏

𝐬)

H 6 Find ℒ (

e−bt − e−at

t).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 (𝐬 + 𝐚

𝐬 + 𝐛 )

C 7 Find the Laplace transform of

1 − cos2t

t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 (√𝐬𝟐 + 𝟒

𝐬)

W – 17

H 8 Find the Laplace transform of

1 − cos t

t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠 (√𝐬𝟐 + 𝟏

𝐬)



T 9 Find the Laplace transform of

cos at − cos bt

t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠√𝐬𝟐 + 𝐛𝟐

𝐬𝟐 + 𝐚𝟐

S – 16


sin2 t

t2 .


𝟐𝐥𝐨𝐠 (

𝐬𝟐 + 𝟒

𝐬𝟐)

THEOREM. LAPLACE TRANSFORM OF INTEGRATION OF A FUNCTION

Statement: If 𝐋{𝐟(𝐭)} = 𝐅(𝐬), then 𝓛 {∫ 𝐟(𝐭)𝐭

𝟎 𝐝𝐭} =

𝐅(𝐬)

𝐬.

Proof: By definition ℒ (∫ f(u) dut

0) = ∫ e−st {∫ f(u)

t

0du}

∞

0 dt,

Suppose that U = ∫ f(u)t

0du then we have,

ℒ(U) = ∫ e−st U

∞

0

dt

= [U ∙ (e−st

−s) − ∫ {

dU

dt∙ (e−st

−s)} dt]

0

∞

(By Integration by parts)

= [U (e−st

−s)]

0

∞

−∫ {dU

dt(e−st

−s)} dt

∞

0

………… [1]

𝐈𝐧 𝐟𝐢𝐫𝐬𝐭 𝐬𝐭𝐞𝐩,when t → ∞ then e−st → 0 and t → 0 then U = ∫ f(u)t

0du → 0 .

𝐈𝐧 𝐬𝐞𝐜𝐨𝐧𝐝 𝐬𝐭𝐞𝐩, By Fundamental Theorem of Calculus,dU

dt=

d

dt{∫ f(u)

t

0 du} = f(t)

Now by [1] we have,

ℒ(U) = 0 −1

s∫ e−st f(t) dt

∞

0

Thus, ℒ {∫ f(t)

t

0

dt} =1

sF(s).



Remark:

Similarly,we can prove that ∫ ∫ f(t)t

0

t

0

dt dt =1

s2F(S).

METHOD – 6: EXAMPLE ON INTEGRATION OF A FUNCTION

H 1 Find ℒ {∫ e−x cos x dx

t

0

} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬 + 𝟏

𝐬[(𝐬 + 𝟏)𝟐 + 𝟏]

H 2 Find ℒ {∫ ∫ sin au

t

0

t

0

du du} .


𝐬𝟐𝐚

(𝐬𝟐 + 𝐚𝟐)

C 3

Find ℒ {∫eu(u + sin u)du

t

0

}.


𝐬{

𝟏

𝐬𝟐 − 𝟐𝐬 + 𝟏+

𝟏

𝐬𝟐 − 𝟐𝐬 + 𝟐}

W – 15

C 4 Find the Laplace transformation of f(t) = e−3t∫ tsin 3t dt

t

0

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔

(𝐬𝟐 + 𝟔𝐬 + 𝟏𝟖)𝟐

H 5 Find the Laplace transformation of ∫ e−t t cos t dt

t

0

.

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐬 + 𝟏)𝟐 − 𝟏

𝐬[(𝐬 + 𝟏)𝟐 + 𝟏]𝟐

C 6 Find the Laplace transformation of f(t) = ∫

et sint

t

t

0

dt .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐜𝐨𝐭−𝟏(𝐬 − 𝟏)

𝐬

W – 16



C 7 Evaluate: ∫ e−2t

∞

0

t cos t dt .


𝟐𝟓

T 8

Evaluate: ∫ e−t∞

0

(∫sin u

u

t

0

du) dt .


𝟒

THEOREM. LAPLACE TRANSFORM OF PERIODIC FUNCTION:

Statement: The Laplace transform of a piecewise continuous periodic function 𝐟(𝐭) having

period "𝐩" is

F(s) = ℒ{f(t)} =1

1 − e−ps∫ e−stf(t)dt

p

0

,Where s > 0.

Proof: By definition, ℒ{f(t)} = ∫ e−stf(t) dt∞

0

ℒ{f(t)} = ∫ e−stf(t) dtp

0

+∫ e−stf(t) dt∞

p

……… (A)

Now, ∫ e−stf(t) dt∞

p

= ∫ e−s(u+p)f(u + p) du∞

0

Since f(u) is Periodic. i. e. f(u) = f(u + p)

= ∫ e−s(u+p)f(u) du∞

0

= e−sp∫ e−suf(u) du∞

0

= e−spF(s)

By eqn. …(A)

ℒ{f(t)} = ∫ e−stf(t) dtp

0

+∫ e−stf(t) dt∞

p

Let, t = u + p ⟹ dt = du

When t ⟶ p ⟹ u⟶ 0

t ⟶ ∞ ⟹ u ⟶ ∞.



⟹ F(s) = ∫ e−stf(t) dtp

0

+ e−spF(s)

⟹ (1− e−sp)F(s) = ∫ e−stf(t) dtp

0

⟹ F(s) =1

1 − e−sp∫ e−stf(t) dt

p

0

Thus, ℒ{f(t)} =1

1 − e−sp∫ e−stf(t)dt

p

0

; Where s > 0

METHOD – 7: EXAMPLE ON L. T. OF PERIODIC FUNCTIONS

H 1 Find the Laplace transformation of f(t) = {

1 ; 0 ≤ t < a

−1 ; a < t < 2a and f(t) is

a period of 2a.


𝐬𝐭𝐚𝐧𝐡 (

𝐚𝐬

𝟐)

H 2 Find the Laplace transform of the periodic function defined by

f(t) =t

2, 0 < t < 3, f(t + 3) = f(t).


𝟐𝐬𝟐[𝟏 −

𝟑𝐬

𝐞𝟑𝐬 − 𝟏]

C 3 Find the Laplace transform of the periodic function

f(t) = {3t ; 0 ≤ t < 2

6 ; 2 < t < 4


𝟏 − 𝐞−𝟒𝐬[𝟑

𝐬𝟐−𝟑𝐞−𝟐𝐬

𝐬𝟐−𝟔𝐞−𝟒𝐬

𝐬]

C 4 Find the Laplace transformation of the periodic function of the waveform

f(t) =2t

3 , 0 < t < 3 ; if f(t) = f(t + 3).


𝟑𝐬𝟐−

𝟐𝐞−𝟑𝐬

𝐬(𝟏 − 𝐞−𝟑𝐬)

W – 17




t

T , 0 < t < T ; if f(t) = f(t + T).


𝐓𝐬𝟐−

𝐞−𝐓𝐬

𝐬(𝟏 − 𝐞−𝐓𝐬)

H 6 Find the Laplace transformation of f(t) = t2 , 0 < t < 2 if f(t) = f(t + 2).


(𝟏 − 𝐞−𝟐𝐬)𝐬𝟑(𝟐 − 𝟐𝐞−𝟐𝐬 − 𝟒𝐬𝐞−𝟐𝐬 − 𝟒𝐬𝟐𝐞−𝟐𝐬)

H 7 Find the Laplace transformation of f(t) = {

t ; 0 < t < a

2a − t ; a < t < 2a

if f(t) = f(t + 2a).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭𝐚𝐧𝐡 (

𝐚𝐬𝟐)

𝐬𝟐

C 8 Find the Laplace transform of the half wave rectifier

f(t) = {

sin ωt , 0 < t <π

ω

0, π

ω< t <

2π

ω

and f(t) = f (t +2π

ω).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝛚

((𝐬𝟐 +𝛚𝟐)(𝟏 − 𝐞−𝛑𝐬

𝛚⁄ ))

T 9 Find the Laplace transform of f(t) = |sin wt|, t ≥ 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝛚(𝟏 + 𝐞

−𝛑𝐬𝛚⁄ )

(𝐬𝟐 + 𝐰𝟐)(𝟏 − 𝐞−𝛑𝐬

𝛚⁄ )

LAPLACE TRANSFORM OF UNIT STEP FUNCTION:

Statement: Show that 𝐋{𝐮(𝐭 − 𝐚)} =𝐞−𝐚𝐬

𝐬


0

Now, ℒ{u(t − a)} = ∫ e−stu(t − a) dt∞

0

= ∫ e−stu(t − a) dta

0

+∫ e−stu(t − a) dt∞

a

We know that, u(t − a) = {0, 0 < x < a1, x ≥ a



= ∫ e−st dt∞

a

= [−e−st

s]a

∞

= [−0 − e−as

s] =

e−as

s

Thus, ℒ{u(t − a)} =e−as

s

Note: Instead of 𝐮(𝐭 − 𝐚) , we can also write 𝐇(𝐭 − 𝐚), which is referred as Heaviside’s unit

step function.

THEOREM. SECOND SHIFTING THEOREM:

Statement: If 𝓛{𝐟(𝐭)} = 𝐅(𝐬), then 𝓛{𝐟(𝐭 − 𝐚)𝐮(𝐭 − 𝐚)} = 𝐞−𝐚𝐬𝐅(𝐬)

Proof: By definition, L{f(t)} = ∫ e−stf(t) dt∞

0

Now, ℒ{f(t − a)u(t − a)}

= ∫ e−stf(t − a)u(t − a) dt∞

0

= ∫ e−stf(t − a)u(t − a) dta

0

+∫ e−stf(t − a)u(t − a) dt∞

a

We know that, 𝐮(𝐭 − 𝐚) = {𝟎, 𝟎 < 𝐱 < 𝐚𝟏, 𝐱 ≥ 𝐚

= ∫ e−stf(t − a) dt∞

a

Let, 𝐭 − 𝐚 = 𝐮 ⟹ 𝐝𝐭 = 𝐝𝐮. When 𝐭 ⟶ 𝐚⟹ 𝐮⟶ 𝟎 and 𝐭 ⟶ ∞ ⟹ 𝐮 ⟶ ∞.

= ∫ e−s(a+u)f(u) du∞

0

= e−sa∫ e−suf(u) du∞

0

= e−asF(s)

Hence, ℒ{f(t − a) ∙ u(t − a)} = e−asF(s)

NOTE: ℒ{f(t) ∙ u(t − a)} = e−as ℒ{f(t + a)}



METHOD – 8: EXAMPLE ON SECOND SHIFTING THEOREM

C 1 Find the Laplace transform of e−3t u(t − 2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟐𝐬𝐞−𝟔

𝐬 + 𝟑

W – 17

H 2 Find the Laplace transform of et u(t − 2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟐𝐬+𝟐

𝐬 − 𝟏

C 3 Find the Laplace transform of t2 u(t − 2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟐𝐬 (𝟐!

𝐬𝟑+

𝟒

𝐬𝟐+𝟒

𝐬)

W – 14

H 4 Find the Laplace transform of (t − 1)2 u(t − 1).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐞−𝐬

𝐬𝟑

H 5 Find the Laplace transform of cost u(t − π).

𝐀𝐧𝐬𝐰𝐞𝐫: −𝐬𝐞−𝛑𝐬

𝐬𝟐 + 𝟏

C 6 Find ℒ (e−t sint u(t − π)).

𝐀𝐧𝐬𝐰𝐞𝐫: −𝐞−𝛑(𝐬+𝟏)

𝐬𝟐 + 𝟐𝐬 + 𝟐

C 7 Express the following in terms of unit step function and hence find F(s).

f(t) = {t − 1 ; 1 < t < 2

3 − t ; 2 < t < 3

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝐬

𝐬𝟐−𝟐𝐞−𝟐𝐬

𝐬𝟐+𝐞−𝟑𝐬

𝐬𝟐

H 8 Express the following in terms of unit step function and hence find F(s).

f(t) = {sin 2t ; 2π < t < 4π

0 ; otherwise

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐(𝐞−𝟐𝛑𝐬 − 𝐞−𝟒𝛑𝐬)

𝐬𝟐 + 𝟒

T 9 Express the following in terms of unit step function and hence find F(s).

f(t) = {t2 ; 0 < t < π

e−2t ; π < t < 2πcos 3t ; t > 2π

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐!

𝐬𝟑−𝐞−𝛑𝐬 (

𝟐!

𝐬𝟑+𝟐𝛑

𝐬𝟐+𝛑𝟐

𝐬) − 𝐞−𝛑𝐬

𝐞−𝟐𝛑

𝐬 + 𝟐− 𝐞−𝛑𝐬

𝐞−𝟒𝛑

𝐬 + 𝟐+ 𝐞−𝟐𝛑𝐬

𝐬

𝐬𝟐 +𝟗



LAPLACE INVERSE TRANSFORM

ℒ−1 {1

s} = 1 ℒ−1 {

1

sn} =

tn−1

(n − 1)! 𝐎𝐑 t

n−1

n⁄

ℒ−1 {1

s − a} = eat ℒ−1 {

1

s + a} = e−at

ℒ−1 {1

s2 + a2} =

1

asin at ℒ−1 {

s

s2 + a2} = cos at

ℒ−1 {1

s2 − a2} =

1

asinh at ℒ−1 {

s

s2 − a2} = cosh at

METHOD – 9: EXAMPLE ON LAPLACE INVERSE TRANSFORM

C 1 Find ℒ−1 {

6s

s2 − 16} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔 𝐜𝐨𝐬𝐡 𝟒𝐭

T 2 Find ℒ−1 {

2s − 5

s2 − 4} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐 𝐜𝐨𝐬𝐡 𝟐𝐭 −𝟓

𝟐𝐬𝐢𝐧𝐡 𝟐𝐭

C 3 Find ℒ−1 {

3s − 8

4s2 + 25} .


𝟒𝐜𝐨𝐬

𝟓

𝟐𝐭 −

𝟒

𝟓𝐬𝐢𝐧

𝟓

𝟐𝐭

H 4 Find ℒ−1 {

3(s2 − 1)2

2s5} .


𝟐[𝟏 − 𝐭𝟐 +

𝐭𝟒

𝟒!]

C 5 Find ℒ−1 {

s3 + 2s2 + 2

s3(s2 + 1)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭𝟐 + 𝐬𝐢𝐧𝐭

H 6 Find ℒ−1 (

4s + 15

16s2 − 25) .


𝟒𝐜𝐨𝐬 𝐡

𝟓

𝟒𝐭 +

𝟑

𝟒𝐬𝐢𝐧 𝐡

𝟓

𝟒𝐭



C 7 Find ℒ−1 (

√s − 1

s)

2

.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 + 𝐭 − 𝟒√𝐭

𝛑

REMARK: (CHANGE OF SCALE PROPERTY)

Statement: If ℒ−1{F(s)} = f(t), then ℒ−1{F(bs)} =1

bf (

t

b).

Example 1: Find ℒ−1 (1

5s+1) using change of scale property.

Solution:

Since, ℒ−1 {1

s + 1} = e−t .

So, ℒ−1 (1

5s + 1) =

1

5e−

t5

Example 2: Find ℒ−1 (2

9s2−4) using change of scale property.

Solution:

ℒ−1 (2

9s2 − 4)= 2 ℒ−1 (

1

(3s)2 − 4)

⟹ ℒ−1 (2

9s2 −4) = 2 (

1

3)(

1

2) sinh 2 (

t

3) (∵ ℒ−1 {

1

s2 − 4} =

1

2sinh 2t)

⟹ ℒ−1 (2

9s2 −4) =

1

3sinh

2t

3 .

FIRST SHIFTING THEOREM:

Statement: If ℒ−1{F(s)} = f(t), then ℒ−1{F(s − a)} = eatf(t).



METHOD – 10: EXAMPLE ON FIRST SHIFTING THEOREM

C 1 Find ℒ−1 {

10

(s − 2)4} .

𝐀𝐧𝐬𝐰𝐞𝐫: 5e2tt3

3

T 2 Find ℒ−1 (

1

√2s + 3) .

𝐀𝐧𝐬𝐰𝐞𝐫: 1

√2πt−

12e−

3t2

T 3 Find ℒ−1 (

3s + 1

(s + 1)4) .


2e−tt2 −

1

3e−tt3

H 4 Find ℒ−1 (s

(s + 2)4) .

𝐀𝐧𝐬𝐰𝐞𝐫: e−2t(t2

2−t3

3)

C 5 Find ℒ−1 (s

(s + 2)2 + 1) .

𝐀𝐧𝐬𝐰𝐞𝐫: e−2t cos t − 2e−2t sin t

C 6 Find ℒ−1 {

3

s2 + 6s + 18} .

𝐀𝐧𝐬𝐰𝐞𝐫: e−3tsin3t

H 7 Find ℒ−1 (

2s + 3

s2 − 4s + 13) .


3e2t(6 cos3t + 7 sin 3t)

H 8 Find ℒ−1 (

1

s2 + s + 1) .

𝐀𝐧𝐬𝐰𝐞𝐫: √2

3 e−

t2 sin (√

3

2t)

H 9 Find ℒ−1 (s

s2 + s + 1) .

𝐀𝐧𝐬𝐰𝐞𝐫: e−t2(cos √

3

2t −

1

√6sin (√

3

2t))



H 10 Find ℒ−1 (

s + 7

s2 + 8s + 25) .

𝐀𝐧𝐬𝐰𝐞𝐫: e−4t(sin 3t + cos3t)

S – 17

C 11 Find the inverse Laplace transform of

6 + s

s2 + 6s + 13 , use shifting theorem.

𝐀𝐧𝐬𝐰𝐞𝐫: e−3tcos2t +3

2e−3t sin 2t

PARTIAL FRACTION METHOD

Case 1 : If the denominator has non-repeated linear factors (s − a), (s − b), (s − c), then

f(s)

(s − a)(s − b)(s − c)=

A

(s − a)+

B

(s − b)+

C

(s − c)

Case 2 : If the denominator has repeated linear factors (s − a), (n times), then

f(s)

(s − a)n=

A1

(s − a)+

A2

(s − a)2+

A3

(s − a)3+⋯+

An

(s − a)n

Case 3 : If the denominator has non-repeated quadratic factors (s2 + as + b),

(s2 + cs + d) ,

f(s)

(s2 + as + b)(s2 + cs + d)=

As + B

(s2 + as + b)+

Cs + D

(s2 + cs + d)

Case 4 : If the denominator has repeated quadratic factors (s2 + as + b), (n times), then

f(s)

(s2 + as + b)n=

As + B

(s2 + as + b)+

Cs + D

(s2 + as + b)2+⋯(n times)

METHOD – 11: EXAMPLE ON PARTIAL FRACTION METHOD

H 1 Find ℒ−1 {

1

s(s + 1)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 − 𝐞−𝐭

H 2 Find ℒ−1 {

1

(s + 1)(s + 2)} .


𝟓[𝐞−𝐭 − 𝐞−𝟐𝐭]

S – 18



H 3 Find ℒ−1 {

1

(s − 2)(s + 3)} .


𝟓[𝐞𝟐𝐭 − 𝐞−𝟑𝐭]

S – 15

T 4 Find ℒ−1 {

1

(s + √2)(s − √3)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞√𝟑𝐭 − 𝐞−√𝟐𝐭

√𝟑 + √𝟐

S – 16

C 5 Find ℒ−1 {−

s + 10

s2 − s − 2} .

𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟒𝐞𝟐𝐭 + 𝟑𝐞−𝐭

C 6 Find ℒ−1 {

5s2 + 3s − 16

(s − 1)(s + 3)(s − 2)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐞𝐭 + 𝐞−𝟑𝐭 + 𝟐𝐞𝟐𝐭

H 7 Find ℒ−1 {

3s2 + 2

(s + 1)(s + 2)(s + 3)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓

𝟐𝐞−𝐭 − 𝟏𝟒𝐞−𝟐𝐭 +

𝟐𝟗

𝟐𝐞−𝟑𝐭

H 8 Find ℒ−1 (

2s2 − 4

(s + 1)(s − 2)(s − 3)) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟕

𝟐𝐞𝟑𝐭 −

𝟏

𝟔𝐞−𝐭 −

𝟒

𝟑𝐞𝟐𝐭

T 9 Find ℒ−1 (

s + 2

s (s + 1)(s + 3)) .


𝟑−𝟏

𝟐𝐞−𝐭 −

𝟏

𝟔𝐞−𝟑𝐭

C 10 Find ℒ−1 (

2s + 3

(s + 2) (s + 1)2) .

𝐀𝐧𝐬𝐰𝐞𝐫: − 𝐞−𝟐𝐭 + 𝐞−𝐭 + 𝐭𝐞−𝐭

H 11 Find ℒ−1 (

4s + 5

(s − 1)2(s + 2)) .


𝟑𝐞𝐭 + 𝟑𝐭𝐞𝐭 +

𝟏

𝟑𝐞−𝟐𝐭

C 12 Find ℒ−1 (

3s + 1

(s + 1) (s2 + 2)) .

𝐀𝐧𝐬𝐰𝐞𝐫: −𝟐

𝟑𝐞−𝐭 +

𝟐

𝟑𝐜𝐨𝐬√𝟐 𝐭 +

𝟕

𝟑√𝟐𝐬𝐢𝐧√𝟐 𝐭



C 13 Find ℒ−1 {

1

s(s2 − 3s + 3)} .


𝟑+ 𝐞

𝟑𝐭𝟐 [

𝟏

√𝟑𝐬𝐢𝐧 (

√𝟑𝐭

𝟐) −

𝟏

𝟑𝐜𝐨𝐬 (

√𝟑𝐭

𝟐)]

W – 15

H 14 Find the inverse Laplace transform of

5s + 3

(s2 + 2s + 5)(s − 1) .

𝐀𝐧𝐬𝐰𝐞𝐫: − 𝐞−𝐭 𝐜𝐨𝐬 𝟐𝐭 +𝟑

𝟐𝐞−𝐭 𝐬𝐢𝐧 𝟐𝐭 + 𝐞𝐭

T 15 Find ℒ−1 (

s + 4

s (s − 1) (s2 + 4)) .

𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟏 + 𝐞𝐭 −𝟏

𝟐𝐬𝐢𝐧𝟐𝐭

C 16 Find ℒ−1 (

1

(s2 + 2) (s2 − 3)) .


𝟓√𝟑𝐬𝐢𝐧𝐡√𝟑𝐭 −

𝟏

𝟓√𝟐𝐬𝐢𝐧√𝟐𝐭

H 17 Find ℒ−1 (s

(s2 + 1) (s2 + 4)) .


𝟑(𝐜𝐨𝐬𝐭 − 𝐜𝐨𝐬𝟐𝐭)

C 18 Find ℒ−1 (

2s2 − 1

(s2 + 1) (s2 + 4)) .


𝟐𝐬𝐢𝐧 𝟐𝐭 − 𝐬𝐢𝐧 𝐭

W – 16

H 19 Find ℒ−1 (

s2

(s2 + a2) (s2 + b2)) .


𝐚𝟐 − 𝐛𝟐(𝐚𝐬𝐢𝐧 𝐚𝐭 − 𝐛𝐬𝐢𝐧 𝐛𝐭)

C 20 Find ℒ−1 {

s3

s4 − 81} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐜𝐨𝐬𝟑𝐭 + 𝐜𝐨𝐬𝐡𝟑𝐭

𝟐

W – 17

H 21 Find ℒ−1 {

1

s4 − 81} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐬𝐢𝐧𝐡 𝟑𝐭 − 𝐬𝐢𝐧 𝟑𝐭

𝟓𝟒

S – 16



T 22 Find ℒ−1 {s

s4 + 64} .


𝟏𝟔[𝐞𝟐𝐭 𝐬𝐢𝐧 𝟐𝐭 − 𝐞−𝟐𝐭 𝐬𝐢𝐧 𝟐𝐭]

SECOND SHIFTING THEOREM:

Statement: If ℒ−1{F(s)} = f(t), then ℒ−1{e−asF(s)} = f(t − a) ∙ H(t − a).

METHOD – 12: EXAMPLE ON SECOND SHIFTING THEOREM

H 1 Find ℒ−1 (

e−as

s) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐭 − 𝐚)

C

2 Find the inverse Laplace transform of

se−2s

s2 + π2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐜𝐨𝐬 𝛑(𝐭 − 𝟐) ∙ 𝐮(𝐭 − 𝟐)

H 3 Find ℒ−1 {

se−πs

s2 − π2} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐜𝐨𝐬𝐡(𝐭 − 𝛑) ∙ 𝐮(𝐭 − 𝛑)

T 4 Find ℒ−1 (e−s {

√s − 1

s}

2

) .

𝐀𝐧𝐬𝐰𝐞𝐫: {𝟏 + (𝐭 − 𝟏) − 𝟒√(𝐭 − 𝟏)

𝛑}𝐮(𝐭 − 𝟏)

C 5 Find the inverse Laplace transform of

e−4s(s + 2)

s2 + 4s + 5 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟐(𝐭−𝟒) 𝐜𝐨𝐬(𝐭 − 𝟒) ∙ 𝐮(𝐭 − 𝟒)

H 6 Find ℒ−1 (

e−2s

(s + 2)(s + 3) ) .

𝐀𝐧𝐬𝐰𝐞𝐫: {𝐞−𝟐(𝐭−𝟐) − 𝐞−𝟑(𝐭−𝟐)} ∙ 𝐮(𝐭 − 𝟐)

H 7 Find ℒ−1 (

e−2s

s2 + 8s + 25 ) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟒(𝐭−𝟐)

𝟑𝐬𝐢𝐧 𝟑(𝐭 − 𝟐) ∙ 𝐮(𝐭 − 𝟐)

W – 16



C 8 Find ℒ−1 {

e−2s

(s2 + 2)(s2 − 3)}.


𝟓[𝟏

√𝟑𝐬𝐢𝐧𝐡√𝟑(𝐭 − 𝟐) −

𝟏

√𝟐𝐬𝐢𝐧√𝟐(𝐭 − 𝟐)]𝐇(𝐭 − 𝟐)

W – 15

INVERSE LAPLACE TRANSFORM OF DERIVATIVES:

Statement: If ℒ−1{F(s)} = f(t), then ℒ−1{F′(s)} = −t ∙ f(t)

METHOD – 13: EXAMPLE ON INVERSE LAPLACE TRANSFORM OF DERIVATIVES

H 1 Find ℒ−1 {log

1

s} .


𝐭

H 2 Find ℒ−1 {log (

1 + s

s)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 − 𝐞−𝐭

𝐭

C 3 Find the inverse transform of the function ln (1 +

w2

s2) .


𝐭(𝟏 − 𝐜𝐨𝐬𝐰𝐭)

T 4 Find ℒ−1 {log

s + a

s + b} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝐛𝐭 − 𝐞−𝐚𝐭

𝐭

H 5 Find ℒ−1 {log

s + 1

s − 1} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐭 − 𝐞−𝐭

𝐭

C 6 Find ℒ−1 {log (

s + 4

s + 3)}.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟑𝐭 − 𝐞−𝟒𝐭

𝐭



H 7 Find ℒ−1{tan−1(s + 1)} .

𝐀𝐧𝐬𝐰𝐞𝐫: −𝟏

𝐭𝐞−𝐭 𝐬𝐢𝐧 𝐭

C 8 Find ℒ−1{tan−1

2

s} .


𝐭𝐬𝐢𝐧 𝟐𝐭

W – 17

C 9 Find ℒ−1 {cot−1 (

s + a

b)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐛𝐞−𝐚𝐭 𝐬𝐢𝐧 𝐛𝐭

𝐭

H 10 Find ℒ−1{cot−1(αs)} .


𝐭𝐬𝐢𝐧

𝐭

𝐚

DEFINITION: CONVOLUTION PRODUCT:

The convolution of f and g is denoted by f ∗ g and is defined as

f ∗ g = ∫ f(u) ∙ g(t − u) dut

0

METHOD – 14: EXAMPLE ON CONVOLUTION PRODUCT

H 1 Find the value of 1 ∗ 1. Where " ∗ " denote convolution product.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭

C 2 Evaluate 1 ∗ e2t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝟐𝐭 − 𝟏

C 3 Evaluate t ∗ et .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝐭 − 𝐭 − 𝟏 S – 15

H 4 Find the convolution of e−t and sin t .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝐭 + 𝐬𝐢𝐧 𝐭 − 𝐜𝐨𝐬 𝐭

𝟐



𝒕

THEOREM. CONVOLUTION THEOREM

Statement: If ℒ−1{F(s)} = f(t) and ℒ−1{G(s)} = g(t), then

ℒ−1{F(s) ∙ G(s)} = ∫ f(u) ∙ g(t − u) dut

0

= f ∗ g

Proof: Let us suppose F(t) = ∫ f(u) ∙ g(t − u) dut

0

Now, ℒ(F(t)) = ∫ e−st (∫ f(u) ∙ g(t − u)dut

0

) dt∞

0

= ∫ ∫ f(u) ∙ g(t − u)t

0

e−stdu dt∞

0

Here, region of integration is entire area

lying between the lines u = 0 and u = t

which is part of the first quadrant.

Changing the order of integration, we have

ℒ(F(t)) = ∫ ∫ e−stf(u) ∙ g(t − u) dt∞

u

du ∞

0

= (∫ e−suf(u)du∞

0

) (∫ e−s(t−u)g(t − u)∞

u

dt)

Let, t − u = v ⟹ dt = dv. When t ⟶ u ⟹ v ⟶ 0andt ⟶ ∞ ⟹ v ⟶ ∞.

= (∫ e−suf(u)du∞

0

) (∫ e−svg(v)∞

0

dv)

= F(s) ∙ G(s)

Thus, ℒ(F(t)) = F(s) ∙ G(s) ⟹ F(t) = ℒ−1{F(s) ∙ G(s)}

⟹ ℒ−1{F(s) ∙ G(s)} = ∫ f(u)g(t − u)dut

0

Hence, f ∗ g = ℒ−1{F(s) ∙ G(s)} = ∫ f(u) ∙ g(t − u) dut

0 is convolution product of f & g.

𝐭

𝐮 u = t

u = 0

t = 0



METHOD – 15: EXAMPLE ON CONVOLUTION THEROREM

C 1 State convolution theorem and using it find ℒ−1 {

1

(s + 1)(s + 3)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝐭 − 𝐞−𝟑𝐭

𝟐

C 2 Using the convolution theorem, obtain the value of ℒ−1 {

1

s(s2 + 4)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 − 𝐜𝐨𝐬 𝟐𝐭

𝟒

S – 15

H 3 Use convolution theorem to find ℒ−1 {

1

s(s2 + a2)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 − 𝐜𝐨𝐬 𝐚𝐭

𝐚𝟐

T 4 State convolution theorem and use it to evaluate ℒ−1 {a

s2(s2 + a2)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚𝐭 − 𝐬𝐢𝐧 𝐚𝐭

𝐚𝟐

C 5 Apply convolution theorem to Evaluate ℒ−1 {s

(s2 + a2)2} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭 𝐬𝐢𝐧 𝐚𝐭

𝟐𝐚

W – 16

H 6 Find the inverse Laplace transform of s

(s2 + 1)2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭 𝐬𝐢𝐧 𝐭

𝟐

W – 14

C 7 State Convolution Theorem and Use to it Evaluate ℒ−1 {

1

(s2 + a2)2} .


𝟐𝐚𝟐(𝐬𝐢𝐧 𝐚𝐭

𝟐𝐚− 𝐭 𝐜𝐨𝐬 𝐚𝐭)

S – 16

H 8 Using convolution theorem, obtain ℒ−1 {

1

(s2 + 4)2} .


𝟖(𝐬𝐢𝐧𝟐𝐭

𝟒− 𝐭 𝐜𝐨𝐬𝟐𝐭)

S – 17 S – 18

H 9 Using convolution theorem find ℒ−1 (

2

(s2 + 1)(s2 + 4)) .


𝟑(𝐬𝐢𝐧 𝐭 −

𝟏

𝟐𝐬𝐢𝐧 𝟐𝐭)


U N I T - 5 » L a p l a c e T r a n s f o r m a n d I t ’ s A p p l i c a t i o n [ 1 0 0 ]

C 10 State convolution theorem and using it find ℒ−1 {

s2

(s2 + a2)(s2 + b2)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 𝐬𝐢𝐧𝐚𝐭 − 𝐛 𝐬𝐢𝐧 𝐛𝐭

𝐚𝟐 − 𝐛𝟐

W – 17

H 11 State convolution theorem and using it to find ℒ−1 {

s2

(s2 + 4)(s2 + 9)} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑 𝐬𝐢𝐧 𝟑𝐭 − 𝟐 𝐬𝐢𝐧 𝟐𝐭

𝟓

T 12 Find ℒ−1 {

s + 2

(s2 + 4s + 5)2} .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟐𝐭 𝐭 𝐬𝐢𝐧 𝐭

𝟐

C 13 Find the inverse Laplace transform of s

(s + 1) (s − 1)2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐭 𝐞𝐭

𝟐+𝐞𝐭

𝟒−𝐞−𝐭

𝟒

W – 14

H 14 Find ℒ−1 (

1

(s − 2)(s + 2)2) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞𝟐𝐭 − 𝐞−𝟐𝐭 − 𝟒𝐭 𝐞−𝟐𝐭

𝟏𝟔

C 15 Find ℒ−1 (

1

(s − 2)4(s + 3)) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−𝟑𝐭

𝟔𝟐𝟓+𝐞𝟐𝐭

𝟔(𝐭𝟑

𝟓−𝟑𝐭𝟐

𝟐𝟓+

𝟔𝐭

𝟏𝟐𝟓−

𝟔

𝟔𝟐𝟓 )

T 16 Find ℒ−1 {

1

s(s + a)3} .


𝟐[−

𝐭𝟐𝐞−𝐚𝐭

𝐚−𝟐𝐭𝐞−𝐚𝐭

𝐚𝟐−𝟐𝐞−𝐚𝐭

𝐚𝟑+

𝟐

𝐚𝟑]

C 17 State the convolution theorem and verify it for f(t) = t and g(t) = e2t .


𝐬𝟐(𝐬 − 𝟐)

W – 15

H 18 State the convolution theorem and verify it for f(t) = 1 and g(t) = sin t .


𝐬(𝐬𝟐 + 𝟏)



NOTE:

(1) If ℒ−1{F(s)} = f(t) then ℒ−1 {F(s)

s} = ∫ f(t)

t

0dt .

(2) If ℒ−1{F(s)} = f(t) then ℒ−1 {F(s)

sn} = (∫ ∫ ……n times) f(t)

t

0

t

0(dt)n ; n ∈ N

THEOREM. DERIVATIVE OF LAPLACE TRANSFORM

Statement: If 𝓛{𝐟(𝐭)} = 𝐅(𝐬), then 𝓛{𝐟′(𝐭)} = 𝐬𝐅(𝐬) − 𝐟(𝟎).


0

⟹ ℒ{f ′(t)} = ∫ e−stf′(t) dt∞

0

⟹ ℒ{f ′(t)} = [e−st∫ f ′(t)dt − ∫ {(d

dte−st) ∙ ∫ f ′(t)dt} dt]

0

∞

⟹ ℒ{f ′(t)} = [e−stf(t) − ∫(−s)e−stf(t)dt]0

∞

⟹ ℒ{f ′(t)} = 0 − f(0) + s∫ e−stf(t) dt∞

0

= sF(s) − f(0)

⟹ ℒ{f ′(t)} = sF(s) − f(0).

NOTE:

ℒ{y′(t)} = sℒ{y(t)} − y(0).

ℒ{y′′(t)} = s2ℒ{y(t)} − s y(0) − y′(0).

ℒ{y′′′(t)} = s3ℒ{y(t)} − s2y(0) − sy′(0) − y′′(0).

METHOD – 16: EXAMPLE ON APPLICATION OF LAPLACE TRANSFORM

T 1 Solve by Laplace transform: dy

dt− 2y = 4, given that t = 0, y = 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = 𝟑𝐞𝟐𝐭 − 𝟐

H 2 Solve by Laplace transform y′′ + 6y = 1, y(0) = 2, y′(0) = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) =𝟏𝟏

𝟔𝐜𝐨𝐬 √𝟔𝐭 +

𝟏

𝟔



C 3 Solve IVP using Laplace transform y′′ + 4y = 0, y(0) = 1, y′(0) = 6.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = 𝐜𝐨𝐬 𝟐𝐭 + 𝟑 𝐬𝐢𝐧 𝟐𝐭

T 4 By using the method of Laplace transform solve the IVP :

y′′ + 2y′ + y = e−t, y(0) = −1 and y′(0) = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) =𝐞−𝐭 𝐭𝟐

𝟐! − 𝐞−𝐭

H 5 By using the method of Laplace transform solve the IVP :

y′′ + 4y′ + 3y = e−t, y(0) = 1 and y′(0) = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = (𝐭

𝟐+𝟗

𝟒) 𝐞−𝐭 −

𝟓

𝟒𝐞−𝟑𝐭

H 6 By using the method of Laplace transform solve the IVP :

y′′ + 3y′ + 2y = et, y(0) = 1 and y′(0) = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) =𝟏

𝟔𝐞𝐭 −

𝟓

𝟐𝐞−𝐭 +

𝟒

𝟑𝐞−𝟐𝐭

S – 15

C 7 Use the Laplace transform to solve the following initial value problem:

y" − 3y′ + 2y = 12e−2t , y(0) = 2 and y′(0) = 6 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = 𝟏

𝟔𝐞−𝟐𝐭 +

𝟗

𝟐𝐞𝟐𝐭 −

𝟖

𝟑𝐞𝐭

S – 17

C 8 By using the method of Laplace transform solve the IVP :

y′′ − 4y′ + 3y = 6t − 8, y(0) = 0 and y′(0) = 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = 𝟐𝐭 + 𝐞𝐭 − 𝐞𝟑𝐭

C 9 Using Laplace transform solve the IVP

y′′ + y = sin 2t , y(0) = 2, y′(0) = 1.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) =𝟓

𝟑𝐬𝐢𝐧 𝐭 −

𝟏

𝟑𝐬𝐢𝐧 𝟐𝐭 + 𝟐 𝐜𝐨𝐬 𝐭

W – 14 S – 18

T 10 By Laplace transform solve y′′ + a2y = K sin at.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = (𝐤

𝟐𝐚𝟐+𝐁

𝐚) 𝐬𝐢𝐧𝐚 𝐭 + (𝐀 −

𝐤

𝟐𝐚𝐭 ) 𝐜𝐨𝐬 𝐚𝐭

T 11 Using Laplace transform solve the differential equation

d2x

dt2+ 2

dx

dt+ 5x = e−t sin t ,where x(0) = 0, x′(0) = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱(𝐭) =𝐞−𝐭(𝐬𝐢𝐧 𝐭 + 𝐬𝐢𝐧 𝟐𝐭)

𝟑



T 12 Solve the initial value problem: y" − 2y′ = et sint , y(0) = y′(0) = 0, using

Laplace transform.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = −𝟏

𝟒+𝟏

𝟒𝐞𝟐𝐭 −

𝟏

𝟐𝐞𝐭 𝐬𝐢𝐧 𝐭

W – 15

C 13 Solve the differential equation using Laplace Transformation method

y" − 3y′ + 2y = 4t + e3t , y(0) = 1 and y′(0) = −1 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) = 𝟑 + 𝟐𝐭 +𝟏

𝟐(𝐞𝟑𝐭 − 𝐞𝐭) − 𝟐𝐞𝟑𝐭

S – 16 W – 16

T 14 Solve using Laplace transforms,

y′′′ + 2y′′ − y′ − 2y = 0; where, y(0) = 1, y′(0) = 2, y′′(0) = 2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲(𝐭) =𝟏

𝟑[𝟓𝐞𝐭 + 𝐞−𝟐𝐭] − 𝐞−𝐭

W – 17

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆


U N I T - 6 » P a r t i a l D i f f e r e n t i a l E q u a t i o n a n d I t ’ s A p p l i c a t i o n [ 1 0 4 ]

UNIT-6 » PARTIAL DIFFERENTIAL EQUATION AND IT’S APPLICATION

INTRODUCTION:

A partial differential equation is a mathematical equation involving two or more

independent variables, unknown function and its partial derivative with respect to

independent variables.

Partial differential equations are used to formulate the problems containing functions of

several variables, such as propagation of heat or sound, fluid flow, electrodynamics etc.

DEFINITION: PARTIAL DIFFERENTIAL EQUATION:

An equation which involves function of two or more variables and partial derivatives of that

function then it is called Partial Differential Equation.

e.g. ∂y

∂x+

∂y

∂t= 0.

DEFINITION: ORDER OF DIFFERENTIAL EQUATION:

The order of highest derivative which appears in differential equation is “Order of D.E”.

e.g. (∂y

∂x)2+

∂y

∂t+ 5y = 0 has order 1.

DEFINITION: DEGREE OF DIFFERENTIAL EQUATION:

When a D.E. is in a polynomial form of derivatives, the highest power of highest order

derivative occurring in D.E. is called a “Degree Of D.E.”.

e.g. (∂y

∂x)2+

∂y

∂t+ 5y = 0 has degree 2.

NOTATION:

Suppose z = f(x, y). For that , we shall use ∂z

∂x= p ,

∂z

∂y= q ,

∂2z

∂x2= r,

∂2z

∂x∂y= s,

∂2z

∂y2= t.

FORMATION OF PARTIAL DIFFERENTIAL EQUATION:

By Eliminating Arbitrary Constants

o Consider the function f(x, y, z, a, b) = 0. Where, a & b are independent arbitrary

constants.

o Step 1: f(x, y, z, a, b) = 0. ……(1)



o Step 2: fx(x, y, z, a, b) = 0. ……(2) and fy(x, y, z, a, b) = 0. ……(3)

o Step 3: Eliminate a & b from eq. (1), eq. (2) & eq. (3).

o We get partial differential equation of the form F(x, y, z, p, q) = 0

By Eliminating Arbitrary Functions

Type 1: Consider, the function (u, v) = 0 ; u and v are functions of x and y

o Step 1: Let, u = F(v).

o Step 2: Find ux & uy.

o Step 3: Eliminate the function F from ux & uy .

o Note: In such case, for elimination of function, substitution method is used.

Type 2: Consider, the function z = f(x, y)

o Step 1: Find zx & zy.

o Step 2: Eliminate the function f from zx & zy.

o Note: In such case, for elimination of function, division of zx & zy is used.

METHOD – 1: EXAMPLE ON FORMATION OF PARTIAL DIFFERENTIAL EQUATION

H 1 Form the partial differential equation z = ax + by + ct .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐩𝐱 + 𝐪𝐲 + 𝐭𝛛𝐳

𝝏𝒕

S – 17

C 2 Form the partial differential equation z = (x − 2)2 + (y − 3)2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝐳 = 𝐩𝟐 + 𝐪𝟐

C 3 Form the partial differential equation for the equation

(x − a)(y − b) − z2 = x2 + y2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒(𝐱 + 𝐩𝐳)(𝐲 + 𝐪𝐳)−𝐳𝟐 = 𝐱𝟐 + 𝐲𝟐

W – 15

H 4 Eliminate the function f from the relation f(xy + z2, x + y + z) = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 + 𝟏

𝐪 + 𝟏=𝐲 + 𝟐𝐳𝐩

𝐱 + 𝟐𝐳𝐪



C 5 Form the partial differential equation of f(x + y + z, x2+y2 + z2) = 0.


𝐪 + 𝟏=𝐱 + 𝐳𝐩

𝐲 + 𝐳𝐪

S – 15

T 6 From a partial differential equation by eliminating the arbitrary function ∅

from ∅(x + y + z, x2 + y2 − z2) = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: (𝐲 + 𝐳)𝐩 − (𝐱 + 𝐳)𝐪 = 𝐱 − 𝐲

H 7 Form the partial differential equation f(x2 − y2, xyz) = 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲𝐳 + 𝐱𝐲𝐩

𝐱𝐳 + 𝐱𝐲𝐪= −

𝐱

𝐲

W – 14

C 8 Form the partial differential equations by eliminating the arbitrary

function from f(x2 + y2, z − xy) = 0.


𝐲=𝐩 − 𝐲

𝐪 − 𝐱

W – 17

C 9 Form partial differential equation by eliminating the arbitrary function

from xyz = Ф(x + y + z).


𝐪 + 𝟏=𝐲𝐳 + 𝐱𝐲𝐩

𝐱𝐳 + 𝐱𝐲𝐪

H 10 Form the partial differential equation of z = f (𝑥

𝑦).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩

𝐪= −

𝐲

𝐱

S – 17

H 11 Form the partial differential equation by eliminating the arbitrary function

from z = f(x2 − y2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩

𝐪= −

𝐱

𝐲

T 12 Form the partial differential equation of z = xy + f(x2 + y2).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 − 𝐲

𝐪 − 𝐱=𝐱

𝐲

C 13 Form partial differential equation of z = f(x + iy) + g(x − iy).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝛛𝟐𝐳

𝛛𝐱𝟐+𝛛𝟐𝐳

𝛛𝐲𝟐= 𝟎



METHOD – 2: EXAMPLE ON DIRECT INTEGRATION

C 1 Solve

∂2u

∂x ∂y= x3+y3 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) =𝐱𝟒𝐲

𝟒+𝐱𝐲𝟒

𝟒+ 𝐅(𝐲) + 𝐠(𝐱)

H 2 Solve

∂2u

∂x ∂t= e−t cos x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = −𝐞−𝐭 𝐬𝐢𝐧 𝐱 + 𝐅(𝐭) + 𝐠(𝐱)

C 3 Solve

∂3u

∂x2 ∂y= cos(2x + 3y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = − 𝐬𝐢𝐧(𝟐𝐱 + 𝟑𝐲)

𝟏𝟐+ 𝐱𝐅(𝐲) + 𝐆(𝐲) + 𝐡(𝐱)

H 4 Solve ∂2z

∂x∂y= sin x sin y , given that

∂z

∂y= −2 sin y ,when x = 0 & z = 0,

when y is an odd multiple of π

2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳(𝐱, 𝐲) = 𝐜𝐨𝐬 𝐱 𝐜𝐨𝐬 𝐲 + 𝐜𝐨𝐬 𝐲

LINEAR PDE WITH CONSTANT CO-EFFICIENT:

The nth order linear partial differential equation with constant co-efficient is

a0∂nz

∂xn+ a1

∂nz

∂xn−1 ∂y+⋯an

∂nz

∂yn= F(x, y)……… (A)

Where, a0, a1, … , an are constants.

NOTATIONS:

Replacing ∂

∂x= D and

∂

∂y= D′ in Eq. (A) , it can be written in operator form as below,

a0Dnz + a1D

n−1D′z + ⋯+ anD′nz = F(x, y) 𝐎𝐑 [f(D, D′)]z = F(x, y)

AUXILIARY EQUATION:

The auxiliary equation for nth order PDE a0Dnz + a1D

n−1D′z + ⋯+ anD′nz = F(x, y)

is derived by replacing D by m , D’ by 1 and F(x, y) by 0.

COMPLEMENTARY FUNCTION (C.F.--𝐳𝐜):

A general solution of [f(D, D′)]z = 0 is called complementary function of [f(D, D′)]z = F(x, y).



PARTICULAR INTEGRAL (P.I.--𝐳𝐩):

A particular integral of [f(D, D′)]z = F(x, y) is P. I. =1

f(D,D′)F(x, y).

GENERAL SOLUTION OF PDE:

G. S. = C. F.+P. I = zc + zp

METHOD FOR FINDING C.F. OF PARTIAL DIFFERENTIAL EQUATION:

Consider, a0Dnz + a1D

n−1D′z + ⋯+ anD′nz = F(x, y)

The Auxiliary equation is a0mnz + a1m

n−1z + ⋯+ anz = 0.

Let m1 , m2 , … be the roots of auxiliary equation.

Case Nature of the “n” roots General Solutions

1. m1 ≠ m2 ≠ m3 ≠ m4 ≠ ⋯ z = ϕ1(y + m1x) + ϕ2(y + m2x) + ϕ3(y + m3x) + ⋯

2. m1 = m2 = m

m3 ≠ m4 ≠ ⋯ z = ϕ1(y + mx) + xϕ2(y + mx) + ϕ3(y + m3x) + ⋯

3. m1 = m2 = m3 = m

m4 ≠ m5 , …

z = ϕ1(y + mx) + xϕ2(y + mx)

+x2ϕ3(y + mx) + ϕ4(y + m4x) + ⋯

METHOD FOR FINDING PARTICULAR INTEGRAL:

For partial differential equation the value of Particular integral can be find by following

methods.

(1) General Method

(2) Short-cut Method

GENERAL METHOD

o Consider the partial differential equation f(D, D′)z = F(x, y)

o Particular integral P. I. =1

f(D,D′)F(x, y)

o Suppose, f(x, y) is factorized into n linear factors.



P. I. =1

f(D, D′)F(x, y) =

1

(D −m1D′)(D − m2D

′)… (D − mnD′)F(x, y)

Which can be evaluated by

1

D −mD′F(x, y) = ∫F(x, c − mx)dx

o Where, c is replaced by y + mx after integration.

SHORTCUT METHOD

Case-1 F(x, y) = eax+by

P. I. =1

f(D, D′)eax+by =

1

f(a, b)eax+by, if f(a, b) ≠ 0

If f(a, b) = 0 then m =a

b is a root of auxiliary equation repeated r times.

f(D, D′) = (D −a

bD′)

r

g(D, D′)

P. I. =1

(D −abD′)

rg(D, D′)

eax+by =xr

r!

1

g(a, b)eax+by, g(a, b) ≠ 0

Case-2 F(x, y) = sin(ax + by)

P. I. =1

f(D2, DD′, D′2)sin(ax + by) =

1

f(−a2, −ab,−b2)sin(ax + by)

Where, f(−a2, −ab,−b2) ≠ 0

If f(−a2, −ab, −b2) = 0, then use general method for finding P.I.

Case-3 F(x, y) = cos(ax + by)

P. I. =1

f(D2, DD′, D′2)cos(ax + by) =

1

f(−a2, −ab,−b2)cos(ax + by)

Where, f(−a2, −ab,−b2) ≠ 0

If f(−a2, −ab, −b2) = 0, then use general method for finding P.I.



Case-4 F(x, y) = xmyn

P. I. =1

f(D, D′)xmyn = [f(D, D′)]−1xmyn

Expand [f(D, D′)]−1 by using binomial expansion according to the following rules:

o If n < m, expand in power of D′

D.

o If m < n, expand in power of D

D′.

Case-5 f(x, y) = eax+by V(x, y)

P. I. =1

f(D, D′)eax+by V(x, y) = eax+by

1

f(D + a, D′ + b) V(x, y)

METHOD – 3: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE

C 1 Solve

∂2z

∂x2= z .

𝐀𝐧𝐬𝐰𝐞𝐫: z(x, y) = f(y)ex + g(y)e−x W – 14

H 2 Solve

∂2z

∂x2+ z = 0 , given that when x = 0, z = ey and

∂z

∂x= 1.

𝐀𝐧𝐬𝐰𝐞𝐫: z = ey cos x + sin x

C 3 Solve

∂3z

∂x3− 3

∂3z

∂x2 ∂y+ 2

∂3z

∂y3= 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 + 𝐱) + 𝛟𝟐[𝐲 + (𝟏 + √𝟑)𝐱] + 𝛟𝟑[𝐲 + (𝟏 − √𝟑)𝐱]

W – 14

H 4 Solve

∂2z

∂x2−

∂2z

∂x∂y− 6

∂2z

∂y2= 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 + 𝟑𝐱)

H 5 Solve

∂3z

∂x3− 2

∂3z

∂x2 ∂y= 2e2x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲) + 𝐱𝛟𝟐(𝐲) + 𝛟𝟑(𝐲 + 𝟐𝐱) +𝟏

𝟒𝐞𝟐𝐱

W – 16



C 6 Find complete solution𝜕3𝑧

𝜕𝑥3− 3

𝜕3𝑧

𝜕2𝑥𝜕𝑦+ 4

𝜕3𝑧

𝜕𝑦3= 𝑒𝑥+2𝑦

𝐀𝐧𝐬𝐰𝐞𝐫:𝛟𝟏(𝒚 − 𝒙) + 𝛟𝟐(𝒚 + 𝟐𝒙) + 𝒙𝛟𝟑(𝒚 + 𝟐𝒙) +𝒆𝒙+𝟐𝒚

11

W – 17

H 7 Solve 𝜕2𝑧

𝜕𝑥2− 4

𝜕2𝑧

𝜕 𝑥𝜕𝑦+ 4

𝜕2𝑧

𝜕𝑦2= 𝑒2𝑥+3𝑦

𝐀𝐧𝐬𝐰𝐞𝐫: 𝛟𝟏(𝒚 + 𝟐𝒙) + 𝒙𝛟𝟐(𝒚 + 𝟐𝒙) +𝒆𝟐𝒙+𝟑𝒚

16

S – 18

H 8 Solve (D2 − 3DD′ + 2D′2)z = cos(x + 2y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 − 𝐱) −𝟏

𝟑𝐜𝐨𝐬(𝐱 + 𝟐𝐲)

C 9 Solve

𝜕3𝑧

𝜕2𝑥𝜕𝑦= cos(2𝑥 + 3𝑦).

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = −𝟏

𝟏𝟐𝐬𝐢𝐧(𝟐𝐱 + 𝟑𝐲) + 𝐱𝐅(𝐲) + 𝐆(𝐲) + 𝛟(𝐱)

S – 18

H 10 Solve

∂2z

∂x2−

∂2z

∂x ∂y− 6

∂2z

∂y2= sin(2x + y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 + 𝟑𝐱) +𝟏

𝟒𝐬𝐢𝐧(𝟐𝐱 + 𝐲)

C 11 Solve

∂2z

∂x2+ 3

∂2z

∂x∂y+ 2

∂2z

∂y2= x + y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 − 𝐱) −𝐱𝟑

𝟑+𝐱𝟐𝐲

𝟐

S – 15 W – 17

LAGRANGE’S DIFFERENTIAL EQUATION:

A partial differential equation of the form Pp + Qq = R where P, Q and R are functions of

x, y, z, or constant is called lagrange linear equation of the first order.

METHOD FOR OBTAINING GENERAL SOLUTION OF 𝐏𝐩 + 𝐐𝐪 = 𝐑:

Step-1: From the A.E. dx

P=

dy

Q=

dz

R.

Step-2: Solve this A.E. by the method of grouping or by the method of multiples or both to

get two independent solution u(x, y, z) = c1 and v(x, y, z) = c2.



Step-3: The form F(u, v) = 0 or u = f(v) & v = f(u) is the general solution Pp + Qq = R .

FOLLOWING TWO METHODS WILL BE USED TO SOLVE LANGRAGE’S LINEAR EQUATION

Grouping Method

o This method is applicable only if the third variable z is absent in dx

P=

dy

Q or it is

possible to eliminate z from dx

P=

dy

Q.

o Similarly, if the variable x is absent in last two fractions or it is possible to eliminate

x from last two fractions dy

Q=

dz

R, then we can apply grouping method.

Multipliers Method

o In this method, we require two sets of multiplier l,m, n and l′, m′ , n′ .

o By appropriate selection multiplier l,m, n (either constants or functions of x, y, z)

we may write

dx

P=dy

Q=dz

R=ldx + mdy + ndz

lP + mQ + nRSuch that, lP + mQ + nR = 0 .

o This implies ldx + mdy + ndz = 0

o Solving it we get u(x, y, z) = c1 … (1)

o Again we may find another set of multipliers l′, m′, n′ . So that, l′P + m′Q + n′R = 0

o This gives, l′dx + m′dy + n′dz = 0

o Solving it we get v(x, y, z) = c2 … (2)

o From (1) and (2), we get the general solution as F(u, v) = 0.

METHOD – 4: EXAMPLE ON LAGRANGE’S DIFFERENTIAL EQUATION

H 1 Solve x2p + y2q = z2 .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝟏

𝐲−𝟏

𝐱,𝟏

𝐳−𝟏

𝐲) = 𝟎

S – 16

H 2 Solve y2p − xyq = x(z − 2y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱𝟐 + 𝐲𝟐, 𝐲𝐳 − 𝐲𝟐) = 𝟎 S – 17



H 3 Solve xp + yq = 3z.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱

𝐲,𝐲𝟑

𝐳) = 𝟎

S – 18

H 4 Find the general solution to the P.D.E. xp + yq = x − y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱

𝐲, 𝐱 − 𝐲 − 𝐳) = 𝟎 W – 15

C 5 Solve (z − y)p + (x − z)q = y − x .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱 + 𝐲 + 𝐳, 𝐱𝟐+𝐲𝟐 + 𝐳𝟐) = 𝟎 S – 15

H 6 Solve x(y − z)p + y(z − x)q = z(x − y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱 + 𝐲 + 𝐳, 𝐱𝐲𝐳) = 𝟎

C 7 Solve (x2 − y2 − z2)p + 2xyq = 2xz .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐲

𝐳,𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐

𝐳) = 𝟎

W – 16

T 8 Solve (y + z)p + (x + z)q = x + y .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱 − 𝐲

𝐲 − 𝐳, (𝐱 + 𝐲 + 𝐳)(𝐱 − 𝐲)𝟐) = 𝟎

T 9 Solve x2(y − z)p + y2(z − x)q = z2(x − y) .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱𝐲𝐳,𝟏

𝐱+𝟏

𝐲+𝟏

𝐳) = 𝟎

C 10 Solve (x2 − yz)p + (y2 − zx)q = z2 − xy .

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱 − 𝐲

𝐲 − 𝐳,𝐲 − 𝐳

𝐳 − 𝐱) = 𝟎 W – 17

NON LINEAR PARTIAL DIFFERENTIAL EQUATION OF FIRST ORDER:

A partial differential equation in which p & q occur in more than one order is known as Non

Linear Partial Differential Equation.

Type 1: Equation Of the form f(p, q) = 0.

o Step 1: Substitute p = a & q = b.

o Step 2: Convert b = g(a).

o Step 3: Complete Solution : z = ax + by + c ⟹ z = ax + g(a)y + c

Type 2: Equation Of the form f(x, p) = g(y, q).

o Step 1: f(x, p) = g(y, q) = a



o Step 2: Solving equations for p & q. Assume p = F(x) & q = G(y).

o Step 3: Complete Solution : z = ∫ F(x) dx + ∫G(y) dy + b.

Type 3: Equation Of the form z = px + qy + f(p, q) (Clairaut’s form. ) W-15

o Step 1: Substitute p = a & q = b.

o Step 2: Complete Solution : z = ax + by + f(a, b).

Type 4: Equation Of the form f(z, p, q) = 0.

o Step 1: Assume q = ap

o Step 2: Solve the Equation in dz = p dx + q dy

CHARPIT’S METHOD:

Consider, f(x, y, z, p, q) = 0.

o Step 1: Find value of p & q by using the relation

dx

∂f∂p

=dy

∂f∂q

=dz

p∂f∂p

+ q∂f∂q

=dp

−(∂f∂x

+ p∂f∂z)=

dq

−(∂f∂y

+ q∂f∂z)( lagrange − Charpit eqn )

o Step 2: Find value of p & q.

o Step 3: Complete Solution : z = ∫p dx + ∫q dy + c.

METHOD – 5: EXAMPLE ON NON-LINEAR PDE

C 1 Solve p2 + q2 = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax ± (√1 − a2) y + c S – 18

H 2 Solve √p + √q = 1 .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax + (1 − √a)2y + c

W – 14

H 3 Find the complete integral of q = pq + p2 .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax +a2

1 − ay + c ; a ≠ 1



C 4 Solve p2+q2 = npq .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax +na ± a√n2 − 4

2y + c

T 5 Find the complete integral of p2 = q + x .

𝐀𝐧𝐬𝐰𝐞𝐫: z =2

3(x + a)

32 + ay + b

C 6 Solve p2 + q2 = x + y .


3(a + x)

32 +

2

3(y − a)

32 + b

T 7 Solve p2 − q2 = x − y .


3(a + x)

32 +

2

3(a + y)

32 + b

W – 14 S – 17 S – 18

H 8 Solve p − x2 = q + y2 .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax +x3

3+ ay −

y3

3+ b

S – 15

C 9 Solve z = px + qy + p2q2 .

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax + by + a2b2

H 10 Solve qz = p(1 + q) .

𝐀𝐧𝐬𝐰𝐞𝐫: log(az − 1) = x + ay + b

C 11 Solve pq = 4z .

𝐀𝐧𝐬𝐰𝐞𝐫: az = (x + ay + b)2

T 12 Using Charpit’s method solve z = px + qy + p2 + q2.

𝐀𝐧𝐬𝐰𝐞𝐫: z = ax + by + a2 + b2

C 13 Solve py = 2 yx + log q .

𝐀𝐧𝐬𝐰𝐞𝐫: az = ax2 + a2x + eay + ab

H 14 Solve by charpit’s method px + qy = pq .

𝐀𝐧𝐬𝐰𝐞𝐫: 2az = (ax + y)2 + b S – 16

H 15 Solve by charpit’s method p = (z + qy)2.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳𝐲 = 𝐚𝐱 + 𝐳√𝐚√𝐲 + 𝐛 S – 18

METHOD OF SEPARATION OF VARIABLES:

Step 1: Let u(x, y) = X(x) ∙ Y(y)



Step 2: Find ∂u

∂x,∂u

∂y,∂2u

∂x2,∂2u

∂x∂y,∂2u

∂y2 as requirement and substitute in given Partial Differential

Eqn.

Step 3: Convert it into Separable Variable equation and equate with constant say k

individually.

Step 4: Solve each Ordinary Differential Equation.

Step 5: Put value of X(x) & Y(y) in equation u(x, y) = X(x) ∙ Y(y).

METHOD – 6: EXAMPLE ON SEPARATION OF VARIABLES

H 1 Solve the equation by method of separation of variables ∂u

∂x= 4

∂u

∂y ,

where u(0, y) = 8e−3y.

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = 8 e−12x−3y

C 2 Solve ∂u

∂x= 2

∂u

∂t+ u subject to the condition u(x, 0) = 6e−3x

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, t) = 6 e−3x−2t

S – 15

S – 17

H 3 Solve the equation ux = 2ut + u given u(x, 0) = 4e−4x by the method of

separation of variable.

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, t) = 4e−4x−52t

S – 16

C 4 Using method of separation of variables solve ∂u

∂x+

∂u

∂y= 2(x + y)u.

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = c1c2ex2+y2+kx−ky

H 5 Solve x∂u

∂x− 2y

∂u

∂y= 0 using method of separation variables.

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = c1 c2 xk y

k2

W – 16

H 6 Using the method of separation variables, solve the partial differential

equation uxx = 16uy .

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = (c1e√kx + c2e

−√kx) c3 eky16

H 7 Using method of separation of variables solve ∂2u

∂x2=

∂u

∂y+ 2u.


−√kx) c3 e(k−2)y



C 8 Solve two dimensional Laplace’s equation ∂2u

∂x2+

∂2u

∂y2= 0,using the method

separation of variables.


−√kx) (c3 cos √ky + c4 sin √ky)

H 9 Using the method of separation of variables, solve the partial differential

equation ∂2u

∂x2= 16

∂2u

∂y2.


−√kx) (c3e√k4 y + c4e

−√k4 y)

W – 14

H 10 Solve the method of separation of variables ∂2u

∂x2− 4

∂u

∂x+

∂u

∂y= 0 .

𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = (c1e(2+√4+k)x + c2e

(2−√4+k)x) c3e−ky

C 11 Solve ∂2z

∂x2− 2

∂z

∂x+

∂z

∂y= 0 by the method of separation variable.

𝐀𝐧𝐬𝐰𝐞𝐫: z(x, y) = (c1e(1+√1+k)x+ c2e

(1−√1+k)x) c3e−ky

W – 17

CLASSIFICATION OF SECOND ORDER PARTIAL DIFFERENTIAL EQUATION:

The general form of a non-homogeneous second order P.D.E.

A(x, y)∂2z

∂x2+ B(x, y)

∂2z

∂x ∂y+ C(x, y)

∂2z

∂y2+ f (x, y, z,

∂z

∂x,∂z

∂y) = F(x, y)…… (1)

Equation (1) is said to be

Elliptic, If B2 − 4AC < 0 Parabolic, If B2 − 4AC = 0 Hyperbolic, If B2 − 4AC > 0

METHOD – 7: EXAMPLE ON CLASSIFICATION OF 2ND ORDER PDE

C 1 Classify the 2nd order P.D.E. t∂2u

∂t2+ 3

∂2u

∂x∂t+ x

∂2u

∂x2+ 17

∂u

∂x= 0.

𝐀𝐧𝐬𝐰𝐞𝐫: Hyperbolic, if 𝐱𝐭 >𝟗

𝟒 ; Parabolic, if 𝐱𝐭 =

𝟗

𝟒 ; Elliptical, if 𝐱𝐭 <

𝟗

𝟒

H 2 Classify the 2nd order P.D.E. 4∂2u

∂t2− 9

∂2u

∂t∂x+ 5

∂2u

∂x2= 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐇𝐲𝐩𝐞𝐫𝐛𝐨𝐥𝐢𝐜



H 3 Classify the 2nd order P.D.E. ∂2u

∂t2+ 2

∂2u

∂x∂t+ 2

∂2u

∂x2= 0.

𝐀𝐧𝐬𝐰𝐞𝐫: 𝐄𝐥𝐥𝐢𝐩𝐭𝐢𝐜

⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

GUJARAT TECHNOLOGICAL UNIVERSITY ADVANCED ENGINEERING MATHEMATICS

SUBJECT CODE: 2130002

B.E. 3rd

SEMESTER

Type of course: Engineering Mathematics

Prerequisite: The course follows from Calculus, Linear algebra

Rationale: Mathematics is a language of Science and Engineering

Teaching and Examination Scheme:

Content:

Sr.

No. Topics

Teaching

Hrs.

Module

Weightage

1

Introduction to Some Special Functions: Gamma function, Beta function, Bessel function, Error function and

complementary Error function, Heaviside’s function, pulse unit height

and duration function, Sinusoidal Pulse function, Rectangle function,

Gate function, Dirac’s Delta function, Signum function, Saw tooth wave

function, Triangular wave function, Halfwave rectified sinusoidal

function, Full rectified sine wave, Square wave function.

02

4

2

Fourier Series and Fourier integral: Periodic function, Trigonometric series, Fourier series, Functions of any

period, Even and odd functions, Half-range Expansion, Forced

oscillations, Fourier integral

05

10

3

Ordinary Differential Equations and Applications: First order differential equations: basic concepts, Geometric meaning of

y’ = f(x,y) Direction fields, Exact differential equations, Integrating

factor, Linear differential equations, Bernoulli equations, Modeling ,

Orthogonal trajectories of curves.Linear differential equations of second

and higher order: Homogeneous linear differential equations of second

order, Modeling: Free Oscillations, Euler- Cauchy Equations,

Wronskian, Non homogeneous equations, Solution by undetermined

coefficients, Solution by variation of parameters, Modeling: free

Oscillations resonance and Electric circuits, Higher order linear

differential equations, Higher order homogeneous with constant

coefficient, Higher order non homogeneous equations. Solution by

[1/f(D)] r(x) method for finding particular integral.

11

20

Teaching Scheme Credits Examination Marks Total

Marks L T P C Theory Marks Practical Marks

ESE

(E)

PA (M) PA (V) PA

(I) PA ALA ESE OEP

3 2 0 5 70 20 10 30 0 20 150

4

Series Solution of Differential Equations: Power series method, Theory of power series methods, Frobenius

method.

03

6

5

Laplace Transforms and Applications: Definition of the Laplace transform, Inverse Laplace transform,

Linearity, Shifting theorem, Transforms of derivatives and integrals

Differential equations, Unit step function Second shifting theorem,

09

15

Dirac’s delta function, Differentiation and integration of transforms,

Convolution and integral equations, Partial fraction differential

equations, Systems of differential equations

6

Partial Differential Equations and Applications:

Formation PDEs, Solution of Partial Differential equations f(x,y,z,p,q) = 0, Nonlinear PDEs first order, Some standard forms of nonlinear

PDE, Linear PDEs with constant coefficients,Equations reducible to

Homogeneous linear form, Classification of second order linear

PDEs.Separation of variables use of Fourier series, D’Alembert’s

solution of the wave equation,Heat equation: Solution by Fourier series

and Fourier integral

12

15

Reference Books:

1. Advanced Engineering Mathematics (8th Edition), by E. Kreyszig, Wiley-India (2007). 2. Engineering Mathematics Vol 2, by Baburam, Pearson

3. W. E. Boyce and R. DiPrima, Elementary Differential Equations (8th Edition), John Wiley (2005)

4. R. V. Churchill and J. W. Brown, Fourier series and boundary value problems (7th Edition),

McGraw-Hill (2006).

5. T.M.Apostol, Calculus , Volume-2 ( 2nd Edition ), Wiley Eastern , 1980

Course Outcome:

After learning the course the students should be able to

1. Fourier Series and Fourier Integral

o Identify functions that are periodic. Determine their periods.

o Find the Fourier series for a function defined on a closed interval.

o Find the Fourier series for a periodic function.

o Recall and apply the convergence theorem for Fourier series.

o Determine whether a given function is even, odd or neither.

o Sketch the even and odd extensions of a function defined on the interval [0,L].

o Find the Fourier sine and cosine series for the function defined on [0,L]

2. Ordinary Differential Equations and Their Applications

o Model physical processes using differential equations.

o Solve basic initial value problems, obtain explicit solutions if possible.

o Characterize the solutions of a differential equation with respect to initial values.

o Use the solution of an initial value problem to answer questions about a physical system. o Determine the order of an ordinary differential equation. Classify an ordinary differential equation as

linear or nonlinear.

o Verify solutions to ordinary differential equations.

o Identify and solve first order linear equations.

o Analyze the behavior of solutions.

o Analyze the models to answer questions about the physical system modeled.

o Recall and apply the existence and uniqueness theorem for first order linear differential equations.

o Identify whether or not a differential equation is exact.

o Use integrating factors to convert a differential equation to an exact equation and then solve. o Solve second order linear differential equations with constant coefficients that have a characteristic equation

with real and distinct roots.

o Describe the behavior of solutions. o Recall and verify the principal of superposition for solutions of second order linear differential

equations. o Evaluate the Wronskian of two functions.

GUJARAT TECHNOLOGICAL UNIVERSITY B E - S E M E S T E R - I I I • EXAMINATION - W I N T E R • 2014

Subject Code: 2130002 Date: 06-01-2015 Subject Name: Advanced Engineering Mathematics Time: 02.30 pm - 05.30 pm Total Marks: 70 Instructions:

1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Q . l (a) Define the terms: (1) Unit Step Function (2) Dirac Delta Function 04 (b) Form partial differential equation by eliminating arbitrary function from 03

F ( . Y 2 - / , x v z ) = 0

( c ) Find the series solution of (1 - x 2 ) y - 2xv + 2y = 0 .

, (n-xV Q.2 (a) Obtain the Fourier Series o f / ( x ) = in the interval 0 < x < In. Hence 07

V 2 , ^r- 1 1 1

deduce that — = — — r + - r • • 12 1- 2 3

(b) Find the Fourier Series of / (x ) = x + |x| in the interval - n < x < n . 07

O R (b) Obtain Half range cosine series of / ( - * ) = s inx in the interval 0 < x < n . Hence 07

, ^r2 1 1 1 deduce that — = — — r + — .

12 l2 2 3

Q.3 (a) ( I ) Define Laplace Transform. Prove that I ( e " " ) = -s + a

s > -a 03

(2) Find the Laplace Transform of (i) e 2 ,(sin 4/ + / 2) (ii) 04

(b) Find the inverse Laplace Transform of (i) — (ii) — j 07 (5 + 1 ) ( J -1 ) " ( , r + l)

O R Q.3 (a) Do as Directed.

( 1 ) F i n d l ( / 2 » ( / - 2 ) )

(2) Find the Laplace Transform of (i) r cosher (ii) sin2 3/ (b) Solve the initial value problem by using Laplace Transforms: 07

y +y = s i n2 / , v(0) = 2, >' (0) = 1

03

04

03

04

Q- 4 (a) (1) Solve: (x 1 + 3x 3 ' 2 )dx + (3x2 v + / ) dy = 0

(2) Solve: x2ydx-(x3 + xy'^dy = 0

(b) Solve: v" + 2 y + 3y = 2x2 07 OR

Q.4 (a) ( l ) Solve the initial value problem y -4y +4y = 0 >'(0) = 3, y (0 ) = 1

(2) Solve: (x2y2 + 2)ydx+(l-x2y2)xdy = 0

(b) y"'- 3 v"+ 3 v v = 4e'

Q.5 (a) (i) Solve: J p + 4 q = l

d2z (ii) Solve:

a r (b) Using the method of separation of variables, solve the partial differential 07

d'u ,, d'u equation — - - 1 6 — -

dx~ ch'' O R

Q.5 (a) (i) Solve: p2-q2=x-y 0 , a 3 z „ d'z „ d*z n (n) Solve: - - 3 — + 2 — r = 0

dx* d2xdy d\>}

(b) Find the Fourier Integral representation of f ( x ) (l if |x| < 1

| 0 if l.vl > 1

03

04

07

kkkkkkkkkkkki

1

1

Seat No.: ________ Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER– III (NEW)EXAMINATION – SUMMER 2015

Subject Code:2130002 Date: 06/06/2015

Subject Name:Advanced Engineering Mathematics

Time: 02.30pm-05.30pm Total Marks: 70 Instructions:

1. Attempt all questions.

2. Make suitable assumptions wherever necessary.

3. Figures to the right indicate full marks.

Q.1

(a) (1) Solve the differential equation2

1

x

e

xdx

dy y

. 04

(2) Solve the differential equation . 0)2( dyeydxye xx

03

(b)

Find the series solution of 09'")1( 2 yxyyx . 07

Q.2 (a) (1)Solve the differential equation using the method of variation of parameter

xyy 3sec 9 . 04

(2) Solve the differential equationxeyDD 10)12( 2 . 03

(b) Using the method of separation of variables, solve

xexuut

u

x

u 36)0,(;2

. 07

OR

(b) Find the series solution of 0;0)1()1(2 0 xyyxyxx 07

Q.3 (a) Find the Fourier Series for

xx

xxxf

0;

0;)( 07

(b) (1) Find the Half range Cosine Series for .10;)1()( 2 xxxf 04

(2) Find the Fourier sine series for .0;)( xexf x 03

OR

Q.3 (a) Find the Fourier Series for

xx

xxf

0;

0;)( . 07

(b) (1) Find the Fourier cosine series for .0;)( 2 xxxf 04

(2) Find the Fourier sine series for 10;2)( xxxf . 03

Q.4 (a) (1) Prove that (i) as

aseL at

;

1)( (ii)

22)(sinh

as

aatL

. 04

(2) Find the Laplace transform of tt 2sin . 03

(b) (1) Using convolution theorem, obtain the value of

4

12

1

ssL . 04

(2) Find the inverse Laplace transform of

32

1

ss. 03

OR

Q.4 (a) Solve the initial value problem using Laplace transform:

00,10,23 yyeyyy t.

07

(b) (1) Find the Laplace transform of

tt

ttf

;sin

0;0. 04

2

(2) Evaluatetet . 03

Q.5 (a) Using Fourier integral representation prove that

0

2

0

02

00

1

sincos

xife

xif

xif

dxx

x

. 07

(b) (1) Form the partial differential equation by eliminating the arbitrary functions from

0, 222 zyxzyxf . 04

(2) Solve the following partial differential equation xyqzxpyz .

03

OR

Q.5 (a) A homogeneous rod of conducting material of length 100 cm has its ends kept

at zero temperature and the temperature initially is

10050;100

500;)0,(

xx

xxxu

Find the temperature ),( txu at any time.

07

(b) (1) Solve .23

2

22

2

2

yxy

z

yx

z

x

z

04

(2) Solve 22 yqxp . 03

*************

1


GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (New) EXAMINATION – WINTER 2015

Subject Code:2130002 Date:31/12/2015

Subject Name: Advanced Engineering Mathematics

Time: 2:30pm to 5:30pm Total Marks: 70 Instructions:




Q.1 Answer the following one mark each questions: 14

1 Find Γ (13

2)

2 State relationship between beta and gamma functions.

3 Represent graphically the given saw-tooth function 𝑓(𝑥) = 2𝑥, 0 ≤𝑥 < 2 and 𝑓(𝑥 + 2) = 𝑓(𝑥) for all 𝑥.

4 For a periodic function 𝑓 with fundamental period p, state the formula to

find Laplace transform of 𝑓.

5 Find 𝐿(𝑒−3𝑡𝑓(𝑡)), if 𝐿(𝑓(𝑡)) =𝑠

(𝑠−3)2.

6 Find 𝐿[(2𝑡 − 1)2].

7 Find the extension of the function 𝑓(𝑥) = 𝑥 + 1, define over (0,1] to [−1, 1] − {0} which is an odd function.

8 Is the function 𝑓(𝑥) = {

𝑥, 0 ≤ 𝑥 ≤ 2

𝑥2, 2 < 𝑥 ≤ 4; continuous on [0,4]? Give

reason.

9 Is the differential equation 𝑑𝑦

𝑑𝑥=

𝑦

𝑥 exact? Give reason.

10 Give the differential equation of the orthogonal trajectory to the equation

𝑦 = 𝑐𝑥2.

11 If 𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2 = 𝑒𝑥(𝑐1 cos 𝑥 + 𝑐2 sin 𝑥) is a complementary

function of a second order differential equation, find the Wronskian

𝑊(𝑦1, 𝑦2).

12 Solve (𝐷2 + 𝐷 + 1)𝑦 = 0; where 𝐷 =𝑑

𝑑𝑡.

13 Is 𝑢(𝑡, 𝑥) = 50𝑒(𝑡−𝑥) 2⁄ , a solution to 𝜕𝑢

𝜕𝑡=

𝜕𝑢

𝜕𝑥+ 𝑢?

14 Give an example of a first order partial differential equation of Clairaut’s

form.

Q.2 (a) Solve: 𝑑𝑦

𝑑𝑥=

𝑥2−𝑥−𝑦2

2𝑥𝑦. 03

2

(b) Solve: 𝑑𝑦

𝑑𝑥+

1

𝑥𝑦 = 𝑥3𝑦3. 04

(c) Find the series solution of (𝑥 − 2)𝑑2𝑦

𝑑𝑥2− 𝑥2 𝑑𝑦

𝑑𝑥+ 9𝑦 = 0 about 𝑥0 = 0. 07

OR

(c) Explain regular-singular point of a second order differential equation and

find the roots of the indicial equation to 𝑥2𝑦′′ + 𝑥𝑦′ − (2 − 𝑥)𝑦 = 0.

07

Q.3 (a) Find the complete solution of 𝑑3𝑦

𝑑𝑥3 + 8𝑦 = cosh(2𝑥). 03

(b) Find solution of 𝑑2𝑦

𝑑𝑥2 + 9𝑦 = tan 3𝑥, using the method of variation of

parameters.

04

(c) Using separable variable technique find the acceptable general solution to

the one-dimensional heat equation 𝜕𝑢

𝜕𝑡= 𝑐2

𝜕2𝑢

𝜕𝑥2 and find the solution

satisfying the conditions 𝑢(0, 𝑡) = 𝑢(𝜋, 𝑡) = 0 for 𝑡 > 0 and 𝑢(𝑥, 0) =𝜋 − 𝑥, 0 < 𝑥 < 𝜋.

07

OR

Q.3 (a) Solve completely, the differential equation 𝑑2𝑦

𝑑𝑥2 − 6𝑑𝑦

𝑑𝑥+ 9𝑦 = cos(2𝑥) sin 𝑥.

03

(b) Solve completely the differential equation

𝑥2 𝑑2𝑦

𝑑𝑥2 − 6𝑥𝑑𝑦

𝑑𝑥+ 6𝑦 = 𝑥−3 log 𝑥.

04

(c) (i) Form the partial differential equation for the equation (𝑥 − 𝑎)(𝑦 −𝑏) − 𝑧2 = 𝑥2 + 𝑦2.

(ii) Find the general solution to the partial differential equation 𝑥𝑝 +𝑦𝑞 = 𝑥 − 𝑦.

07

Q.4 (a) Find the Fourier cosine integral of 𝑓(𝑥) =𝜋

2𝑒−𝑥, 𝑥 ≥ 0. 03

(b) For the function 𝑓(𝑥) = cos 2𝑥, find its Fourier sine series over [0, 𝜋]. 04

(c) For the function 𝑓(𝑥) = {

𝑥; 0 ≤ 𝑥 ≤ 2 4 − 𝑥; 2 ≤ 𝑥 ≤ 4

, find its Fourier series.

Hence show that 1

12 +1

32 +1

52 + ⋯ =𝜋2

16.

07

OR

Q.4 (a) Find the Fourier cosine series of 𝑓(𝑥) = 𝑒−𝑥, where 0 ≤ 𝑥 ≤ 𝜋. 03

(b) Show that ∫𝜆3 sin 𝜆𝑥

𝜆4+4

∞

0𝑑𝜆 =

𝜋

2𝑒−𝑥 cos 𝑥, 𝑥 > 0. 04

(c) Is the function 𝑓(𝑥) = 𝑥 + |𝑥|, -𝜋 ≤ 𝑥 ≤ 𝜋 even or odd? Find its Fourier

series over the interval mentioned.

07

Q.5 (a) Find 𝐿 {∫ 𝑒𝑢(𝑢 + sin 𝑢)𝑑𝑢𝑡

0}. 03

(b) Find 𝐿−1 {1

𝑠(𝑠2−3𝑠+3)}. 04

(c) Solve the initial value problem: 𝑦′′ − 2𝑦′ = 𝑒𝑡 sin 𝑡, 𝑦(0) = 𝑦′(0) = 0,

using Laplace transform.

07

OR

Q.5 (a) Find 𝐿{𝑡(sin 𝑡 − 𝑡 cos 𝑡 )}. 03

(b) Find 𝐿−1 {𝑒−2𝑠

(𝑠2+2)(𝑠2−3)}. 04

(c) State the convolution theorem and verify it for 𝑓(𝑡) = 𝑡 and 𝑔(𝑡) = 𝑒2𝑡. 07

**********

1


GUJARAT TECHNOLOGICAL UNIVERSITY

BE - SEMESTER–III(New) EXAMINATION – SUMMER 2016


Subject Name:Advanced Engineering Mathematics Time:10:30 AM to 01:30 PM Total Marks: 70 Instructions:


2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Q.1 Answer the following one mark each questions : 14

1 Integreating factor of the differential equation

𝑑𝑥

𝑑𝑦+

3𝑥

𝑦=

1

𝑦2 is _________

Type equation here.

2 The general solution of the differential equation 𝑑𝑦

𝑑𝑥+

𝑦

𝑥

=tan2x____________.

3 The orthogonal trajectory of the family of curve 𝑥2 +𝑦2 = 𝑐2 is _________

4 Particular integral of (𝐷2 + 4)𝑦 = cos 2𝑥 is _________

5 X=0 is a regular singular point of 2𝑥2𝑦′′ + 3𝑥𝑦′(𝑥2 − 4)𝑦 = 0 say true or false.

6 The solution of

(𝑦 − 𝑧)𝑝 + (𝑧 − 𝑥)𝑞 = 𝑥 − 𝑦 𝑖𝑠_________

7 State the type ,order and degree of differential equation

(𝑑𝑥

𝑑𝑦)2 + 5𝑦

1

3 = x is __________

8 Solve (D+𝐷′)z= cos x

9 Is the partial differential equation

2𝜕2𝑢

𝜕𝑥2+4𝜕2𝑢

𝜕𝑥𝜕𝑦+ 3

𝜕2𝑢

𝜕𝑦2 = 6 elliptic?

10 𝐿−1 (

1

(𝑠 + 𝑎)2) = ____________

11 If f(t) is a periodic function with period t then

L[𝑓(𝑡)] = _________

12 Laplace transform of f(t) is defined for +ve and –ve

values of t. Say true or false.

13 State Duplication (Legendre) formula.

14 Find B ( 9 ,

2

7

2 )

Q.2 (a) Solve : 9y 𝑦 ˊ + 4𝑥 = 0 03

2

(b) Solve : 𝑑𝑦

𝑑𝑥+ 𝑦 cot 𝑥 = 2 cos 𝑥 04

(c) Find series solution of 𝑦 ˊˊ + 𝑥𝑦 = 0 07

OR

(c) Determine the value of (a) J1

2 (𝑥) (𝑏) J3

2 (𝑥) 07

Q.3 (a) Solve (𝐷2 + 9)𝑦 = 2sin 3𝑥 + cos 3𝑥 03

(b) Solve 𝑦′′ + 4𝑦′ = 8𝑥2 by the method of

undetermined coefficients.

04

(c) (i) Solve 𝑥2𝑝 + 𝑦2𝑞 = 𝑧2

(ii) Solve by charpit’s method px+qy = pq

07

OR

Q.3 (a) Solve 𝑦′′ + 4𝑦′ + 4 = 0 , 𝑦(0) = 1 , 𝑦′(0) = 1 03

(b) Find the solution of 𝑦′′ + 𝑎2𝑦′ = tan 𝑎𝑥 , by the

method of variation of parameters.

04

(c) Solve the equation ux = 2ut + u given u(x,0)= 4𝑒−4𝑥 by

the method of separation of variable. 07

Q.4 (a) Find the fourier transform of the function f(x) = 𝑒−𝑎𝑥2

03

(b) Obtain fourier series to represent f(x) =x2 in the interval

-𝜋 < 𝑥 < 𝜋 .Deduce that ∑1

𝑛2∞𝑛=1 =

𝜋2

6

04

(c) Find Half-Range cosine series for

F(x) = kx , 0≤ 𝑥 ≤𝑙

2

= k(𝑙-x ) , 𝑙

2 ≤ 𝑥 ≤ 𝑙

Also prove that ∑1

(2𝑛−1)2∞𝑛=1 =

𝜋2

8

07

OR

Q.4 (a) Expres the function

F(x)= 2 , |x| < 2

= 0 , |x| > 2 as Fourier integral.

03

(b) Find the fourier series expansion of the function

F(x) = -𝜋 − 𝜋 < 𝑥 < 0

= x 0 < 𝑥 < 𝜋

04

(c) Find fourier series to represent the function

F(x) = 2x-x2 in 0 < 𝑥 < 3

07

Q.5 (a) 𝐹𝑖𝑛𝑑 𝐿−1 {1

(𝑠+√2)(𝑠−√3)}

03

(b) Find the laplace transform of

(i) 𝑐𝑜𝑠𝑎𝑡−𝑐𝑜𝑠𝑏𝑡

𝑡

(ii) tsinat

04

(c) State convolution theorem and use to it evaluate

𝐿−1 {1

(𝑠2+𝑎2)2}

07

3

OR

Q.5 (a) L {𝑡2 cos ℎ3𝑡} 03

(b) Find 𝐿−1 {1

𝑠4−81} 04

(c) Solve the equation 𝑦′′ − 3𝑦′ + 2𝑦 = 4𝑡 + 𝑒3𝑡,when

y(0)=1 , 𝑦′(0) = −1

07

*************

1


GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III(New) • EXAMINATION – WINTER 2016


Subject Name:Advanced Engineering Mathematics

Time: 10:30 AM to 01:30 PM Total Marks: 70 Instructions:




MARKS

Q.1 Answer the following one mark questions 14

1 Find ⌈(1

2).

2 State relation between beta and gamma function.

3 Define Heaviside’s unit step function.

4 Define Laplace transform of f (t), t ≥ 0.

5 Find Laplace transform of 𝑡−1

2 .

6 Find L { 𝑠𝑖𝑛𝑎𝑡

𝑡 }, given that L {

𝑠𝑖𝑛𝑡

𝑡 } = 𝑡𝑎𝑛−1{

1

𝑠 }.

7 Find the continuous extension of the function f(x) = 𝑥2+𝑥−2

𝑥2−1 to x = 1

8 Is the function f(x) = 1

𝑥 continuous on [-1, 1] ? Give reason.

9 Solve 𝑑𝑦

𝑑𝑥 = 𝑒3𝑥−2𝑦 + 𝑥2𝑒−2𝑦.

10 Give the differential equation of the orthogonal trajectory of the

family of circles 𝑥2 + 𝑦2 = 𝑎2.

11 Find the Wronskian of the two function sin2x and cos2x .

12 Solve ( 𝐷2 + 6𝐷 + 9) x = 0; D = 𝑑

𝑑𝑡.

13 To solve heat equation 𝜕𝑢

𝜕𝑡= 𝑐2 𝜕2𝑢

𝜕𝑥2 how many initial and

boundary conditions are required.

14 Form the partial differential equations from z = f(x + at) + g(x –at).

Q.2 (a) Solve : (x+1)𝑑𝑦

𝑑𝑥− 𝑦 = 𝑒3𝑥(𝑥 + 1)2. 03

(b) Solve : 𝑑𝑦

𝑑𝑥+

𝑦𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑦 +𝑦

𝑠𝑖𝑛𝑥 + 𝑥𝑐𝑜𝑠𝑦 + 𝑥= 0 04

(c) Find the series solution of 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑦 = 0. 07

OR

(c) Find the general solution of 2𝑥2𝑦′′ + 𝑥𝑦′ + (𝑥2 − 1)𝑦 = 0 by using frobenius method.

07

Q.3 (a) Solve : (𝐷3 − 3𝐷2 + 9𝐷 − 27)y = cos3x. 03

(b) Solve : 𝑥2 𝑑2𝑦

𝑑𝑥2 + 4𝑥𝑑𝑦

𝑑𝑥+ 2𝑦 = 𝑥2 sin(𝑙𝑛𝑥). 04

(c) (i) Solve : 𝜕3𝑧

𝜕𝑥3 − 2𝜕3𝑧

𝜕𝑥2𝜕𝑦= 2𝑒2𝑥.

(ii) find the general solution to the partial differential equation

03

04

2

(𝑥2 − 𝑦2 − 𝑧2)𝑝 + 2𝑥𝑦𝑞 = 2𝑥𝑧.

OR

Q.3 (a) Solve : (𝐷3 − 𝐷)𝑦 = 𝑥3 . 03

(b) Find the solution of 𝑦′′ − 3𝑦′ + 2y = 𝑒𝑥 , using the method of


04

(c) Solve 𝑥𝜕𝑢

𝜕𝑥 - 2y

𝜕𝑢

𝜕𝑦= 0 using method of separation of variables. 07

Q.4 (a) Find the Fourier cosine integral of 𝑓(𝑥) = 𝑒−𝑘𝑥, 𝑥 > 0, 𝑘 > 0 03

(b) Express f(x) = |x|, −𝜋 < 𝑥 < 𝜋 as fouries series. 04

(c) Find Fourier Series for the function f(x) given by

𝑓(𝑥) = {1 +

2𝑥

𝜋; −𝜋 ≤ 𝑥 ≤ 0

1 −2𝑥

𝜋 ; 0 ≤ 𝑥 ≤ 𝜋

Hence deduce that 1

12 + 1

32 +1

52 + … . =𝜋2

8 .

07

OR

Q.4 (a) Obtain the Fourier Series of periodic function function

f(x) =2x, -1 < x < 1, p = 2L =2

03

(b) Show that ∫𝑠𝑖𝑛𝜆𝑐𝑜𝑠𝜆

𝜆

∞

0𝑑𝜆 = 0, if x ˃ 1. 04

(c) Expand f(x) in Fourier series in the interval (0, 2𝜋) if

𝑓(𝑥) = {−𝜋 ; 0 < 𝑥 < 𝜋

𝑥 − 𝜋 ; 𝜋 < 𝑥 < 2𝜋

and hence show that ∑1

(2𝑟+1)2 =𝜋2

8.∞

𝑟=0

07

Q.5 (a) Find L{∫ 𝑒𝑡𝑡

0

𝑠𝑖𝑛𝑡

𝑡𝑑𝑡}. 03

(b) Find 𝐿−1{ 2𝑠2−1

(𝑠2+1)(𝑠2+4) }. 04

(c) Solve initial value problem : 𝑦′′ − 3𝑦′ + 2y = 4t + 𝑒3𝑡 ,y(0) = 1 and

𝑦′(0) = −1 ,using Laplace transform.

07

OR

Q.5 (a) Find L {tsin3tcos2t}. 03

(b) Find 𝐿−1{ 𝑒−3𝑠

𝑠2+8𝑠+25 }. 04

(c) State the convolution theorem and apply it to evaluate 𝐿−1 { 𝑠

(𝑠2+𝑎2)2 }. 07

*************

1


GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) - EXAMINATION – SUMMER 2017

Subject Code: 2130002 Date: 25/05/2017 Subject Name: Advanced Engineering Mathematics Time: 10:30 AM to 01:30 PM Total Marks: 70 Instructions:


MARKS

Q.1 Short Questions 14

1 What are the order and the degree of the differential

equation xyy cos33"2 .

2 What is the integrating factor of the linear differential

equation: 2)/1(' xyxy

3 Is the differential equation 0)2( dyeydxyexx is

exact? Justify.

4 Solve: 010'11" yyy .

5 Find particular integral of : xeyy

2''"

6 If xexccy )( 21 is a complementary function of a

second order differential equation, find the Wronskian

).,( 21 yyW

7 Find the value of

2

7

8 What is the value of the Fourier coefficients a0 and bn for

.11,)(2

xxxf

9 Find 33 teL

10 Find

)9(

14

22

1

ss

L

11 Find the singular point of the differential equation

0)1('2")1(2

ynnxyyx

12 Obtain the general integral of 0

3

3

x

z

13 Obtain the general integral of zqp

14 State the relationship between beta and gamma function.

Q.2 (a) Solve: 02)3(22

xydydxyx 03

(b) Solve: 0)0(,2sin)(tan yxyx

dx

dy

04

(c) dxdDwherexeyD

x/,)16(

424 07

OR

(c) Use the method of variation of parameters to find the 07

2

general solution of x

eyyy

x2

4'4"

Q.3 (a) Find half range sine series of xxxf 0,)(3 03

(b) Find the Fourier integral representation of the function

2

2

,0

,2)(

x

xxf

04

(c) Find the Fourier series expansion for the 2 - periodic

function 2

)( xxxf in the interval x and show

that 12

...

4

1

3

1

2

1

1

12

2222

07

OR

Q.3 (a) Discuss about ordinary point, singular point, regular

singular point and irregular singular point for the

differential equation: 07')1(3")1(3

xyyxyxx

03

(b) Use the method of undetermined coefficients to solve the

differential equation 229" xyy

04

(c) Find the series solution of 0'")1(2

xyxyyx

about 00 x .

07

Q.4 (a) Solve: dxdDwherexeyDx

/,)1(2

03

(b) Solve: )sin(ln

2

22

xydx

dyx

dx

ydx

04

(c) Use Laplace Transform to solve the following initial

value problem:

6)0(',2)0(,122'3"2

yyeyyyt

07

OR

Q.4 (a) Obtain teLt 22

sin 03

(b) Find

258

7

2

1

ss

sL

04

(c) Using Convolution theorem, obtain

22

1

)4(

1

s

L 07

Q.5 (a) Find the Laplace Transform of tett

2cos4

03

(b) Form the partial differential equation from the following:

1) ctbyaxz 2)

y

xfz

04

(c) Using the method of separation of variables solve,

ut

u

x

u

2 where

xexu

36)0,(

07

OR

Q.5 (a) Obtain the solution of the partial differential equation:

,yxqp 22

where y

zq

x

zp

,

03

(b) Solve: )yz(xxyqpy 2

2 ,where

y

zq,

x

zp

04

(c) Find the solution of the wave equation 07

3

Lxucu xxtt 0,2

satisfying the conditions:

LxL

xxuxutLutu t 0,)0,(,0)0,(,0),(),0(

*************

1


GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) EXAMINATION – WINTER 2017

Subject Code: 2130002 Date:06/11/2017 Subject Name: Advanced Engineering Mathematics Time: 10:30 AM to 01:30 PM Total Marks: 70 Instructions:




Q.1 (a) Solve the following differential equation using variable separable method

0sec1tan3 2 dyyedxye xx

03

(b) Find the Laplace transform of tt 3sin 2 . 04

(c) Given that 2xxxf for , x find the Fourier expression of .xf

Deduce that ............4

1

3

1

2

11

6 222

2

07

Q.2 (a) Define rectangle function and saw-tooth wave function. Also sketch the graphs. 03

(b) Find the general solution of the following differential equation :

xeydx

dy

dx

yd x cos423

3

04

(c) Find the power series solution of 0221 2 yyxyx about the ordinary

point 0x

07

OR

(c) Find the power series solution of 0232

2

ydx

dy

dx

ydx about the point

0x , using Frobenius method.

07

Q.3

(a) Express

xfor

xforxf

,0

0,1

As a Fourier sine integral and hence evaluate

dxsin

cos1

0

03

(b) Check whether the given differential equations is exact or not

0sin422 32424 dyyyxyxdxyyxx

Hence find the general solution.

04

(c) Solve the following differential equation using the method of undetermined

coefficient : xexydx

dy

dx

yd 3242 2

2

2

07

OR

Q.3 (a) Find the cosine series for xxf in the interval x,0 . 03

(b) Solve the following differential equation xydx

ydsin

2

2

using the method of


04

(c) Solve the following Cauchy-Euler equation

xxydx

dyx

dx

ydx logsin.log

2

22

07

2

Q.4 (a) Find the orthogonal trajectory of the cardioids cos1 ar 03

(b) Find the Laplace transforms of :

(i) 23 tue t

(ii) t

t2cos1

04

(c) Solve the equation 022

2

y

z

x

z

x

z by the method of separation of variables.

07

OR

Q.4 (a) Solve the following Bernoulli’s equation:

2

2

x

y

x

y

dx

dy

03

(b) Find the inverse Laplace transforms of :

(i)

s

2tan 1

(ii) 44

3

as

s

04

(c) Find the complete solution of the following partial differential equations:

(i) yxey

z

yx

z

x

z 2

3

3

2

3

3

3

43

(ii) yxy

z

yx

z

x

z

2

22

2

2

23

07

Q.5 (a) Form the partial differential equations by eliminating the arbitrary function

from 0,22 xyzyxf

03

(b) Solve the following Lagrange’s linear differential equation:

xyzqzxypyzx 222

04

(c) Solve the following initial value problem using the method of Laplace

transforms

022 yyyy given that 20,20,10 yyy

07

OR

Q.5 (a) Find the Laplace transform of the periodic function of the waveform

tftftt

tf 3,30,3

2

03

(b) Using the convolution theorem, find

ba

bsas

sL

,2222

21

04

(c) A tightly stretched string of length l with fixed ends is initially in equilibrium

position. It is set vibrating by giving each point a velocity l

xv

3

0 sin .find the

displacement .,txy

07

*************

1


GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) - EXAMINATION – SUMMER 2018

Subject Code:2130002 Date:16/05/2018 Subject Name:Advanced Engineering Mathematics Time:10:30 AM to 01:30 PM Total Marks: 70 Instructions:


Q.1 (a) Solve y2yx exe

dx

dy by variable separable method. 03

(b) Solve xcosexsiny

dx

dy

04

(c)

State convolution theorem and hence find

22

1

4s

1L

07

Q.2 (a) Solve 3" 3 ' 2 xy y y e 03

(b) Find Fourier series for 2x)x(f ; x 04

(c) Find Fourier series in the interval 2,0 if

2xx

x0)x(f

and hence show that 81n2

1 2

1n2

07

OR

(c) Find the series solution of 2" 0y x y about an ordinary point

0x

07

Q.3 (a) Solve ''' 6 '' 11 ' 6 0y y y y 03

(b) Solve x4cosy9D2 04

(c) Solve " 4 4tan 2y y x by method of variation parameter. 07

OR

Q.3 (a) Find

2s1s

1L 1

03

(b) Solve x6x6x5y5'y2''y 23 by method of undetermined

coefficients.

04

(c) Find the series solution of 28 " 10 ' 1 0x y xy x y 07

Q.4 (a) Solve 0dyxydxyx 344 03

(b) Express sinx as cosine series in x0 04

(c) Find Fourier series for axe)x(f in 2,0 ; 0a 07

OR

Q.4 (a) Find tcosL 2 03

2

(b) Find t3sineL t2 04

(c) Solve ;t2siny"y with ,2)0(y 10'y by using Laplace

transform.

07

Q.5 (a) Solve 1qp 22 03

(b) Solve z3yqxp 04

(c) Solve e

y3x2

2

2

2

2

y

z4

yx

z4

x

z

07

OR

Q.5 (a) Solve yxqp 22 03

(b) Solve y3x2cos

yx

z2

3

04

(c) Solve by Charpit’s method 2qyzp 07

*************

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