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1. INTRODUCTION TO SOME SPECIAL FUNCTIONS

PAGE | 1

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 1

If m > 0, n > 0, then Beta function is defined by the integral ∫0 x m−1 (1 − x)n−1 dx and is denoted by β(m, n). 𝟏

𝐢. 𝐞. 𝛃(𝐦, 𝐧) = ∫ 𝐱 𝐦−𝟏 (𝟏 − 𝐱)𝐧−𝟏 𝐝𝐱 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 ∫02 sin2m−1 θ cos2n−1 θdθ



∫02 sinp θ cosq θdθ = 2 B (

π

π

1

p+1 q+1 2

,

)

2

2. Gamma Function ∞

If n > 0, then Gamma function is defined by the integral ∫0 e−x x n−1 dx and is denoted by ⌈n. ∞

𝐢. 𝐞. ⌈𝐧 = ∫ 𝐞−𝐱 𝐱 𝐧−𝟏 𝐝𝐱 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 ∫0 e−x x 2m−1 dx = 2 ⌈m



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



∫02 sinp θ cosq θdθ = 2

∞

2

1

⌈m⌈n

p+1

π

q+1

1 ⌈( 2 )⌈( 2 ) p+q+2 ⌈( ) 2

1



⌈ 2 = √π



⌈

n+1 2

=

(2n)!√π n!4n

for n = 0,1,2,3, …

Special cases 1

For n = 0, ⌈2 = √π

A.E.M.(2130002)

3

For n = 1, ⌈2 =

√π 2

5

For n = 2, ⌈2 =

3√π 4

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

1. INTRODUCTION TO SOME SPECIAL FUNCTIONS

PAGE | 2

3. Error Function and Complementary Error Function The error function of x is defined by the integral

x

2

2

∫ e−t dt, where x may be real or complex √π 0

variable and is denoted by erf(x). i. e. erf(x) =

x

2

2

√π

∫ e−t dt 0

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

2 √π

∞ 2

∫ e−t dt x

Properties 

erf(0) = 0



erf(∞) = 1



erf(x) + erfc (x) = 1



erf(−x) = −erf(x)

4. Unit Step Function u(x-a)

The Unit Step Function is defined by u(x − a) = {

1, for x ≥ a , where a ≥ 0. 0, for x < a

1

0

x a

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

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

1. INTRODUCTION TO SOME SPECIAL FUNCTIONS 6. Sinusoidal Pulse Function

PAGE | 3

f(x)

The sinusoidal pulse function is defined by f(x) = {

sin ax , for 0 ≤ x ≤ 0, for x >

π

π

a

.

a

π a

0

f(x)

7. Rectangle Function 1

A Rectangular function f(x) defined on ℝ as f(x) = {

1, for a ≤ x ≤ b . 0, otherwise a

0

x

b

f(x)

8. Gate Function A Gate function fa (x) defined on ℝ as fa (x) = {

1, for |x| ≤ a . 0, for |x| > a 1

Note that gate function is symmetric about axis of co-domain. Gate function is also a rectangle function. 0

-a

x

a

fε (x)

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

1⁄ ε

1 fε (x) = { ε , for 0 ≤ x ≤ ε . 0, for x > ε ε

A.E.M.(2130002)

x

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

1. INTRODUCTION TO SOME SPECIAL FUNCTIONS

PAGE | 4 f(x)

10. Signum Function The Signum function is defined by f(x) = {

1

1, for x > 0 . −1, for x < 0 0 −1

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. f(x)

12. Square Wave Function

1

A square wave function f(x) of period 2a is defined by 3a

a

1, for 0 < x < a f(x) = { . −1, for a < x < 2a

0

-a

2a

−1

f(x)

13. Saw Tooth Wave Function A saw tooth wave function f(x) with period a is defined as f(x) = x ; 0 ≤ x < a.

1

0

A.E.M.(2130002)

a

2a

3a

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

1. INTRODUCTION TO SOME SPECIAL FUNCTIONS

PAGE | 5

14. Triangular Wave Function A Triangular wave function f(x) having period 2a

f(x)

is defined by f(x) = {

x ; 0≤x 0, k > 0). ∞

Ans. f(x) = ∫ 0

H

5.

PAGE | 15 Mar-10

2k cos ωx dω + ω2 )

π(k 2

x; 0 π.

7.

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

Mar-10

∞

2 a) f(x) = ∫ (sin ω − sin 2ω) cos ωx dω πω T

0 ∞

Ans. b) f(x) =

2 ∫ (cos 2ω − cos ω) sin ωx dω πω 0

T

8.

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

a) f(x) = ∫ Ans.

0 ∞

b) f(x) = ∫ 0

A.E.M.(2130002)

sin x ; 0 ≤ x ≤ π 0 ; x > π.

2(1 + cos ωπ) cos ωx dω ; A(1) = 0 π (1 − ω2 ) 2 sin ωπ sin ωx dω ; B(1) = 1 π (1 − ω2 )

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 1

Definition: Differential Equation An eqn which involves differential co-efficient is called Differential Equation. e.g.

d2 y dx2

dy

+ x 2 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”. dy 2

dy

e.g. (dx) + 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.”. dy 2

dy

e.g. (dx) + dx + 5y = 0 Has degree 2. Exercise-1

Order And Degree Differential Equation 1

C

1.

d2 y dx2

dy 2 4

= [y + (dx) ] . 1 2

dy

2.

[dx + y] = sin x.

C

3.

y = x dx +

C

4.

(dx2 ) = [x + sin (dx)] .

T

dy

H

5.

A.E.M.(2130002)

d2 y

𝑑2 𝑦 𝑑𝑥 2

x dy dx

.

3

𝑑𝑦

May-11

[𝟏, 𝟏]

May-11

[𝟏, 𝟐] dy

= ln (𝑑𝑥 ) + 𝑦.

[𝟐, 𝟒]

2

[𝟐, 𝐔𝐧𝐝𝐞𝐟𝐢𝐧𝐞𝐝] [𝟐, 𝐔𝐧𝐝𝐞𝐟𝐢𝐧𝐞𝐝]

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 2

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

6.

d3 y

d2 y

of differential equation √x 2 dx2 + 2y = 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. 2. 3.

d2 y dx2 d2 y dx2 d2 y dx2

dy

+ x 2 dx + y = 0 Is linear. dy

+ y dx + y = 0 Is non-linear. dy 2

+ x 2 (dx) + y = 0 Is non-linear.

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

i.e. dx + P(x)y = Q(x) + c OR

A.E.M.(2130002)

dx dy

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

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 3

Type Of First Order Differential Equation I. II. III. IV. V.

Variable Separable Equation Homogeneous Differential Equation Exact Differential Equation Linear(Leibnitz’s) Differential Equation Bernoulli’s Equation

Variable Separable Equation dy

If differential equation of type 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.

H

2.

ex dx − ey dy = 0

C

3.

C

4.

y ′ = ex−y + xe−y

C

5.

xy ′ + y = 0 ; y(2) = −2.

C

6.

L

H

7.

(1 + x)ydx + (1 − y)xdy = 0.

[𝐥𝐨𝐠(𝐱𝐲) + 𝐱 − 𝐲 = 𝐜]

T

8.

ex tanydx + (1 − ex )sec 2ydy = 0.

[(𝟏 − 𝐞𝐱 )−𝟏 𝐭𝐚𝐧 𝐲 = 𝐜]

C

9.

xy ′ = y 2 + y.

dy

= e2x+3y dx

A.E.M.(2130002)

dI

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

[𝟗𝐲 𝟐 + 𝟒𝐱 𝟐 = 𝐜]

Dec-11

[𝐞𝐱 = 𝐞𝐲 + 𝐜] 𝐞−𝟑𝐲

[ −𝟑 =

𝐞𝟐𝐱

+ 𝐜]

Jun-14

+ 𝐜]

Jun-15

[𝐱𝐲 = −𝟒]

Dec-11

[𝐞𝐲 = 𝐞𝐱 +

𝟐 𝐱𝟐 𝟐

𝐑

[𝐈 = 𝐈𝟎 𝐞− 𝐋 𝐭 ]

Dec-10

Jun-13

𝐲

[𝐲+𝟏 = 𝐱𝐜]

Dec-10

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER dy

[𝐲 − 𝐥𝐨𝐠(𝟏 + 𝐲) = 𝐥𝐨𝐠 𝐱 + 𝐱 + 𝐜]

H

10.

xy

C

11.

tany dx = sin(x + y) + sin(x − y).

C

12.

1 + dx = ex+y .

H

13.

T

14.

(x + y)2 dx = a2 .

C

15.

x dx = y + xe x .

T

16.

H

17.

(x + y)dx + (y − x)dy = 0.

C

18.

[1 + ey ] dx + ey [1 − y] dy = 0.

C

19.

(x + y)2 [x

dx

PAGE | 4

= 1 + x + y + xy. dy

[𝐬𝐞𝐜 𝐲 = −𝟐 𝐜𝐨𝐬 𝐱 + 𝐜]

dy

[−(𝐞−𝐱−𝐲 ) = 𝐱 + 𝐜]

dy

= cosx cosy − sinx siny. dx

[𝐭𝐚𝐧 (

dy

dy dx

y

𝐱+𝐲

𝐚

= 𝐜]

𝐲

[𝐬𝐢𝐧 𝐱 = 𝐱𝐜]

x

dx

𝐱+𝐲

Jan- 13

𝐲

y

dy

) = 𝐱 + 𝐜]

[− (𝐞−𝐱 ) = 𝐥𝐨𝐠 𝐱 + 𝐜]

= x + tan x.

x

𝟐

[𝐲 − 𝐚 𝐭𝐚𝐧−𝟏

y

dy

May-12

𝐲

[𝐥𝐨𝐠(𝐱 𝟐 + 𝐲 𝟐 ) = 𝟐𝐭𝐚𝐧−𝟏 (𝐱) + 𝐜]

Jan-15 Jun-15

𝐱

x

[𝐱 + 𝐲𝐞𝐲 = 𝐜]

dy

Jun-15

𝟏

+ y] = xy [1 + dx].

[𝐥𝐨𝐠 𝐱𝐲 = − 𝐱+𝐲 + 𝐜]

May-12

Leibnitz’s Linear Differential Equation 𝐝𝐲

A differential equation of the form 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.

H

2.

C

3.

H

4.

A.E.M.(2130002)

dy dx dy dx

− y = e2x .

[𝐲𝐞−𝐱 = 𝐞𝐱 + 𝐜]

+ 2xy = e−x

2.

y ′ + y sin x = ecos x . dy

1

1

+ x2 y = 6ex . dx

Dec-09

𝟐

[𝐲𝐞𝐱 = 𝐱 + 𝐜] [𝐲𝐞− 𝐜𝐨𝐬 𝐱 = 𝐱 + 𝐜] −𝟏

[𝐲𝐞 𝐱 = 𝟔𝐱 + 𝐜]

Dec-11 Jan-13 Jun-15

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER e−2x

3

5.

C

6.

(x + 1)

C

7.

H

8.

T

9.

C

10.

H

11.

C

12.

C

13.

+ x2 +1 y = (x2 +1)3. dx

C

14.

(1 + y 2 ) dy = tan−1 y − x.

T

15.

ydx − xdy + (logx)dx = 0.

T

′ −1 16. y − (1 + 3x )y = x + 2, y(1) = e − 1.

+ dx

dy dx

2y x

dy dx

𝐲

− y = (x + 1)2 e3x . dx

[𝐱+𝟏 =

Jun-14

[𝐱𝐲ex = 𝐱 𝟑 ex − 𝟑𝐱 𝟐 ex + 𝟔𝐱ex − 𝟔ex + 𝐜]

Jun-13

[𝐲 𝐬𝐞𝐜 𝐱 = 𝐱 + 𝟐]

Jan-15

[𝐲 𝐬𝐞𝐜 𝟐 𝐱 = 𝐬𝐞𝐜 𝐱 + 𝐜]

Jan-15

+ 2 y tanx = sinx.

dy

+ y cot x = 2 cos x. dx

[𝐲𝐬𝐢𝐧 𝐱 = −

1

𝐜𝐨𝐬 𝟐𝐱 𝟐

+ 𝐜]

[𝐲(𝐱 𝟐 + 𝟏)𝟐 = 𝐭𝐚𝐧−𝟏 𝐱 + 𝐜]

dx

1

+ 𝐜]

Dec-09

+ (tan x)y = cos x ; y(0) = 2.

4x

𝟑

[𝐲𝐞𝐱 = 𝐞𝐱 𝐱 − 𝐞𝐱 + 𝐜]

+ y = x.

dy

𝐞𝟑𝐱

May-10 Mar-10

[𝐲𝐱 𝟐 = −𝐱 𝟐 𝐜𝐨𝐬 𝐱 + 𝟐𝐱 𝐬𝐢𝐧 𝐱 + 𝟐 𝐜𝐨𝐬 𝐱 + 𝐜]

= sin x.

dy

dx

𝐱

dy

x dx + (1 + x)y = x 3 . dy

[𝐲𝐞𝟐𝐱 = (𝟏 − )]

, y(1) = 0.

x2

𝟏

𝟑

T

y ′ + 6x 2 y =

dy

PAGE | 5

−𝟏 𝐲

[𝐱𝐞𝐭𝐚𝐧

= 𝐞𝐭𝐚𝐧

−𝟏 𝐲

(𝐭𝐚𝐧−𝟏 𝐲 − 𝟏) + 𝐜]

Dec-13 Jun-13

[𝐲 + 𝐥𝐨𝐠 𝐱 + 𝟏 = 𝐜𝐱] [𝐲 = −𝐱 + 𝐱 𝟑 𝐞𝐱 ]

Dec-10

𝟏 𝐱 𝐞𝟑 {−𝟔𝐱𝟓 + 𝟗𝟑𝐱 𝟒 − 𝟏𝟏𝟏𝟔𝐱 𝟑 + 𝟏𝟎𝟎𝟒𝟒𝐱 𝟐 − 𝟔𝟎𝟐𝟔𝟒𝐱 + 𝟏𝟖𝟎𝟕𝟗𝟐 + 𝐜}] 𝟑

Dec-10

1

y ′ + 3 y = 3 (1 − 2x)x 4 . H

17.

𝐱

[𝐲𝐞𝟑 =

Bernoulli’s Differential Equation 𝐝𝐲

𝐝𝐱

A differential equation of the form 𝐝𝐱 + 𝐏(𝐱)𝐲 = 𝐐(𝐱)𝐲 𝐧 OR𝐝𝐲 + 𝐏(𝐲)𝐱 = 𝐐(𝐲)𝐱 𝐧 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)𝐲 𝐧 _____(1)

Dividing both sides of equation (1) by 𝐲 𝐧 ,

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 6

𝐝𝐲

We get, 𝐲 −𝐧 𝐝𝐱 + 𝐏(𝐱)𝐲 𝟏−𝐧 = 𝐐(𝐱)____(2) Let 𝐲 𝟏−𝐧 = 𝐯 𝐝𝐲

𝐝𝐯

𝐝𝐲

𝟏

 (𝟏 − 𝐧)𝐲 −𝐧 𝐝𝐱 = 𝐝𝐱 ⟹ 𝐲 −𝐧 𝐝𝐱 = (𝟏−𝐧) 𝟏

Equation (2) becomes (𝟏−𝐧)

𝐝𝐯

𝐝𝐯 𝐝𝐱 𝐝𝐯

+ 𝐏(𝐱)𝐯 = 𝐐(𝐱) ⟹ 𝐝𝐱 + 𝐏(𝐱)(𝟏 − 𝐧)𝐯 = 𝐐(𝐱)(𝟏 − 𝐧) 𝐝𝐱

Which is Linear Differential equation and accordingly can be solved. 

𝐝𝐲

CASE 2 : A differential of form 𝐝𝐱 + 𝐏(𝐱)𝐟(𝐲) = 𝐐(𝐱)𝐠(𝐲) ……(3)

Dividing both sides of equation (3) by "y" , 𝟏 𝐝𝐲

𝐟(𝐲)

We get, 𝐠(𝐲) 𝐝𝐱 + 𝐏(𝐱) 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 dy

x

C

1.

+ y = − y. dx

C

2.

H

3.

dy

H

4.

x dx + ylogy = x y ex .

C

5.

T

6.

C

7.

dy dx

[𝐲 𝟐 𝐞𝟐𝐱 = −𝐱𝐞𝟐𝐱 +

y

dy dx

−

ey

tan y 1+x

+ xsin2y = x 3 cos 2 y dx

dy dx

[

𝐞−𝐲 𝐱

8.

A.E.M.(2130002)

y + xy)dx = dy.

May-11 Jun-15

𝐬𝐢𝐧 𝐲

𝟐

[𝐞𝐱 𝐭𝐚𝐧 𝐲 = [

𝐬𝐞𝐱 𝟐 𝐱 𝐲

(𝐱 𝟐 −𝟏)𝐞 𝐱

[

𝐞𝟐 𝐲

𝟐

𝟐

=−

𝐱𝟐

C

𝟏

= 𝟐𝐱𝟐 + 𝐜]

[ 𝟏+𝐱 = 𝐞𝐱 + 𝐜]

− 2y tan x = y 2 tan2 x

(x 3 2

Dec-09

[𝐱 𝐥𝐨𝐠𝐲 = 𝐞𝐱 𝐱 − 𝐞𝐱 + 𝐜]

= (1 + x)ex sec y.

dy

+ 𝐜]

[𝐲 −𝟓 𝐱 −𝟓 = 𝟐 𝐱 −𝟐 + 𝐜]

+ x = x2 . dx dy

𝟐

𝟓

+ x = x2 y6 1

𝐞𝟐𝐱

= (𝟐 − 𝐱

𝐭𝐚𝐧𝟑 𝐱 𝟑 𝐱𝟐

𝟐 )𝐞 𝟐

+ 𝐜]

Jan-15

+ 𝐜]

+ 𝐜]

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 7

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

𝐲=𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

Where, c is an arbitrary constant Exercise-5

Exact Differential Equation C

1.

(x 3 + 3xy 2 )dx + (y 3 + 3x 2 y)dy = 0

H

2.

(x 2 + y 2 )dx + 2xydy = 0.

H

3.

2xydx + x 2 dy = 0.

T

4.

yex dx + (2y + ex )dy = 0; y(0) = −1.

T

5.

(ey + 1) cos x dx + ey sin x dy = 0.

𝐱𝟒

[𝟒 +

𝟑𝐱 𝟐 𝐲 𝟐 𝟐

+

𝐲𝟒 𝟒

= 𝐜]

𝐱𝟑

[ 𝟑 + 𝐲 𝟐 𝐱 = 𝐜] [𝐱 𝟐 𝐲 = 𝐜]

Dec-09

[𝐲𝐞𝐱 + 𝐲 𝟐 = 𝟎]

Mar-10 Jun-15

[(𝐞𝐲 + 𝟏) 𝐬𝐢𝐧 𝐱 = 𝐜]

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

6.

C

7.

dy

+ dx

ycosx+siny+y sinx+xcosy+x

Jan-15

= 0.

[𝐱(𝐞𝐱 − 𝐞𝐲 ) = 𝐞 − 𝟏]

Jun-14 Dec -11

[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 ∂M ∂N Differential Equation. i.e. if ∂y ≠ ∂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.

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 8

Some Standard Rules for Finding I.F. 1

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

Non-Homogeneous), then I. F. = Mx−Ny . 1

∂M

∂N

1

∂N

∂M

3. If N ( ∂y − ∂x ) = f(x) (i. e. function of only x), then I. F. = e∫ f(x) dx 4. If M ( ∂x −

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

𝐲=𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

Where, c is an arbitrary constant. Exercise-6

Non-Exact Differential Equation State the necessary & sufficient condition to be exact differential equation. C

1.

And using it Solve x 2 y dx − (x 3 + xy 2 )dy = 0. 𝐱𝟐

Jan-13 Jan-15

Jan-15

[− 𝟐𝐲𝟐 + logy = 𝐜] 𝐱

C

2.

(x 2 y − 2xy 2 )dx − (x 3 − 3x 2 y)dy = 0

[𝐲 − 𝟐 𝐥𝐨𝐠 𝐱 + 𝟑𝐥𝐨𝐠𝐲 = 𝐜]

H

3.

(xy − 2y 2 )dx − (x 2 − 3xy)dy = 0.

[ 𝐲 − 𝟐𝐥𝐨𝐠 𝐱 + 𝟑 𝐥𝐨𝐠 𝐲 = 𝐜]

T

4.

(x 3 +y 3 )dx − xy 2 dy = 0.

[𝐥𝐨𝐠𝐱 − 𝟑𝐱𝟑 = 𝐜]

C

5.

(x 2 +y 2 )dx − 2xydy = 0.

[𝐱 𝟐 − 𝐲 𝟐 = 𝐜𝐱]

C

6.

(x 2 y 2 + 2)ydx + (2 − x 2 y 2 )xdy = 0.

C

7.

y(1 + xy)dx + x(1 + xy + x 2 y 2 )dy = 0.

H

8.

y(xy + 2x 2 y 2 )dx + x(xy − x 2 y 2 )xdy = 0

C

9.

(x 2 +y 2 + x)dx + xydy = 0.

A.E.M.(2130002)

𝐱

𝐲𝟑

𝐱

𝟏

[𝐥𝐨𝐠 𝐲 − 𝐱𝟐 𝐲𝟐 = 𝐜] 𝟏

May-12 Jan-15

𝟏

[𝟐𝐱𝟐 𝐲𝟐 + 𝐱𝐲 − 𝐥𝐨𝐠𝐲 = 𝐜] 𝟏

[− 𝐱𝐲 + 𝐥𝐨𝐠

𝐱𝟐 𝐲

= 𝐜]

[𝟑𝐱 𝟒 + 𝟔𝐱 𝟐 𝐲 𝟐 + 𝟒𝐱 𝟑 = 𝟏𝟐𝐜]

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3A. DIFFERENTIAL EQUATION OF FIRST ORDER

PAGE | 9

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 dy

dx

III.

Replace dx by − 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 dr

dθ

III.

Replace dθ by −r 2 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 = x 2 + c.

H

2.

y 2 + (x − a)2 = a2

T

3.

x 2 = 4b(y + b).

A.E.M.(2130002)

𝟏

𝐜

[𝐲 = − 𝟐 𝐥𝐨𝐠 𝐱 + 𝟐]

Dec-10

[(𝐲 − 𝐛)𝟐 + 𝐱 𝟐 = 𝐛𝟐 ] [𝐱 𝟐 = 𝟒𝐛(𝐲 + 𝐛)]

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

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

dn y dn−1 y dn−2 y + P + P + ⋯ + Pn y = f(x) … … … (A) 1 2 dx n dx n−1 dx n−2

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

dn y dn−1 y dn−2 y + a + a + ⋯ + an y = f(x) … … … (B) 1 2 dx n dx n−1 dx n−2

Where, a0 , a1 , a2 , … are constants.

Notations Eq. (B) can be written in operator form as below, a0 Dn y + a1 Dn−1 y + a2 Dn−2 y + ⋯ + an y = 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 a0 Dn y + a1 Dn−1 y + a2 Dn−2 y + ⋯ + an y = f(x) is derived by replacing D by m and equating with 0. i. e. a0 mn y + a1 mn−1 y + a2 mn−2 y + ⋯ + an y = 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 ) 1

A particular integral of [g(D)]y = f(x) is y = 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 A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 2

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, a0 Dn y + a1 Dn−1 y + a2 Dn−2 y + ⋯ + an y = f(x) The Auxiliary equation is a0 mn y + a1 mn−1 y + a2 mn−2 y + ⋯ + an y = 0 Let, m1 , m2 , m3 , … be the roots of auxiliary equation. Case

1.

2.

3.

Nature of the “n” roots m1 ≠ m2 ≠ m3 ≠ m4 ≠⋯ m1 = m2 = m m3 ≠ m4 ≠ ⋯

emx , xemx , em3 x , em4 x , … emx , xemx , x 2 emx ,

m4 ≠ m5 , …

em4 x , em5 x , …

m2 = p − iq m3 ≠ m4 , …

5.

em1 x , em2 x , em3 x , …

m1 = m2 = m3 = m

m1 = p + iq 4.

L.I. solutions

epx cos qx , epx sin qx, em3 x , em4 x , …

General Solutions y = c1 em1 x + c2 em2 x + c3 em3 x +⋯ y = (c1 + c2 x)emx + c3 em3 x + c4 em4 x + ⋯ y = (c1 + c2 x + c3 x 2 )emx + c4 em4 x + c5 em5 x +⋯ y = epx (c1 cos qx + c2 sin qx) + c2 em3 x + c3 em4 x +⋯

m1 = m2 = p + iq

epx cos qx , xepx cos qx,

y = epx [(c1 + c2 x) cos qx

m3 = m4 = p − iq

epx sin qx , xepx sin qx,

+(c3 + c4 x) sin qx]

m5 ≠ m6 , …

em5 x , em6 x , …

A.E.M.(2130002)

+c5 em5 x + c6 em6 x + ⋯

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 3

Exercise-1

Solution Of Homogeneous Differential Equation C

1.

H

2.

C

3.

T

4.

T

5.

H

6.

T

7.

T

8.

C

9.

H

10.

T

11.

C

12.

H C

13.

y ′′ + y ′ − 2y = 0 [𝐜𝟏 𝐞−𝟐𝐱 + 𝐜𝟐 𝐞𝐱 ]

Dec-09

[𝐜𝟏 𝐞−𝟗𝐱 + 𝐜𝟐 𝐞𝟐𝐱 ]

Jan-15

y ′′ + 7y ′ − 18y = 0 y ′′ + y ′ − 2y = 0, y(0) = 4, y ′ (0) = −5.

[𝟑𝐞−𝟐𝐱 + 𝐞𝐱 ]

Dec-09 Jan-15

[𝟔 𝐞𝟑𝐱 + 𝟔 𝐞−𝟑𝐱 ]

𝟕

Jan-15

[𝐞𝟑𝐱−𝟏 ]

May-12

y ′′ − 9y = 0 ; y(0) = 2, y ′ (0) = −1. 𝟓

y ′′ − 5y ′ + 6y = 0; y(1) = e2 , y ′ (1) = 3e2 y ′′ − 2√2y ′ + 2y = 0.

Jan-15

[(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞√𝟐𝐱 ] y ′′ + 4y ′ + 4y = 0; y(0) = 1, y ′ (0) = 1 . [(𝟏 + 𝟑𝐱)𝐞−𝟐𝐱 ]

Dec-11

[(𝟑 − 𝟓𝐱)𝐞𝟐𝐱 ]

Jan-15

[(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞𝟑𝐱 + (𝐜𝟑 + 𝐜𝟒 𝐱)𝐞−𝟑𝐱 ]

May-12

y ′′ − 4y ′ + 4y = 0; y(0) = 3, y ′ (0) = 1 d4 y dx4

d2 y

− 18 dx2 + 81y = 0 .

(D2 + 1)y = 0 .

[(𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱)]

16y ′′ − 8y ′ + 5y = 0.

𝐱

𝐱

𝐱

[𝐞𝟒 (𝐜𝟏 𝐜𝐨𝐬 𝟐 + 𝐜𝟐 𝐬𝐢𝐧 𝟐)]

(D4 − 1)y = 0.

[𝐜𝟏 𝐞−𝐱 + 𝐜𝟐 𝐞𝐱 + (𝐜𝟑 𝐜𝐨𝐬𝐱 + 𝐜𝟒 𝐬𝐢𝐧𝐱)]

y ′′′ − y = 0.

𝟏

√𝟑

√𝟑

[𝐜𝟏 𝐞𝐱 + 𝐞−𝟐𝐱 (𝐜𝟐 𝐜𝐨𝐬 𝟐 𝐱 + 𝐜𝟑 𝐬𝐢𝐧 𝟐 𝐱)] y ′′′ − y ′′ + 100y ′ − 100y = 0; y(0) = 4, y ′ (0) = 11, y ′′ (0) = −299 14. [𝐞𝐱 + 𝐬𝐢𝐧 𝟏𝟎𝐱 + 𝟑 𝐜𝐨𝐬 𝟏𝟎𝐱]

Dec-11 Dec-11 Jun-14 Jun-13 Jun-15 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. B. C. D.

General Methods Short-cut Methods involving operators Method of Undetermined Co-efficient Method of Variation of parameters

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 4

A. General Methods Consider the differential equation a0 Dn y + a1 Dn−1 y + a2 Dn−2 y + ⋯ + an y = X It may be written as f(D)y = X 1 ∴ Particular Integral = X f(D) Particular Integral may be obtained by following two ways:

1. Method Of Factors 1

The operator f(D) X may be factorized into n linear factors; then the particular integral will be P. I. =

1 1 X= X (D − m1 )(D − m2 ) … … … … . (D − mn ) f(D)

Now, we know that, 1 X = emn x ∫ X e−mnx dx D − mn On opening with the first symbolic factor, beginning at the right, the particular integral will have form P. I. =

1 emn x ∫ X e−mn x dx (D − m1 )(D − m2 ) … … … … . (D − mn−1 )

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 1

The operator f(D) X may be factorized into n linear factors; then the particular integral will be P. I. = 1

1 A1 A2 An X=( + + ⋯+ )X f(D) D − m1 D − m2 D − mn 1

1

= A1 D−m X + A2 D−m X + ⋯ + An D−m X 1

Using

2

n

1 X = emn x ∫ X e−mn x dx , we get D − mn

P. I. = A1 em1 x ∫ X e−m1 x dx + A2 em2 x ∫ X e−m2 x dx + ⋯ + An emn x ∫ X e−mn x dx Out of these two methods, this method is generally preferred.

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 5

B. Shortcut Method 1.

F(x) = eax P. I. =

1 ax 1 ax e = e , if f(a) ≠ 0 f(D) f(a)

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

1 ax x ax e = e , if f ′ (a) ≠ 0 f(D) f′(a)

In general, If f n−1 (a) = 0 , 1 ax x n ax P. I. = e = n e , if f n (a) ≠ 0 f(D) f (a) 2.

F(x) = sin(ax + b) P. I. =

1 1 sin(ax + b) = sin(ax + b) , if f(−a2 ) ≠ 0 2 f(D ) f(−a2 )

If f(−a2 ) = 0 , 1 x P. I. = sin(ax + b) = sin(ax + b) , if f′(−a2 ) ≠ 0 2 2 f(D ) f′(−a ) If f ′ (−a2 ) = 0 , 1 x2 P. I. = sin(ax + b) = sin(ax + b) , if f"(−a2 ) ≠ 0 f(D2 ) f"(−a2 ) 3.

and so on …

F(x) = cos(ax + b) P. I. =

1 1 cos(ax + b) = cos(ax + b) , if f(−a2 ) ≠ 0 2 f(D ) f(−a2 )

If f(−a2 ) = 0 , 1 x P. I. = cos(ax + b) = cos(ax + b) , if f′(−a2 ) ≠ 0 2 f(D ) f′(−a2 ) If f ′ (−a2 ) = 0 , 1 x2 P. I. = cos(ax + b) = cos(ax + b) , if f"(−a2 ) ≠ 0 f(D2 ) f"(−a2 )

and so on …

4. F(x) = x m ; m > 0 In this case convert f (D) in the form of 1 + ϕ(D) or 1 − ϕ(D) form so that we get 1 m 1 P. I. = x = x m = {1 − ϕ(D) + [ϕ(D)]2 − . . . } x m f(D) 1 + ϕ(D) (Using Binomial Theorem) ax 5. F(x) = e V(X) ,Where V(X) is a function of x. 1 ax 1 P. I. = e V(x) = eax V(x) f(D) f(D + a)

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 6

Exercise-2

Solution Of Non-Homogeneous Differential Equation d2 y

C

1.

H

2.

H

3.

dx2

4.

H

5.

H

6.

𝟏

[(𝐜𝟏 𝐞−𝟒𝐱 + 𝐜𝟐 𝐞𝟑𝐱 ) + 𝟑𝟎 𝐞𝟔𝐱 ]

(D2 + 5D + 6)y = ex

𝐞𝐱

[𝐜𝟏 𝐞−𝟑𝐱 + 𝐜𝟐 𝐞−𝟐𝐱 + 𝟏𝟐]

y ′′ − 5y ′ + 6y = 3e−2x

𝟑

d2 y

H

dy

+ dx − 12y = e6x

dx2

[(𝐜𝟏 𝐞𝟐𝐱 + 𝐜𝟐 𝐞𝟑𝐱 ) + 𝟐𝟎 𝐞−𝟐𝐱 ]

7.

C

8.

T T H

9. 10. 11.

C

12.

C

13.

T

14.

H

15.

T

16.

Jun-14 Jan-15

dy

− 5 dx + 6y = e4x [𝐜𝟏 𝐞𝟐𝐱 + 𝐜𝟐 𝐞𝟑𝐱 +

y ′′ − 3y ′ + 2y = ex (D2 − 3D + 2)y = 2ex . (D3 − 7D + 6)y = e2x .

H

Jan-13

(D2 − 2D + 1)y = 10ex . y ′′′ − 3y ′′ + 3y ′ − y = 4et

𝐞𝟒𝐱 𝟐

]

[𝐜𝟏 𝐞𝐱 + 𝐜𝟐 𝐞𝟐𝐱 − 𝐱𝐞𝐱 ]

Dec-11

[𝐜𝟏 𝐞𝐱 + 𝐜𝟐 𝐞𝟐𝐱 − 𝟐𝐱𝐞𝐱 ]

May-12

𝐱 [𝐜𝟏 𝐞𝐱 + 𝐜𝟐 𝐞𝟐𝐱 + 𝐜𝟑 𝐞−𝟑𝐱 + 𝐞𝟐𝐱 ] 𝟓

Jun-15

[(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞𝐱 + 𝟓𝐱 𝟐 𝐞𝐱 ]

Jun-15

𝟐

y ′′′ − 3y ′′ + 3y ′ − y = e−x

Dec-13

[(𝐜𝟏 + 𝐜𝟐 𝐭 + 𝐜𝟑 𝐭 𝟐 )𝐞𝐭 + 𝟑 𝐭 𝟑 𝐞𝐭 ]

Jan-15

𝟏

[(𝐜𝟏 + 𝐜𝟐 𝐱 + 𝐜𝟑 𝐱 𝟐 )𝐞𝐱 − 𝟖 𝐞−𝐱 ]

(D2 − 49)y = sinh 3x

𝟏

[𝐜𝟏 𝐞−𝟕𝐱 + 𝐜𝟐 𝐞𝟕𝐱 − 𝟒𝟎 𝐬𝐢𝐧𝐡 𝟑𝐱]

y ′′ + 2y ′ + 2y = sinh x

Jun-15

𝟏

[𝐞−𝐱 (𝐜𝟏 𝐜𝐨𝐬𝐱 + 𝐜𝟐 𝐬𝐢𝐧𝐱) + 𝟏𝟎 (𝐞𝐱 − 𝟓𝐞−𝐱 )]

(D2 − 25)y = cos 5x

𝟏

[𝐜𝟏 𝐞−𝟓𝐱 + 𝐜𝟐 𝐞𝟓𝐱 − 𝟓𝟎 𝐜𝐨𝐬𝟓𝐱] Find the steady state oscillation of the mass-spring system governed by the equation: y ′′ + 3y ′ + 2y = 20 cos 2t . [𝟑𝐬𝐢𝐧𝟐𝐭 − 𝐜𝐨𝐬𝟐𝐭] (D2 + 9)y = cos 2x + sin 2x 𝟏 𝟏 [𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 + 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱] 𝟓 𝟓 (D2 − 4D + 3)y = sin 3x cos 2x. 𝐬𝐢𝐧 𝐱+𝟐 𝐜𝐨𝐬 𝐱 𝟏𝟎 𝐜𝐨𝐬 𝟓𝐱−𝟏𝟏 𝐬𝐢𝐧 𝟓𝐱 [𝐜𝟏 𝐞𝐱 + 𝐜𝟐 𝐞𝟑𝐱 + − ] 𝟐𝟎 𝟖𝟖𝟒

Jun-14 Dec-09

Jun-15 Dec-13

(D4 + 2a2 D2 + a4 )y = cos ax. T

17.

A.E.M.(2130002)

[(𝐜𝟏 + 𝐜𝟐 𝐱) 𝐜𝐨𝐬 𝐚𝐱 + (𝐜𝟑 + 𝐜𝟒 𝐱) 𝐬𝐢𝐧 𝐚𝐱 −

𝐱𝟐 𝐜𝐨𝐬 𝐚𝐱] 𝟖𝐚𝟐

May-11

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 7

(D2 + D − 6)y = e2x sin3x T

C

19.

T

20.

H

21.

C

𝐞𝟐𝐱 (𝟏𝟓 𝐜𝐨𝐬 𝟑𝐱 + 𝟗 𝐬𝐢𝐧 𝟑𝐱)] 𝟑𝟎𝟔

Jan-13

(D3 − D2 + 3D + 5)y = ex cos 3x. 𝐞𝐱 [𝐜𝟏 𝐞−𝐱 + 𝐞𝐱 (𝐜𝟐 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟑 𝐬𝐢𝐧 𝟐𝐱) − (𝟑 𝐬𝐢𝐧 𝟑𝐱 + 𝟐 𝐜𝐨𝐬 𝟑𝐱)]

May-12

18.

22.

[𝐜𝟏 𝐞−𝟑𝐱 + 𝐜𝟐 𝐞𝟐𝐱 −

y ′′ + 2y ′ + 3y = 2x 2

𝟐

y ′′ + 2y ′ + 10y = 25x 2 + 3

23.

𝟐𝟖

[𝐞−𝐱 (𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱) + (𝟐 𝐱 𝟐 − 𝐱)]

(D3 − D2 − 6D)y = x 2 + 1

𝐱𝟑

𝐱𝟐

𝟐𝟓𝐱

[𝐜𝟏 + 𝐜𝟐 𝐞𝟑𝐱 − 𝟏𝟖 + 𝟑𝟔 − 𝟏𝟎𝟖] dt4

Jan-15 Dec-10 Jan-13

d2 y

− 2 dt2 + y = cos t + e2t + et [(𝐜𝟏 + 𝐜𝟐 𝐭)𝐞−𝐭 + (𝐜𝟑 + 𝐜𝟒 𝐭)𝐞𝐭 + (

′′

y − 4y = e H

𝟖

[𝐞−𝐱 (𝐜𝟏 𝐜𝐨𝐬√𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧√𝟐𝐱) + ( 𝟑 𝐱 𝟐 − 𝟗 𝐱 − 𝟐𝟕)] 𝟓

d4 y

T

𝟔𝟓

−2x

− 2x, y(0) = 0, y

′ (0)

𝐜𝐨𝐬𝐭 𝟒

+

𝐞𝟐𝐭 𝟗

+

𝐭 𝟐 𝐞𝐭 𝟖

)]

May-12

=0 𝟏 𝟏 𝟏 [− 𝐬𝐢𝐧𝐡 𝟐𝐱 + 𝐱 − 𝐱𝐞−𝟐𝐱 ] 𝟖 𝟐 𝟒

24. (D2 + 16)y = x 4 + e3x + cos3x

C

25.

H

26.

𝟏 𝟒 𝟑 𝟐 𝟑 𝟏 𝟑𝐱 𝟏 [𝐱 − 𝐱 + ] + 𝐞 + 𝐜𝐨𝐬 𝟑𝐱] 𝟏𝟔 𝟒 𝟑𝟐 𝟐𝟓 𝟕 ′′ −2x 2 ′ (0) y + 4y = 8e + 4x + 2; y(0) = 2, y = 2. [𝐜𝐨𝐬𝟐𝐱 + 𝟐𝐬𝐢𝐧𝟐𝐱 + 𝐞−𝟐𝐱 + 𝐱 𝟐 ] [(𝐜𝟏 𝐜𝐨𝐬 𝟒𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟒𝐱) +

Jan-15 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

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 8

Here we guess the form of YP depending on X as per the following table. RHS of 𝐟(𝐃)𝐲 = 𝐗 1. 2.

3.

4.

X = eax X = sin ax X = cos ax X = a + bx + cx 2 + dx 3 X = ax 2 + bx X = ax + b X=c X = eax sin bx ax

X = e cos bx

Form of Trial Solution YP = Aeax YP = A sin ax + B cos ax YP YP YP YP

= A + Bx + Cx 2 + Dx 3 = A + Bx + Cx 2 = A + Bx =A

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

5.

X = xeax X = x 2 eax

YP = eax (A + Bx) YP = eax (A + Bx + Cx 2 )

6.

X = x sin ax X = x 2 cos ax

YP = sin ax (A + Bx) + cos ax (C + Dx) YP = sin ax (A + Bx + Cx 2 ) + cos ax (D + Ex + Fx 2 )

X = e2x X = e2x − 3e−x X = cos 3x X = 2 sin(4x − 5)

YP YP YP YP

7. 8.

= Ae2x = Ae2x + Be−x = A sin 3x + B cos 3x = A sin(4x − 5) + B cos(4x − 5)

Exercise-3

Solution By Method Of Undetermined Co-Efficient C

1.

H

2.

T

3.

y ′′ + 4y = 2sin3x 𝟐

[𝐜𝟏 𝐜𝐨𝐬𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧𝟐𝐱 − 𝟓 𝐬𝐢𝐧𝟑𝐱]

Find particular Integral of Differential Equation y ′′ + 9y = cos 5x 𝟏 [− 𝟏𝟔 𝐜𝐨𝐬𝟓𝐱] y ′′ + 4y = 8x 2 . d2 y

T

4.

C

5.

T

6.

H

7.

C

8.

dx2

[𝐜𝟏 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟐𝐱 + 𝟐𝐱 𝟐 − 𝟏]

Jun-15 Dec-09

dy

+ dx − 6y = 6x + 3x 2 − 6x 3 .

[𝐜𝟏 𝐞−𝟑𝐱 + 𝐜𝟐 𝐞𝟐𝐱 + 𝐱 𝟑 ] y ′′ + 1.5y ′ − y = 12x 2 + 6x 3 − x 4 , y(0) = 4, y ′ (0) = −8 [𝟒𝐞−𝟐𝐱 + 𝐱 𝟒 ] ′′′ ′′ ′ −x ′ (0) ′′ (0) y + 3y + 3y + y = 30e ; y(0) = 3, y = −3, y = −47 . [(𝟑 − 𝟐𝟓𝐱 𝟐 + 𝟓𝐱 𝟑 )𝐞−𝐱 ] y ′′ + 1.2y ′ + 0.36y = 4e−0.6x , y(0) = 0, y ′ (0) = 1 [(𝐱 + 𝟐𝐱 𝟐 )𝐞−𝟎.𝟔𝐱 ] y ′′ − 2y ′ + y = ex + x

A.E.M.(2130002)

Mar-10 Jun-14

[(𝐜𝟏 + 𝐱𝐜𝟐 )𝐞𝐱 + [

𝐱𝟐 𝐞𝐱 𝟐

Jan-13

May-11

+ (𝐱 + 𝟐)]]

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 9

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

W(𝐲𝟏 , 𝐲𝟐 , … 𝐲𝐧 ) = |

(𝒏−𝟏) 𝐲𝟏

𝐲𝟐 𝐲𝟐′ ⋮ (𝒏−𝟏) 𝐲𝟐

𝐲𝐧 𝐲′𝐧 ⋮ |

⋯ ⋱ ⋯

(𝒏−𝟏)

𝐲𝐧

Theorem: Let y1 , y2 , … yn be differentiable functions defined on some interval I. Then 1. 2.

y1 , y2 , … yn Are linearly independent on I if and only if W(y1 , y2 , … yn ) ≠ 0 for all x ∈ I. 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.

C

2.

x, log x , x(log x)2 [𝐋. 𝐈] ex , e−x [𝐋. 𝐈]

May-11 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 X

X

Where yc = c1 y1 + c2 y2 and yp (x) = −y1 ∫ y2 W dx + y2 ∫ y1 W dx y1 Where y1 and y2 are the solutions of y" + py′ + qy = 0, and W = |y′

1

y2 y′2 | = y1 y′2 − y2 y′1 ≠ 0.

The general solution of third order differential equation by the method of variation of parameters is y(x) = yc + yp Where,yc = c1 y1 + c2 y2 + c3 y3 and yp = P(X)y1 + Q(X)y2 + R(X)y3.

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 10

Where, P(X) = ∫(y2 y3′ − y3 y2′ ) y1 ′ Where, w = | y1 y1′′

y2 y2′ y2′′

X dx w

Q(X) = ∫(y3 y1′ − y1 y3′ )

X dx w

R(X) = ∫(y1 y2′ − y2 y1′ )

X dx w

y3 y3′ | ≠ 0. y3′′

Exercise-5

Solution By Variation OF Parameter C

T

H

1.

2.

3.

(D2 − 4D + 4)y =

(D2 + 4D + 4)y =

e2x x5

𝟏 𝐞𝟐𝐱

[(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞𝟐𝐱 + 𝟏𝟐

e−2x x2

May-12

] 𝐱𝟑 Mar-10

[(𝐜𝟏 + 𝐜𝟐 𝐱 − (𝟏 + 𝐥𝐨𝐠(𝐱)))𝐞−𝟐𝐱 ]

d2 y dy ex − 2 + y = dx 2 dx x2

[(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞𝐱 − 𝐞𝐱 𝐥𝐨𝐠 𝐱 − 𝐞𝐱 ]

Jan-13

3

T

4.

T

5.

T

6.

(D2 − 2D + 1)y = 3x 2 ex 𝟏𝟐 𝟕 𝐱 [(𝐜𝟏 + 𝐜𝟐 𝐱 + 𝐱𝟐) 𝐞 ] 𝟑𝟓

y ′′ + 2y ′ + y = e−x cos x (D2 − 4D + 4)y =

[(𝐜𝟏 + 𝐜𝟐 𝐱 − 𝐜𝐨𝐬 𝐱)𝐞−𝐱 ]

e2x 1 + x2 [(𝐜𝟏 + 𝐜𝟐 𝐱)𝐞

C

7.

H

8.

(D2 − 3D + 2)y =

ex 1 + ex

𝟐𝐱

𝟏 −𝐞 𝐥𝐨𝐠(𝟏 + 𝐱 𝟐 ) + 𝐱𝐞𝟐𝐱 (𝐭𝐚𝐧−𝟏 𝐱) ] 𝟐

9.

C

10.

H

11.

[𝐜𝟏 𝐞𝐱 + 𝐜𝟐 𝐞𝟐𝐱 + 𝐞𝐱 [(𝟏 + 𝐞𝐱 ) 𝐥𝐨𝐠(𝟏 + 𝐞−𝐱 ) − 𝟏]] y ′′ + y = sec x . [𝐜𝟏 𝐜𝐨𝐬 𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝐱 + 𝐱 𝐬𝐢𝐧 𝐱 + 𝐜𝐨𝐬 𝐱 𝐥𝐨𝐠(𝐜𝐨𝐬 𝐱)] 𝟏 𝟏 [𝐲 = 𝐜𝟏 𝐜𝐨𝐬 𝟑𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟑𝐱 + 𝐱 𝐬𝐢𝐧 𝟑𝐱 + 𝐜𝐨𝐬 𝟑𝐱 𝐥𝐨𝐠 𝐜𝐨𝐬 𝟑𝐱] 𝟑 𝟗 ′′ y + 4y = tan 2x. 𝟏 [𝐜𝟏 𝐜𝐨𝐬 𝟐𝐱 + 𝐜𝟐 𝐬𝐢𝐧 𝟐𝐱 − 𝟒 𝐜𝐨𝐬 𝟐𝐱 𝐥𝐨𝐠(𝐬𝐞𝐜 𝟐𝐱 + 𝐭𝐚𝐧 𝟐𝐱) ] y ′′ + y = cotx .

A.E.M.(2130002)

May-12

𝟐𝐱

y ′′ + 9y = sec 3x. H

Dec-10 Jun-13

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

Jan-13

Dec-09 Mar-10 Jan-15 Jun-15 Jun-14

Jun-15

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

H

12.

(D2 + a2 )y = cosec ax [𝐜𝟏 𝐜𝐨𝐬𝐚𝐱 + 𝐜𝟐 𝐬𝐢𝐧𝐚𝐱 + ( d3 y

C

13.

PAGE | 11

dx3

−𝐱 𝐜𝐨𝐬 𝐚𝐱 𝐚

𝟏

+ 𝐚𝟐 𝐬𝐢𝐧 𝐚𝐱 𝐥𝐨𝐠|𝐬𝐢𝐧 𝐱|)]

dy

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

May-11 Jan-13 May-12

Cauchy – Euler Equation An equation of the form 𝑥𝑛

𝑑𝑛 𝑦 𝑑 𝑛−1 𝑦 𝑑 𝑛−2 𝑦 𝑑𝑦 𝑛−1 𝑛−2 + 𝑎 𝑥 + 𝑎 𝑥 + ⋯ + 𝑎𝑛−1 𝑥 + 𝑎𝑛 𝑦 = 𝑋 1 2 𝑛 𝑛−2 𝑛−2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥

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 = 𝑒 𝑧 so that 𝑧 = 𝑙𝑜𝑔𝑥. 𝑁𝑜𝑤, 𝑧 = 𝑙𝑜𝑔 𝑥 ⟹ 𝑁𝑜𝑤, ⟹𝑥

𝑑𝑦 𝑑𝑦 𝑑𝑧 = 𝑑𝑥 𝑑𝑧 𝑑𝑥

𝑑𝑧 1 = 𝑑𝑥 𝑥 ⟹

𝑑𝑦 1 𝑑𝑦 = 𝑑𝑥 𝑥 𝑑𝑧

𝑑𝑦 𝑑𝑦 𝑑 = = 𝐷𝑦, 𝑤ℎ𝑒𝑟𝑒 𝐷 = 𝑑𝑥 𝑑𝑧 𝑑𝑧

𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑥 2

𝑑2𝑦 𝑑3𝑦 3 = 𝐷(𝐷 − 1)𝑦 & 𝑥 = 𝐷(𝐷 − 1)(𝐷 − 2)𝑦 𝑑𝑥 2 𝑑𝑥 3

Using this transformation, the given equation reduces to [𝐷(𝐷 − 1)(D − 2) … (𝐷 − 𝑛 + 1) + 𝑎1 𝐷(𝐷 − 1) … (𝐷 − 𝑛 + 2) + ⋯ + 𝑎𝑛−1 𝐷 + 𝑎𝑛 ]𝑦 = 𝑓(𝑒 𝑧 ) This is a linear equation with constant coefficients, which can be solved by the methods discussed earlier. Exercise-6

Cauchy – Euler Equation H

1.

T

2.

( 𝑥 2 𝐷2 − 3𝑥𝐷 + 4)𝑦 = 0; 𝑦(1) = 0, 𝑦 ′ (1) = 3

Dec-11

𝟐

[𝟑𝒙 𝒍𝒐𝒈 𝒙] 𝑥 2 𝑦 ′′ − 4𝑥𝑦 ′ + 6𝑦 = 21𝑥 −4 .

A.E.M.(2130002)

𝟏

[𝒄𝟏 𝒙𝟐 + 𝒄𝟐 𝒙𝟑 + 𝟐 𝒙−𝟒 ]

May-11

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

3B. DIFFERENTIAL EQUATION OF HIGHER ORDER

PAGE | 12

(𝑥 2 𝐷2 − 3𝑥𝐷 + 4)𝑦 = 𝑥 2 ; 𝑦(1) = 1, 𝑦 ′ (1) = 0 C

3.

C

4.

H

5.

T

6.

𝟏 [(𝟏 − 𝟐 𝒍𝒐𝒈 𝒙)𝒙𝟐 + 𝒙𝟐 (𝒍𝒐𝒈 𝒙)𝟐 ] 𝟐

May-12

𝑥 3 𝑦 ′′ ′ + 2𝑥 2 𝑦 ′′ + 2𝑦 = 10 (𝑥 + 𝑥). [𝒄𝟏 𝒙−𝟏 + 𝒙{𝒄𝟐 𝒄𝒐𝒔(𝒍𝒐𝒈 𝒙) + 𝒄𝟑 𝒔𝒊𝒏(𝒍𝒐𝒈 𝒙)} + 𝟓𝒙 + 𝟐𝒙−𝟏 𝒍𝒐𝒈 𝒙]

May-11 Jan-13 Jun-14

1

(𝑥 2 𝐷2 − 3𝑥𝐷 + 3)𝑦 = 3 𝑙𝑛 𝑥 − 4 [𝒄𝟏 𝒙 + 𝒄𝟐 𝒙𝟑 + 𝒍𝒏 𝒙] 𝑥 2 𝐷2 𝑦 − 3𝑥𝐷𝑦 + 5𝑦 = 𝑥 2 𝑠𝑖𝑛(𝑙𝑜𝑔𝑥) [𝒙𝟐 {𝒄𝟏 𝒄𝒐𝒔(𝒍𝒐𝒈 𝒙) + 𝒄𝟐 𝒔𝒊𝒏(𝒍𝒐𝒈 𝒙)} − 𝒙𝟐 𝒍𝒐𝒈𝒙𝒄𝒐𝒔(𝒍𝒐𝒈𝒙)]

Mar-10

Jun-15

Solution Of Differential Equation By One Of Its Solution 𝑑2 𝑦

𝑑𝑦

Step-1

Convert given D.E. into 𝑑𝑥 2 + 𝑃(𝑥) 𝑑𝑥 + Q(𝑥)𝑦 = 0 and find 𝑃(𝑥) & 𝑄(𝑥).

Step-2

Find 𝑈. 𝑈=

Step-3

1 − ∫ 𝑃 𝑑𝑥 𝑒 𝑦12

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.

𝑥 2 𝑦 ′′ − 𝑥𝑦 ′ + 𝑦 = 0; 𝑦1 = 𝑥. 2 ′′

T H

2. 3.

′

Dec-10 2

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

sin x . x

[𝒚𝟐 = 𝒙𝒍𝒐𝒈𝒙] [𝒚𝟐 = 𝒙𝟑 ]

[𝐲𝟐 = − H

4.

y ′′ + 4y ′ + 4y = 0, y1 = e−2x .

A.E.M.(2130002)

𝐜𝐨𝐬𝐱 𝐱

Jan-13 May-11

]

[𝐲𝟐 = 𝐱𝐞−𝟐𝐱 ]

Jun-15

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4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

PAGE | 1

Definition: Power Series An infinite series of the form ∞

∑ ak (x − x0 )k = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + ⋯ k=0

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

d2 y

dy

P0 (x) dx2 + P1 (x) dx + P2 (x)y = 0 be the given differential equation with variable co-

efficient. Dividing by P0 (x), d2 y P1 (x) dy P2 (x) + + y=0 dx 2 P0 (x) dx P0 (x) P (x)

P (x)

Let, P(x) = P1 (x) & Q(x) = P2 (x) 0

∴

0

d2 y dy + P(x) + Q(x)y = 0 2 dx dx

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 )2 Q(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.

C

2.

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

Ans.

A.E.M.(2130002)

Jun-14 Jun-15

No Singular Points.

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4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS H

3.

PAGE | 2

y ′′ + ex y ′ + sin(x 2 )y = 0.

Ans. No Singular Points. T

4.

x 3 y ′′ + 5xy ′ + 3y = 0

Jan-15

Ans. x = 0 is Irregular Singular Point. T

5.

(1 − x 2 )y ′′ − 2xy ′ + n(n + 1)y = 0.

Dec-11

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

6.

(x 2 + 1)y ′′ + xy ′ − xy = 0.

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

7.

2x(x − 2)2 y ′′ + 3xy ′ + (x − 2)y = 0.

May-12

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

8.

d2 y

Dec-12

dy

2x 2 dx2 + 6x dx + (x + 3)y = 0.

Ans. x = 0 is Regular Singular Point. C

9.

x(x + 1)2 y ′′ + (2x − 1)y ′ + x 2 y = 0.

Jun-15

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

10. Discuss singularities of x 3 (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 d2 y

dy

A power series solution of a differential equation P0 (x) dx2 + P1 (x) dx + P2 (x)y = 0 at an ordinary point x0 can be obtained using the following steps. k 2 3 4 5  STEP-1: Assume that y = ∑∞ k=0 a k (x − x0 ) = a 0 + a1 x + a 2 x + a 3 x + a 4 x + a 5 x + ⋯ is the solution of the given differential equation.

Differentiating with respect to y we get, 𝑑𝑦

𝑦 ′ = 𝑑𝑥 = a1 + 2a2 x + 3a3 x 2 + 4a4 x 3 + 5a5 x 4 +. . . ∞ 𝑦 ′′ =

𝑑𝑦 𝑑𝑥

= 2a2 + 6a3 x + 12a4 x 2 + 20a5 x 3 +. . . ∞ dy

 STEP-2: Substitute the expressions of y, dx , and

d2 y 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 + a1 x + a2 + a3 x 3 + a4 x 4 + a5 x 5 +. .. which will be the required solution.

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4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

PAGE | 3

Exercise - 2

Solution By Power Series Method C

1. Ans.

H

2. Ans.

T

3.

Ans. C

4. Ans.

H

5. Ans.

H

6. Ans.

H

7. Ans.

C

C

y ′ + 2xy = 0.

Dec-11

1 1 y = a0 − a0 x 2 + a0 x 4 − a0 x 6 + ⋯ 2 6 May-11 Jun-15

y ′ = 2xy. 1 1 y = a0 + a0 x 2 + a0 x 4 + a0 x 6 + ⋯ 2 6

Dec 09,11,12,13 Jan-13 Jun-14,15

y′′ + y = 0.

y = a 0 + a1 x −

1 1 1 1 a 0 x 2 − a1 x 3 + a0 x 4 + a x5 + ⋯ 2 6 24 120 1

1 1 1 y = a 0 + a1 x − a 0 x 3 − a1 x 4 + a x6 + ⋯ 6 12 180 0 Dec-13 Jan-15

y′′ + x 2 y = 0. y = a 0 + a1 x −

1 1 1 1 a 0 x 4 − a1 x 5 + a0 x 8 + a x9 + ⋯ 12 20 672 1440 1

y′′ = y′. y = a 0 + a1 x +

Mar-10 1 1 1 1 a1 x 2 + a1 x 3 + a1 x 4 + a x5 + ⋯ 2 6 24 120 1

y ′′ = 2y ′ in powers of x.

y′′ − 2xy′ + 2py = 0.

Ans.

y = a 0 + a1 x − p a 0 x 2 +

Ans.

Jan-13

2 1 2 y = a 0 + a1 x + a1 x 2 + a1 x 3 + a1 x 4 + a1 x 5 + ⋯ 3 3 15

8.

9.

Jun-14 May-13

y ′′ + xy = 0 in powers of x.

(1 − p) (1 − p) (3 − p) p (2 − p) a1 x 3 − a0 x 4 + a1 x 5 + ⋯ 3 6 30

Jun-13 Dec-13 Jan-15

(1 − x 2 )y ′′ − 2xy ′ + 2y = 0. 1 1 y = a 0 + a1 x − a 0 x 2 − a 0 x 4 − a 0 x 6 + ⋯ 3 5

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4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

T

d2 y dy (1 − x 2 ) − x + py = 0. 2 dx dx

10. Ans.

H

May-11

(1 − p) (9 − p)(1 − p) p p(4 − p) a0 x 2 + a1 x 3 − a0 x 4 + a1 x 5 + ⋯ 2 6 24 120

May-12 Jun-15

(1 + x 2 )y ′′ + xy ′ − 9y = 0.

11. Ans.

C

y = a 0 + a1 x −

9 4 15 7 y = a 0 + a1 x + a 0 x 2 + a1 x 3 + a 0 x 4 − a 0 x 6 + ⋯ 2 3 8 16 May-11,13 Dec-12 Jan-13 Jun-13

(x 2 + 1)y ′′ + xy ′ − xy = 0 near x = 0.

12.

Ans.

PAGE | 4

x3 x3 a1 3 y = a 0 + a1 x + a 0 − a1 + ( ) x 4 − ( ) a0 x 5 +.. 6 6 12 40

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 d2 y m+k−1 (m + k)(m + k − 1)ak (x − x0 )m+k−2 = ∑ (m + k)ak (x − x0 ) & = ∑ dx dx 2 k=1 k=2

dy

 STEP-2: Substitute the expressions of y , dx , and

d2 y 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 m1 and 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 = c1 y1 + c2 y2 .

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4. SERIES SOLUTION OF DIFFERENTIAL EQUATIONS

PAGE | 5

CASE-II Equal roots m+k In this case, we will have only one series solution. i.e. y = y1 = ∑∞ k=0 a k (x − x0 ) dy

In terms of a0 and the variable m. The general solution is y = c1 (y1 )m + c2 ( dm1 )

m

CASE-III Distinct roots differing by an integer When m1 − m2 ∈ Z, i.e. difference of m1 & m2 is 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 . 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

2.

dy

modified form of the series for y and in dm obtained from this modified form. Exercise - 3

Solution By Frobenius Method C

1.

4x

d2 y dy + 2 + y = 0. 2 dx dx

Jan-15

H

2.

(x 2 − 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.

x 2 y′′ + xy′ + (x 2 − 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.

x 2 y ′′ + x 3 y ′ + (x 2 − 2)y = 0.

Dec- 10

A.E.M.(2130002)

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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−𝐬𝐭 𝐟(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)} = ∫0 e−st f(t) dt Now, ℒ{α ∙ f(t) + β ∙ g(t)} ∞

= ∫ e−st [α ∙ f(t) + β ∙ g(t)] dt 0 ∞

∞

= ∫ e−st ∙ α ∙ f(t) dt + ∫ e−st ∙ β ∙ g(t) dt 0

0 ∞

=α∙∫ e

∞ −st

f(t) dt + β ∙ ∫ e−st g(t) dt

0

0

= α ∙ ℒ{f(t)} + β ∙ ℒ{g(t)}

Laplace Transform of some Standard Function 𝟏

1. 𝓛{𝐭 𝐧 } = 𝐬𝐧+𝟏 n  1 ; 𝐧 > −𝟏.

Jun-13 ; Dec-13 ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

ℒ{t

n}

= ∫ e−st t n dt 0

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

⟹ ℒ{t

n}

=∫ e 0

A.E.M.(2130002)

−x

x n dx sn s

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5. LAPLACE TRANSFORM AND APPLICATIONS 1

=

s

PAGE | 2

∞

∫ e−x x n dx n+1 0

∞

1

= sn+1 n  1 ( By definition of Gamma function n = ∫0 e−x x n−1 dx ) If n is a positive integer, then n! = n  1 n!

⟹ ℒ{t n } =

s n+1 𝟏

2. 𝓛{𝟏} = 𝐬 .

Dec-12 ; Jun-14 ; Jan-15 ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

ℒ{1} = ∫ e−st dt 0 ∞

e−st =[ ] −s 0 =

0−1 1 = −s s 𝟏

2. 𝓛{𝐞𝐚𝐭 } = 𝐬−𝐚 , 𝐬 > 𝐚

Jun-15 ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

ℒ{eat } = ∫ e−st eat dt 0 ∞

= ∫ e−(s−a)t dt 0 ∞

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 } =

A.E.M.(2130002)

1 s−a

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5. LAPLACE TRANSFORM AND APPLICATIONS 𝟏

3. 𝓛{𝐞−𝐚𝐭 } = 𝐬+𝐚 , 𝐬 > −𝐚

PAGE | 3 Jun-13

∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

ℒ{e−at } = ∫ e−st e−at dt 0 ∞

= ∫ e−(s+a)t dt 0 ∞

e−(s+a)t =[ ] −(s + a) 0

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

0−1 1 ]= −(s + a) s+a 1 s+a

⟹ ℒ{e−at } =

𝐚

4. 𝓛{𝐬𝐢𝐧 𝐚𝐭} = 𝐬𝟐 +𝐚𝟐 , 𝐬 > 0 and 𝐚 is a constant. ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

ℒ{sin at} = ∫ e−st sin at dt 0 ∞

e−st =[ 2 (−s sin at − a cos at)] s + a2 0 [When t ⟶ ∞ ⟹ e−st ⟶ 0(∵ s > 0)] =0− ⟹ ℒ{sin at} =

s2

s2

1 (−a) + a2

a + a2 𝐬

5. 𝓛{𝐜𝐨𝐬 𝐚𝐭} = 𝐬𝟐 +𝐚𝟐 , 𝐬 > 0 and 𝐚 is a constant. ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt

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5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 4

∞

ℒ{cos at} = ∫ e−st cos at dt 0 ∞

e−st =[ 2 (−s cos at + a sin at)] s + a2 0 [ When t ⟶ ∞ ⟹ e−st ⟶ 0 (∵ s > 0) ] 1 (−s) s 2 + a2

=0− ⟹ ℒ{cos at} =

s s 2 + a2 𝐚

6. 𝓛{𝐬𝐢𝐧𝐡 𝐚𝐭} = 𝐬𝟐 −𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐 (𝐬 > |𝐚|) eat − e−at

Proof: ℒ{sinh at} = ℒ {

2

}

=

1 [ℒ{eat } − ℒ{e−at }] 2

=

1 1 1 [ − ] 2 s−a s+a

=

1 s+a−s+a [ ] 2 s 2 − a2

⟹ ℒ{sinh at} =

Dec-11;Dec-12 ; Jun-14 ; Jan-15 ; Jun-15

a s 2 − a2 𝐬

7. 𝓛{𝐜𝐨𝐬𝐡 𝐚𝐭} = 𝐬𝟐 −𝐚𝟐 , 𝐬𝟐 > 𝐚𝟐 (𝐬 > |𝐚|) Proof: ℒ{cosh at} = ℒ {

eat + e−at 2

Dec-13

}

1 = [ℒ{eat } + ℒ{e−at }] 2 1 1 1 = [ + ] 2 s−a s+a 1 s+a+s−a = [ ] 2 s 2 − a2 ⟹ ℒ{cosh at} =

A.E.M.(2130002)

s2

s − a2

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5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 5

Exercise-1

Laplace Transform C

1.

0,0 < t < π Find the Laplace transform of f(t) = { . sin t , t > π −𝛑𝐬

[

H

H

T

2.

3.

4.

−𝐞 ] 𝐬𝟐 + 𝟏

0, 0 ≤ t < 2 Find the Laplace transform of (t) = { . 3, when t ≥ 2

May-12 Jun-15

𝟑𝐞−𝟐𝐬 [ ] 𝐬

Dec-11

𝟒𝐞−𝟑𝐬 [ ] 𝐬

Jan-15

−𝐞−𝟐𝐬 𝟏 𝟏 [ 𝟐 + + 𝟐] 𝐬 𝐬 𝐬

Dec-13

0, 0 ≤ t < 3 Find the Laplace transform of (t) = { . 4, when t ≥ 3

t + 1,0 ≤ t ≤ 2 Given that f(t) = { . Find ℒ{f(t)}. 3, t ≥ 2

1

Find the Laplace transform of t 3 + e−3t + t 2 . T

5.

𝟑!

𝟏

[𝐬𝟒 + 𝐬+𝟑 +

√𝛑 𝟑 𝟐𝐬 ⁄𝟐

]

Jun-13

4

Find the Laplace transform of 2t 3 + e−2t + t 3 . C

6.

𝟒 1/ 3 𝟏𝟐 𝟏 [ 𝟒+ + ] 𝐬 𝐬 + 𝟐 𝟗𝐬𝟕⁄𝟑

Dec-13

𝟓! 𝟏 𝐬 + + ] 𝐬𝟔 𝐬 + 𝟏𝟎𝟎 𝐬𝟐 + 𝟐𝟓

Jun-14

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

C

7.

8.

[

Find the Laplace transform of 100t + 2t10 + sin 10t. 𝟏 𝟐 ∙ 𝟏𝟎 ! 𝟏𝟎 [ + + ] 𝐬 − 𝐥𝐨𝐠 𝐞 𝟏𝟎𝟎 𝐬𝟏𝟏 𝐬𝟐 + 𝟏𝟎𝟎 Find ℒ{sin 2t cos 2t}.

C

𝟐 [ 𝟐 ] 𝐬 + 𝟏𝟔

9.

Jun-15

Dec-09 Jan-15

Find ℒ{sin 2t sin 3t}. T

10.

[

𝟏𝟐𝐬 ] (𝐬𝟐 + 𝟐𝟓)(𝐬𝟐 + 𝟏)

Jun-14

𝐬𝟐 + 𝟖 [ 𝟐 ] 𝐬(𝐬 + 𝟏𝟔)

Dec-12

Find the Laplace transform of cos2 2t. H

11.

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5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 6

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

C

H

T

𝐬𝟐 + 𝟐𝐚𝟐 [ 𝟐 ] 𝐬(𝐬 + 𝟒𝐚𝟐 )

12.

Find the Laplace transform of following functions: (i) cos3 t. (ii) sin2 t. 13. 𝐬(𝐬𝟐 + 𝟕) 𝟐 [(𝐢) 𝟐 , (𝐢𝐢) ] 𝟐 𝟐 (𝐬 + 𝟏)(𝐬 + 𝟗) 𝐬(𝐬 + 𝟒) Find the Laplace transform of following functions: (i) sin3 2t. (ii) sin2 2t. 14. 𝟒𝟖 𝟖 (𝐢𝐢) [(𝐢) 𝟐 , ] (𝐬 + 𝟒)(𝐬𝟐 + 𝟑𝟔) 𝐬(𝐬𝟐 + 𝟏𝟔) Find the Laplace transform of (i) cos3 2t (ii) sin2 3t. 𝐬𝟑 + 𝟐𝟖𝐬 𝟏𝟖 15. [(𝐢) 𝟐 , (𝐢𝐢) ] 𝟐 𝟐 (𝐬 + 𝟒)(𝐬 + 𝟑𝟔) 𝐬(𝐬 + 𝟑𝟔)

Dec-11

Jun-14

Jan-15

Theorem. First Shifting Theorem Statement: If ℒ{f(t)} = F(s), then show that ℒ{eat f(t)} = F(s − a). ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt ∞

Now, ℒ{eat f(t)} = ∫0 e−st eat f(t) dt ∞

= ∫0 e−(s−a)t f(t) dt Since, s and a are constants. s − a is also a constant. Thus, ℒ{eat f(t)} = F(s − a) Corollary: If 𝓛{𝐟(𝐭)} = 𝐅(𝐬), then show that 𝓛{𝐞−𝐚𝐭 𝐟(𝐭)} = 𝐅(𝐬 + 𝐚). Exercise-2

First Shifting Theorem H

1.

T

2.

C

3.

H

4.

By using first shifting theorem, obtain the value of L{(t + 1)2 et }. 𝟐 𝟐 𝟏 [ + + ] 𝟑 𝟐 (𝐬 − 𝟏) (𝐬 − 𝟏) 𝐬−𝟏 Find Laplace transform of e−2t sin2 2t, where a is a constant. 𝟖 [ ] (𝐬 + 𝟐)[(𝐬 + 𝟐)𝟐 + 𝟏𝟔] Find Laplace transform of e−2t (sin2 4t + t 2 ). 𝟑𝟐 𝟐 [( + )] (𝐬 + 𝟐)(𝐬𝟐 + 𝟒𝐬 + 𝟔𝟖) (𝐬 + 𝟐)𝟑 Find Laplace transform of e−3t (2 cos 5t − 3 sin 5t). 𝟐𝐬 − 𝟗 [ ] (𝐬 + 𝟑)𝟐 + 𝟐𝟓

A.E.M.(2130002)

Dec-09

Jun-13

Jan-15

Jun-14

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS T

5.

PAGE | 7

Find Laplace transform of e4t (sin 2t cos t). 𝟏 𝟑 𝟏 [ 𝟐 + 𝟐 ] 𝟐 (𝐬 − 𝟖𝐬 + 𝟐𝟓) (𝐬 − 𝟖𝐬 + 𝟏𝟕)

Jun-15

Theorem. Differentiation of Laplace Transform dn

Statement: If ℒ{f(t)} = F(s), then show that ℒ{t n f(t)} = (−1)n dsn F(s), n = 1,2,3, … ∞

Proof: By definition, F(s) = ∫0 e−st f(t) dt ⟹

dn dn ∞ −st F(s) = ∫ e f(t) dt dsn ds n 0 ∞

=∫ [ 0

∂n −st e ] f(t) dt ∂s n

∞

∂n−1 −st = ∫ (−1)(t) n−1 e f(t) dt ∂s 0 ∞

= ∫ (−1)2 (t)2 0

∂n−2 −st e f(t) dt ∂s n−2

Continuing in this way, we have ∞

=

(−1)n

∫ e−st (t)n f(t) dt 0

= (−1)n ℒ{t n f(t)} dn

Thus,ℒ{t n f(t)} = (−1)n dsn F(s), n = 1,2,3, … Exercise-3

Differentiation Of Laplace Transform (Multiplication by tn ) Find the value of ℒ{t sin ωt}. T

𝟐𝛚𝐬 ] (𝐬𝟐 + 𝛚𝟐 )𝟐

Dec-09

𝟐𝐬 𝟒𝐬 , ] (𝐬𝟐 + 𝟏𝟐 )𝟐 (𝐬𝟐 + 𝟒)𝟐

Jun-15

Find the Laplace transform of t 2 sin πt & t 2 sin 2t. 𝟐𝛑(𝟑𝐬𝟐 − 𝛑𝟐 ) 𝟒(𝟑𝐬𝟐 − 𝟒) [ , 𝟐 ] (𝛑𝟐 + 𝐬𝟐 )𝟑 (𝐬 + 𝟒)𝟑

Dec-10 Jan-15 Jun-13

1.

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

H

H

2.

3.

[

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 8 d

H

4.

C

5.

If ℒ{f(t)} = F(s), then show that ℒ{tf(t)} = − ds F(s) use this result to obtained ℒ{eat t sin at}. 𝟐𝐚(𝐬 − 𝐚) [ ] [(𝐬 − 𝐚)𝟐 + 𝐚𝟐 ]𝟐 Find the value of ℒ{t cosh t}. 𝟏 + 𝐬𝟐 [ 𝟐 ] (𝐬 − 𝟏)𝟐

Dec-13

Jun-14

Find the Laplace transform of f(t) = t 2 sinh at. C

6.

H

[

𝟏 𝟏 − ] 𝟑 (𝐬 − 𝐚) (𝐬 + 𝐚)𝟑

Find the Laplace transform of f(t) = t 2 cos h πt & f(t) = t 2 cos h 3t 𝟏 𝟏 𝟏 𝟏 [ + , + ] (𝐬 − 𝛑)𝟑 (𝐬 + 𝛑)𝟑 (𝐬 − 𝟑)𝟑 (𝐬 + 𝟑)𝟑

7.

May-12

Jan-15 Jun-15

Find the Laplace transform of t 3 cosh 2t. H

8.

[

𝟑 𝟑 + ] 𝟒 (𝐬 − 𝟐) (𝐬 + 𝟐)𝟒

Dec-12

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

∞

} = ∫s F(s)ds.

∞

Proof: By definition, F(s) = ∫0 e−st f(t) dt Integrating both sides with respect to "s" in the range s to ∞. ∞

∞

∞

∫ F(s)ds = ∫ e−st (∫ e−st ds) f(t) dt s

0

s ∞

∞

∞ e−st 0 − e−st =∫ ( ) f(t) dt = ∫ ( ) f(t) dt −t s −t 0 0 ∞ f(t) f(t) = ∫ e−st [ ] dt = ℒ { } t t 0 f(t)

Thus, ℒ {

t

∞

} = ∫s F(s)ds.

Exercise-4

Integration of Laplace Transform (Division by t) sin 2t

Find ℒ { H

1.

A.E.M.(2130002)

t

}. 𝛑 𝐬 [ − 𝐭𝐚𝐧−𝟏 ( )] 𝟐 𝟐

Dec-12 Jun-14

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS 1−et

Find ℒ { C

2.

t

}

Find the Laplace transform of T

1−cos t t

4.

sin wt t

𝐬−𝟏 [𝐥𝐨𝐠 ( )] 𝐬

Dec-13

√𝐬𝟐 + 𝟏 [𝐥𝐨𝐠 ( )] 𝐬

May-12

𝛑 𝐬 [ − 𝐭𝐚𝐧−𝟏 ( )] 𝟐 𝛚

May-11

.

3. Find the Laplace transform of [

H

PAGE | 9

].

Theorem. Laplace Transform of integration of a function t

Statement: If ℒ{f(t)} = F(s), then ℒ {∫0 f(t) dt} = t

∞

F(s) s

.

t

Proof: By definition ℒ (∫0 f(u) du) = ∫0 e−st {∫0 f(u) du} dt, t

Suppose that, U = ∫0 f(u) du ∞

=∫e

∞

−st

0

e−st dU e−st U dt = [U ∙ ( ) − ∫{ ∙ ( )} dt] −s dt −s 0

By Integration by parts, ∞

∞

e−st dU e−st = [U ( )] − ∫ { ( )} dt −s 0 dt −s 0

In first step, t

When t → 0 ⟹ U = ∫0 f(u) du → 0 & t → ∞ , ⟹ e−st → 0. In second step, By Fundamental theorem of calculus,

dU dt

d

t

= dt {∫0 f(u) du} = f(t).

t

1 ∞ −st ⟹ ℒ {∫ f(t) dt} = ∫ e f(t) dt s 0 0

t

1

Thus, ℒ {∫0 f(t) dt} = s F(s).

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 10

Exercise-5

Laplace Transform of integration of a function t

Find ℒ {∫0 e−x cos x dx}. C

1.

𝐬+𝟏 [ ] 𝐬[(𝐬 + 𝟏)𝟐 + 𝟏] t

Jun-13 May-12

t

Find ℒ {∫0 ∫0 sin au du du}. C

2.

[

𝟏 𝐚 ] 𝐬𝟐 (𝐬𝟐 + 𝐚𝟐 )

Jun-13

Theorem. Laplace Transform of Periodic Function Statement: The Laplace transform of a piecewise continuous periodic function f(t) having p

1

period "p" is F(s) = ℒ{f(t)} = 1−e−ps ∫0 e−st f(t)dt , where s > 0. ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt1 p

=∫ e

∞ −st

f(t) dt + ∫ e−st f(t) dt … (A)

0

p

∞

Now, ∫p e−st f(t) dt Let, t = u + p ⟹ dt = du When t ⟶ p ⟹ u ⟶ 0 and t ⟶ ∞ ⟹ u ⟶ ∞. ∞

⟹∫ e

∞ −st

f(t) dt = ∫ e−s(u+p) f(u + p) du

p

0

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

= ∫ e−s(u+p) f(u) du 0 ∞

=e

−sp

∫ e−su f(u) du = e−sp F(s) 0

By eqn. …(A) p

∞

ℒ{f(t)} = ∫ e−st f(t) dt + ∫ e−st f(t) dt 0

A.E.M.(2130002)

p

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 11

p

⟹ F(s) = ∫ e−st f(t) dt + e−sp F(s) 0 p

⟹ (1 − e

−sp )F(s)

= ∫ e−st f(t) dt 0

p 1 ∫ e−st f(t) dt 1 − e−sp 0

⟹ F(s) =

p

1

Thus, ℒ{f(t)} = 1−e−sp ∫0 e−st f(t)dt, where s > 0 is a Laplace Transform of periodic function f(t) of period p. Exercise-6

Laplace Transform of Periodic Function

T

1.

Find the Laplace transform of the half wave rectifier π sin ωt , 0 < t < ω 2π f(t) = { π 2π and f(t) = f (t + ω ). 0, ω < t < ω [

Dec-12 Mar-10 𝛚

((𝐬𝟐 + 𝐰 𝟐 )(𝟏 − 𝐞

−𝛑𝐬⁄ 𝛚 ))

]

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

2.

[

𝛚(𝟏+𝐞

−𝛑𝐬⁄ 𝛚)

−𝛑𝐬 (𝐬 𝟐 +𝐰 𝟐 )(𝟏−𝐞 ⁄𝛚 )

May-11

]

Laplace Transform of Unit Step Function Show that 𝓛{𝐮(𝐭 − 𝐚)} =

𝐞−𝐚𝐬 𝐬 ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt Now, ∞

ℒ{u(t − a)} = ∫ e−st u(t − a) dt 0 a

∞

= ∫ e−st u(t − a) dt + ∫ e−st u(t − a) dt 0

We know that, u(t − a) = {

A.E.M.(2130002)

a

0, 0 < x < a 1, x ≥ a

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS ∞

∞

=∫ e

−st

a

= [−

PAGE | 12

e−st dt = [− ] s a

0 − e−sa e−sa ]= s 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−as F(s) ∞

Proof: By definition, ℒ{f(t)} = ∫0 e−st f(t) dt Now, ℒ{f(t − a)u(t − a)} ∞

= ∫ e−st f(t − a)u(t − a) dt 0 a

∞

= ∫ e−st f(t − a)u(t − a) dt + ∫ e−st f(t − a)u(t − a) dt 0

a

We know that, 𝑢(𝑡 − 𝑎) = {

0, 0 < 𝑥 < 𝑎 1, 𝑥 ≥ 𝑎

∞

= ∫ 𝑒 −𝑠𝑡 𝑓(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 −𝑠𝑢 𝑓(𝑢) 𝑑𝑢 = 𝑒 −𝑎𝑠 𝐹(𝑠) 0

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

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 13

Exercise-7

Second Shifting Theorem Find the Laplace transform of 𝑒 −3𝑡 𝑢(𝑡 − 2). H

1.

𝒆−𝟐𝒔 𝒆−𝟔 [ ] 𝒔+𝟑

May-12

𝒆−𝟐𝒔+𝟐 [ ] 𝒔−𝟏

Dec-10

𝟐! 𝟒 𝟒 [𝒆−𝟐𝒔 ( 𝟑 + 𝟐 + )] 𝒔 𝒔 𝒔

Jan-15

𝟐𝒆−𝒔 ] 𝒔𝟑

Jun-15

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

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

C

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

T

4.

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

T

5.

[

−𝒔𝒆−𝜋𝒔 ] 𝒔𝟐 + 𝟏

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

𝐵

𝐶

Partial fractions = (𝑠−𝑎) + (𝑠−𝑏) + (𝑠−𝑐) Case 2 : If the denominator has repeated linear factors (𝑠 − 𝑎), (n times), then 𝐴

𝐴

𝐴

𝐴

1 2 3 𝑛 Partial fractions = (𝑠−𝑎) + (𝑠−𝑎) 2 + (𝑠−𝑎)3 + ⋯ + (𝑠−𝑎)𝑛

Case 3 : If the denominator has non-repeated quadratic factors (𝑠 2 + 𝑎𝑠 + 𝑏),(𝑠 2 + 𝑐𝑠 + 𝑑), then 𝐴𝑠+𝐵

𝐶𝑠+𝐷

Partial fractions = (𝑠2 +𝑎𝑠+𝑏) + (𝑠2 +𝑐𝑠+𝑑) Case 4 : If the denominator has repeated quadratic factors (𝑠 2 + 𝑎𝑠 + 𝑏), (n times), then 𝐴𝑠+𝐵

𝐶𝑠+𝐷

Partial fractions = (𝑠2 +𝑎𝑠+𝑏) + (𝑠2 +𝑎𝑠+𝑏)2 + ⋯(n times)

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 14

Exercise-8

Inverse Laplace Transform 6𝑠

H

1.

Find ℒ −1 {𝑠2 −16}. [𝟔𝒄𝒐𝒔𝒉𝟒𝒕] Find ℒ −1 {

C

3(𝑠2 −1)

2

2𝑠5

}.

2.

𝟑 𝒕𝟒 [ [𝟏 − 𝒕𝟐 + ]] 𝟐 𝟒! 𝑠3 +2𝑠2 +2

T

Dec-12

3.

Find ℒ −1 { 𝑠3 (𝑠2 +1) }.

4.

Find ℒ −1 {𝑠(𝑠+1)}.

5.

Find ℒ −1 {𝑠2 −3𝑠+2}.

[𝒕𝟐 + 𝒔𝒊𝒏𝒕]

Dec-13

Mar-10 Jun-13

1

T

[𝟏 − 𝒆−𝒕 ]

Jan-15

𝑠

H

[𝟐𝒆𝟐𝒕 − 𝒆𝒕 ] Find ℒ −1 {(𝑠+

C

1 √2)(𝑠−√3)

}.

6. [

𝒆√𝟑𝒕 − 𝒆−√𝟐𝒕 √𝟑 + √𝟐

]

Jun-14

Dec-10 Dec-09 Jun-15

1

H

7.

Find ℒ −1 {(𝑠−2)(𝑠+3)}. 𝟏 𝟐𝒕 [𝒆 − 𝒆−𝟑𝒕 ] 𝟓 5𝑠2 +3𝑠−16

C

8.

Find ℒ −1 {(𝑠−1)(𝑠+3)(𝑠−2)}. [𝟐𝒆𝒕 + 𝒆−𝟑𝒕 + 𝟐𝒆𝟐𝒕 ]

Jun-15 Dec-12 Dec-11

3𝑠2 +2

T

9.

Find ℒ −1 {(𝑠+1)(𝑠+2)(𝑠+3)}. 𝟓 𝟐𝟗 −𝟑𝒕 [ 𝒆−𝒕 − 𝟏𝟒𝒆−𝟐𝒕 + 𝒆 ] 𝟐 𝟐

Jun-13 Jan-15

𝑠3

C

10.

Find ℒ −1 {𝑠4 −81}. [

𝒄𝒐𝒔𝟑𝒕 + 𝒄𝒐𝒔𝒉𝟑𝒕 ] 𝟐

Dec-12 Mar-10

𝑠+10

H

11.

Find ℒ −1 {− 𝑠2 −𝑠−2}.

A.E.M.(2130002)

[−𝟒𝒆𝟐𝒕 + 𝟑𝒆−𝒕 ]

Mar-10

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 15

Exercise-9

First Shifting Theorem If 𝓛

−𝟏 {𝑭(𝒔)}

= 𝒇(𝒕), then 𝓛−𝟏 {𝑭(𝒔 − 𝒂)} = 𝒆𝒂𝒕 𝒇(𝒕)

10

Find ℒ −1 {(𝑠−2)4 }. H

1.

[

𝟓𝒆𝟐𝒕 𝒕𝟑 ] 𝟑

Dec-12

3

C

2.

Evaluate ℒ −1 {𝑠2 +6𝑠+18}. [𝒆−𝟑𝒕 𝒔𝒊𝒏𝟑𝒕] Find the inverse Laplace transform of

T

3.

6+𝑠

Dec-09

, use shifting theorem.

𝑠2 +6𝑠+13

𝟑 [𝒆−𝟑𝒕 𝒄𝒐𝒔𝟐𝒕 + 𝒆−𝟑𝒕 𝒔𝒊𝒏 𝟐𝒕] 𝟐

Dec-11

𝟏 [ 𝟐 [𝒆𝒂𝒕 𝒔𝒊𝒏 𝒂𝒕 − 𝒆−𝒂𝒕 𝒔𝒊𝒏 𝒂𝒕]] 𝟒𝒂

Dec-13

𝑠

Find ℒ −1 {𝑠4 +4𝑎4 }. T

4.

5𝑠+3

C

5.

Find the inverse Laplace transform of (𝑠2 +2𝑠+5)(𝑠−1). 𝟑 [−𝒆−𝒕 𝒄𝒐𝒔 𝟐𝒕 + 𝒆−𝒕 𝒔𝒊𝒏 𝟐𝒕 + 𝒆𝒕 ] 𝟐

May-12 Jun-14 Jan-15

Exercise-10

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

C

1.

2.

Find the inverse Laplace transform of

𝑠𝑒 −2𝑠 𝑠2 +𝜋 2

. [𝒄𝒐𝒔 𝝅(𝒕 − 𝟐) ∙ 𝒖(𝒕 − 𝟐)]

Find the inverse Laplace transform of

Dec-10 Dec-09

𝑒 −4𝑠 (𝑠+2) 𝑠2 +4𝑠+5

.

[𝒆−𝟐(𝒕−𝟒) 𝒄𝒐𝒔(𝒕 − 𝟒) ∙ 𝒖(𝒕 − 𝟒)]

Jun-13

Exercise-11

Inverse Laplace Transform Of Derivatives If 𝓛−𝟏 {𝑭(𝒔)} = 𝒇(𝒕), then 𝓛−𝟏 {𝑭′(𝒔)} = −𝒕𝒇(𝒕) 𝑠+𝑎

Find ℒ −1 {𝑙𝑜𝑔 𝑠+𝑏}. T

1.

𝒆−𝒃𝒕 − 𝒆−𝒂𝒕 [ ] 𝒕

Jun-13 Dec-10

𝑠+1

Find ℒ −1 {𝑙𝑜𝑔 𝑠−1}. H

2.

A.E.M.(2130002)

𝒆𝒕 − 𝒆−𝒕 [ ] 𝒕

Jun-13

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 16

1

H

3.

Obtain ℒ −1 {𝑙𝑜𝑔 𝑠 }. 𝟏 [ ] 𝒕 Find the inverse transform of the function 𝑙𝑛 (1 +

C

4.

𝑤2 𝑠2

). 𝟐 [ (𝟏 − 𝒄𝒐𝒔 𝒘𝒕)] 𝒕

May-11

Mar-10 Jun-14

Definition: Convolution Product The convolution of 𝑓 and 𝑔 is denoted by 𝑓 ∗ 𝑔 and is defined as 𝑡

𝑓 ∗ 𝑔 = ∫ 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 0

Exercise-12

Convolution Product T

1.

C

2.

Find the value of 1 ∗ 1. Where " ∗ " denote convolution product. [𝒕] Find the convolution of 𝑡 and 𝑒 𝑡 . [𝒆𝒕 − 𝒕 − 𝟏]

Dec-09 Jun-15 Dec-10 Jun-15

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

Then ℒ −1 {𝐹(𝑠) ∙ 𝐺(𝑠)} = ∫0 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 = 𝑓 ∗ 𝑔 𝑡

Proof: Let us suppose 𝐹(𝑡) = ∫0 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 ∞

𝑡

∞

𝑡

Now, ℒ(𝐹(𝑡)) = ∫0 𝑒 −𝑠𝑡 (∫0 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢) 𝑑𝑡 = ∫0 ∫0 𝑓(𝑢)𝑔(𝑡 − 𝑢) 𝑒 −𝑠𝑡 𝑑𝑢 𝑑𝑡

𝑡=𝑢 Here, region of integration is entire area lying between the lines 𝑢 =

𝑡⟶∞ 𝑢=0

A.E.M.(2130002)

0 and 𝑢 = 𝑡 which is part of the first quadrant .

𝑡

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 17

Changing the order of integration, we have ∞

∞

ℒ(𝐹(𝑡)) = ∫ ∫ 𝑒 −𝑠𝑡 𝑓(𝑢)𝑔(𝑡 − 𝑢) 𝑑𝑡 𝑑𝑢 0

𝑢 ∞

∞

= (∫ 𝑒 −𝑠𝑢 𝑓(𝑢)𝑑𝑢) (∫ 𝑒 −𝑠(𝑡−𝑢) 𝑔(𝑡 − 𝑢) 𝑑𝑡) 0

𝑢

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

∞

= (∫ 𝑒 −𝑠𝑢 𝑓(u)𝑑𝑢) (∫ 𝑒 −𝑠𝑣 𝑔(𝑣) 𝑑𝑣) 0

0

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

⟹ ℒ −1 {𝐹(𝑠) ∙ 𝐺(𝑠)} = ∫ 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 0 𝑡

Hence, 𝑓 ∗ 𝑔 = ℒ −1 {𝐹(𝑠) ∙ 𝐺(𝑠)} = ∫0 𝑓(𝑢)𝑔(𝑡 − 𝑢)𝑑𝑢 is convolution product of function 𝑓 & 𝑔.

Exercise-13

Convolution Theorem Using convolution theorem, find the inverse Laplace transform of H

1.

𝟏 𝒔𝒊𝒏 𝒂𝒕 [ 𝟐( − 𝒕 𝒄𝒐𝒔 𝒂𝒕)] 𝟐𝒂 𝒂 ℒ −1 {

T

2.

ℒ −1 { T

1 . (𝑠2 +𝑎2 )2

3.

(𝑠 2

𝑠 } + 1)2

𝑠 } (𝑠 + 1) (𝑠 − 1)2

May-11 Dec-10

𝟏 [ 𝒕 𝒔𝒊𝒏 𝒕] 𝟐

Jan-15

𝟏 𝒕 𝒆𝒕 𝒆−𝒕 [ 𝒕𝒆 + − ] 𝟐 𝟒 𝟒

Jan-15

1

State convolution theorem and using it find ℒ −1 {(𝑠+1)(𝑠+3)} . H

4.

A.E.M.(2130002)

𝒆−𝒕 − 𝒆−𝟑𝒕 [ ] 𝟐

Dec-13

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

H

5.

PAGE | 18

State convolution theorem and using it find ℒ −1 {(𝑠2

𝑠2

}.

+4)(𝑠2 +9)

𝟏 [ (𝟑 𝒔𝒊𝒏 𝟑𝒕 − 𝟐 𝒔𝒊𝒏 𝟐𝒕)] 𝟓

Jun-14

𝑠2

C

6.

State convolution theorem and using it find ℒ −1 {(𝑠2 +𝑎2 )(𝑠2 +𝑏2 )} . 𝟏 (𝒂 𝒔𝒊𝒏 𝒂𝒕 − 𝒃 𝒔𝒊𝒏 𝒃𝒕)] [ 𝟐 (𝒂 − 𝒃𝟐 )

Jun-15

𝑠+2

Find ℒ −1 {(𝑠2 +4𝑠+5)2 }. T

H

7.

8.

𝒆−𝟐𝒕 𝒕 𝒔𝒊𝒏 𝒕 [ ] 𝟐 State the convolution theorem on Laplace transform and using it find ℒ

−1

{

1

}.

𝑠(𝑠2 +4)

[

𝟏 − 𝒄𝒐𝒔 𝟐𝒕 ] 𝟒

Jun-13 Dec-12 Dec-09 Jan-15 Jun-15

𝑠

H

9.

Apply convolution theorem to Evaluate ℒ −1 {(𝑠2 +𝑎2 )2 } . 𝒕 𝒔𝒊𝒏 𝒂𝒕 [ ] 𝟐𝒂

1

Jun-14 Jan-15

Find ℒ −1 {𝑠(𝑠+𝑎)3 } . T

10.

𝟏 𝒕𝟐 𝒆−𝒂𝒕 𝟐𝒕𝒆−𝒂𝒕 𝟐𝒆−𝒂𝒕 𝟐 [ [− − − + 𝟑 ]] 𝟐 𝒂 𝒂𝟐 𝒂𝟑 𝒂

Dec-13

𝑎

State convolution theorem and use to evaluate Laplace inverse of 𝑠2 (𝑠2 +𝑎2 ) C

11.

[

𝒂𝒕 − 𝒔𝒊𝒏 𝒂𝒕 ] 𝒂𝟐

Mar-10

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)} = ∫0 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡 ∞

⟹ ℒ{𝑓 ′ (𝑡)} = ∫ 𝑒 −𝑠𝑡 𝑓′(𝑡) 𝑑𝑡 0

= [𝑒

−𝑠𝑡

= [𝑒

−𝑠𝑡

∫𝑓

′ (𝑡)𝑑𝑡

∞ 𝑑 −𝑠𝑡 ′ (𝑡)𝑑𝑡} − ∫ {( 𝑒 ) ∙ ∫ 𝑓 𝑑𝑡] 𝑑𝑡 0 ∞

𝑓(𝑡) − ∫(−𝑠)𝑒

−𝑠𝑡

𝑓(𝑡)𝑑𝑡] 0

∞

= 0 − 𝑓(0) + 𝑠 ∫ 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = s𝐹(𝑠) − 𝑓(0) 0

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 19

Thus, ℒ{𝑓 ′ (𝑡)} = 𝑠𝐹(𝑠) − 𝑓(0). ℒ{𝑓 ′′ (𝑡)} = 𝑠 2 𝐹(𝑠) − s𝑓(0) − 𝑓 ′ (0) In general, 𝓛{𝒇𝒏 (𝒕)} = 𝒔𝒏 𝑭(𝒔) − 𝒔𝒏−𝟏 𝒇(𝟎) − 𝒔𝒏−𝟐 𝒇′(𝟎) − 𝒔𝒏−𝟑 𝒇” (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 𝟏 [ 𝒄𝒐𝒔 √𝟔𝒕 + ] 𝟔 𝟔 Using Laplace transform solve the differential equation 𝑑2 𝑥

T

2.

𝑑𝑡 2

Dec-12 Jun-15

𝑑𝑥

+ 2 𝑑𝑡 + 5𝑥 = 𝑒 −𝑡 𝑠𝑖𝑛 𝑡 , 𝑤ℎ𝑒𝑟𝑒 𝑥(0) = 0, 𝑥 ′ (0) = 1. 𝒆−𝒕 (𝒔𝒊𝒏 𝒕 + 𝒔𝒊𝒏 𝟐𝒕) [ ] 𝟑

May-12

𝑑2 𝑦

Solve the differential equation 𝑑𝑡 2 + 4𝑦 = 𝑓(𝑡), 𝑦(0) = 0, 𝑦 ′ (0) = 1 by Laplace transform where (i)𝑓(𝑡) = {

1,0 < 𝑡 < 1 , (ii)𝑓(𝑡) = 𝐻(𝑡 − 2). 0, 𝑡 > 1 May-12

C

3.

C

4.

Solve IVP using Laplace transform 𝑦 ′′ + 4𝑦 = 0, 𝑦(0) = 1, 𝑦 ′ (0) = 6. [𝒄𝒐𝒔 𝟐𝒕 + 𝟑 𝒔𝒊𝒏 𝟐𝒕]

Dec-11

5.

Using Laplace transform solve the IVP 𝑦 ′′ + 𝑦 = 𝑠𝑖𝑛 2𝑡 , 𝑦(0) = 2, 𝑦 ′ (0) = 1. 𝟓 𝟏 [𝒚 = 𝒔𝒊𝒏 𝒕 − 𝒔𝒊𝒏 𝟐𝒕 + 𝟐 𝒄𝒐𝒔 𝒕] 𝟑 𝟑

Dec-10 Jan-15

6.

By Laplace transform solve, 𝑦 ′′ + 𝑎2 𝑦 = 𝐾 𝑠𝑖𝑛 𝑎𝑡. 𝒌 𝑩 𝒌 [𝒚 = ( 𝟐 + ) 𝒔𝒊𝒏𝒂 𝒕 + (𝑨 − 𝒕 ) 𝒄𝒐𝒔 𝒂𝒕] 𝟐𝒂 𝒂 𝟐𝒂

Mar-10 Jun-14

H

C

C

H

7.

8.

𝒔𝒊𝒏 𝟐𝒕 𝟏 − 𝒄𝒐𝒔 𝟐𝒕 [𝟏 − 𝒄𝒐𝒔 𝟐(𝒕 − 𝟏)]𝑯(𝒕 − 𝟏) (𝒊)𝒚 = + − 𝟐 𝟒 𝟒 [ ] 𝒔𝒊𝒏 𝟐𝒕 𝑯(𝒕 − 𝟐) − 𝒄𝒐𝒔 𝟐(𝒕 − 𝟐) 𝑯(𝒕 − 𝟐) (𝒊𝒊)𝒚 = + 𝟐 𝟒

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 [ − 𝒆−𝒕 ] 𝟐! By using the method of Laplace transform solve the IVP : 𝑦 ′′ + 5𝑦 ′ + 6𝑦 = 𝑒 −𝑡 , 𝑦(0) = 0 𝑎𝑛𝑑 𝑦 ′ (0) = −1 . 𝟏 𝟑 [ 𝒆−𝒕 − 𝟐𝒆−𝟐𝒕 + 𝒆−𝟑𝒕 ] 𝟐 𝟐

A.E.M.(2130002)

Dec-09

Jun-14

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 20

T

9.

By using the method of Laplace transform solve the IVP : 𝑦 ′′ + 4𝑦 ′ + 3𝑦 = 𝑒 −𝑡 , 𝑦(0) = 1 𝑎𝑛𝑑 𝑦 ′ (0) = 1 . 𝒕 𝟕 𝟑 [( + ) 𝒆−𝒕 − 𝒆−𝟑𝒕 ] 𝟐 𝟒 𝟒

Dec-13

H

By using the method of Laplace transform solve the IVP : 𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 𝑒 𝑡 , 𝑦(0) = 1 𝑎𝑛𝑑 𝑦 ′ (0) = 0 . 10. 𝟏 𝟑 𝟐 [ 𝒆𝒕 + 𝒆−𝒕 − 𝒆−𝟐𝒕 ] 𝟔 𝟐 𝟑

Jun-15

H

Solve using Laplace transforms, 𝑦 ′′′ + 2𝑦 ′′ − 𝑦 ′ − 2𝑦 = 0; where, 𝑦(0) = 1, 𝑦 ′ (0) = 2, 𝑦 ′′ (0) = 2 11. 𝟏 [ [𝟓𝒆𝒕 + 𝒆−𝟐𝒕 ] − 𝒆−𝒕 ] 𝟑

Jan-15

T

By using the method of Laplace transform solve the IVP : 12. 𝑦 ′′ − 4𝑦 ′ + 3𝑦 = 6𝑡 − 8, 𝑦(0) = 0 𝑎𝑛𝑑 𝑦 ′ (0) = 0 . [𝒚(𝒕) = 𝟐𝒕 + 𝒆𝒕 − 𝒆𝟑𝒕 ]

Jun-15

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

5. LAPLACE TRANSFORM AND APPLICATIONS

PAGE | 21

Laplace Transform Of Some Standard Functions 1.

ℒ{1} =

1 𝑠

1.

−1

1 𝑡 𝑛−1 𝑡 𝑛−1 { 𝑛} = 𝑶𝑹 (𝑛 − 1)! 𝑠 n

2.

ℒ

3.

ℒ −1 {

1 } = 𝑒 𝑎𝑡 𝑠−𝑎

4.

ℒ −1 {

1 } = 𝑒 −𝑎𝑡 𝑠+𝑎

𝑎 + 𝑎2

5.

ℒ −1 {

𝑠 𝑠 2 + 𝑎2

6.

𝑠 ℒ −1 { 2 } = 𝑐𝑜𝑠 𝑎𝑡 𝑠 + 𝑎2

𝑎 − 𝑎2

7.

ℒ −1 {

𝑠 𝑠 2 − 𝑎2

8.

𝑠 ℒ −1 { 2 } = 𝑐𝑜𝑠ℎ 𝑎𝑡 𝑠 − 𝑎2

1

2.

ℒ{𝑡 𝑛 } =

3.

ℒ{𝑒 𝑎𝑡 } =

4.

ℒ{𝑒 −𝑎𝑡 } =

5.

ℒ{𝑠𝑖𝑛 𝑎𝑡} =

6.

ℒ{𝑐𝑜𝑠 𝑎𝑡} =

7.

ℒ{𝑠𝑖𝑛ℎ 𝑎𝑡} =

8.

ℒ{𝑐𝑜𝑠ℎ 𝑎𝑡} =

9.

1 ℒ −1 { } = 1 𝑠

n  1 𝑶𝑹

𝑠 𝑛+1

𝑛! 𝑠 𝑛+1

1 𝑠−𝑎 1 𝑠+𝑎 𝑠2

𝑠2

n 1

𝑛! ℒ{𝑒 𝑎𝑡 𝑡 𝑛 } = 𝑶𝑹 (𝑠 − 𝑎)𝑛+1 (𝑠 − 𝑎)𝑛+1

10. ℒ{eat sin bt} =

b (s − a)2 + b 2

11. ℒ{eat cos bt} =

s−a (s − a)2 + b 2

−1

𝑠2

𝑠2

1 1 } = 𝑠𝑖𝑛 𝑎𝑡 2 +𝑎 𝑎

1 1 } = 𝑠𝑖𝑛ℎ 𝑎𝑡 2 −𝑎 𝑎

1 𝑒 𝑎𝑡 𝑡 𝑛−1 𝑒 𝑎𝑡 𝑡 𝑛−1 { }= 𝑶𝑹 (𝑠 − 𝑎)𝑛 (n − 1)! n

9.

ℒ

10.

ℒ −1 {

1 1 at } = e sin bt (s − a)2 + b 2 a

11.

ℒ −1 {

s−a } = eat cos bt (s − a)2 + b 2

12. ℒ{eat sinh bt} =

b (s − a)2 − b 2

12.

ℒ −1 {

1 1 at } = e sinh bt (s − a)2 − b 2 b

13. ℒ{eat cosh bt} =

s−a (s − a)2 − a2

13.

ℒ −1 {

s−a } = eat cosh bt (s − a)2 − b 2

s 1 } = t sin at (s2 + a2 )2 2a

14.

ℒ −1 {

1 1 } = (sin at − at cos at) (s2 + a2 )2 2a3

14. ℒ −1 {

A.E.M.(2130002)

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

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”. ∂y 2

∂y

e.g. (∂x) + ∂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.”. ∂y 2

∂y

e.g. (∂x) + ∂x + 5y = 0 has degree 2.

Notation ∂z

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

∂z

=q, ∂y

∂2 z ∂x2

= r,

∂2 z

= s, ∂x ∂y

∂2 z ∂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.

A.E.M.(2130002)

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PAGE | 2

Exercise-1

Formation Of Partial Differential Equation C

1.

C

2.

T

C

Form the partial differential equation z = (x − 2)2 + (y − 3)2 . [𝟒𝐳 = 𝐩𝟐 + 𝐪𝟐 ]

Form the partial differential equation z = (x + a)(y + b). [𝐳 = 𝐩𝐪]

Jun-15

3.

Eliminate the function f from the relation f(xy + z 2 , x + y + z) = 0. 𝐩 + 𝟏 𝐲 + 𝟐𝐳𝐩 [ = ] 𝐪 + 𝟏 𝐱 + 𝟐𝐳𝐪

Jun-13

4.

Form the partial differential equation of f(x + y + z, x 2 +y 2 + z 2 ) = 0. 𝐩 + 𝟏 𝐱 + 𝐳𝐩 [ = ] 𝐪 + 𝟏 𝐲 + 𝐳𝐪

Jan-13 Jun-14 Jun-15

Form the partial differential equation f(x 2 − y 2 , xyz) = 0. T

Jan-13 Jun-14

5.

H

6.

H

7.

−[

𝐲𝐳 + 𝐱𝐲𝐩 𝐱 =− ] 𝐱𝐳 + 𝐱𝐲𝐪 𝐲

Form partial differential equation by eliminating the arbitrary function from xyz = Ф(x + y + z). 𝐩 + 𝟏 𝐲𝐳 + 𝐱𝐲𝐩 [ = ] 𝐪 + 𝟏 𝐱𝐳 + 𝐱𝐲𝐪 Form the partial differential equation by eliminating the arbitrary function from z = f(x 2 − y 2 ). 𝐩 𝐱 [ =− ] 𝐪 𝐲

Jan-15

Dec-13

Dec-13

x

C

8.

Form the partial differential equation of z = f (y).

Form the partial differential equation of z = xy + f(x 2 + y 2 ).

𝐩 𝐲 [ =− ] 𝐪 𝐱

Jan-13

𝐩−𝐲 𝐱 = ] 𝐪−𝐱 𝐲

Jan-15

T

9.

C

Form the partial differential equation of y = f(x − at) + F(x + at). 𝛛𝟐 𝐲 𝛛𝟐 𝐲 10. 𝟐 =𝐚 𝛛𝐭 𝟐 𝛛𝐱 𝟐

A.E.M.(2130002)

[

Jun-15

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PAGE | 3

Exercise-2

Solution Of Partial Differential Equation ∂2 u

C

1.

Solve ∂x ∂y = x 3 +y 3 . 𝐱 𝟒 𝐲 𝐱𝐲 𝟒 [𝐮(𝐱, 𝐲) = + + 𝐅(𝐲) + 𝐠(𝐱)] 𝟒 𝟒

Jun-14

∂2 u

C

2.

Solve ∂x ∂t = e−t cosx.

[𝐮(𝐱, 𝐲) = −𝐞−𝐭 𝐬𝐢𝐧 𝐱 + 𝐅(𝐭) + 𝐠(𝐱)]

Dec-13

∂3 u

H

3.

Solve ∂x2 ∂y = cos(2x + 3y). [𝐮(𝐱, 𝐲) = − ∂2 z

T

4.

𝐬𝐢𝐧(𝟐𝐱 + 𝟑𝐲) + 𝐱𝐅(𝐲) + 𝐆(𝐲) + 𝐡(𝐱)] 𝟏𝟐

∂z

Solve ∂x ∂y = sin x sin y, given that ∂y = −2 sin y,when x = 0 & z = 0, π

when y is an odd multiple of 2 .

5.

Jun-13

[𝐳(𝐱, 𝐲) = 𝐜𝐨𝐬 𝐱 𝐜𝐨𝐬 𝐲 + 𝐜𝐨𝐬 𝐲]

∂2 z

C

Jan-15

Solve ∂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 dx dy dz  Step-1: From the A.E. P = Q = 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 eliminate z from

dx P

=

dx P

=

dy Q

or it is possible to

dy Q

.

Similarly, if the variable x is absent in last two fractions or it is possible to eliminate x from last dx dy two fractions Q = R , then we can apply grouping method.

A.E.M.(2130002)

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6.PARTIAL DIFFERENTIAL EQUATION AND APPLICATIONS

PAGE | 4

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 dy dz ldx + mdy + ndz = = = P Q R 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.

C

2.

Solve (z − y)p + (x − z)q = y − x. [𝐅(𝐱 + 𝐲 + 𝐳, 𝐱 𝟐 +𝐲 𝟐 + 𝐳 𝟐 ) = 𝟎] Solve x(y − z)p + y(z − x)q = z(x − y). [𝐅(𝐱 + 𝐲 + 𝐳, 𝐱𝐲𝐳) = 𝟎]

Jun-15

Jun-13

Solve(x 2 − y 2 − z 2 )p + 2xyq = 2xz. T

3. Solve (y + z)p + (x + z)q = x + y.

T

4.

A.E.M.(2130002)

[𝐅 (

𝐲 𝐱𝟐 + 𝐲𝟐 + 𝐳𝟐 [𝐅 ( , ) = 𝟎] 𝐳 𝐳

Jun-14

𝐱−𝐲 , (𝐱 + 𝐲 + 𝐳)(𝐱 − 𝐲)𝟐 ) = 𝟎] 𝐲−𝐳

Jan-15

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6.PARTIAL DIFFERENTIAL EQUATION AND APPLICATIONS

PAGE | 5

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 dy dz dp dq = = = = ( lagrange − Charpit eqn ) ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f − − −p − q +p +q ∂p ∂q ∂p ∂q ∂x ∂z ∂y ∂z  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.

H

2.

Solve p + q2 = 1. [𝐳 = 𝐚𝐱 ± (√𝟏 − 𝐚)𝐲 + 𝐜] Solve √p + √q = 1 𝟐

[𝐳 = 𝐚𝐱 + (𝟏 − √𝐚) 𝐲 + 𝐜]

Jan-13 Jan-15

Solve p2 +q2 = npq. T

3.

𝐧𝐚 ± 𝐚√𝐧𝟐 − 𝟒 [𝐳 = 𝐚𝐱 + 𝐲 + 𝐜] 𝟐

Jan-15

𝟑 𝟑 𝟐 𝟐 (𝐚 + 𝐱)𝟐 + (𝐲 − 𝐚)𝟐 + 𝐛] 𝟑 𝟑

Jan-14

Solve p2 + q2 = x + y. C

4.

A.E.M.(2130002)

[𝐳 =

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PAGE | 6

Solve p2 − q2 = x − y. H

5.

[𝐳 = Solve p − x 2 = q + y 2 .

T

6.

T

7.

C

8.

H

9.

C

10.

𝟑 𝟑 𝟐 𝟐 (𝐚 + 𝐱)𝟐 + (𝐚 + 𝐲)𝟐 + 𝐛] 𝟑 𝟑

Jan-15

𝐱𝟑 𝐲𝟑 + 𝐚𝐲 − + 𝐛] 𝟑 𝟑

Jun-15

[𝐳 = 𝐚𝐱 + 𝐛𝐲 + 𝐚𝟐 𝐛𝟐 ]

Jun-13

[𝐳 = 𝐚𝐱 + Solve z = px + qy + p2 q2 . Solve z = px + qy − 2√pq.

[𝐳 = 𝐚𝐱 + 𝐛𝐲 − 𝟐√𝐚𝐛] Solve qz = p(1 + q).

Dec-13

[𝐛𝐳 − 𝟏 = 𝐜 𝐞𝐱+𝐚𝐲 ]

Jun-14

[𝟒𝐛𝐳 = (𝟐𝐱 + 𝟐𝐛𝐲 + 𝐛)𝟐 ]

Jun-15

Solve pq = 4z.

Method Of Separation Of Variables  Step 1: Let u(x, y) = X(x) ∙ Y(y) ∂u ∂u ∂2 u

∂2 u

∂2 u

 Step 2: Find ∂x , ∂y , ∂x2 , ∂x ∂y , ∂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 Solve the equation by method of separation of variables C

1.

∂u ∂x

∂u

= 4 ∂y ,

[𝐮(𝐱, 𝐲) = 𝟖 𝐞 ∂u

C

∂u

−𝟏𝟐𝐱−𝟑𝐲 ]

2.

Solve 2 ∂x =

+ u subject to the condition u(x, 0) = 4e−3x [𝐮(𝐱, 𝐭) = 𝟒 𝐞−𝟑𝐱−𝟕𝐭 ]

Jan-13

3.

Solve ∂x = 2 ∂t + u subject to the condition u(x, 0) = 6e−3x [𝐮(𝐱, 𝐭) = 𝟔 𝐞−𝟑𝐱−𝟐𝐭 ]

Jun-15

∂u

H

Dec-13

where u(0, y) = 8e−3y .

A.E.M.(2130002)

∂t

∂u

DARSHAN INSTITUTE OF ENGINEERING AND TECHNOLOGY

6.PARTIAL DIFFERENTIAL EQUATION AND APPLICATIONS

∂u

C

T

∂u

4.

Using method of separation of variables solve ∂x + ∂y = 2(x + y)u.

5.

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

[𝐮(𝐱, 𝐲) = 𝐞

[𝐮(𝐱, 𝐲) = (𝐜𝟏 𝐞

H

6.

Using method of separation of variables solve

7.

∂2 u ∂x2

√𝐤𝐱

𝐱 𝟐 +𝐲 𝟐 +𝐤𝐱−𝐤𝐲+𝐜

+ 𝐜𝟐 𝐞

−√𝐤𝐱

)

]

𝐤𝐲 𝐜𝟑 𝐞𝟏𝟔 ]

∂z

∂z

Solve ∂x2 − 2 ∂x + ∂y = 0 by the method of separation variables. [𝐳(𝐱, 𝐲) = (𝐜𝟏 𝐞

(𝟏+√𝟏+𝐤)𝐱

∂2 u

+ 𝐜𝟐 𝐞

(𝟏−√𝟏+𝐤)𝐱

) 𝐜𝟑 𝐞

−𝐤𝐲

]

8.

Dec-10

Jun-13

May-12 Jun-14

∂2 u

Solve two dimensional Laplace’s equation ∂x2 + ∂y2 = 0,using the method C

Jan-15 Jun-15

∂u

= ∂y + 2u.

[𝐮(𝐱, 𝐲) = (𝐜𝟏 𝐞√𝐤𝐱 + 𝐜𝟐 𝐞−√𝐤𝐱 ) 𝐜𝟑 𝐞(𝐤−𝟐)𝐲 ] ∂2 z

C

PAGE | 7

separation of variables. [𝐮(𝐱, 𝐲) = (𝐜𝟏 𝐞√𝐤𝐱 + 𝐜𝟐 𝐞−√𝐤𝐱 ) (𝐜𝟑 𝐜𝐨𝐬 √𝐤𝐲 + 𝐜𝟒 𝐬𝐢𝐧 √𝐤𝐲)]

Jan-13 Dec-09

Using the method of separation of variables, solve the partial differential ∂2 u

T

9.

∂2 u

equation ∂x2 = 16 ∂y2 . [𝐮(𝐱, 𝐲) = (𝐜𝟏 𝐞√𝐤𝐱 + 𝐜𝟐 𝐞−√𝐤𝐱 ) ∂u

C

10.

√𝐤 (𝐜𝟑 𝐞 𝟒 𝐲

+ 𝐜𝟒 𝐞

−

√𝐤 𝐲 𝟒 )]

∂u

Solve x ∂x − 2y ∂y = 0 using method of separation variables.

A.E.M.(2130002)

Jan-15

𝐤

[𝐮(𝐱, 𝐲) = 𝐜𝟏 𝐜𝟐 𝐱 𝐤 𝐲 𝟐 ]

Jun-13

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