1. INTRODUCTION TO SOME SPECIAL FUNCTIONS … · 1. INTRODUCTION TO SOME SPECIAL FUNCTIONS PAGE | 1...

1. INTRODUCTION TO SOME SPECIAL FUNCTIONS PAGE | 1

A.E.M.(2130002) DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

Beta and Gamma Function

The name gamma function and its symbol were introduced by Adrien-marie Legendre in 1811.

It is found that some specific definite integrals can be conveniently used as Beta and Gamma

function. The gamma and beta functions have wide applications in the area of quantum physics,

Fluid dynamics, Engineering and statistics.

1. 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).

𝐢. 𝐞. 𝛃(𝐦, 𝐧) = ∫ 𝐱𝐦−𝟏(𝟏 − 𝐱)𝐧−𝟏𝐝𝐱

𝟏

𝟎

Properties

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

B(m, n) = 2 ∫ sin2m−1 θ cos2n−1 θdθπ

20

∫ sinp θ cosq θdθ =1

2B (

p+1

2,

q+1

2)

π

20

2. Gamma Function

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

0 and is denoted by ⌈n.

𝐢. 𝐞. ⌈𝐧 = ∫ 𝐞−𝐱𝐱𝐧−𝟏𝐝𝐱

∞

𝟎

Properties

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

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

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

1

2

∞

0⌈m

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

⌈(m+n)

∫ sinp θ cosq θdθ =1

2

⌈(p+1

2)⌈(

q+1

2)

⌈(p+q+2

2)

π

20

⌈1

2= √π

⌈n+1

2=

(2n)!√π

n!4n for n = 0,1,2,3, …

Special cases

For n = 0, ⌈1

2= √π For n = 1, ⌈

3

2=

√π

2 For n = 2, ⌈

5

2=

3√π

4



3. Error Function and Complementary Error Function

The error function of x is defined by the integral 2

√π∫ e−t2

dt,x

0 where x may be real or complex

variable and is denoted by erf(x).

i. e. erf(x) =2

√π∫ e−t2

dt

x

0

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

erfc(x) =2

√π∫ e−t2

dt

∞

x

Properties

erf(0) = 0

erf(∞) = 1

erf(x) + erfc(x) = 1

erf(−x) = −erf(x)

4. Unit Step Function

The Unit Step Function is defined by

u(x − a) = {1, for x ≥ a0, for x < a

, where a ≥ 0.

5. Pulse of unit Height

The pulse of unit height of duration T is defined by f(x) = {1, for 0 ≤ x ≤ T0, for T < x

.

0 a

1

u(x-a)

x



6. Sinusoidal Pulse Function

The sinusoidal pulse function is defined by

f(x) = {sin ax , for 0 ≤ x ≤

π

a

0, for x >π

a

.

7. Rectangle Function

A Rectangular function f(x) defined on ℝ as

f(x) = {1, for a ≤ x ≤ b0, otherwise

.

8. Gate Function

A Gate function fa(x) defined on ℝ as

fa(x) = {1, for |x| ≤ a0, for |x| > a

.

Note that gate function is symmetric about axis

of co-domain. Gate function is also a rectangle function.

9. Dirac Delta Function

A Dirac delta Function fε(x) defined of ℝ as

fε(x) = {1

ε, for 0 ≤ x ≤ ε

0, for x > ε .

f(x)

0 π

a

0 a

1

f(x)

x b

0 -a

1

f(x)

x a

1ε⁄

fε(x)

x ε



10. Signum Function

The Signum function is defined by

f(x) = { 1, for x > 0−1, for x < 0

.

11. 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).

Note

Constant function is periodic without Fundamental period.

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

12. Square Wave Function

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

f(x) = { 1, for 0 < x < a

−1, for a < x < 2a.

13. Saw Tooth Wave Function

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

f(x) = x ; 0 ≤ x < a.

0

−1

1

f(x)

0

−1

1

f(x)

a

-a 2a

3a

0

1

f(x)

a 2a 3a



14. 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.

15. Full Rectified Sine Wave Function

A full rectified sine wave function with

period π is defined as

f(t) = sin t ; 0 ≤ t < π and f(t + π) = f(t).

16. Half Rectified Sine Wave Function

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

f(x) = {sin x , for 0 ≤ x < π 0, for π ≤ x < 2π

.

`

f(x)

π

`

`

2π 3π −π 0 4π −2π

0

1

f(x)

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

`

f(x)

π

`

`

2π −π 0



17. Bessel’s Function

A Bessel’s function of order n is defined by

Jn(x) =xn

2n⌈(n + 1)[1 −

x2

2(n + 2)+

x4

2 ∙ 4(2n + 2)(2n + 4)− ⋯ ] = ∑

(−1)k

k! ⌈(n + k + 1)(

x

2)

n+2k∞

k=0

Properties

J−n(x) = (−1)nJn(x),If n is a positive integer.

Jn+1(x) =2n

xJn(x) − Jn−1(x)

d

dx(xnJn(x)) = xnJn−1(x)

2. FOURIER SERIES AND FOURIER INTEGRAL PAGE | 1


Fourier Series in the interval (𝐜, 𝐜 + 𝟐𝐥)

The Fourier series for the function f(x) in the interval (0,2l) 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

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

FORMULAE

1. Leibnitz’s Formula

∫ 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, A means Algebraic Function

L means Logarithmic Function T means Trigonometric Function

I means Invertible Function E means Exponential Function

2. ∫ 𝐞𝐚𝐱 𝐬𝐢𝐧 𝐛𝐱 𝐝𝐱 =𝐞𝐚𝐱

𝐚𝟐+𝐛𝟐[𝐚 𝐬𝐢𝐧 𝐛𝐱 − 𝐛 𝐜𝐨𝐬 𝐛𝐱] + 𝐜

3. ∫ 𝐞𝐚𝐱 𝐜𝐨𝐬 𝐛𝐱 𝐝𝐱 =𝐞𝐚𝐱

𝐚𝟐+𝐛𝟐[𝐚 𝐜𝐨𝐬 𝐛𝐱 + 𝐛 𝐬𝐢𝐧 𝐛𝐱] + 𝐜



Fourier series in the interval (𝟎, 𝟐𝐥)

The Fourier series for the function f(x) in the interval (0,2l) is defined by

𝐟(𝐱) =𝐚𝟎

𝟐+ ∑ (𝐚𝐧 𝐜𝐨𝐬

𝐧𝛑𝐱

𝐥+ 𝐛𝐧 𝐬𝐢𝐧

𝐧𝛑𝐱

𝐥)

∞

𝐧=𝟏


𝐚𝟎 =𝟏


𝟐𝐥

𝟎 𝐚𝐧 =

𝟏


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝟐𝐥

𝟎 𝐛𝐧 =

𝟏


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝟐𝐥

𝟎

Exercise-1

Fourier Series In Arbitrary Period [𝟎, 𝟐𝐥]

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

Ans. f(x) =1

3+ ∑ [

1

n2π2cos(2nπx) −

1

nπsin(2nπx)]

∞

n=1

C 2. Find the Fourier series to represent f(x) = 2x − x2 in (0,3). May-12

Ans. f(x) = ∑ [−9

n2π2cos (

2nπx

3) +

3

nπsin (

2nπx

3)]

∞

n=1

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

Ans. f(x) =(1 − e−2)

2+ ∑ [

(1 − e−2)

n2π2 + 1cos(nπx) +

nπ(1 − e−2)

n2π2 + 1sin(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. Dec-10

Ans. f(x) = 2 + ∑4

1 − 4n2cos(nπx)

∞

n=1

H 5. Develop f(x) in a Fourier series in the interval (0,2) if

f(x) = {x, 0 < x < 10, 1 < x < 2.

Dec-13

Ans. f(x) =1

4+ ∑ [

(−1)n − 1

n2π2cos(nπx) +

(−1)n+1

nπsin(nπx)]

∞

n=1



C 6. Find the Fourier series for periodic function with period 2 of

f(x) = {π, 0 ≤ x ≤ 1

π (2 − x), 1 ≤ x ≤ 2.

Jun-13

Ans. f(x) =3π

4+ ∑ [

(−1)n − 1

πn2cos(nπx) +

1

nsin(nπx)]

∞

n=1

T 7. Find the Fourier series for periodic function with period 2 of

f(x) = {πx, 0 ≤ x ≤ 1

π (2 − x), 1 ≤ x ≤ 2.

Jun-15

Ans. f(x) =π

2+ ∑

2[(−1)n − 1]

πn2cos(nπx)

∞

n=1

Fourier series in the interval (𝟎, 𝟐𝛑)

The Fourier series for the function f(x) in the interval (0,2π) is defined by

𝐟(𝐱) =𝐚𝟎

𝟐+ ∑(𝐚𝐧 𝐜𝐨𝐬 𝐧𝐱 + 𝐛𝐧 𝐬𝐢𝐧 𝐧𝐱)

∞

𝐧=𝟏


𝐚𝟎 =𝟏

𝛑∫ 𝐟(𝐱) 𝐝𝐱

𝟐𝛑

𝟎 𝐚𝐧 =

𝟏

𝛑∫ 𝐟(𝐱) 𝐜𝐨𝐬 𝐧𝐱 𝐝𝐱

𝟐𝛑

𝟎 𝐛𝐧 =

𝟏

𝛑∫ 𝐟(𝐱) 𝐬𝐢𝐧 𝐧𝐱 𝐝𝐱

𝟐𝛑

𝟎

Exercise-2

Fourier Series In [𝟎, 𝟐𝛑]

H 1. Find Fourier Series for f(x) = x2 ; where 0 ≤ x ≤ 2π. Jan-13

Ans. f(x) =4π2

3+ ∑ [

4

n2cos nx −

4π

nsin nx]

∞

n=1

T 2. Express f(x) =π−x

2 in a Fourier series in interval 0 < x < 2π. Dec-13

Ans. f(x) = ∑1

nsin nx

∞

n=1

C 3. Show that, π − x =π

2+ ∑

sin 2nx

n∞n=1 , when 0 < x < π .

C 4. Obtain the Fourier series for f(x) = (

π−x

2)

2

in the interval 0 < x < 2π.

Hence prove that ∑(−1)n+1

n2∞n=1 =

π2

12.

Jan-15 Jun-15



Ans. f(x) =π2

12+ ∑

1

n2cos nx

∞

n=1

C 5. Find Fourier Series for f(x) = e−x where0 < x < 2π. Jun-14

Ans. f(x) =1 − e−2π

2π+ ∑ [

1 − e−2π

π(n2 + 1)cos nx +

n(1 − e−2π)

π(n2 + 1)sin nx]

∞

n=1

C 6. Find the Fourier series of f(x) = {x2 ; 0 < x < π

0 ; π < x < 2π. Jan-13

Ans. f(x) =

π2

6+ ∑ [

2(−1)n

n2cos nx +

1

π{−

π2(−1)n

n+

2(−1)n

n3−

2

n3} sin nx]

∞

n=1

Fourier series in the interval (−𝐥, 𝐥)

The Fourier series for the function f(x) in the interval (−l, l) is defined by

𝐟(𝐱) =𝐚𝟎

𝟐+ ∑ (𝐚𝐧 𝐜𝐨𝐬

𝐧𝛑𝐱

𝐥+ 𝐛𝐧 𝐬𝐢𝐧

𝐧𝛑𝐱

𝐥)

∞

𝐧=𝟏


𝐚𝟎 =𝟏


𝐥

−𝐥 𝐚𝐧 =

𝟏


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝐥

−𝐥 𝐛𝐧 =

𝟏


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝐥

−𝐥

Definition: Odd Function & Even Function

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

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

Fourier Series For Odd & Even Function

Let, f(x) be a periodic function defined in (−𝐥, 𝐥)

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

f(x) =a0

2+ ∑ an cos (

nπx

l)

∞

n=1

f(x) = ∑ bn sin (nπx

l)

∞

n=1



Exercise-3

Fourier Series In Arbitrary Period [−𝐥, 𝐥]

H 1. Expand f(x) = x in – l < x < l the Fourier series.

Ans. f(x) = ∑2 l (−1)n+1

nπsin (

nπx

l)

∞

n=1

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

Where−1 < x < 1 , p = 2l = 2. Dec-09

Ans. f(x) = ∑4(−1)n+1

nπsin(nπx)

∞

n=1

T 3. Find the Fourier series for f(x) = x2 in −2 < x < 2. Dec-11

Ans. f(x) =4

3+ ∑

16(−1)n

n2π2cos (

nπx

2)

∞

n=1

H 4. Find the Fourier series of (i) f(x) = x; – π < x < π, f(x) = f(x + 2π). (ii) f(x) = x2; – l < x < l.

Jun-14

Ans.

(i)f(x) = ∑2(−1)n+1

nsin nx

∞

n=1

(ii)f(x) =l2

3+ ∑

4l2(−1)n

n2π2cos nx

∞

n=1

H 5. Expand f(x) = x2 − 2 in – 2 < x < 2 the Fourier series. Jan-15

Ans. f(x) = −2

3+

16

π2∑

(−1)n+1

n2cos (

nπx

2)

∞

n=1

T 6. Find the Fourier series of f(x) = x2 + x Where −2 < x < 2. Jan-13

Ans. f(x) =4

3+ ∑ [

16(−1)n

n2π2cos (

nπx

l) +

4(−1)n+1

nπsin (

nπx

l)]

∞

n=1

H 7. Find the Fourier expansion for function f(x) = x − x3 in −1 < x < 1.

Ans. f(x) = ∑12(−1)n+1

n3π3sin nπx

∞

n=1

C 8. Find the Fourier expansion for function f(x) = x − x2 in −1 < x < 1. Jun-15



Ans. f(x) = −1

3+ ∑ [

4(−1)n+1

n2π2cos(nπx) +

2(−1)n+1

nπsin(nπx)]

∞

n=1

T 9. Find the Fourier series for periodic function with period 2, which is

given below f(x) = {0, −1 < x < 0x, 0 < x < 1

. Jun-13

Ans. f(x) =1

4+ ∑ [

(−1)n − 1

n2π2cos nx +

(−1)n+1

nπsin nx]

∞

n=1

H 10. Find the Fourier series of f(x) = {x , −1 < x < 02, 0 < x < 1

. Jan-13

Ans. f(x) =3

4+ ∑ [

1 − (−1)n

n2π2cos 𝑛𝑥 +

2 − 3(−1)n

nπsin nx]

∞

n=1

C 11. Find the Fourier series for periodic function f(x) with period 2

Where f(x) = {−1, −1 < x < 0

1, 0 < x < 1 .

Jan-13

Ans. f(x) = ∑2 − 2(−1)n

πnsin nx

∞

n=1

Fourier series in the interval (−𝛑, 𝛑)

The Fourier series for the function f(x) in the interval (−π, π) is defined by

𝐟(𝐱) =𝐚𝟎

𝟐+ ∑(𝐚𝐧 𝐜𝐨𝐬 𝐧𝐱 + 𝐛𝐧 𝐬𝐢𝐧 𝐧𝐱)

∞

𝐧=𝟏

𝐚𝟎 =𝟏

𝛑∫ 𝐟(𝐱) 𝐝𝐱

𝛑

−𝛑 𝐚𝐧 =

𝟏

𝛑∫ 𝐟(𝐱) 𝐜𝐨𝐬 𝐧𝐱 𝐝𝐱

𝛑

−𝛑 𝐛𝐧 =

𝟏

𝛑∫ 𝐟(𝐱) 𝐬𝐢𝐧 𝐧𝐱 𝐝𝐱

𝛑

−𝛑

Exercise-4

Fourier Series In [−𝛑, 𝛑]

H 1. Find the Fourier series expansion of f(x) = x; – π < x < π . Dec-11

Jun-14

Ans. f(x) = ∑ 2(−1)n+1

nsin nx

∞

n=1

T 2. Find the Fourier series expansion of f(x) = |x|; – π < x < π . Jun-15

Ans. f(x) =π

2+ ∑

2 [(−1)n − 1]

πn2cos nx

∞

n=1



C 3.

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. (iii) ∑

1

(2n − 1)2

∞

n=1

=π2

8.

Dec-09

Jan-15

Ans. f(x) =π2

3+ ∑

4(−1)n

n2cos nx

∞

n=1

H 4. Find the Fourier series of f(x) =x2

2; – π < x < π. Mar-10

Ans. f(x) =π2

6+ ∑

2(−1)n

n2cos nx

∞

n=1

H 5. Find the Fourier series of f(x) = x3 ; x ∈ (−π, π). Jun-13

Ans. f(x) = ∑2(−1)n+1

n[π2 −

6

n2] sin nx

∞

n=1

C 6. Sketch the function f(x) = x + π ; – π < x < π. Where f(x) = f(x + 2π) and Find the Fourier series.

Mar-10

Ans. f(x) = π + ∑2(−1)n+1

nsin nx

∞

n=1

T 7. Find the Fourier series of f(x) = x − x2; – π < x < π.

Deduce that: 1

12−

1

22+

1

32−

1

42+ ⋯ =

π2

12.

Jun-14

Ans. f(x) = −π2

3+ ∑ [

4(−1)n+1

n2cos nx +

2(−1)n+1

nsin nx]

∞

n=1

H 8. Find the Fourier series of f(x) = x + x2; – π < x < π. Jun-15

Ans. f(x) =π2

3+ ∑ [

4(−1)n

n2cos nx +

2(−1)n+1

nsin nx]

∞

n=1

T 9. Find the Fourier series of f(x) = x + |x|; – π < x < π.

Mar-10

Jun-13

Jan-15

Jan-15*

Ans. f(x) =π

2+ ∑ [

2[(−1)n − 1]

πn2cos nx +

2(−1)n+1

nsin nx]

∞

n=1



T 10. Find the Fourier series to representation ex in the the interval(−π, π). Dec-13

Ans.

f(x) =eπ − e−π

2𝜋+ ∑ [

[eπ − e−π] (−1)n

π(n2 + 1)cos nx +

n [e−π − eπ] (−1)n

π(n2 + 1)sin nx]

∞

n=1

C 11. Find the Fourier series for f(x) = |sin x| in– π < x < π. May-11

Ans. f(x) =2

π+ ∑

2 [(−1)n + 1]

π(1 − n2)cos nx ; a1 = 0

∞

n=2

T 12. Find the Fourier series expansion of f(x) = √1 − cos x in the interval,

(i)– π < x < π. (ii) 0 ≤ x ≤ 2π.

Ans. f(x) =2√2

π+ ∑

4√2

π(1 − 4n2)cos nx

∞

n=1

(same answer in both intervals)

H 13. Find the Fourier series of the periodic function f(x) with period 2π

defined as follows f(x) = {0 ; – π < x < 0x ; 0 < x < π.

Jun-13

Ans. f(x) =π

4+ ∑ [

(−1)n − 1

πn2cos nx +

(−1)n+1

nsin nx]

∞

n=1

H 14. 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.

Ans. f(x) =π2

6+ ∑ [

2(−1)n

n2cos nx +

1

π{−

π2(−1)n

n+

2(−1)n

n3−

2

n3} sin nx]

∞

n=1

H 15. 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.

May-12

Jun-14

Ans. f(x) = −π

4+ ∑ [

(−1)n − 1

πn2cos nx +

1 − 2(−1)n

nsin nx]

∞

n=1

C 16. Find Fourier series for 2π periodic function f(x) = {

−k; if − π < x < 0k; if 0 < x < π

Hence deduce that 1 −1

3+

1

5−

1

7+ ⋯ =

π

4.

Jan-15



Ans. f(x) = ∑2k[1 − (−1)n]

nπsin nx

n=1

= ∑4k

nπsin nx

n=odd

T 17. 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.

Jan-13 Jun-15

Ans. f(x) =π

2+ ∑

2

πn2[1 − (−1)n] cos nx

∞

n=1

H 18.

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

f(x) = {−π , −π < x < 0

x − π , 0 < x < π

Jun-15

Ans. f(x) = −3π

4+ ∑ [

(−1)n − 1

πn2cos nx +

(−1)n+1

nsin nx]

∞

n=1

C 19.

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.

May-11

Jan-15

Ans. 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, π) instead of (0,2π), 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

𝐚𝟎 =𝟐


𝐥

𝟎 𝐚𝐧 =

𝟐


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝐥

𝟎



Exercise-5

Half Range Cosine Series

C 1. Find Fourier cosine series of the periodic function f(x) = x, (0 < x < L), p = 2L. also sketch f(x) and its periodic extension.

Dec-10

Ans. f(x) =L

2+ ∑

2L[(−1)n − 1]

n2π2cos (

nπx

L)

∞

n=1

H 2. Find the Half range cosine series for f(x) = x, 0 < x < 3. Jan-13

Ans. f(x) =3

2+ ∑

6[(−1)n − 1]

n2π2cos (

nπx

3)

∞

n=1

H 3. Find Fourier cosine series for f(x) = x2; 0 < x ≤ c . Also sketch f(x). Jun-13

Ans. f(x) =c2

3+ ∑

4c2(−1)n

n2π2 cos (

nπx

c)

∞

n=1

H 4. Find Half-range cosine series for f(x) = x2 in 0 < x < π. Jun-15

Ans. f(x) =π2

3+ ∑

4(−1)n

n2cos nx

∞

n=1

T 5. Find Half-range cosine series for f(x) = (x − 1)2 in 0 < x < 1. Jun-15

Ans. f(x) =1

3+ ∑

4

n2π2cos(nπx)

∞

n=1

T 6. Find a cosine series for f(x) = ex in 0 < x < L. Mar-10

Jan-15

Ans. f(x) =eL − 1

𝐿+ ∑

2L[eL(−1) n − 1]

n2π2 + L2cos (

nπx

L)

∞

n=1

H 7. Find Half-range cosine series for f(x) = ex in (0,1). May-12

Ans. f(x) = (e − 1) + ∑2[e(−1)n − 1]

n2π2 + 1cos(nπx)

∞

n=1



H 8. Find Half-range cosine series for f(x) = ex in 0 < x < 2. Jun-14

Ans. f(x) =e2 − 1

2+ ∑

4[e2(−1)n − 1]

n2π2 + 4cos (

nπx

2)

∞

n=1

H 9. Find Half-range cosine series for f(x) = ex in 0 < x < π. Jun-15

Ans. f(x) =eπ − 1

π+ ∑

2(eπ(−1)n − 1)

π(1 + n2)cos nx

∞

n=1

C 10. Fine Half range cosine series for sin x in (0, π) and show that

1 −1

3+

1

5− ⋯ =

π

4.

May-12

Jan-15

Ans. f(x) =2

π+ ∑ −

4

π(n2 − 1)cos nx

∞

n=2

, a1 = 0

Half Range Sine Series In The Interval (𝟎, 𝐥)

f(x) = ∑ bn sin (nπx

l)

∞

n=1

𝐛𝐧 =𝟐


𝐧𝛑𝐱

𝐥) 𝐝𝐱

𝐥

𝟎

Exercise-6

Half Range Sine Series

H 1. Express f(x) = x as a Half range sine series in 0 < x < 2 Half range cosine series in 0 < x < 2.

Jun-14

Ans.

(i)f(x) = ∑4(−1)n+1

nπsin (

nπx

2)

∞

n=1

(ii)f(x) = 1 + ∑ (2

nπ)

2

[(−1)n − 1] cos (nπx

2)

∞

n=1

H 2. Find the Half range sine series for f(x) = 2x, 0 < x < 1. Jun-15



Ans. f(x) = ∑4

nπ(−1)n+1 sin nπx

∞

n=1

H 3. Find the Half range sine series for f(x) = π − x, 0 < x < π. Jan-13

Ans. f(x) = ∑2

nsin nx

∞

n=1

T 4. Find Half-range sine series for f(x) = ex in 0 < x < π. Jun-15

Ans. f(x) = ∑2n[eπ(−1)n+1 + 1]

π(1 + n2)sin nx

∞

n=1

C 5. Expand πx − x2 in a half-range sine series in the interval (0, π) up to first three terms.

Jan-15 Jun-15

Ans. f(x) =8

π(sin x +

sin 3x

33+

sin 5x

53+ ⋯ )

C 6. Find the sine series f(x) = {x ; 0 < x <

π

2

π − x ;π

2< x < π

. May-11

Ans. f(x) = ∑4

n2πsin (

nπ

2) sin nx

∞

n=1

T 7. If f(x) = {

mx ; 0 < x <π

2

m(π − x) ;π

2< x < π

then, show that

f(x) =4m

π{sin x

12−

sin 3x

32−

sin 5x

52− ⋯ }

May-12



Some Useful Integrals And Formulae

Consider m and n be positive integers or zero.

1. ∫ sin nx dx

2π

0

= 0; n ≠ 0 2. ∫ cos nx dx

2π

0

= 0; n ≠ 0

3. ∫ cos mx cos nx dx

2π

0

= {0 ; m ≠ n π ; m = n ≠ 0

4. ∫ sin mx sin nx dx

2π

0

= {0 ; m ≠ n π ; m = n ≠ 0

5. ∫ sin mx cos nx dx

2π

0

= 0, ∀m, n

Let m, n ∈ ℤ.

1. sin nπ = 0 ; ∀n 2. cos nπ = (−1)n ; ∀n

3. sin(2n + 1)π

2= (−1)n 4. cos(2n + 1)

π

2= 0

5. ∫ ex[f(x) + f ′(x)] dx = ex f(x) + c 6. ∫[f(x)]n

f ′(x)dx =

[f(x)]n+1

n + 1+ c

If f(x) is an odd function defined in (−a, a), then ∫ f(x)a

−a dx = 0.

If f(x) is a even function defined in (−a, a), then ∫ f(x)a

−a dx = 2 ∫ f(x)

a

0 dx.



Fourier Integrals

Fourier Integral of f(x) is given by

f(x) = ∫[A(ω) cos ωx + B(ω) sin ωx]

∞

0

dω

Where, A(ω) =1

π∫ f(v) cos ωv

∞

−∞

dv & B(ω) =1

π∫ f(v) sin ωv

∞

−∞

dv

Fourier Cosine Integral

f(x) = ∫ A(ω) cos ωx

∞

0

dω ; Where, A(ω) =2

π∫ f(v) cos ωv

∞

0

dv

Fourier Sine Integral

f(x) = ∫ B(ω) sin ωx

∞

0

dω ; Where, B(ω) =2

π∫ f(v) sin ωv

∞

0

dv

Exercise-6

FOURIER INTEGRAL

C 1.

Using Fourier integral prove that

∫cos ωx + ωsin ωx

1 + ω2dω

∞

0

= {

0 ; x < 0π

2⁄ ; x = 0

πe−x ; x > 0.

Dec-10

Jan-15

Jun-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

Dec-13

Jan-15

Jun-15

Ans. f(x) = ∫2sin ω

πωcos ωx

∞

0

dω a) {

π

2 ; |x| < 1

0 ; |x| > 1 b)

π

2

T 3. Find the Fourier integral representation of f(x) = {2 ; |x| < 2

0 ; |x| > 2.

Jan-13 Jun-14

Ans. f(x) = ∫4 sin 2ω cos ωx

πω

∞

0

dω



H 4. Find the Fourier cosine integral of f(x) = e−kx (x > 0, k > 0). Mar-10

Ans. f(x) = ∫2k

π(k2 + ω2)cos ωx

∞

0

dω

H 5. Find Fourier cosine integral of f(x) = {x ; 0 < x < a0 ; x > a.

Jun-13

Ans. f(x) =2

π∫ [

a sin ωa

ω+

cos ωa

ω2−

1

ω2] cos ωx

∞

0

dω

H 6.

Using Fourier integral prove that

∫1 − cos ωπ

ω

∞

0

sin ωx dω = {

π

2 ; 0 < x < π

0 ; x > π.

Mar-10

7. Find Fourier cosine and sine integral of f(x) = {0 ; 0 ≤ x < 1−1 ; 1 < x < 20 ; 2 < x < ∞

T Ans.

a) f(x) =2

πω∫ (sin ω − sin 2ω) cos ωx

∞

0

dω

b) f(x) =2

πω∫ (cos 2ω − cos ω) sin ωx dω

∞

0

T 8. Find Fourier cosine and sine integral of f(x) = {sin x ; 0 ≤ x ≤ π

0 ; x > π.

Ans.

a) f(x) = ∫2(1 + cos ωπ)

π (1 − ω2)cos ωx

∞

0

dω ; A(1) = 0

b) f(x) = ∫2 sin ωπ

π (1 − ω2)sin ωx dω

∞

0

; B(1) = 1

3A. DIFFERENTIAL EQUATION OF FIRST ORDER PAGE | 1


Definition: Differential Equation

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

e.g. dy

dx+ y = 0

Definition: Partial Differential Equation

An eqn which involves function of two or more variable 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 appeared in differential equation is “Order of D.E”.

e.g. (dy

dx)

2

+dy

dx+ 5y = 0 Has order 1.

Definition: Degree Of Differential Equation

When a D.E. is in a polynomial form of derivatives, the highest power of highest order derivative occurring in D.E. is called a “Degree of D.E.”.

e.g. (dy

dx)

2

+dy

dx+ 5y = 0 Has degree 2.

Exercise-1

Order And Degree Differential Equation

C 1.

d2y

dx2 = [y + (dy

dx)

2

]

1

4

. [𝟐, 𝟒] May-11

T 2. [dy

dx+ y]

1

2= sin x. [𝟏, 𝟏] May-11

C 3. y = x

dy

dx+

xdy

dx

. [𝟏, 𝟐]

C 4. (d2y

dx2)3

= [x + sin (dy

dx)]

2

. [𝟐, 𝐔𝐧𝐝𝐞𝐟𝐢𝐧𝐞𝐝]

H 5. 𝑑2𝑦

𝑑𝑥2 = ln (𝑑𝑦

𝑑𝑥) + 𝑦. [𝟐, 𝐔𝐧𝐝𝐞𝐟𝐢𝐧𝐞𝐝]



T 6.

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

of differential equation √x2 d2y

dx2+ 2y =

d3y

dx3. [𝟑, 𝟐]

Jan-15

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.

1. 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.

2. Particular Solution The solution obtained from the general solution by giving a particular value to the arbitrary constants is called a particular solution.

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

Definition: Linear Differential Equation

A differential equation is called “LINEAR DIFFERENTIAL EQUATION” if

1. dependent variable and its all derivative are of first degree 2. dependent variable and its derivative are not multiplied together

If one of above condition is not satisfy, then it is called “NON-LINEAR DIFFERENTIAL EQUATION”.

e.g.

1. d2y

dx2 + x2 dy

dx+ y = 0 Is linear.

2. d2y

dx2+ y

dy

dx+ y = 0 Is non-linear.

3. d2y

dx2+ x2 (

dy

dx)

2

+ y = 0 Is non-linear.

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

i.e. dy

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

dx

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



Type Of First Order Differential Equation

I. Variable Separable Equation II. Homogeneous Differential Equation

III. Exact Differential Equation IV. Linear(Leibnitz’s) Differential Equation V. Bernoulli’s Equation

Variable Separable Equation

If differential equation of type dy

dx= f(x, y) can convert into M(x)dx = N(y)dy, then it is known

as Variable Separable Equation.

The general solution of Variable Separable Equation is

∫ 𝐌(𝐱)𝐝𝐱 = ∫ 𝐍(𝐲)𝐝𝐲 + 𝐜

Where, c is a arbitrary constant.

Homogeneous Differential Equation

A differential equation of the form 𝐌(𝐱, 𝐲)𝐝𝐱 + 𝐍(𝐱, 𝐲)𝐝𝐲 = 𝟎 is said to be Homogeneous Differential Equation if M(x, y) & N(x, y) are homogeneous function of same degree.

Such differential equation is solved by the substitution 𝐲 = 𝐯𝐱.

Exercise-2

Separable method

C 1. 9yy′ + 4x = 0. [𝟗𝐲𝟐 + 𝟒𝐱𝟐 = 𝐜] Dec-11

H 2. exdx − eydy = 0 [𝐞𝐱 = 𝐞𝐲 + 𝐜]

C 3. dy

dx= e2x+3y [

𝐞−𝟑𝐲

−𝟑=

𝐞𝟐𝐱

𝟐+ 𝐜] Jun-14

C 4. y′ = ex−y + xe−y [𝐞𝐲 = 𝐞𝐱 +𝐱𝟐

𝟐+ 𝐜] Jun-15

C 5. xy′ + y = 0 ; y(2) = −2. [𝐱𝐲 = −𝟒] Dec-11

C 6. L dI

dt+ RI = 0, I(0) = I0. [𝐈 = 𝐈𝟎𝐞−

𝐑

𝐋𝐭] Dec-10

H 7. (1 + x)ydx + (1 − y)xdy = 0. [𝐥𝐨𝐠(𝐱𝐲) + 𝐱 − 𝐲 = 𝐜]

T 8. extanydx + (1 − ex)sec2ydy = 0. [(𝟏 − 𝐞𝐱)−𝟏 𝐭𝐚𝐧 𝐲 = 𝐜] Jun-13

C 9. xy′ = y2 + y. [𝐲

𝐲+𝟏= 𝐱𝐜] Dec-10



H 10. xydy

dx= 1 + x + y + xy. [𝐲 − 𝐥𝐨𝐠(𝟏 + 𝐲) = 𝐥𝐨𝐠 𝐱 + 𝐱 + 𝐜] May-12

C 11. tanydy

dx= sin(x + y) + sin(x − y). [𝐬𝐞𝐜 𝐲 = −𝟐 𝐜𝐨𝐬 𝐱 + 𝐜]

C 12. 1 +dy

dx= ex+y. [−(𝐞−𝐱−𝐲) = 𝐱 + 𝐜]

H 13. dy

dx= cosx cosy − sinx siny. [𝐭𝐚𝐧 (

𝐱+𝐲

𝟐) = 𝐱 + 𝐜] Jan- 13

T 14. (x + y)2 dy

dx= a2. [𝐲 − 𝐚 𝐭𝐚𝐧−𝟏 𝐱+𝐲

𝐚= 𝐜]

C 15. xdy

dx= y + xe

y

x . [− (𝐞−𝐲

𝐱) = 𝐥𝐨𝐠 𝐱 + 𝐜]

T 16. dy

dx=

y

x+ tan

y

x. [𝐬𝐢𝐧

𝐲

𝐱= 𝐱𝐜] Jan-15

H 17. (x + y)dx + (y − x)dy = 0. [𝐥𝐨𝐠(𝐱𝟐 + 𝐲𝟐) = 𝟐𝐭𝐚𝐧−𝟏 (𝐲

𝐱) + 𝐜] Jun-15

C 18. [1 + ex

y] dx + ex

y [1 −x

y] dy = 0. [𝐱 + 𝐲𝐞

𝐱

𝐲 = 𝐜] Jun-15

C 19. (x + y)2 [xdy

dx+ y] = xy [1 +

dy

dx]. [𝐥𝐨𝐠 𝐱𝐲 = −

𝟏

𝐱+𝐲+ 𝐜] May-12

Leibnitz’s Linear Differential Equation

A differential equation of the form 𝐝𝐲

𝐝𝐱+ 𝐏(𝐱)𝐲 = 𝐐(𝐱) “OR”

𝐝𝐱

𝐝𝐲+ 𝐏(𝐲)𝐱 = 𝐐(𝐲) is known as

Linear Differential Equation.

The general solution of Linear Differential Equation is

𝐲(𝐈. 𝐅. ) = ∫ 𝐐(𝐱)(𝐈. 𝐅. )𝐝𝐱 + 𝐜 “OR” 𝐱(𝐈. 𝐅. ) = ∫ 𝐐(𝐲)(𝐈. 𝐅. )𝐝𝐲 + 𝐜

Where, 𝐈. 𝐅. = 𝐞∫ 𝐩 𝐝𝐱 “OR” 𝐈. 𝐅. = 𝐞∫ 𝐩 𝐝𝐲

Exercise-3

Linear Differential Equation

C 1. dy

dx− y = e2x. [𝐲𝐞−𝐱 = 𝐞𝐱 + 𝐜] Dec-09

H 2. dy

dx+ 2xy = e−x2. [𝐲𝐞𝐱𝟐

= 𝐱 + 𝐜]

C 3. y′ + y sin x = ecos x. [𝐲𝐞− 𝐜𝐨𝐬 𝐱 = 𝐱 + 𝐜] Dec-11

H 4. dy

dx+

1

x2y = 6e

1

x. [𝐲𝐞−𝟏

𝐱 = 𝟔𝐱 + 𝐜] Jan-13 Jun-15



T 5. y′ + 6x2y =e−2x3

x2, y(1) = 0. [𝐲𝐞𝟐𝐱𝟑

= (𝟏 −𝟏

𝐱)] May-10

Mar-10

C 6. (x + 1)dy

dx− y = (x + 1)2e3x. [

𝐲

𝐱+𝟏=

𝐞𝟑𝐱

𝟑+ 𝐜] Jun-14

C 7. dy

dx+

2y

x= sin x. [𝐲𝐱𝟐 = −𝐱𝟐 𝐜𝐨𝐬 𝐱 + 𝟐𝐱 𝐬𝐢𝐧 𝐱 + 𝟐 𝐜𝐨𝐬 𝐱 + 𝐜]

H 8. dy

dx+ y = x. [𝐲𝐞𝐱 = 𝐞𝐱𝐱 − 𝐞𝐱 + 𝐜] Dec-09

T 9. xdy

dx+ (1 + x)y = x3. [𝐱𝐲ex = 𝐱𝟑ex − 𝟑𝐱𝟐ex + 𝟔𝐱ex − 𝟔ex + 𝐜] Jun-13

C 10. dy

dx+ (tan x)y = cos x ; y(0) = 2. [𝐲 𝐬𝐞𝐜 𝐱 = 𝐱 + 𝟐] Jan-15

H 11. dy

dx+ 2 y tanx = sinx. [𝐲 𝐬𝐞𝐜𝟐𝐱 = 𝐬𝐞𝐜 𝐱 + 𝐜] Jan-15

C 12. dy

dx+ y cot x = 2 cos x. [𝐲𝐬𝐢𝐧 𝐱 = −

𝐜𝐨𝐬 𝟐𝐱

𝟐+ 𝐜]

C 13. dy

dx+

4x

x2+1y =

1

(x2+1)3. [𝐲(𝐱𝟐 + 𝟏)𝟐 = 𝐭𝐚𝐧−𝟏 𝐱 + 𝐜] Dec-13

C 14. (1 + y2)dx

dy= tan−1y − x. [𝐱𝐞𝐭𝐚𝐧−𝟏 𝐲 = 𝐞𝐭𝐚𝐧−𝟏 𝐲(𝐭𝐚𝐧−𝟏 𝐲 − 𝟏) + 𝐜] Jun-13

T 15. ydx − xdy + (logx)dx = 0. [𝐲 + 𝐥𝐨𝐠 𝐱 + 𝟏 = 𝐜𝐱]

T 16. y′ − (1 + 3x−1)y = x + 2, y(1) = e − 1. [𝐲 = −𝐱 + 𝐱𝟑𝐞𝐱] Dec-10

H 17.

y′ +1

3y =

1

3(1 − 2x)x4.

[𝐲𝐞𝐱𝟑 =

𝟏

𝟑𝐞

𝐱𝟑{−𝟔𝐱𝟓 + 𝟗𝟑𝐱𝟒 − 𝟏𝟏𝟏𝟔𝐱𝟑 + 𝟏𝟎𝟎𝟒𝟒𝐱𝟐 − 𝟔𝟎𝟐𝟔𝟒𝐱 + 𝟏𝟖𝟎𝟕𝟗𝟐 + 𝐜}]

Dec-10

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 is real number (except for n = 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−𝐧 𝐝𝐲

𝐝𝐱+ 𝐏(𝐱)𝐲𝟏−𝐧 = 𝐐(𝐱)____(2)

Let 𝐲𝟏−𝐧 = 𝐯

(𝟏 − 𝐧)𝐲−𝐧 𝐝𝐲

𝐝𝐱=

𝐝𝐯

𝐝𝐱⟹ 𝐲−𝐧 𝐝𝐲

𝐝𝐱=

𝟏

(𝟏−𝐧) 𝐝𝐯

𝐝𝐱

Equation (2) becomes 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 "y" ,

We get, 𝟏

𝐠(𝐲)

𝐝𝐲

𝐝𝐱+ 𝐏(𝐱)

𝐟(𝐲)

𝐠(𝐲)= 𝐐(𝐱) ……(4)

Let 𝐟(𝐲)

𝐠(𝐲)= 𝐯

Differentiate with respect to x both the side,

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

Exercise-4

Bernoulli’s Differential Equation

C 1. dy

dx+ y = −

x

y. [𝐲𝟐𝐞𝟐𝐱 = −𝐱𝐞𝟐𝐱 +

𝐞𝟐𝐱

𝟐+ 𝐜] Dec-09

C 2. dy

dx+

y

x= x2y6 [𝐲−𝟓𝐱−𝟓 =

𝟓

𝟐𝐱−𝟐 + 𝐜]

H 3. dy

dx+

1

x=

ey

x2. [𝐞−𝐲

𝐱=

𝟏

𝟐𝐱𝟐 + 𝐜] May-11 Jun-15

H 4. xdy

dx+ ylogy = x y ex. [𝐱 𝐥𝐨𝐠𝐲 = 𝐞𝐱𝐱 − 𝐞𝐱 + 𝐜]

C 5. dy

dx−

tan y

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

𝐬𝐢𝐧 𝐲

𝟏+𝐱= 𝐞𝐱 + 𝐜]

T 6. dy

dx+ xsin2y = x3 cos2 y [𝐞𝐱𝟐

𝐭𝐚𝐧 𝐲 =(𝐱𝟐−𝟏)𝐞𝐱𝟐

𝟐+ 𝐜] Jan-15

C 7. dy

dx− 2y tan x = y2tan2x [

𝐬𝐞𝐱𝟐𝐱

𝐲= −

𝐭𝐚𝐧𝟑𝐱

𝟑+ 𝐜]

C 8. (x3y2 + xy)dx = dy. [𝐞

𝐱𝟐

𝟐

𝐲= (𝟐 − 𝐱𝟐)𝐞

𝐱𝟐

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 i.e. 𝛛𝐌

𝛛𝐲=

𝛛𝐍

𝛛𝐱.

The general solution of Exact Differential Equation is

∫ 𝐌(𝐱)𝐝𝐱𝐲=𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

+ ∫(𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝐍 𝐧𝐨𝐭 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐢𝐧𝐠 𝐱)𝐝𝐲 = 𝐜

Where, c is an arbitrary constant

Exercise-5

Exact Differential Equation

C 1. (x3 + 3xy2)dx + (y3 + 3x2y)dy = 0 [𝐱𝟒

𝟒+

𝟑𝐱𝟐𝐲𝟐

𝟐+

𝐲𝟒

𝟒= 𝐜] Jan-15

H 2. (x2 + y2)dx + 2xydy = 0. [𝐱𝟑

𝟑+ 𝐲𝟐𝐱 = 𝐜]

H 3. 2xydx + x2dy = 0. [𝐱𝟐𝐲 = 𝐜] Dec-09

T 4. yexdx + (2y + ex)dy = 0; y(0) = −1. [𝐲𝐞𝐱 + 𝐲𝟐 = 𝟎] Mar-10 Jun-15

T 5. (ey + 1) cos x dx + ey sin x dy = 0. [(𝐞𝐲 + 𝟏) 𝐬𝐢𝐧 𝐱 = 𝐜]

C 6. Test for exactness and solve :[(x + 1)ex − ey]dx − xeydy = 0, y(1) = 0.

[𝐱(𝐞𝐱 − 𝐞𝐲) = 𝐞 − 𝟏] Jun-14 Dec -11

C 7. dy

dx+

ycosx+siny+y

sinx+xcosy+x= 0. [𝐲 𝐬𝐢𝐧 𝐱 + 𝐱 𝐬𝐢𝐧 𝐲 + 𝐱𝐲 = 𝐜] Jun-13

Definition: 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.

Definition: Integrating Factor

A differential equation which is not exact be made 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∫ f(y) dy

Then find, M∗ = M(I. F. ) & N∗ = N(I. F. ) and solution is

∫ 𝐌∗𝐝𝐱𝐲=𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

+ ∫(𝐭𝐞𝐫𝐦𝐬 𝐨𝐟 𝐍∗ 𝐧𝐨𝐭 𝐜𝐨𝐧𝐭𝐚𝐢𝐧𝐢𝐧𝐠 𝐱)𝐝𝐲 = 𝐜;

Where, c is an arbitrary constant.

Exercise-6

Non-Exact Differential Equation

C 1.

State the necessary & sufficient condition to be exact differential equation.

And using it Solve x2y dx − (x3 + xy2)dy = 0.

[−𝐱𝟐

𝟐𝐲𝟐 + logy = 𝐜]

Jan-13 Jan-15

C 2. (x2y − 2xy2)dx − (x3 − 3x2y)dy = 0 [𝐱

𝐲− 𝟐 𝐥𝐨𝐠 𝐱 + 𝟑𝐥𝐨𝐠𝐲 = 𝐜] Jan-15

H 3. (xy − 2y2)dx − (x2 − 3xy)dy = 0. [𝐱

𝐲− 𝟐𝐥𝐨𝐠 𝐱 + 𝟑 𝐥𝐨𝐠 𝐲 = 𝐜]

T 4. (x3+y3)dx − xy2dy = 0. [𝐥𝐨𝐠𝐱 −𝐲𝟑

𝟑𝐱𝟑 = 𝐜]

C 5. (x2+y2)dx − 2xydy = 0. [𝐱𝟐 − 𝐲𝟐 = 𝐜𝐱]

C 6. (x2y2 + 2)ydx + (2 − x2y2)xdy = 0. [𝐥𝐨𝐠𝐱

𝐲−

𝟏

𝐱𝟐𝐲𝟐 = 𝐜] May-12 Jan-15

C 7. y(1 + xy)dx + x(1 + xy + x2y2)dy = 0. [𝟏

𝟐𝐱𝟐𝐲𝟐 +𝟏

𝐱𝐲− 𝐥𝐨𝐠𝐲 = 𝐜]

H 8. y(xy + 2x2y2)dx + x(xy − x2y2)xdy = 0 [−𝟏

𝐱𝐲+ 𝐥𝐨𝐠

𝐱𝟐

𝐲= 𝐜]

C 9. (x2+y2 + x)dx + xydy = 0. [𝟑𝐱𝟒 + 𝟔𝐱𝟐𝐲𝟐 + 𝟒𝐱𝟑 = 𝟏𝟐𝐜]



Definition: Orthogonal Trajectory

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

Trajectory.

1. Methods of finding orthogonal trajectory of 𝐟(𝐱, 𝐲, 𝐜) = 𝟎

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

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

III. Replace dy

dx by −

dx

dy. This will give you differential equation of the orthogonal trajectories.

IV. Solve the differential equation to get the equation of the orthogonal trajectories.

2. Methods of finding orthogonal trajectory of 𝐟(𝐫, 𝛉, 𝐜) = 𝟎

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

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

III. Replace dr

dθ by −r2 dθ

dr. This will give you differential eqn of the orthogonal trajectories.

IV. Solve the differential equation to get the equation of the orthogonal trajectories.

Exercise-7

Orthogonal Trajectory

C 1. y = x2 + c. [𝐲 = −𝟏

𝟐𝐥𝐨𝐠 𝐱 +

𝐜

𝟐] Dec-10

H 2. y2 + (x − a)2 = a2 [(𝐲 − 𝐛)𝟐 + 𝐱𝟐 = 𝐛𝟐]

T 3. x2 = 4b(y + b). [𝐱𝟐 = 𝟒𝐛(𝐲 + 𝐛)]

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER PAGE | 1


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

P0

dny

dxn+ P1

dn−1y

dxn−1+ P2

dn−2y

dxn−2+ ⋯ + Pny = f(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

a0

dny

dxn+ a1

dn−1y

dxn−1+ a2

dn−2y

dxn−2+ ⋯ + any = f(x) … … … (B)

Where, a0, a1, a2, … are constants.

Notations

Eq. (B) can be written in operator form as below,

a0Dny + a1Dn−1y + a2Dn−2y + ⋯ + any = f(x) … … … (C)

OR

[g(D)]y = f(x) … … … (D)

Note

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

Auxiliary Equation

The auxiliary equation for nth order linear differential equation

a0Dny + a1Dn−1y + a2Dn−2y + ⋯ + any = f(x)

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

i. e. a0mny + a1mn−1y + a2mn−2y + ⋯ + any = 0

Complimentary Function (C.F.) i.e.(𝐲𝐜)

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

Particular Integral (P.I.) i.e. (𝐲𝐏)

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

g(D)f(x).

General Solution [𝐲 (𝐱)] Of Higher Order Linear Differential Equation

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



Note

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

general solution.

Method For Finding C.F. Of Higher Order Differential Equation

Consider,

a0Dny + a1Dn−1y + a2Dn−2y + ⋯ + any = f(x)

The Auxiliary equation is

a0mny + a1mn−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 ≠ m4

≠ ⋯ em1x, em2x, em3x, …

y = c1em1x + c2em2x + c3em3x

+ ⋯

2. m1 = m2 = m

m3 ≠ m4 ≠ ⋯ emx, xemx, em3x, em4x, …

y = (c1 + c2x)emx + c3em3x

+ c4em4x + ⋯

3. m1 = m2 = m3 = m

m4 ≠ m5, …

emx, xemx, x2emx,

em4x, em5x, …

y = (c1 + c2x + c3x2)emx

+ c4em4x + c5em5x

+ ⋯

4.

m1 = p + iq

m2 = p − iq

m3 ≠ m4, …

epx cos qx , epx sin qx,

em3x, em4x, …

y = epx(c1 cos qx + c2 sin qx)+ c2em3x + c3em4x

+ ⋯

5.

m1 = m2 = p + iq

m3 = m4 = p − iq

m5 ≠ m6, …

epx cos qx , xepx cos qx,

epx sin qx , xepx sin qx,

em5x, em6x, …

y = epx[(c1 + c2x) cos qx

+(c3 + c4x) sin qx]

+c5em5x + c6em6x + ⋯



Exercise-1

Solution Of Homogeneous Differential Equation

C 1. y′′ + y′ − 2y = 0

[𝐜𝟏𝐞−𝟐𝐱 + 𝐜𝟐𝐞𝐱] Dec-09

H 2. y′′ + 7y′ − 18y = 0

[𝐜𝟏𝐞−𝟗𝐱 + 𝐜𝟐𝐞𝟐𝐱] Jan-15

C 3. y′′ + y′ − 2y = 0, y(0) = 4, y′(0) = −5.

[𝟑𝐞−𝟐𝐱 + 𝐞𝐱] Dec-09 Jan-15

T 4. y′′ − 9y = 0 ; y(0) = 2, y′(0) = −1.

[𝟓

𝟔𝐞𝟑𝐱 +

𝟕

𝟔𝐞−𝟑𝐱]

Jan-15

T 5. y′′ − 5y′ + 6y = 0; y(1) = e2, y′(1) = 3e2

[𝐞𝟑𝐱−𝟏] May-12

H 6. y′′ − 2√2y′ + 2y = 0.

[(𝐜𝟏 + 𝐜𝟐𝐱)𝐞√𝟐𝐱] Jan-15

T 7. y′′ + 4y′ + 4y = 0; y(0) = 1, y′(0) = 1 .

[(𝟏 + 𝟑𝐱)𝐞−𝟐𝐱] Dec-11

T 8. y′′ − 4y′ + 4y = 0; y(0) = 3, y′(0) = 1

[(𝟑 − 𝟓𝐱)𝐞𝟐𝐱] Jan-15

C 9. d4y

dx4 − 18d2y

dx2 + 81y = 0 . [(𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟑𝐱 + (𝐜𝟑 + 𝐜𝟒𝐱)𝐞−𝟑𝐱] May-12

H 10. (D2 + 1)y = 0 .

[(𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱)]

Dec-11

T 11. 16y′′ − 8y′ + 5y = 0.

[𝐞𝐱

𝟒 (𝐜𝟏 𝐜𝐨𝐬𝐱

𝟐+ 𝐜𝟐 𝐬𝐢𝐧

𝐱

𝟐)]

Dec-11

C 12. (D4 − 1)y = 0.

[𝐜𝟏𝐞−𝐱 + 𝐜𝟐𝐞𝐱 + (𝐜𝟑𝐜𝐨𝐬𝐱 + 𝐜𝟒𝐬𝐢𝐧𝐱)]

Jun-14

H 13. y′′′ − y = 0.

[𝐜𝟏𝐞𝐱 + 𝐞−𝟏

𝟐𝐱 (𝐜𝟐 𝐜𝐨𝐬

√𝟑

𝟐𝐱 + 𝐜𝟑 𝐬𝐢𝐧

√𝟑

𝟐𝐱)]

Jun-13 Jun-15

C 14. y′′′ − y′′ + 100y′ − 100y = 0; y(0) = 4, y′(0) = 11, y′′(0) = −299

[𝐞𝐱 + 𝐬𝐢𝐧 𝟏𝟎𝐱 + 𝟑 𝐜𝐨𝐬 𝟏𝟎𝐱] Dec-11

Method Of Finding The Particular Integral

There are many methods of finding the particular integral 1

f(D) X, We shall discuss

following four main methods,

A. General Methods B. Short-cut Methods involving operators

C. Method of Undetermined Co-efficient

D. Method of Variation of parameters



A. General Methods Consider the differential equation

a0Dny + a1Dn−1y + a2Dn−2y + ⋯ + any = X

It may be written as f(D)y = X

∴ Particular Integral =1

f(D) X

Particular Integral may be obtained by following two ways:

1. Method Of Factors

The operator 1

f(D) X may be factorized into n linear factors; then the particular

integral will be

P. I. =1

f(D) X =

1

(D − m1)(D − m2) … … … … . (D − mn)X

Now, we know that,

1

D − mnX = emnx ∫ X e−mnxdx

On opening 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 ∫ 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) X may be factorized into n linear factors; then the particular

integral will be

P. I. =1

f(D) X = (

A1

D − m1+

A2

D − m2+ ⋯ +

An

D − mn) X

= A11

D−m1X + A2

1

D−m2X + ⋯ + An

1

D−mnX

Using 1

D − mnX = emnx ∫ X e−mnxdx , we get

P. I. = A1em1x ∫ X e−m1xdx + A2em2x ∫ X e−m2xdx + ⋯ + Anemnx ∫ X e−mnxdx

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



B. Shortcut Method

1. F(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 f n−1(a) = 0 ,

P. I. =1

f(D)eax =

xn

f n(a)eax, if f n(a) ≠ 0

2. F(x) = sin(ax + b)

P. I. =1

f(D2)sin(ax + b) =

1

f(−a2)sin(ax + b) , if f(−a2) ≠ 0

If f(−a2) = 0 ,

P. I. =1

f(D2)sin(ax + b) =

x

f′(−a2)sin(ax + b) , if f′(−a2) ≠ 0

If f ′(−a2) = 0 ,

P. I. =1

f(D2)sin(ax + b) =

x2

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

3. F(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 …

4. F(x) = xm; m > 0 In this case convert f (D) in the form of 1 + ϕ(D) or 1 − ϕ(D) form so that we get

P. I. =1

f(D)xm =

1

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

(Using Binomial Theorem) 5. F(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)



Exercise-2

Solution Of Non-Homogeneous Differential Equation

C 1.

d2y

dx2 +dy

dx− 12y = e6x

[(𝐜𝟏𝐞−𝟒𝐱 + 𝐜𝟐𝐞𝟑𝐱) +𝟏

𝟑𝟎𝐞𝟔𝐱]

Jan-13

H 2. (D2 + 5D + 6)y = ex

[𝐜𝟏𝐞−𝟑𝐱 + 𝐜𝟐𝐞−𝟐𝐱 +𝐞𝐱

𝟏𝟐]

Jun-14

H 3. y′′ − 5y′ + 6y = 3e−2x

[(𝐜𝟏𝐞𝟐𝐱 + 𝐜𝟐𝐞𝟑𝐱) +𝟑

𝟐𝟎𝐞−𝟐𝐱]

Jan-15

H 4.

d2y

dx2 − 5dy

dx+ 6y = e4x

[𝐜𝟏𝐞𝟐𝐱 + 𝐜𝟐𝐞𝟑𝐱 +𝐞𝟒𝐱

𝟐]

Dec-13

H 5. y′′ − 3y′ + 2y = ex

[𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞𝟐𝐱 − 𝐱𝐞𝐱] Dec-11

H 6. (D2 − 3D + 2)y = 2ex.

[𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞𝟐𝐱 − 𝟐𝐱𝐞𝐱] May-12

H 7. (D3 − 7D + 6)y = e2x.

[𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞𝟐𝐱 + 𝐜𝟑𝐞−𝟑𝐱 +𝐱

𝟓𝐞𝟐𝐱 ]

Jun-15

C 8. (D2 − 2D + 1)y = 10ex.

[(𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝐱 + 𝟓𝐱𝟐𝐞𝐱] Jun-15

T 9. y′′′ − 3y′′ + 3y′ − y = 4et

[(𝐜𝟏 + 𝐜𝟐𝐭 + 𝐜𝟑𝐭𝟐)𝐞𝐭 +𝟐

𝟑𝐭𝟑𝐞𝐭]

Jan-15

T 10. y′′′ − 3y′′ + 3y′ − y = e−x

[(𝐜𝟏 + 𝐜𝟐𝐱 + 𝐜𝟑𝐱𝟐)𝐞𝐱 −𝟏

𝟖𝐞−𝐱]

H 11. (D2 − 49)y = sinh 3x

[𝐜𝟏𝐞−𝟕𝐱 + 𝐜𝟐𝐞𝟕𝐱 −𝟏

𝟒𝟎𝐬𝐢𝐧𝐡 𝟑𝐱]

Jun-15

C 12. y′′ + 2y′ + 2y = sinh x

[𝐞−𝐱(𝐜𝟏𝐜𝐨𝐬𝐱 + 𝐜𝟐𝐬𝐢𝐧𝐱) +𝟏

𝟏𝟎(𝐞𝐱 − 𝟓𝐞−𝐱)]

C 13. (D2 − 25)y = cos 5x

[𝐜𝟏𝐞−𝟓𝐱 + 𝐜𝟐𝐞𝟓𝐱 −𝟏

𝟓𝟎𝐜𝐨𝐬𝟓𝐱]

Jun-14

T 14. Find the steady state oscillation of the mass-spring system governed by the equation: y′′ + 3y′ + 2y = 20 cos 2t .

[𝟑𝐬𝐢𝐧𝟐𝐭 − 𝐜𝐨𝐬𝟐𝐭] Dec-09

H 15. (D2 + 9)y = cos 2x + sin 2x

[𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 +𝟏

𝟓𝐜𝐨𝐬𝟐𝐱 +

𝟏

𝟓𝐬𝐢𝐧𝟐𝐱]

Jun-15

T 16. (D2 − 4D + 3)y = sin 3x cos 2x.

[𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞𝟑𝐱 +𝐬𝐢𝐧 𝐱+𝟐 𝐜𝐨𝐬 𝐱

𝟐𝟎−

𝟏𝟎 𝐜𝐨𝐬 𝟓𝐱−𝟏𝟏 𝐬𝐢𝐧 𝟓𝐱

𝟖𝟖𝟒]

Dec-13

T 17.

(D4 + 2a2D2 + a4)y = cos ax.

[(𝐜𝟏 + 𝐜𝟐𝐱) 𝐜𝐨𝐬 𝐚𝐱 + (𝐜𝟑 + 𝐜𝟒𝐱) 𝐬𝐢𝐧 𝐚𝐱 −𝐱𝟐

𝟖𝐚𝟐𝐜𝐨𝐬 𝐚𝐱]

May-11



T 18.

(D2 + D − 6)y = e2xsin3x

[𝐜𝟏𝐞−𝟑𝐱 + 𝐜𝟐𝐞𝟐𝐱 −𝐞𝟐𝐱

𝟑𝟎𝟔(𝟏𝟓 𝐜𝐨𝐬 𝟑𝐱 + 𝟗 𝐬𝐢𝐧 𝟑𝐱)]

Jan-13

C 19. (D3 − D2 + 3D + 5)y = ex cos 3x.

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

𝟔𝟓(𝟑 𝐬𝐢𝐧 𝟑𝐱 + 𝟐 𝐜𝐨𝐬 𝟑𝐱)]

May-12

T 20. y′′ + 2y′ + 3y = 2x2

[𝐞−𝐱(𝐜𝟏𝐜𝐨𝐬√𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧√𝟐𝐱) + ( 𝟐

𝟑𝐱𝟐 −

𝟖

𝟗𝐱 −

𝟐𝟖

𝟐𝟕)]

Jan-15

H 21. y′′ + 2y′ + 10y = 25x2 + 3

[𝐞−𝐱(𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱) + (𝟓

𝟐𝐱𝟐 − 𝐱)]

Dec-10

C 22. (D3 − D2 − 6D)y = x2 + 1

[𝐜𝟏 + 𝐜𝟐𝐞𝟑𝐱 −𝐱𝟑

𝟏𝟖+

𝐱𝟐

𝟑𝟔−

𝟐𝟓𝐱

𝟏𝟎𝟖]

Jan-13

T 23.

d4y

dt4 − 2d2y

dt2 + y = cos t + e2t + et

[(𝐜𝟏 + 𝐜𝟐𝐭)𝐞−𝐭 + (𝐜𝟑 + 𝐜𝟒𝐭)𝐞𝐭 + (𝐜𝐨𝐬𝐭

𝟒+

𝐞𝟐𝐭

𝟗+

𝐭𝟐𝐞𝐭

𝟖)]

May-12

H 24. y′′ − 4y = e−2x − 2x, y(0) = 0, y′(0) = 0

[−𝟏

𝟖𝐬𝐢𝐧𝐡 𝟐𝐱 +

𝟏

𝟐𝐱 −

𝟏

𝟒𝐱𝐞−𝟐𝐱]

C 25. (D2 + 16)y = x4 + e3x + cos3x

[(𝐜𝟏 𝐜𝐨𝐬 𝟒𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟒𝐱) +𝟏

𝟏𝟔[𝐱𝟒 −

𝟑

𝟒𝐱𝟐 +

𝟑

𝟑𝟐] +

𝟏

𝟐𝟓𝐞𝟑𝐱 +

𝟏

𝟕𝐜𝐨𝐬 𝟑𝐱]

Jan-15

H 26. y′′ + 4y = 8e−2x + 4x2 + 2; y(0) = 2, y′(0) = 2.

[𝐜𝐨𝐬𝟐𝐱 + 𝟐𝐬𝐢𝐧𝟐𝐱 + 𝐞−𝟐𝐱 + 𝐱𝟐] Mar-10

C. Method of Undetermined Co-efficient

This method determines P.I. of f(D)y = X. In this method we will assume a trial solution containing unknown constants, which will be obtained by substitution in f(D)y = X. The trial solution depends upon X (the RHS of the given equationf(D)y = X).

Let the given equation be f(D)y = 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 = 𝐗 Form of Trial Solution

1. X = eax YP = Aeax

2. X = sin ax

YP = A sin ax + B cos ax X = cos ax

3.

X = a + bx + cx2 + dx3 X = ax2 + bx X = ax + b X = c

YP = A + Bx + Cx2 + Dx3 YP = A + Bx + Cx2 YP = A + Bx YP = A

4. X = eax sin bx

YP = eax(A sin bx + B cos bx) X = eax cos bx

5. X = xeax X = x2eax

YP = eax(A + Bx) YP = eax(A + Bx + Cx2)

6. X = x sin ax X = x2 cos ax

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

7. X = e2x X = e2x − 3e−x

YP = Ae2x YP = Ae2x + Be−x

8. X = cos 3x X = 2 sin(4x − 5)

YP = A sin 3x + B cos 3x YP = A sin(4x − 5) + B cos(4x − 5)

Exercise-3

Solution By Method Of Undetermined Co-Efficient

C 1. y′′ + 4y = 2sin3x

[𝐜𝟏𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐𝐬𝐢𝐧𝟐𝐱 −𝟐

𝟓𝐬𝐢𝐧𝟑𝐱]

Mar-10 Jun-14

H 2. Find particular Integral of Differential Equation y′′ + 9y = cos 5x

[−𝟏

𝟏𝟔𝐜𝐨𝐬𝟓𝐱]

Jun-15

T 3. y′′ + 4y = 8x2 .

[𝐜𝟏 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟐𝐱 + 𝟐𝐱𝟐 − 𝟏] Dec-09

T 4. d2y

dx2 +dy

dx− 6y = 6x + 3x2 − 6x3.

[𝐜𝟏𝐞−𝟑𝐱 + 𝐜𝟐𝐞𝟐𝐱 + 𝐱𝟑]

Jan-13

C 5. y′′ + 1.5y′ − y = 12x2 + 6x3 − x4 , y(0) = 4, y′(0) = −8

[𝟒𝐞−𝟐𝐱 + 𝐱𝟒]

T 6. y′′′ + 3y′′ + 3y′ + y = 30e−x; y(0) = 3, y′(0) = −3, y′′(0) = −47 .

[(𝟑 − 𝟐𝟓𝐱𝟐 + 𝟓𝐱𝟑)𝐞−𝐱] May-11

H 7. y′′ + 1.2y′ + 0.36y = 4e−0.6x, y(0) = 0, y′(0) = 1 [(𝐱 + 𝟐𝐱𝟐)𝐞−𝟎.𝟔𝐱]

C 8.

y′′ − 2y′ + y = ex + x

[(𝐜𝟏 + 𝐱𝐜𝟐)𝐞𝐱 + [𝐱𝟐𝐞𝐱

𝟐+ (𝐱 + 𝟐)]]



Definition: Wronskian

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

W(𝐲𝟏, 𝐲𝟐, … 𝐲𝐧) = ||

𝐲𝟏 𝐲𝟐

𝐲𝟏′ 𝐲𝟐

′ ⋯𝐲𝐧

𝐲′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 for all x ∈ I.

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

Exercise-4

Check Whether L.D. Or L.I.

T 1. x, log x , x(log x)2

[𝐋. 𝐈] May-11

C 2. ex, e−x

[𝐋. 𝐈] Dec-09

D. 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 = X. Where, q and X are the functions of x.

The general solution of second order differential equation by the method of variation of parameters is

y(x) = yc + yp

Where yc = c1y1 + c2y2 and yp(x) = −y1 ∫ y2X

Wdx + y2 ∫ y1

X

Wdx

Where y1and y2 are the solutions of y" + py′ + qy = 0, and W = |y1 y2

y′1 y′2| = y1y′2 − y2y′1 ≠ 0.

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 = P(X)y1 + Q(X)y2 + R(X)y3.



Where,

P(X) = ∫(y2y3′ − y3y2

′ ) X

w dx Q(X) = ∫(y3y1

′ − y1y3′ )

X

w dx R(X) = ∫(y1y2

′ − y2y1′ )

X

w dx

Where, w = |

y1 y2 y3

y1′ y2

′ y3′

y1′′ y2

′′ y3′′

| ≠ 0.

Exercise-5

Solution By Variation OF Parameter

C 1. (D2 − 4D + 4)y =

e2x

x5

[(𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟐𝐱 +𝟏

𝟏𝟐

𝐞𝟐𝐱

𝐱𝟑 ] May-12

T 2. (D2 + 4D + 4)y =

e−2x

x2

[(𝐜𝟏 + 𝐜𝟐𝐱 − (𝟏 + 𝐥𝐨𝐠(𝐱)))𝐞−𝟐𝐱] Mar-10

H 3.

d2y

dx2− 2

dy

dx+ y =

ex

x2

[(𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝐱 − 𝐞𝐱 𝐥𝐨𝐠 𝐱 − 𝐞𝐱] Jan-13

T 4. (D2 − 2D + 1)y = 3x

32ex

[(𝐜𝟏 + 𝐜𝟐𝐱 +𝟏𝟐

𝟑𝟓𝐱

𝟕𝟐) 𝐞𝐱]

Dec-10 Jun-13

T 5. y′′ + 2y′ + y = e−x cos x

[(𝐜𝟏 + 𝐜𝟐𝐱 − 𝐜𝐨𝐬 𝐱)𝐞−𝐱]

T 6. (D2 − 4D + 4)y =

e2x

1 + x2

[(𝐜𝟏 + 𝐜𝟐𝐱)𝐞𝟐𝐱 − 𝐞𝟐𝐱𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝐱𝟐) + 𝐱𝐞𝟐𝐱(𝐭𝐚𝐧−𝟏 𝐱) ]

May-12

C 7. (D2 − 3D + 2)y =

ex

1 + ex

[𝐜𝟏𝐞𝐱 + 𝐜𝟐𝐞𝟐𝐱 + 𝐞𝐱[(𝟏 + 𝐞𝐱) 𝐥𝐨𝐠(𝟏 + 𝐞−𝐱) − 𝟏]] Jan-13

H 8. y′′ + y = sec x .

[𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱 + 𝐱 𝐬𝐢𝐧 𝐱 + 𝐜𝐨𝐬 𝐱 𝐥𝐨𝐠(𝐜𝐨𝐬 𝐱)] Dec-09

H 9. y′′ + 9y = sec 3x.

[𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 +𝟏

𝟑𝐱 𝐬𝐢𝐧 𝟑𝐱 +

𝟏

𝟗𝐜𝐨𝐬 𝟑𝐱 𝐥𝐨𝐠 𝐜𝐨𝐬 𝟑𝐱]

Mar-10 Jan-15 Jun-15

C 10. y′′ + 4y = tan 2x.

[𝐜𝟏 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟐𝐱 −𝟏

𝟒𝐜𝐨𝐬 𝟐𝐱 𝐥𝐨𝐠(𝐬𝐞𝐜 𝟐𝐱 + 𝐭𝐚𝐧 𝟐𝐱) ]

Jun-14

H 11. y′′ + y = cotx .

[𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱 + 𝐬𝐢𝐧 𝐱 𝐥𝐨𝐠(𝐜𝐨𝐬𝐞𝐜𝐱 − 𝐜𝐨𝐭𝐱)] Jun-15



H 12. (D2 + a2)y = cosec ax

[𝐜𝟏𝐜𝐨𝐬𝐚𝐱 + 𝐜𝟐𝐬𝐢𝐧𝐚𝐱 + (−𝐱 𝐜𝐨𝐬 𝐚𝐱

𝐚+

𝟏

𝐚𝟐 𝐬𝐢𝐧 𝐚𝐱 𝐥𝐨𝐠|𝐬𝐢𝐧 𝐱|)] May-11

C 13. d3y

dx3 +dy

dx= cosecx

𝐲 = [𝐥𝐨𝐠(𝐜𝐨𝐬𝐞𝐜 𝐱 − 𝐜𝐨𝐭 𝐱) + 𝐜𝐨𝐬 𝐱 (− 𝐥𝐨𝐠 𝐬𝐢𝐧 𝐱) − 𝐱 𝐬𝐢𝐧 𝐱]

Jan-13 May-12

Cauchy – Euler Equation

An equation of the form

𝑥𝑛𝑑𝑛𝑦

𝑑𝑥𝑛+ 𝑎1𝑥𝑛−1

𝑑𝑛−1𝑦

𝑑𝑥𝑛−2+ 𝑎2𝑥𝑛−2

𝑑𝑛−2𝑦

𝑑𝑥𝑛−2+ ⋯ + 𝑎𝑛−1𝑥

𝑑𝑦

𝑑𝑥+ 𝑎𝑛𝑦 = 𝑋

Where 𝑎1, 𝑎2 , … , 𝑎𝑛 are constants and 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= 𝐷𝑦, 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.

Exercise-6

Cauchy – Euler Equation

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

[𝟑𝒙𝟐 𝒍𝒐𝒈 𝒙] Dec-11

T 2. 𝑥2𝑦′′ − 4𝑥𝑦′ + 6𝑦 = 21𝑥−4 .

[𝒄𝟏𝒙𝟐 + 𝒄𝟐𝒙𝟑 +𝟏

𝟐𝒙−𝟒]

May-11



C 3. (𝑥2𝐷2 − 3𝑥𝐷 + 4)𝑦 = 𝑥2; 𝑦(1) = 1, 𝑦′(1) = 0

[(𝟏 − 𝟐 𝒍𝒐𝒈 𝒙)𝒙𝟐 +𝟏

𝟐𝒙𝟐(𝒍𝒐𝒈 𝒙)𝟐]

May-12

C 4. 𝑥3𝑦′′′ + 2𝑥2𝑦′′ + 2𝑦 = 10 (𝑥 +

1

𝑥).

[𝒄𝟏 𝒙−𝟏 + 𝒙{𝒄𝟐 𝒄𝒐𝒔(𝒍𝒐𝒈 𝒙) + 𝒄𝟑 𝒔𝒊𝒏(𝒍𝒐𝒈 𝒙)} + 𝟓𝒙 + 𝟐𝒙−𝟏 𝒍𝒐𝒈 𝒙]

May-11 Jan-13 Jun-14

H 5. (𝑥2𝐷2 − 3𝑥𝐷 + 3)𝑦 = 3 𝑙𝑛 𝑥 − 4

[𝒄𝟏𝒙 + 𝒄𝟐𝒙𝟑 + 𝒍𝒏 𝒙] Mar-10

T 6. 𝑥2𝐷2𝑦 − 3𝑥𝐷𝑦 + 5𝑦 = 𝑥2 𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)

[𝒙𝟐{𝒄𝟏 𝒄𝒐𝒔(𝒍𝒐𝒈 𝒙) + 𝒄𝟐 𝒔𝒊𝒏(𝒍𝒐𝒈 𝒙)} − 𝒙𝟐𝒍𝒐𝒈𝒙𝒄𝒐𝒔(𝒍𝒐𝒈𝒙)] Jun-15

Solution Of Differential Equation By One Of Its Solution

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

Exercise-7

To Find Another Solution Of Differential Equation

C 1. 𝑥2𝑦′′ − 𝑥𝑦′ + 𝑦 = 0; 𝑦1 = 𝑥.

[𝒚𝟐 = 𝒙𝒍𝒐𝒈𝒙] Dec-10

T 2. 𝑥2𝑦′′ − 4𝑥𝑦′ + 6𝑦 = 0 is 𝑦1 = 𝑥2 ; 𝑥 > 0.

[𝒚𝟐 = 𝒙𝟑] Jan-13

H 3. 𝑥𝑦′′ + 2𝑦′ + 𝑥y = 0, y1 =

sin x

x.

[𝐲𝟐 = −𝐜𝐨𝐬𝐱

𝐱]

May-11

H 4. y′′ + 4y′ + 4y = 0, y1 = e−2x.

[𝐲𝟐 = 𝐱𝐞−𝟐𝐱] Jun-15

4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS PAGE | 1


Definition: Power Series

An infinite series of the form

∑ ak(x − x0)k

∞

k=0

= a0 + a1(x − x0) + a2(x − x0)2 + ⋯

is called a power series in (x − x0).

Definition: Analytic Function

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

Definition: Ordinary and Singular Point

Let P0(x)d2y

dx2 + P1(x)dy

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

efficient.

Dividing by P0(x),

d2y

dx2+

P1(x)

P0(x)

dy

dx+

P2(x)

P0(x)y = 0

Let, P(x) =P1(x)

P0(x) & Q(x) =

P2(x)

P0(x)

∴d2y

dx2+ P(x)

dy

dx+ Q(x)y = 0

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 is further classified into regular singular point and irregular singular

point as follows.

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.

Exercise - 1

Classify The Singularities Of Following Differential Equation

C 1. Define Ordinary Point of the differential equation

y′′ + P(x)y′ + Q(x)y = 0. Jun-14 Jun-15

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

Ans. No Singular Points.



H 3. y′′ + exy′ + sin(x2)y = 0.

Ans. No Singular Points.

T 4. x3y′′ + 5xy′ + 3y = 0 Jan-15

Ans. x = 0 is Irregular Singular Point.

T 5. (1 − x2)y′′ − 2xy′ + n(n + 1)y = 0. Dec-11

Ans. x = 1 & − 1 are Regular Singular Point.

C 6. (x2 + 1)y′′ + xy′ − xy = 0.

Ans. x = i , −i are Regular Singular Point.

H 7. 2x(x − 2)2y′′ + 3xy′ + (x − 2)y = 0. May-12

Ans. x = 0 is Regular Singular Point & x = 2 is Irregular Singular Point.

H 8. 2x2 d2y

dx2+ 6x

dy

dx+ (x + 3)y = 0.

Dec-12

Ans. x = 0 is Regular Singular Point.

C 9. x(x + 1)2y′′ + (2x − 1)y′ + x2y = 0. Jun-15

Ans. x = 0 is Regular Singular Point & x = −1 is Irregular Singular Point.

T 10. Discuss singularities of x3(x − 1)y′′ − 3(x − 1)y′ + xy = 0. Jun-13

Ans. x = 1 is Regular Singular Point & x = 0 is Irregular Singular Point.

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 + a2x2 + a3x3 + a4x4 + a5x5 + ⋯ is

the solution of the given differential equation.

Differentiating with respect to y we get,

𝑦′ =𝑑𝑦

𝑑𝑥= a1 + 2a2x + 3a3x2 + 4a4x3 + 5a5x4+. . . ∞

𝑦′′ =𝑑𝑦

𝑑𝑥= 2a2 + 6a3x + 12a4x2 + 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 + a2 + a3x3 + a4x4 + a5x5+. .. which will be the required solution.



Exercise - 2

Solution By Power Series Method

C 1. y′ + 2xy = 0. Dec-11

Ans. y = a0 − a0 x2 +1

2a0x4 −

1

6a0 x6 + ⋯

H 2. y′ = 2xy. May-11

Jun-15

Ans. y = a0 + a0 x2 +1

2a0x4 +

1

6a0 x6 + ⋯

T 3. y′′ + y = 0.

Dec

09,11,12,13

Jan-13

Jun-14,15

Ans. y = a0 + a1 x −1

2 a0 x2 −

1

6 a1 x3 +

1

24 a0 x4 +

1

120 a1 x5 + ⋯

C 4. y′′ + xy = 0 in powers of x. Jun-14 May-13

Ans. y = a0 + a1 x −1

6a0 x3 −

1

12a1 x4 +

1

180a0 x6 + ⋯

H 5. y′′ + x2y = 0. Dec-13

Jan-15

Ans. y = a0 + a1 x −1

12a0 x4 −

1

20a1 x5 +

1

672a0 x8 +

1

1440a1x9 + ⋯

H 6. y′′ = y′. Mar-10

Ans. y = a0 + a1 x +1

2 a1 x2 +

1

6 a1 x3 +

1

24 a1 x4 +

1

120 a1 x5 + ⋯

H 7. y′′ = 2y′ in powers of x. Jan-13

Ans. y = a0 + a1 x + a1 x2 +2

3a1 x3 +

1

3 a1 x4 +

2

15a1 x5 + ⋯

C 8. y′′ − 2xy′ + 2py = 0.

Ans. y = a0 + a1 x − p a0 x2 +(1 − p)

3a1x3 −

p (2 − p)

6a0 x4 +

(1 − p) (3 − p)

30a1x5 + ⋯

C 9. (1 − x2)y′′ − 2xy′ + 2y = 0. Jun-13 Dec-13 Jan-15

Ans. y = a0 + a1 x − a0 x2 −1

3a0 x4 −

1

5a0x6 + ⋯



T 10. d2y

dx2(1 − x2) − x

dy

dx+ py = 0. May-11

Ans. y = a0 + a1x −p

2 a0x2 +

(1 − p)

6a1 x3 −

p(4 − p)

24a0 x4 +

(9 − p)(1 − p)

120a1 x5 + ⋯

H 11. (1 + x2)y′′ + xy′ − 9y = 0. May-12

Jun-15

Ans. y = a0 + a1 x +9

2a0 x2 +

4

3a1 x3 +

15

8a0 x4 −

7

16a0x6 + ⋯

C 12. (x2 + 1)y′′ + xy′ − xy = 0 near x = 0.

May-11,13 Dec-12 Jan-13 Jun-13

Ans. y = a0 + a1 x + a0

x3

6− a1

x3

6+ (

a1

12) x4 − (

3

40) a0 x5+..

Frobenius Method

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

singular point.

The following steps are useful.

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=1 &

d2y

dx2= ∑ (m + k)(m + k − 1)ak(x − x0)m+k−2

∞

k=2

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.

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 = y_1and y = y2, then the general solution is y = c1y1 + c2y2.



CASE-II Equal roots

In this case, we will have only one series solution. i.e. y = y1 = ∑ ak(x − x0)m+k∞k=0

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

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.

1. 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.

2. 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.

Exercise - 3

Solution By Frobenius Method

C 1. 4xd2y

dx2+ 2

dy

dx+ y = 0. Jan-15

H 2. (x2 − x)y′′ − xy′ + y = 0. Mar-10

C 3. xy′′ + y′ + xy = 0. May-11

H 4. xy′′ + 2y′ + xy = 0. Mar-10

C 5. x2y′′ + xy′ + (x2 − 1)y = 0.

T 6. 2x(1 − x)y′′ + (1 − x)y′ + 3y = 0 Jun-13

C 7. 2x(x − 1)y′′ + (1 + x)y′ + y = 0 ;x = 0 Jun-15

T 8. xy′′ + y′ − y = 0. May-12

C 9. x(x − 1)y′′ + (3x − 1)y′ + y = 0. Dec-11

T 10. x2y′′ + x3y′ + (x2 − 2)y = 0. Dec- 10

5. LAPLACE TRANSFORM AND APPLICATIONS PAGE | 1


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

1. 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

Now, 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 ∙ ∫ e−st∞

0

f(t) dt + β ∙ ∫ e−st∞

0

g(t) dt

= α ∙ ℒ{f(t)} + β ∙ ℒ{g(t)}

Laplace Transform of some Standard Function

1. 𝓛{𝐭𝐧} =𝟏

𝐬𝐧+𝟏 1n ; 𝐧 > −𝟏. Jun-13 ; Dec-13


0f(t) dt

ℒ{tn} = ∫ e−st∞

0

tn dt

Let, st = x ⟹ sdt = dx

When t ⟶ 0 ⇒ x → 0 and t ⟶ ∞ ⇒ x → ∞

⟹ ℒ{tn} = ∫ e−x∞

0

xn

sn

dx

s



=1

sn+1∫ e−x

∞

0

xn dx

=1

sn+1 1n ( By definition of Gamma function n = ∫ e−x∞

0xn−1 dx )

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

⟹ ℒ{tn} =n!

sn+1

2. 𝓛{𝟏} =𝟏

𝐬. Dec-12 ; Jun-14 ; Jan-15


0f(t) dt

ℒ{1} = ∫ e−st∞

0

dt

= [e−st

−s]

0

∞

=0 − 1

−s=

1

s

2. 𝓛{𝐞𝐚𝐭} =𝟏

𝐬−𝐚 , 𝐬 > 𝐚 Jun-15


0f(t) dt

ℒ{eat} = ∫ e−st∞

0

eat dt

= ∫ e−(s−a)t∞

0

dt

= [e−(s−a)t

−(s − a)]

0

∞

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

= [0 − 1

−(s − a)]

⟹ ℒ{eat} =1

s − a



3. 𝓛{𝐞−𝐚𝐭} =𝟏

𝐬+𝐚 , 𝐬 > −𝐚 Jun-13


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

∞

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

= [0 − 1

−(s + a)] =

1

s + a

⟹ ℒ{e−at} =1

s + a

4. 𝓛{𝐬𝐢𝐧 𝐚𝐭} =𝐚

𝐬𝟐+𝐚𝟐 , 𝐬 > 0 and 𝐚 is a constant.


0f(t) dt

ℒ{sin at} = ∫ e−st∞

0

sin at dt

= [e−st

s2 + a2(−s sin at − a cos at)]

0

∞

[When t ⟶ ∞ ⟹ e−st ⟶ 0(∵ s > 0)]

= 0 −1

s2 + a2(−a)

⟹ ℒ{sin at} =a

s2 + a2

5. 𝓛{𝐜𝐨𝐬 𝐚𝐭} =𝐬

𝐬𝟐+𝐚𝟐 , 𝐬 > 0 and 𝐚 is a constant.


0f(t) dt



ℒ{cos at} = ∫ e−st∞

0

cos at dt

= [e−st

s2 + a2(−s cos at + a sin at)]

0

∞

[ When t ⟶ ∞ ⟹ e−st ⟶ 0 (∵ s > 0) ]

= 0 −1

s2 + a2(−s)

⟹ ℒ{cos at} =s

s2 + a2

6. 𝓛{𝐬𝐢𝐧𝐡 𝐚𝐭} =𝐚

𝐬𝟐−𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐(𝐬 > |𝐚|) Dec-11;Dec-12 ; Jun-14 ; Jan-15 ; Jun-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

7. 𝓛{𝐜𝐨𝐬𝐡 𝐚𝐭} =𝐬

𝐬𝟐−𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐(𝐬 > |𝐚|) Dec-13

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



Exercise-1

Laplace Transform

C 1.

Find the Laplace transform of f(t) = {0,0 < t < πsin t , t > π

.

[−𝐞−𝛑𝐬

𝐬𝟐 + 𝟏]

May-12

Jun-15

H 2.

Find the Laplace transform of (t) = {0, 0 ≤ t < 23, when t ≥ 2

.

[𝟑𝐞−𝟐𝐬

𝐬]

Dec-11

H 3.

Find the Laplace transform of (t) = {0, 0 ≤ t < 34, when t ≥ 3

.

[𝟒𝐞−𝟑𝐬

𝐬]

Jan-15

T 4.

Given that f(t) = {t + 1,0 ≤ t ≤ 2

3, t ≥ 2. Find ℒ{f(t)}.

[−𝐞−𝟐𝐬

𝐬𝟐+

𝟏

𝐬+

𝟏

𝐬𝟐]

Dec-13

T 5. Find the Laplace transform of t3 + e−3t + t

1

2.

[𝟑!

𝐬𝟒 +𝟏

𝐬+𝟑+

√𝛑

𝟐𝐬𝟑

𝟐⁄]

Jun-13

C 6.

Find the Laplace transform of 2t3 + e−2t + t4

3.

[𝟏𝟐

𝐬𝟒+

𝟏

𝐬 + 𝟐+

𝟒 3/1

𝟗𝐬𝟕

𝟑⁄]

Dec-13

H 7.

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

[𝟓!

𝐬𝟔+

𝟏

𝐬 + 𝟏𝟎𝟎+

𝐬

𝐬𝟐 + 𝟐𝟓]

Jun-14

C 8.

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

[𝟏

𝐬 − 𝐥𝐨𝐠𝐞𝟏𝟎𝟎+

𝟐 ∙ 𝟏𝟎 !

𝐬𝟏𝟏+

𝟏𝟎

𝐬𝟐 + 𝟏𝟎𝟎]

Jun-15

C 9.

Find ℒ{sin 2t cos 2t}.

[𝟐

𝐬𝟐 + 𝟏𝟔]

Dec-09

Jan-15

T 10.

Find ℒ{sin 2t sin 3t}.

[𝟏𝟐𝐬

(𝐬𝟐 + 𝟐𝟓)(𝐬𝟐 + 𝟏)]

Jun-14

H 11.

Find the Laplace transform of cos22t.

[𝐬𝟐 + 𝟖

𝐬(𝐬𝟐 + 𝟏𝟔)]

Dec-12



H 12.

Find Laplace transform of cos2(at), where a is a constant.

[𝐬𝟐 + 𝟐𝐚𝟐

𝐬(𝐬𝟐 + 𝟒𝐚𝟐)]

Dec-11

C 13.

Find the Laplace transform of following functions: (i) cos3t. (ii) sin2t.

[(𝐢)𝐬(𝐬𝟐 + 𝟕)

(𝐬𝟐 + 𝟏)(𝐬𝟐 + 𝟗), (𝐢𝐢)

𝟐

𝐬(𝐬𝟐 + 𝟒)]

Jun-14

H 14.

Find the Laplace transform of following functions: (i) sin32t. (ii) sin22t.

[(𝐢)𝟒𝟖

(𝐬𝟐 + 𝟒)(𝐬𝟐 + 𝟑𝟔), (𝐢𝐢)

𝟖

𝐬(𝐬𝟐 + 𝟏𝟔)]

T 15.

Find the Laplace transform of (i) cos32t (ii) sin23t.

[(𝐢)𝐬𝟑 + 𝟐𝟖𝐬

(𝐬𝟐 + 𝟒)(𝐬𝟐 + 𝟑𝟔), (𝐢𝐢)

𝟏𝟖

𝐬(𝐬𝟐 + 𝟑𝟔)]

Jan-15

Theorem. First Shifting Theorem

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


0f(t) dt

Now, ℒ{eatf(t)} = ∫ e−steat∞

0f(t) dt

= ∫ e−(s−a)t∞

0f(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).

Exercise-2

First Shifting Theorem

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

[𝟐

(𝐬 − 𝟏)𝟑+

𝟐

(𝐬 − 𝟏)𝟐+

𝟏

𝐬 − 𝟏]

Dec-09

T 2. Find Laplace transform of e−2tsin22t, where a is a constant.

[𝟖

(𝐬 + 𝟐)[(𝐬 + 𝟐)𝟐 + 𝟏𝟔]]

Jun-13

C 3. Find Laplace transform of e−2t(sin24t + t2).

[(𝟑𝟐

(𝐬 + 𝟐)(𝐬𝟐 + 𝟒𝐬 + 𝟔𝟖)+

𝟐

(𝐬 + 𝟐)𝟑)]

Jan-15

H 4. Find Laplace transform of e−3t(2 cos 5t − 3 sin 5t).

[𝟐𝐬 − 𝟗

(𝐬 + 𝟑)𝟐 + 𝟐𝟓]

Jun-14



T 5. Find Laplace transform of e4t(sin 2t cos t).

𝟏

𝟐 [

𝟑

(𝐬𝟐 − 𝟖𝐬 + 𝟐𝟓)+

𝟏

(𝐬𝟐 − 𝟖𝐬 + 𝟏𝟕)]

Jun-15

Theorem. Differentiation of Laplace Transform

Statement: If ℒ{f(t)} = F(s), then show that ℒ{tnf(t)} = (−1)n dn

dsn F(s), n = 1,2,3, …

Proof: By definition, F(s) = ∫ e−st∞

0f(t) dt

⟹ dn

dsn F(s) =

dn

dsn ∫ e−st

∞

0

f(t) dt

= ∫ [ ∂n

∂sn e−st]

∞

0

f(t) dt

= ∫ (−1)(t) ∂n−1

∂sn−1 e−stf(t) dt

∞

0

= ∫ (−1)2(t)2 ∂n−2

∂sn−2 e−stf(t) dt

∞

0

Continuing in this way, we have

= (−1)n ∫ e−st(t)nf(t) dt∞

0

= (−1)nℒ{tnf(t)}

Thus,ℒ{tnf(t)} = (−1)n dn

dsn F(s), n = 1,2,3, …

Exercise-3

Differentiation Of Laplace Transform (Multiplication by tn )

T 1.

Find the value of ℒ{t sin ωt}.

[𝟐𝛚𝐬

(𝐬𝟐 + 𝛚𝟐)𝟐]

Dec-09

H 2.

Find the value of ℒ{t sin t} & ℒ{t sin 2t}

[𝟐𝐬

(𝐬𝟐 + 𝟏𝟐)𝟐,

𝟒𝐬

(𝐬𝟐 + 𝟒)𝟐]

Jun-15

H 3.

Find the Laplace transform of t2 sin πt & t2 sin 2t.

[𝟐𝛑(𝟑𝐬𝟐 − 𝛑𝟐)

(𝛑𝟐 + 𝐬𝟐)𝟑,𝟒(𝟑𝐬𝟐 − 𝟒)

(𝐬𝟐 + 𝟒)𝟑]

Dec-10

Jan-15

Jun-13



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 (∫ e−stds∞

s

) f(t) dt∞

0

= ∫ (e−st

−t)

s

∞

f(t) dt∞

0

= ∫ (0 − e−st

−t) f(t) dt

∞

0

= ∫ e−st [f(t)

t] dt

∞

0

= ℒ {f(t)

t}

Thus, ℒ {f(t)

t} = ∫ F(s)ds

∞

s.

Exercise-4

H 4.

If ℒ{f(t)} = F(s), then show that ℒ{tf(t)} = −d

dsF(s) use this result to

obtained ℒ{eatt sin at}.

[𝟐𝐚(𝐬 − 𝐚)

[(𝐬 − 𝐚)𝟐 + 𝐚𝟐]𝟐]

Dec-13

C 5.

Find the value of ℒ{t cosh t}.

[𝟏 + 𝐬𝟐

(𝐬𝟐 − 𝟏)𝟐]

Jun-14

C 6.

Find the Laplace transform of f(t) = t2 sinh at.

[𝟏

(𝐬 − 𝐚)𝟑−

𝟏

(𝐬 + 𝐚)𝟑]

May-12

H 7.

Find the Laplace transform of f(t) = t2cos h πt & f(t) = t2cos h 3t

[𝟏

(𝐬 − 𝛑)𝟑+

𝟏

(𝐬 + 𝛑)𝟑 ,

𝟏

(𝐬 − 𝟑)𝟑+

𝟏

(𝐬 + 𝟑)𝟑 ]

Jan-15 Jun-15

H 8.

Find the Laplace transform of t3 cosh 2t.

[𝟑

(𝐬 − 𝟐)𝟒+

𝟑

(𝐬 + 𝟐)𝟒]

Dec-12

Integration of Laplace Transform (Division by t)

H 1. Find ℒ {

sin 2t

t}.

[𝛑

𝟐− 𝐭𝐚𝐧−𝟏 (

𝐬

𝟐)]

Dec-12

Jun-14



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

= ∫ 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

In first step,

When t → 0 ⟹ U = ∫ f(u)t

0du → 0 & t → ∞ , ⟹ e−st → 0.

In second step,

By Fundamental theorem of calculus, dU

dt=

d

dt{∫ f(u)

t

0 du} = f(t).

⟹ ℒ {∫ f(t)

t

0

dt} =1

s∫ e−st f(t) dt

∞

0

Thus, ℒ {∫ f(t)t

0 dt} =

1

sF(s).

C 2. Find ℒ {

1−et

t}

[𝐥𝐨𝐠 (𝐬 − 𝟏

𝐬)]

Dec-13

T 3.

Find the Laplace transform of 1−cos t

t.

[𝐥𝐨𝐠 (√𝐬𝟐 + 𝟏

𝐬)]

May-12

H 4. Find the Laplace transform of [

sin wt

t].

[𝛑

𝟐− 𝐭𝐚𝐧−𝟏 (

𝐬

𝛚)]

May-11



Exercise-5

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)dtp

0 , where s > 0.

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

01

= ∫ e−stf(t) dtp

0

+ ∫ e−stf(t) dt∞

p

… (A)

Now, ∫ e−stf(t) dt∞

p

Let, t = u + p ⟹ dt = du

When t ⟶ p ⟹ u ⟶ 0 and t ⟶ ∞ ⟹ u ⟶ ∞.

⟹ ∫ 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

Laplace Transform of integration of a function

C 1.

Find ℒ {∫ e−x cos x dxt

0}.

[𝐬 + 𝟏

𝐬[(𝐬 + 𝟏)𝟐 + 𝟏]]

Jun-13

May-12

C 2.

Find ℒ {∫ ∫ sin aut

0

t

0du du}.

[𝟏

𝐬𝟐

𝐚

(𝐬𝟐 + 𝐚𝟐)]

Jun-13



⟹ 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)dtp

0, where s > 0 is a Laplace Transform of periodic function f(t)

of period p.

Exercise-6

Laplace Transform of Unit Step Function

Show that 𝓛{𝐮(𝐭 − 𝐚)} =𝐞−𝐚𝐬

𝐬


0

Now,

ℒ{u(t − a)} = ∫ e−stu(t − a) dt∞

0

= ∫ e−stu(t − a) dta

0

+ ∫ e−stu(t − a) dt∞

a

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

Laplace Transform of Periodic Function

T 1.

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).

[𝛚

((𝐬𝟐 + 𝐰𝟐)(𝟏 − 𝐞−𝛑𝐬

𝛚⁄ ))]

Dec-12

Mar-10

C 2.

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

[𝛚(𝟏+𝐞

−𝛑𝐬𝛚⁄ )

(𝐬𝟐+𝐰𝟐)(𝟏−𝐞−𝛑𝐬

𝛚⁄ )]

May-11



= ∫ e−st dt∞

a

= [−e−st

s]

a

∞

= [−0 − e−sa

s] =

e−sa

s

Thus, ℒ{u(t − a)} =e−sa

s

Note: Instead of 𝒖(𝐭 − 𝐚) , we can also write 𝐇(𝐭 − 𝐚),which will be Heaviside’s unit step

function.

Theorem. Second Shifting Theorem: Jun-13

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


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−𝑎𝑠ℒ{𝑓(𝑡 + 𝑎)}



Exercise-7

Use of partial fractions

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

Partial fractions =𝐴

(𝑠−𝑎)+

𝐵

(𝑠−𝑏)+

𝐶

(𝑠−𝑐)

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

Partial fractions =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),

then

Partial fractions =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

Partial fractions =𝐴𝑠+𝐵

(𝑠2+𝑎𝑠+𝑏)+

𝐶𝑠+𝐷

(𝑠2+𝑎𝑠+𝑏)2 + ⋯(n times)

Second Shifting Theorem

H 1.

Find the Laplace transform of 𝑒−3𝑡𝑢(𝑡 − 2).

[𝒆−𝟐𝒔𝒆−𝟔

𝒔 + 𝟑]

May-12

H 2.

Find the Laplace transform of 𝑒𝑡𝑢(𝑡 − 2).

[𝒆−𝟐𝒔+𝟐

𝒔 − 𝟏]

Dec-10

C 3. Find the Laplace transform of 𝑡2𝑢(𝑡 − 2).

[𝒆−𝟐𝒔 (𝟐!

𝒔𝟑+

𝟒

𝒔𝟐+

𝟒

𝒔)]

Jan-15

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

[𝟐𝒆−𝒔

𝒔𝟑]

Jun-15

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

[−𝒔𝒆−𝜋𝒔

𝒔𝟐 + 𝟏]



Exercise-8

Inverse Laplace Transform

H 1. Find ℒ−1 {

6𝑠

𝑠2−16}.

[𝟔𝒄𝒐𝒔𝒉𝟒𝒕] Dec-12

C 2.

Find ℒ−1 {3(𝑠2−1)

2

2𝑠5 }.

[𝟑

𝟐[𝟏 − 𝒕𝟐 +

𝒕𝟒

𝟒!]]

Dec-13

T 3. Find ℒ−1 {

𝑠3+2𝑠2+2

𝑠3(𝑠2+1)}.

[𝒕𝟐 + 𝒔𝒊𝒏𝒕]

Mar-10

Jun-13

T 4. Find ℒ−1 {

1

𝑠(𝑠+1)}.

[𝟏 − 𝒆−𝒕] Jan-15

H 5. Find ℒ−1 {

𝑠

𝑠2−3𝑠+2}.

[𝟐𝒆𝟐𝒕 − 𝒆𝒕] Jun-14

C 6.

Find ℒ−1 {1

(𝑠+√2)(𝑠−√3)}.

[𝒆√𝟑𝒕 − 𝒆−√𝟐𝒕

√𝟑 + √𝟐]

Dec-10

Dec-09

Jun-15

H 7.

Find ℒ−1 {1

(𝑠−2)(𝑠+3)}.

𝟏

𝟓[𝒆𝟐𝒕 − 𝒆−𝟑𝒕]

Jun-15

C 8. Find ℒ−1 {

5𝑠2+3𝑠−16

(𝑠−1)(𝑠+3)(𝑠−2)}.

[𝟐𝒆𝒕 + 𝒆−𝟑𝒕 + 𝟐𝒆𝟐𝒕]

Dec-12

Dec-11

T 9.

Find ℒ−1 {3𝑠2+2

(𝑠+1)(𝑠+2)(𝑠+3)}.

[𝟓

𝟐𝒆−𝒕 − 𝟏𝟒𝒆−𝟐𝒕 +

𝟐𝟗

𝟐𝒆−𝟑𝒕]

Jun-13 Jan-15

C 10.

Find ℒ−1 {𝑠3

𝑠4−81}.

[𝒄𝒐𝒔𝟑𝒕 + 𝒄𝒐𝒔𝒉𝟑𝒕

𝟐]

Dec-12

Mar-10

H 11. Find ℒ−1 {−

𝑠+10

𝑠2−𝑠−2}.

[−𝟒𝒆𝟐𝒕 + 𝟑𝒆−𝒕] Mar-10



Exercise-9

Exercise-10

Exercise-11

First Shifting Theorem If 𝓛−𝟏{𝑭(𝒔)} = 𝒇(𝒕), then 𝓛−𝟏{𝑭(𝒔 − 𝒂)} = 𝒆𝒂𝒕𝒇(𝒕)

H 1.

Find ℒ−1 {10

(𝑠−2)4}.

[𝟓𝒆𝟐𝒕𝒕𝟑

𝟑]

Dec-12

C 2. Evaluate ℒ−1 {

3

𝑠2+6𝑠+18}.

[𝒆−𝟑𝒕𝒔𝒊𝒏𝟑𝒕] Dec-09

T 3.

Find the inverse Laplace transform of 6+𝑠

𝑠2+6𝑠+13 , use shifting theorem.

[𝒆−𝟑𝒕𝒄𝒐𝒔𝟐𝒕 +𝟑

𝟐𝒆−𝟑𝒕 𝒔𝒊𝒏 𝟐𝒕]

Dec-11

T 4.

Find ℒ−1 {𝑠

𝑠4+4𝑎4}.

[𝟏

𝟒𝒂𝟐[𝒆𝒂𝒕 𝒔𝒊𝒏 𝒂𝒕 − 𝒆−𝒂𝒕 𝒔𝒊𝒏 𝒂𝒕]]

Dec-13

C 5.

Find the inverse Laplace transform of 5𝑠+3

(𝑠2+2𝑠+5)(𝑠−1).

[−𝒆−𝒕 𝒄𝒐𝒔 𝟐𝒕 +𝟑

𝟐𝒆−𝒕 𝒔𝒊𝒏 𝟐𝒕 + 𝒆𝒕]

May-12

Jun-14

Jan-15

Second Shifting Theorem If 𝓛−𝟏{𝑭(𝒔)} = 𝒇(𝒕), then 𝓛−𝟏{𝒆−𝒂𝒔𝑭(𝒔)} = 𝒇(𝒕 − 𝒂)𝑯(𝒕 − 𝒂)

T 1. Find the inverse Laplace transform of

𝑠𝑒−2𝑠

𝑠2+𝜋2.

[𝒄𝒐𝒔 𝝅(𝒕 − 𝟐) ∙ 𝒖(𝒕 − 𝟐)]

Dec-10

Dec-09

C 2. Find the inverse Laplace transform of

𝑒−4𝑠(𝑠+2)

𝑠2+4𝑠+5.

[𝒆−𝟐(𝒕−𝟒) 𝒄𝒐𝒔(𝒕 − 𝟒) ∙ 𝒖(𝒕 − 𝟒)] Jun-13

Inverse Laplace Transform Of Derivatives

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

T 1.

Find ℒ−1 {𝑙𝑜𝑔𝑠+𝑎

𝑠+𝑏}.

[𝒆−𝒃𝒕 − 𝒆−𝒂𝒕

𝒕]

Jun-13

Dec-10

H 2.

Find ℒ−1 {𝑙𝑜𝑔𝑠+1

𝑠−1}.

[ 𝒆𝒕 − 𝒆−𝒕

𝒕]

Jun-13



𝑡 ⟶ ∞

𝑡 = 𝑢

𝑢 = 0 𝑡

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

Exercise-12

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

0𝑒−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 .

H 3.

Obtain ℒ−1 {𝑙𝑜𝑔1

𝑠}.

[𝟏

𝒕]

May-11

C 4.

Find the inverse transform of the function 𝑙𝑛 (1 +𝑤2

𝑠2 ).

[𝟐

𝒕(𝟏 − 𝒄𝒐𝒔 𝒘𝒕)]

Mar-10

Jun-14

Convolution Product

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

[𝒕]

Dec-09

Jun-15

C 2. Find the convolution of 𝑡 and 𝑒𝑡.

[𝒆𝒕 − 𝒕 − 𝟏]

Dec-10

Jun-15



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 function 𝑓 & 𝑔.

Exercise-13

Convolution Theorem

H 1.

Using convolution theorem, find the inverse Laplace transform of 1

(𝑠2+𝑎2)2.

[𝟏

𝟐𝒂𝟐(

𝒔𝒊𝒏 𝒂𝒕

𝒂− 𝒕 𝒄𝒐𝒔 𝒂𝒕)]

May-11

Dec-10

T 2. ℒ−1 {

𝑠

(𝑠2 + 1)2}

[𝟏

𝟐 𝒕 𝒔𝒊𝒏 𝒕]

Jan-15

T 3.

ℒ−1 {𝑠

(𝑠 + 1) (𝑠 − 1)2}

[𝟏

𝟐𝒕 𝒆𝒕 +

𝒆𝒕

𝟒−

𝒆−𝒕

𝟒]

Jan-15

H 4.

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

(𝑠+1)(𝑠+3)} .

[𝒆−𝒕 − 𝒆−𝟑𝒕

𝟐]

Dec-13



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

= [𝑒−𝑠𝑡 ∫ 𝑓′(𝑡)𝑑𝑡 − ∫ {(𝑑

𝑑𝑡𝑒−𝑠𝑡) ∙ ∫ 𝑓′(𝑡)𝑑𝑡} 𝑑𝑡]

0

∞

= [𝑒−𝑠𝑡𝑓(𝑡) − ∫(−𝑠)𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡]0

∞

= 0 − 𝑓(0) + 𝑠 ∫ 𝑒−𝑠𝑡𝑓(𝑡) 𝑑𝑡∞

0

= s𝐹(𝑠) − 𝑓(0)

H 5. State convolution theorem and using it find ℒ−1 {

𝑠2

(𝑠2+4)(𝑠2+9)} .

[𝟏

𝟓(𝟑 𝒔𝒊𝒏 𝟑𝒕 − 𝟐 𝒔𝒊𝒏 𝟐𝒕)]

Jun-14

C 6. State convolution theorem and using it find ℒ−1 {

𝑠2

(𝑠2+𝑎2)(𝑠2+𝑏2)} .

[𝟏

(𝒂𝟐 − 𝒃𝟐)(𝒂 𝒔𝒊𝒏 𝒂𝒕 − 𝒃 𝒔𝒊𝒏 𝒃𝒕)]

Jun-15

T 7.

Find ℒ−1 {𝑠+2

(𝑠2+4𝑠+5)2}.

[𝒆−𝟐𝒕 𝒕 𝒔𝒊𝒏 𝒕

𝟐]

Jun-13

H 8.

State the convolution theorem on Laplace transform and using it

find ℒ−1 {1

𝑠(𝑠2+4)}.

[𝟏 − 𝒄𝒐𝒔 𝟐𝒕

𝟒]

Dec-12

Dec-09

Jan-15

Jun-15

H 9. Apply convolution theorem to Evaluate ℒ−1 {

𝑠

(𝑠2+𝑎2)2} .

[𝒕 𝒔𝒊𝒏 𝒂𝒕

𝟐𝒂]

Jun-14 Jan-15

T 10.

Find ℒ−1 {1

𝑠(𝑠+𝑎)3} .

[𝟏

𝟐[−

𝒕𝟐𝒆−𝒂𝒕

𝒂−

𝟐𝒕𝒆−𝒂𝒕

𝒂𝟐−

𝟐𝒆−𝒂𝒕

𝒂𝟑+

𝟐

𝒂𝟑]]

Dec-13

C 11.

State convolution theorem and use to evaluate Laplace inverse of 𝑎

𝑠2(𝑠2+𝑎2)

[𝒂𝒕 − 𝒔𝒊𝒏 𝒂𝒕

𝒂𝟐]

Mar-10



Thus, ℒ{𝑓′(𝑡)} = 𝑠𝐹(𝑠) − 𝑓(0).

ℒ{𝑓′′(𝑡)} = 𝑠2𝐹(𝑠) − s𝑓(0) − 𝑓′(0)

In general, 𝓛{𝒇𝒏(𝒕)} = 𝒔𝒏𝑭(𝒔) − 𝒔𝒏−𝟏𝒇(𝟎) − 𝒔𝒏−𝟐𝒇′(𝟎) − 𝒔𝒏−𝟑𝒇”(𝟎) − ⋯ − 𝒇𝒏−𝟏(𝟎).

Exercise-14

Application to Laplace Transform

H 1.

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

[𝟏𝟏

𝟔𝒄𝒐𝒔 √𝟔𝒕 +

𝟏

𝟔]

Dec-12

Jun-15

T 2.

Using Laplace transform solve the differential equation 𝑑2𝑥

𝑑𝑡2 + 2𝑑𝑥

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

[𝒆−𝒕(𝒔𝒊𝒏 𝒕 + 𝒔𝒊𝒏 𝟐𝒕)

𝟑]

May-12

C 3.

Solve the differential equation𝑑2𝑦

𝑑𝑡2 + 4𝑦 = 𝑓(𝑡), 𝑦(0) = 0, 𝑦′(0) = 1 by

Laplace transform where (i)𝑓(𝑡) = {1,0 < 𝑡 < 1

0, 𝑡 > 1 , (ii)𝑓(𝑡) = 𝐻(𝑡 − 2).

[(𝒊)𝒚 =

𝒔𝒊𝒏 𝟐𝒕

𝟐+

𝟏 − 𝒄𝒐𝒔 𝟐𝒕

𝟒−

[𝟏 − 𝒄𝒐𝒔 𝟐(𝒕 − 𝟏)]𝑯(𝒕 − 𝟏)

𝟒

(𝒊𝒊)𝒚 =𝒔𝒊𝒏 𝟐𝒕

𝟐+

𝑯(𝒕 − 𝟐) − 𝒄𝒐𝒔 𝟐(𝒕 − 𝟐) 𝑯(𝒕 − 𝟐)

𝟒

]

May-12

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

[𝒄𝒐𝒔 𝟐𝒕 + 𝟑 𝒔𝒊𝒏 𝟐𝒕] Dec-11

H 5.

Using Laplace transform solve the IVP 𝑦′′ + 𝑦 = 𝑠𝑖𝑛 2𝑡 , 𝑦(0) = 2, 𝑦′(0) = 1.

[𝒚 =𝟓

𝟑𝒔𝒊𝒏 𝒕 −

𝟏

𝟑𝒔𝒊𝒏 𝟐𝒕 + 𝟐 𝒄𝒐𝒔 𝒕]

Dec-10

Jan-15

C 6.

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

[𝒚 = (𝒌

𝟐𝒂𝟐+

𝑩

𝒂) 𝒔𝒊𝒏𝒂 𝒕 + (𝑨 −

𝒌

𝟐𝒂𝒕 ) 𝒄𝒐𝒔 𝒂𝒕]

Mar-10

Jun-14

C 7.

By using the method of Laplace transform solve the IVP :

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

[𝒆−𝒕 𝒕𝟐

𝟐! − 𝒆−𝒕]

Dec-09

H 8.


𝑦′′ + 5𝑦′ + 6𝑦 = 𝑒−𝑡, 𝑦(0) = 0 𝑎𝑛𝑑 𝑦′(0) = −1 .

[𝟏

𝟐𝒆−𝒕 − 𝟐𝒆−𝟐𝒕 +

𝟑

𝟐𝒆−𝟑𝒕]

Jun-14



T 9.


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

[(𝒕

𝟐+

𝟕

𝟒) 𝒆−𝒕 −

𝟑

𝟒𝒆−𝟑𝒕]

Dec-13

H 10.


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

[𝟏

𝟔𝒆𝒕 +

𝟑

𝟐𝒆−𝒕 −

𝟐

𝟑𝒆−𝟐𝒕]

Jun-15

H 11.

Solve using Laplace transforms, 𝑦′′′ + 2𝑦′′ − 𝑦′ − 2𝑦 = 0; where, 𝑦(0) = 1,

𝑦′(0) = 2, 𝑦′′(0) = 2

[𝟏

𝟑[𝟓𝒆𝒕 + 𝒆−𝟐𝒕] − 𝒆−𝒕]

Jan-15

T 12.


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

[𝒚(𝒕) = 𝟐𝒕 + 𝒆𝒕 − 𝒆𝟑𝒕]

Jun-15



Laplace Transform Of Some Standard Functions

1. ℒ{1} =1

𝑠 1. ℒ−1 {

1

𝑠} = 1

2. ℒ{𝑡𝑛} =1

𝑠𝑛+1 1n 𝑶𝑹𝑛!

𝑠𝑛+1 2. ℒ−1 {

1

𝑠𝑛} =

𝑡𝑛−1

n 𝑶𝑹

𝑡𝑛−1

(𝑛 − 1)!

3. ℒ{𝑒𝑎𝑡} =1

𝑠 − 𝑎 3. ℒ−1 {

1

𝑠 − 𝑎} = 𝑒𝑎𝑡

4. ℒ{𝑒−𝑎𝑡} =1

𝑠 + 𝑎 4. ℒ−1 {

1

𝑠 + 𝑎} = 𝑒−𝑎𝑡

5. ℒ{𝑠𝑖𝑛 𝑎𝑡} =𝑎

𝑠2 + 𝑎2 5. ℒ−1 {

1

𝑠2 + 𝑎2} =

1

𝑎𝑠𝑖𝑛 𝑎𝑡

6. ℒ{𝑐𝑜𝑠 𝑎𝑡} =𝑠

𝑠2 + 𝑎2 6. ℒ−1 {

𝑠

𝑠2 + 𝑎2} = 𝑐𝑜𝑠 𝑎𝑡

7. ℒ{𝑠𝑖𝑛ℎ 𝑎𝑡} =𝑎

𝑠2 − 𝑎2 7. ℒ−1 {

1

𝑠2 − 𝑎2} =

1

𝑎𝑠𝑖𝑛ℎ 𝑎𝑡

8. ℒ{𝑐𝑜𝑠ℎ 𝑎𝑡} =𝑠

𝑠2 − 𝑎2 8. ℒ−1 {

𝑠

𝑠2 − 𝑎2} = 𝑐𝑜𝑠ℎ 𝑎𝑡

9. ℒ{𝑒𝑎𝑡𝑡𝑛} =1n

(𝑠 − 𝑎)𝑛+1𝑶𝑹

𝑛!

(𝑠 − 𝑎)𝑛+1 9. ℒ−1 {

1

(𝑠 − 𝑎)𝑛} =

𝑒𝑎𝑡 𝑡𝑛−1

n 𝑶𝑹

𝑒𝑎𝑡 𝑡𝑛−1

(n − 1)!

10. ℒ{eat sin bt} =b

(s − a)2 + b2 10. ℒ−1 {

1

(s − a)2 + b2} =

1

aeat sin bt

11. ℒ{eat cos bt} =s − a

(s − a)2 + b2

11. ℒ−1 {s − a

(s − a)2 + b2} = eat cos bt

12. ℒ{eat sinh bt} =b

(s − a)2 − b2 12. ℒ−1 {

1

(s − a)2 − b2} =

1

beat sinh bt

13. ℒ{eatcosh bt} =s − a

(s − a)2 − a2 13. ℒ−1 {

s − a

(s − a)2 − b2} = eat cosh bt

14. ℒ−1 {s

(s2 + a2)2} =

1

2at sin at 14. ℒ−1 {

1

(s2 + a2)2} =

1

2a3(sin at − at cos at)

6.PARTIAL DIFFERENTIAL EQUATION AND APPLICATIONS PAGE | 1


Definition: Partial Differential Equation

An equation which involves function of two or more variable 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 appeared in differential equation is “Order of D.E”.

e.g. (∂y

∂x)

2

+∂y

∂x+ 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

∂x+ 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

1. By Eliminating Arbitrary Constants

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

Step 1: f(x, y, z, a, b) = 0. ……(1) Step 2: fx(x, y, z, a, b) = 0. ……(2) and fy(x, y, z, a, b) = 0. ……(3)

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

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

2. By Eliminating Arbitrary Functions

Type 1: Consider, the function 𝐟(𝐮, 𝐯) = 𝟎

Step 1: Let, u = F(v). Step 2: Find ux & uy.

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

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

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

Step 1: Find zx & zy.

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

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



Exercise-1

Formation Of Partial Differential Equation

C 1. Form the partial differential equation z = (x − 2)2 + (y − 3)2.

[𝟒𝐳 = 𝐩𝟐 + 𝐪𝟐] Jan-13 Jun-14

C 2. Form the partial differential equation z = (x + a)(y + b).

[𝐳 = 𝐩𝐪] Jun-15

T 3. Eliminate the function f from the relation f(xy + z2, x + y + z) = 0.

[𝐩 + 𝟏

𝐪 + 𝟏=

𝐲 + 𝟐𝐳𝐩

𝐱 + 𝟐𝐳𝐪]

Jun-13

C 4. Form the partial differential equation of f(x + y + z, x2+y2 + z2) = 0.

[𝐩 + 𝟏

𝐪 + 𝟏=

𝐱 + 𝐳𝐩

𝐲 + 𝐳𝐪]

Jan-13 Jun-14 Jun-15

T 5. Form the partial differential equation f(x2 − y2, xyz) = 0.

− [𝐲𝐳 + 𝐱𝐲𝐩

𝐱𝐳 + 𝐱𝐲𝐪= −

𝐱

𝐲]

Jan-15

H 6.

Form partial differential equation by eliminating the arbitrary function from xyz = Ф(x + y + z).

[𝐩 + 𝟏

𝐪 + 𝟏=

𝐲𝐳 + 𝐱𝐲𝐩

𝐱𝐳 + 𝐱𝐲𝐪]

Dec-13

H 7.

Form the partial differential equation by eliminating the arbitrary function from z = f(x2 − y2).

[𝐩

𝐪= −

𝐱

𝐲]

Dec-13

C 8. Form the partial differential equation of z = f (

x

y).

[𝐩

𝐪= −

𝐲

𝐱]

Jan-13

T 9. Form the partial differential equation of z = xy + f(x2 + y2).

[𝐩 − 𝐲

𝐪 − 𝐱=

𝐱

𝐲] Jan-15

C 10. Form the partial differential equation of y = f(x − at) + F(x + at).

𝛛𝟐𝐲

𝛛𝐭𝟐= 𝐚𝟐

𝛛𝟐𝐲

𝛛𝐱𝟐

Jun-15



Exercise-2

Solution Of Partial Differential Equation

C 1.

Solve ∂2u

∂x ∂y= x3+y3.

[𝐮(𝐱, 𝐲) =𝐱𝟒𝐲

𝟒+

𝐱𝐲𝟒

𝟒+ 𝐅(𝐲) + 𝐠(𝐱)]

Jun-14

C 2. Solve

∂2u

∂x ∂t= e−tcosx.

[𝐮(𝐱, 𝐲) = −𝐞−𝐭 𝐬𝐢𝐧 𝐱 + 𝐅(𝐭) + 𝐠(𝐱)] Dec-13

H 3.

Solve ∂3u

∂x2 ∂y= cos(2x + 3y).

[𝐮(𝐱, 𝐲) = − 𝐬𝐢𝐧(𝟐𝐱 + 𝟑𝐲)

𝟏𝟐+ 𝐱𝐅(𝐲) + 𝐆(𝐲) + 𝐡(𝐱)]

Jan-15

T 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 .

[𝐳(𝐱, 𝐲) = 𝐜𝐨𝐬 𝐱 𝐜𝐨𝐬 𝐲 + 𝐜𝐨𝐬 𝐲]

Jun-13

C 5. Solve

∂2z

∂x2 = z.

[𝐮(𝐱, 𝐲) = 𝐟(𝐲)𝐞𝐱 + 𝐠(𝐲)𝐞−𝐱]

Jan-13 Jan-15

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.

1. Method obtaining general solution of 𝐏𝐩 + 𝐐𝐪 = 𝐑

Step-1: From the A.E. dx

P=

dy

Q=

dz

R.

Step-2: Solve this A.E. by the method of grouping or by the method of multiples or both to get two independent solution u(x, y, z) = c1 and v(x, y, z) = c2.

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

Following two methods will be used to solve Langrage’s linear Equation

2. Grouping Method

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.

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

two fractions dx

Q=

dy

R, then we can apply grouping method.



3. Multipliers Method

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

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 + nR

Such that, lP + mQ + nR = 0.

This implies ldx + mdy + ndz = 0

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

Again we may find another set of multipliers l′, m′, n′

So that, l′P + m′Q + n′R = 0

This gives, l′dx + m′dy + n′dz = 0

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

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

Exercise-3

Solution Of Lagrange’s Differential Equation

C 1. Solve (z − y)p + (x − z)q = y − x.

[𝐅(𝐱 + 𝐲 + 𝐳, 𝐱𝟐+𝐲𝟐 + 𝐳𝟐) = 𝟎] Jun-15

C 2. Solve x(y − z)p + y(z − x)q = z(x − y).

[𝐅(𝐱 + 𝐲 + 𝐳, 𝐱𝐲𝐳) = 𝟎] Jun-13

T 3.

Solve(x2 − y2 − z2)p + 2xyq = 2xz.

[𝐅 (𝐲

𝐳,𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐

𝐳) = 𝟎]

Jun-14

T 4. Solve (y + z)p + (x + z)q = x + y.

[𝐅 (𝐱 − 𝐲

𝐲 − 𝐳, (𝐱 + 𝐲 + 𝐳)(𝐱 − 𝐲)𝟐) = 𝟎] Jan-15



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.

Step 1: Substitute p = a & q = b. Step 2: Convert b = g(a). Step 3: Complete Solution : z = ax + by + c ⟹ z = ax + g(a) + c

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

Step 1: f(x, p) = g(y, q) = a Step 2: Solving equations for p & q. Assume p = F(x) & q = G(y). Step 3: Complete Solution : z = ∫ F(x) dx + ∫ G(y) dy + b.

Type 3: Equation Of the form 𝐳 = 𝐩𝐱 + 𝐪𝐲 + 𝐟(𝐩, 𝐪)

Step 1: Find value of p & q. Step 2: Complete Solution : z = ax + by + f(a, b).

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

Step 1: Assume q = ap Step 2: Solve the Equation in dz = p dx + q dy

Charpit’s Method

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

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 )

Step 2: Find value of p & q. Step 3: Complete Solution : z = ∫ p dx + ∫ q dy + c.

Exercise-4

Non Linear Equation Of First Order

C 1. Solve p + q2 = 1.

[𝐳 = 𝐚𝐱 ± (√𝟏 − 𝐚)𝐲 + 𝐜] Jan-13

H 2. Solve √p + √q = 1

[𝐳 = 𝐚𝐱 + (𝟏 − √𝐚)𝟐

𝐲 + 𝐜] Jan-15

T 3.

Solve p2+q2 = npq.

[𝐳 = 𝐚𝐱 +𝐧𝐚 ± 𝐚√𝐧𝟐 − 𝟒

𝟐𝐲 + 𝐜]

Jan-15

C 4. Solve p2 + q2 = x + y.

[𝐳 =𝟐

𝟑(𝐚 + 𝐱)

𝟑𝟐 +

𝟐

𝟑(𝐲 − 𝐚)

𝟑𝟐 + 𝐛]

Jan-14



H 5. Solve p2 − q2 = x − y.

[𝐳 =𝟐

𝟑(𝐚 + 𝐱)

𝟑𝟐 +

𝟐

𝟑(𝐚 + 𝐲)

𝟑𝟐 + 𝐛]

Jan-15

T 6.

Solve p − x2 = q + y2.

[𝐳 = 𝐚𝐱 +𝐱𝟑

𝟑+ 𝐚𝐲 −

𝐲𝟑

𝟑+ 𝐛]

Jun-15

T 7. Solve z = px + qy + p2q2.

[𝐳 = 𝐚𝐱 + 𝐛𝐲 + 𝐚𝟐𝐛𝟐] Jun-13

C 8. Solve z = px + qy − 2√pq.

[𝐳 = 𝐚𝐱 + 𝐛𝐲 − 𝟐√𝐚𝐛] Dec-13

H 9. Solve qz = p(1 + q).

[𝐛𝐳 − 𝟏 = 𝐜 𝐞𝐱+𝐚𝐲] Jun-14

C 10. Solve pq = 4z.

[𝟒𝐛𝐳 = (𝟐𝐱 + 𝟐𝐛𝐲 + 𝐛)𝟐] Jun-15

Method Of Separation Of Variables

Step 1: Let u(x, y) = X(x) ∙ Y(y)

Step 2: Find ∂u

∂x,

∂u

∂y,

∂2u

∂x2 ,∂2u

∂x ∂y,

∂2u

∂y2 as requirement and substitute in given Partial Differential

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

individually. Step 4: Solve each Ordinary Differential Equation. Step 5: Put value of X(x) & Y(y) in equation u(x, y) = X(x) ∙ Y(y).

Exercise-5

Method Of Separation Of Variables

C 1.

Solve the equation by method of separation of variables ∂u

∂x= 4

∂u

∂y ,

where u(0, y) = 8e−3y. [𝐮(𝐱, 𝐲) = 𝟖 𝐞−𝟏𝟐𝐱−𝟑𝐲]

Dec-13

C 2. Solve 2

∂u

∂x=

∂u

∂t+ u subject to the condition u(x, 0) = 4e−3x

[𝐮(𝐱, 𝐭) = 𝟒 𝐞−𝟑𝐱−𝟕𝐭] Jan-13

H 3. Solve

∂u

∂x= 2

∂u

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

[𝐮(𝐱, 𝐭) = 𝟔 𝐞−𝟑𝐱−𝟐𝐭] Jun-15



C 4. Using method of separation of variables solve

∂u

∂x+

∂u

∂y= 2(x + y)u.

[𝐮(𝐱, 𝐲) = 𝐞𝐱𝟐+𝐲𝟐+𝐤𝐱−𝐤𝐲+𝐜]

Jan-15 Jun-15

T 5.

Using the method of separation variables, solve the partial differential equation uxx = 16uy .

[𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞√𝐤𝐱 + 𝐜𝟐𝐞−√𝐤𝐱) 𝐜𝟑 𝐞𝐤𝐲𝟏𝟔]

Dec-10

H 6. Using method of separation of variables solve

∂2u

∂x2 =∂u

∂y+ 2u.

[𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞√𝐤𝐱 + 𝐜𝟐𝐞−√𝐤𝐱) 𝐜𝟑 𝐞(𝐤−𝟐)𝐲] Jun-13

C 7. Solve

∂2z

∂x2 − 2∂z

∂x+

∂z

∂y= 0 by the method of separation variables.

[𝐳(𝐱, 𝐲) = (𝐜𝟏𝐞(𝟏+√𝟏+𝐤)𝐱 + 𝐜𝟐𝐞(𝟏−√𝟏+𝐤)𝐱) 𝐜𝟑 𝐞−𝐤𝐲]

May-12 Jun-14

C 8.

Solve two dimensional Laplace’s equation ∂2u

∂x2 +∂2u

∂y2 = 0,using the method

separation of variables.

[𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞√𝐤𝐱 + 𝐜𝟐𝐞−√𝐤𝐱) (𝐜𝟑 𝐜𝐨𝐬 √𝐤𝐲 + 𝐜𝟒 𝐬𝐢𝐧 √𝐤𝐲)]

Jan-13 Dec-09

T 9.

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

equation ∂2u

∂x2 = 16∂2u

∂y2.

[𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞√𝐤𝐱 + 𝐜𝟐𝐞−√𝐤𝐱) (𝐜𝟑𝐞√𝐤𝟒

𝐲 + 𝐜𝟒𝐞−√𝐤𝟒

𝐲)]

Jan-15

C 10. Solve x

∂u

∂x− 2y

∂u

∂y= 0 using method of separation variables.

[𝐮(𝐱, 𝐲) = 𝐜𝟏 𝐜𝟐 𝐱𝐤 𝐲𝐤𝟐]

Jun-13

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