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ADVANCED ENGINEERING MATHEMATICS 2130002 – 5th Edition
Darshan Institute of Engineering and Technology
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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
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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
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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……..……………..…………..………..……………….***
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » 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
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002
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)
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[ 7 ]
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002
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.
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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
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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).
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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 1 ]
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
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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 2 ]
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
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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 3 ]
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
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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 4 ]
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
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002
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 5 ]
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).
𝐀𝐧𝐬𝐰𝐞𝐫:√2
πx cos x
C 3 Determine the value J32
(x).
𝐀𝐧𝐬𝐰𝐞𝐫:√2
πx (sin x
x− cos x)
S – 16
H 4 Determine the value J(−
32)(x).
𝐀𝐧𝐬𝐰𝐞𝐫:√2
πx (cos x
x+ sin x)
T 5 Using Bessel’s function of the first kind, Prove that J0(0) = 1.
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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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
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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 7 ]
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
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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 8 ]
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
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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 9 ]
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
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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 [ 2 0 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑
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
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002
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 [ 2 1 ]
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
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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 [ 2 2 ]
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
𝐧𝟐𝛑𝟐𝐜𝐨𝐬 (
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
𝐧𝟐𝛑𝟐𝐜𝐨𝐬 (
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
𝐧𝛑𝐬𝐢𝐧 (
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑𝟐
𝟑+ ∑
𝟒(−𝟏)𝐧
𝐧𝟐𝐜𝐨𝐬 𝐧𝐱
∞
𝐧=𝟏
S – 16 S – 18
T 8 Find the Fourier series expansion of f(x) = |x|; – π < x < π .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(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
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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 [ 2 3 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑𝟐
𝟑+ ∑[
𝟒(−𝟏)𝐧
𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +
𝟐(−𝟏)𝐧+𝟏
𝐧𝐬𝐢𝐧 𝐧𝐱]
∞
𝐧=𝟏
W – 17
C 12 Find the Fourier series of f(x) = x + |x|; – π < x < π.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑
𝟐+ ∑[
𝟐[(−𝟏)𝐧 − 𝟏]
𝛑𝐧𝟐𝐜𝐨𝐬 𝐧𝐱 +
𝟐(−𝟏)𝐧+𝟏
𝐧𝐬𝐢𝐧 𝐧𝐱]
∞
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
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟏
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
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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 [ 2 4 ]
C 16 Find Fourier series expansion of the function given by
f(x) = { 0, −2 < x < 0
1, 0 < x < 2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝟏
𝟐+∑ [
𝟏 − (−𝟏)𝐧
𝐧𝛑𝐬𝐢𝐧 (
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
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(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
𝐧𝐬𝐢𝐧 𝐧𝐱]
∞
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.
𝐀𝐧𝐬𝐰𝐞𝐫:
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
𝐧𝐬𝐢𝐧 𝐧𝐱]
∞
𝐧=𝟏
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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 [ 2 5 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑
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
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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 [ 2 6 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) =𝛑𝟐
𝟑+ ∑
𝟒(−𝟏)𝐧
𝐧𝟐𝐜𝐨𝐬 𝐧𝐱
∞
𝐧=𝟏
S – 15
H 2 Find Half-range cosine series for f(x) = (x − 1)2 in 0 < x < 1.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(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
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(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
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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 [ 2 7 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟒
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
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱) = ∑𝟒
𝐧𝟐𝛑𝐬𝐢𝐧 (
𝐧𝛑
𝟐) 𝐬𝐢𝐧𝐧𝐱
∞
𝐧=𝟏
W – 14
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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 [ 2 8 ]
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
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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 [ 2 9 ]
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
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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 [ 3 0 ]
T 8 Find Fourier cosine and sine integral of f(x) = {
sin x ; 0 ≤ x ≤ π
0 ; x > π.
𝐀𝐧𝐬𝐰𝐞𝐫:
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
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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[ 3 1 ]
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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:
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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 3 ]
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
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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 4 ]
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
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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 5 ]
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
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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 6 ]
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
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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 7 ]
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
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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 8 ]
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
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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 9 ]
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
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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 [ 4 0 ]
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
Where, c is an arbitrary constant.
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
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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 [ 4 1 ]
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.
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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 [ 4 2 ]
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
Where, c is an arbitrary constant.
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
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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 [ 4 3 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫:𝟏
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 .
𝐀𝐧𝐬𝐰𝐞𝐫:𝟏
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.
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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 [ 4 4 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱
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
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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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.
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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 7 ]
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 + ⋯
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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 8 ]
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)
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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 9 ]
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
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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 [ 5 0 ]
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)
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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 [ 5 1 ]
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)
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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 [ 5 2 ]
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
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞
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) .
𝐀𝐧𝐬𝐰𝐞𝐫:
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
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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 [ 5 3 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐲 = 𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞
−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 .
𝐀𝐧𝐬𝐰𝐞𝐫:
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)
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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 [ 5 4 ]
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
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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 [ 5 5 ]
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
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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 [ 5 6 ]
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
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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 [ 5 7 ]
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
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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 [ 5 8 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫:
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
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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 [ 5 9 ]
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.
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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 [ 6 0 ]
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
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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 [ 6 1 ]
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
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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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 .
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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 3 ]
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
𝐀𝐧𝐬𝐰𝐞𝐫:
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.
𝐀𝐧𝐬𝐰𝐞𝐫:
𝐱 = 𝟎 𝐢𝐬 𝐚 𝐑𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭 & 𝐱 = 𝟐 𝐢𝐬 𝐚𝐧 𝐈𝐫𝐫𝐞𝐠𝐮𝐥𝐚𝐫 𝐒𝐢𝐧𝐠𝐮𝐥𝐚𝐫 𝐏𝐨𝐢𝐧𝐭.
W – 15
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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 4 ]
H 8 Find singularity of x(x + 1)2y′′ + (2x − 1)y′ + x2y = 0.
𝐀𝐧𝐬𝐰𝐞𝐫:
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 +⋯
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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 5 ]
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
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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 6 ]
H 12 (x − 2)y′′ − x2y′ + 9y = 0 .
𝐀𝐧𝐬𝐰𝐞𝐫:
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.
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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 7 ]
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)
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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 8 ]
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
𝐀𝐧𝐬𝐰𝐞𝐫:
𝐲 = 𝐜𝟏𝐚𝟎𝐱𝟏𝟒 (𝟏 +
𝟏
𝟏𝟒𝐱 +
𝟏
𝟔𝟏𝟔𝐱𝟐 +⋯ )+
𝐜𝟐𝐚𝟎
𝐱𝟐(𝟏 +
𝟏
𝟐𝐱 +
𝟏
𝟒𝟎𝐱𝟐 + ⋯)
S – 18
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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[ 6 9 ]
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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)}
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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 1 ]
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
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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 2 ]
⟹ ℒ{tn} =n!
sn+1
(2). 𝓛{𝟏} =𝟏
𝐬.
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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)
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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)
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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
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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 3 ]
⟹ ℒ{e−at} =1
s + a
(5). 𝓛{𝐬𝐢𝐧 𝐚𝐭} =𝐚
𝐬𝟐+𝐚𝟐 , 𝐬 > 𝟎 and 𝐚 is a constant.
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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.
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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
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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 4 ]
(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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬𝟐 + 𝟏(𝐞−𝛑𝐬 (𝐬 − 𝟏) + 𝐬)
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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 5 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬 − 𝐥𝐨𝐠𝐞𝟏𝟎𝟎+𝟐 ∙ 𝟏𝟎 !
𝐬𝟏𝟏+
𝟏𝟎
𝐬𝟐 + 𝟏𝟎𝟎
C 5 Find the Laplace transformation of f(t) = cosh2 3t .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟐𝐬+
𝐬
𝟐(𝐬𝟐 − 𝟑𝟔)
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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 6 ]
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)
𝐀𝐧𝐬𝐰𝐞𝐫:
(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}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
𝐬𝟐 + 𝟏𝟔
H 9 Find ℒ{sin 2t sin 3t}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟒𝐬+
𝐬
𝟒(𝐬𝟐 + 𝟑𝟔)+
𝐬
𝟒(𝐬𝟐 + 𝟏𝟔)+
𝐬
𝟒(𝐬𝟐 + 𝟒)
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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 7 ]
THEOREM. FIRST SHIFTING THEOREM:
Statement: If 𝐋{𝐟(𝐭)} = 𝐅(𝐬), then show that 𝓛{𝐞𝐚𝐭𝐟(𝐭)} = 𝐅(𝐬 − 𝐚).
Proof: By definition, ℒ{f(t)} = ∫ e−st∞
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}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
(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
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 8 ]
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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
𝟐{
𝟏
(𝐬𝟐 + 𝟏𝟒𝐬 + 𝟓𝟖)+
𝟏
(𝐬𝟐 − 𝟐𝐬 + 𝟏𝟎)}
C 12 Find the Laplace transform of f(t) = t2cos h πt.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
(𝐬 − 𝛑)𝟑+
𝟏
(𝐬 + 𝛑)𝟑
W – 14
H 13 Find the Laplace transform of t3 cosh 2t.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
(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
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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 9 ]
= ∫ (−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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬𝟑+𝐬(𝐬𝟐 − 𝟒𝟖)
(𝐬𝟐 + 𝟏𝟔)𝟑
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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 [ 8 0 ]
H 8 Find the Laplace transform of f(t) = t2 cosh 3t.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
(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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 8 1 ]
= ∫ 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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛑
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)
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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 [ 8 2 ]
T 9 Find the Laplace transform of
cos at − cos bt
t .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠√𝐬𝟐 + 𝐛𝟐
𝐬𝟐 + 𝐚𝟐
S – 16
T 10 Find the Laplace transformation of f(t) =
sin2 t
t2 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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).
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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 [ 8 3 ]
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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬𝟐𝐚
(𝐬𝟐 + 𝐚𝟐)
C 3
Find ℒ {∫eu(u + sin u)du
t
0
}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 8 4 ]
C 7 Evaluate: ∫ e−2t
∞
0
t cos t dt .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
𝟐𝟓
T 8
Evaluate: ∫ e−t∞
0
(∫sin u
u
t
0
du) dt .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛑
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 ⟶ ∞.
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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 [ 8 5 ]
⟹ 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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
𝟑𝐬𝟐−
𝟐𝐞−𝟑𝐬
𝐬(𝟏 − 𝐞−𝟑𝐬)
W – 17
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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 [ 8 6 ]
T 5 Find the Laplace transformation of f(t) =
t
T , 0 < t < T ; if f(t) = f(t + T).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
(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
Proof: By definition, ℒ{f(t)} = ∫ e−stf(t) dt∞
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
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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 [ 8 7 ]
= ∫ 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)}
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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 [ 8 8 ]
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
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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 [ 8 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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
𝟒𝐜𝐨𝐬
𝟓
𝟐𝐭 −
𝟒
𝟓𝐬𝐢𝐧
𝟓
𝟐𝐭
H 4 Find ℒ−1 {
3(s2 − 1)2
2s5} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
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) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟒𝐜𝐨𝐬 𝐡
𝟓
𝟒𝐭 +
𝟑
𝟒𝐬𝐢𝐧 𝐡
𝟓
𝟒𝐭
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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 [ 9 0 ]
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).
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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 [ 9 1 ]
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) .
𝐀𝐧𝐬𝐰𝐞𝐫: 3
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) .
𝐀𝐧𝐬𝐰𝐞𝐫: 1
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))
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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 [ 9 2 ]
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)} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟓[𝐞−𝐭 − 𝐞−𝟐𝐭]
S – 18
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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 [ 9 3 ]
H 3 Find ℒ−1 {
1
(s − 2)(s + 3)} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
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)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 9 4 ]
C 13 Find ℒ−1 {
1
s(s2 − 3s + 3)} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟓√𝟑𝐬𝐢𝐧𝐡√𝟑𝐭 −
𝟏
𝟓√𝟐𝐬𝐢𝐧√𝟐𝐭
H 17 Find ℒ−1 (s
(s2 + 1) (s2 + 4)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟑(𝐜𝐨𝐬𝐭 − 𝐜𝐨𝐬𝟐𝐭)
C 18 Find ℒ−1 (
2s2 − 1
(s2 + 1) (s2 + 4)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑
𝟐𝐬𝐢𝐧 𝟐𝐭 − 𝐬𝐢𝐧 𝐭
W – 16
H 19 Find ℒ−1 (
s2
(s2 + a2) (s2 + b2)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 9 5 ]
T 22 Find ℒ−1 {s
s4 + 64} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 9 6 ]
C 8 Find ℒ−1 {
e−2s
(s2 + 2)(s2 − 3)}.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
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
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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 [ 9 7 ]
H 7 Find ℒ−1{tan−1(s + 1)} .
𝐀𝐧𝐬𝐰𝐞𝐫: −𝟏
𝐭𝐞−𝐭 𝐬𝐢𝐧 𝐭
C 8 Find ℒ−1{tan−1
2
s} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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)} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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
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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 [ 9 8 ]
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
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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 [ 9 9 ]
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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟐𝐚𝟐(𝐬𝐢𝐧 𝐚𝐭
𝟐𝐚− 𝐭 𝐜𝐨𝐬 𝐚𝐭)
S – 16
H 8 Using convolution theorem, obtain ℒ−1 {
1
(s2 + 4)2} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝟖(𝐬𝐢𝐧𝟐𝐭
𝟒− 𝐭 𝐜𝐨𝐬𝟐𝐭)
S – 17 S – 18
H 9 Using convolution theorem find ℒ−1 (
2
(s2 + 1)(s2 + 4)) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐
𝟑(𝐬𝐢𝐧 𝐭 −
𝟏
𝟐𝐬𝐢𝐧 𝟐𝐭)
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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} .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬𝟐(𝐬 − 𝟐)
W – 15
H 18 State the convolution theorem and verify it for f(t) = 1 and g(t) = sin t .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏
𝐬(𝐬𝟐 + 𝟏)
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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 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).
Proof: By definition, ℒ{f(t)} = ∫ e−stf(t) dt∞
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
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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 2 ]
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
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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 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
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆
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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)
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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
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C 5 Form the partial differential equation of f(x + y + z, x2+y2 + z2) = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 + 𝟏
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.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐱
𝐲=𝐩 − 𝐲
𝐪 − 𝐱
W – 17
C 9 Form partial differential equation by eliminating the arbitrary function
from xyz = Ф(x + y + z).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐩 + 𝟏
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
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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).
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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.
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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.
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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
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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.
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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
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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
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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
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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 1 5 ]
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 .
𝐀𝐧𝐬𝐰𝐞𝐫: z =2
3(a + x)
32 +
2
3(y − a)
32 + b
T 7 Solve p2 − q2 = x − y .
𝐀𝐧𝐬𝐰𝐞𝐫: z =2
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)
DARSHAN INSTITUTE OF ENGINEERING & TECHNOLOGY » » » AEM - 2130002
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 1 6 ]
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.
𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = (c1e√kx + c2e
−√kx) c3 e(k−2)y
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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 1 7 ]
C 8 Solve two dimensional Laplace’s equation ∂2u
∂x2+
∂2u
∂y2= 0,using the method
separation of variables.
𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = (c1e√kx + c2e
−√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.
𝐀𝐧𝐬𝐰𝐞𝐫: u(x, y) = (c1e√kx + c2e
−√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
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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 1 8 ]
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
Seat No.: ________ Enrolment No.___________
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:
1. Attempt all questions.
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 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
Seat No.: ________ Enrolment No.___________
GUJARAT TECHNOLOGICAL UNIVERSITY
BE - SEMESTER–III(New) EXAMINATION – SUMMER 2016
Subject Code:2130002 Date:07/06/2016
Subject Name:Advanced Engineering Mathematics Time:10:30 AM to 01:30 PM Total Marks: 70 Instructions:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.
Q.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
Seat No.: ________ Enrolment No.___________
GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III(New) • EXAMINATION – WINTER 2016
Subject Code:2130002 Date:30/12/2016
Subject Name:Advanced Engineering Mathematics
Time: 10:30 AM to 01:30 PM Total Marks: 70 Instructions:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary.
3. Figures to the right indicate full marks.
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
variation of parameters.
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
Seat No.: ________ Enrolment No.___________
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:
1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.
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
Seat No.: ________ Enrolment No.___________
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:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary.
3. Figures to the right indicate full marks.
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
variation of parameters.
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
Seat No.: ________ Enrolment No.___________
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:
1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.
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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