Engineering Mathematics - 2 · PDF fileEquations of first order and higher degree (p-y-x equations), Equations solvable for p, y, x. General and singular solutions, Clarauit’s equation

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Common to All Branches

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Engineering Mathematics - 2

[10MAT21]

Chemistry Cycle

2010 Scheme

Branch Name: Common to all branches SEM: 1/2 University: VTU Syllabus: 2010

Table of Contents:

Engineering Mathematics -2 (10MAT21):

Sl. No. Units 1 Differential Equations 1 2 Differential Equations 2 3 Differential Equations 3 4 Partial Differential Equation 5 Integral Calculus 6 Vector Integration 7 Laplace Transforms - 1 8 Laplace Transforms - 2

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SYLLABUS

ENGINEERING MATHEMATICS-II

SUBJECT CODE: 10 MAT 21

PART A

Unit-I: Differential Equations-1

Equations of first order and higher degree (p-y-x equations), Equations solvable for p, y,

x. General and singular solutions, Clarauit’s equation. Applications of differential equations of

first order.

Unit-II: Differential Equations-2

Linear differential equations: Solution of second and higher order equations with constant

coefficients by inverse differential operator method. Simultaneous differential equations of first

order.

Unit-III: Differential Equations-3

Method of variation of parameters, Solution of Cauchy’s and Legendre’s linear

equations, Series solution of equations of second order, Frobenius method-simple problems.

Unit-IV: Partial Differential Equations(PDE)

Formation of partial differential equations (PDE) by elimination of arbitrary constants &

functions. Solution of non-homogeneous PDE by direct integration. Solution of homogeneous

PDE involving derivative with respect to one independent variable only. Solution of Lagrange’s

linear PDE. Solution of PDE by the method of separation of variables (first and second order

equations)

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ENGG. MATHEMATICS-II 10MAT21

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PART-B

Unit-V: Integral Calculus

Multiple Integrals- Evaluation of double integrals and triple integrals. Evaluation of

double integrals over a given region, by change of order of integration, by change of variables.

Applications to area and volume-illustrative examples.

Beta and Gamma Functions- Properties and problems

Unit-VI: Vector Integration:

Line integrals- definition and problems, surface and volume integrals - definition.

Green’s theorem in a plane, Stoke’s and Gauss divergence theorem (statements only).

Unit-VII: Laplace Transforms -I:

Definitions, transforms of elementary functions, properties, periodic function, unit step

function and unit impulse function.

Unit-VIII: Laplace Transforms-II:

Inverse Laplace transforms, Convolution theorem, solution of linear differential

equations using Laplace transforms. Applications.



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Index

Unit 1: Differential Equations – I………………………. ....07-21

Unit 2: Differential Equations – II…………………………21-42

Unit 3: Differential Equations – III…………………….….43-60

Unit 4: Partial Differential Equations ………………….…61-72

Unit 5: Integral Calculus……………………………….….73-92

Unit 6: Vector Integration………………………………...93-102

Unit 7: Laplace Transforms – I………………………….103-138

Unit 8: Laplace Transforms – II……………………...…139-158



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UNIT IDIFFERENTIAL EQUATIONS –I

Introduction:

We are familiar with the solution of differential equations (d.e.) of first order

and first degree in this unit we discuss the solution of differential equations of first

order but not of first degree. In addition to the general solution and particular

solution associated with the d.e, we also introduce singular solution. Thed.es of

first order but not of first degree are also branded as p-y-x equations.

Differential equations of first order and higher degree

If y=f(x), we use the notation pdx

dythroughout this unit.

A differential equation of first order and nth degree is the form

1 2

0 1 2 ....... 0n n n

nA p A p A p A

Where 0 1 2, , ,... nA A A A are functions of x and y. This being a differential equation of

first order, the associated general solution will contain only one arbitrary constant.

We proceed to discuss equations solvable for P or y or x, wherein the problem is

reduced to that of solving one or more differential equations of first order and first

degree. We finally discuss the solution of clairaut’s equation.

Equations solvable for p

Supposing that the LHS of (1) is expressed as a product of n linear factors,

then the equivalent form of (1) is

1 2

1 2

( , ) ( , ) ... ( , ) 0 ....(2)

( , ) 0, ( , ) 0... ( , ) 0

n

n

p f x y p f x y p f x y

p f x y p f x y p f x y

All these are differential equations of first order and first degree. They can

be solved by the known methods. If 1 2( , , ) 0, ( , , ) 0,... ( , , ) 0nF x y c F x y c F x y c

respectively represents the solution of these equations then the general solution is

given by the product of all these solution.

Note: We need to present the general solution with the same arbitrary constant in

each factor.



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1. Solve :

2

0dy dy

y x y xdx dx

>> The given equation is 2

2

( ) 0

( ) ( ) 4

2

( ) ( )

2

.,2 2

., 1 /

,

yp x y p x

x y x y xyp

y

y x x yp

y

y x x y y x x yie p or p

y y

ie p or p x y

We have

2 22 2 2 2

2 2

1 ( ) 0

, 0

., 2 ( ) 02 2

Thus the general solution is given by (y-x-c) (x ) 0

dyy x c or y x c

dx

dy xAlso or ydy xdx y dy x dx k

dx y

y xie k or y x k or x y c

y c

2. Solve : ' 2( ) (2 3 ) 6 0x y x y y y

>> The given equation with the usual notation is, 2

2

3

(2 3 ) 6 0

(2 3 ) (2 3 ) 24

2

(2 3 ) (2 3 ) 32

2

2 2 2 ( 2 ) 0

33 3

., log 3log log log log , log

., log

xp x y p y

x y x y xyp

x

x y x y yp or

x x

We have

dydy dx c or y x c or y x c

dx

dy y dy dx dy dxAlso or k

dx x y x y x

ie y x k or y x c where k x

ie y 3 3 3

3

log ( ) 0

Thus the general solution is (y-2x-c) (y-cx ) 0

cx y cx or y cx



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3) Solve ( ) ( )p p y x x y

Solution: The given equation is, 0)(2 yxxpyp

2

2 2

4 ( )

2

4 4 (2 )

2 2

2( )., ( )

2

,

y y x x yp

y x xy y y x yp

y xie p x or p y x

We have2

2

2

,

., , . ( )

1, ;

., ( ) , int .

(2 ) (

Pdx x

x

x x x

x

dy xx y k

dx

dyAlso y x

dx

dyie y x is a linear d e similar to the previous problem

dx

P Q x e e

Hence ye xe dx c

ie ye xe e c egrating by parts

Thus the general solutionis givenby y x c e 1) 0y x c

Equations solvable for y:

We say that the given differential equation is solvable for y, if it is possible to express y

in terms of x and p explicitly. The method of solving is illustrated stepwise.

Y=f(x, p)

We differentiate (1) w.r.t x to obtain

, ,dy dp

p F x ydx dx

Here it should be noted that there is no need to have the given equation solvable for y in

the explicit form(1).By recognizing that the equation is solvable for y, We can proceed to

differentiate the same w.r.t. x. We notice that (2) is a differential equation of first order in p

and x. We solve the same to obtain the solution in the form. ( , , ) 0x p c

By eliminating p from (1) and (3) we obtain the general solution of the given

differential equation in the form G(x,y,c) =0



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Remark: Suppose we are unable to eliminate p from (1)and (3), we need to solve for x and y

from the same to obtain.

1 2( , ), ( , )x F p c y F p c

Which constitutes the solution of the given equation regarding p as a parameter.

Equations solvable for x

We say that the given equation is solvable for x, if it is possible to express x in terms of y

and p. The method of solving is identical with that of the earlier one and the same is as follows.

X = f(y, p )

Differentiate w.r.t.y to obtain

1, ,

dx dpF x y

dy p dy

(2) being a differential equation of first order in p and y the solution is of the form.

( , , ) 0y p c

By eliminating p from (1) and (3) we obtain the general solution of the given d.e in the form

G(x,y,c) =0

Note: The content of the remark given in the previous article continue to hold good here also.

1. Solve:1 22 tan ( )y px xp

1 2, 2 tan ( )By data y px xp

The equation is of the form y = f (x, p), solvable for y.

Differentiating (1) w.r.t.x,

2

2 4

2

2 4

12 2 .2

1

1., 2 2

1

dp dpp p x x p p

dx x p dx

dp dpie p x xp p

dx x p dx



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2. Obtain the general solution and the singular solution of the equation2 4y px p x

>> The given equation is solvable for y only. 2 4y px p x

Differentiating w.r.t x,

1., 2

2 2

., log log log ( ) log

2 4,

2 2/

2 2 4 4Using (2) in (1) we have, ( / ) ( / )

dp dx dp dx dpie p x or

dx x p x p

ie x p k or x p c x p c

Consider y px p x

x p c or x p c or p c x

y c x x c x x

Thus 2xy c c x is the general solution.

Now, to obtain the singular solution, we differentiate this relation partially

w.r.t c, treating c as a parameter.

2., 2 1

2 4 2 41 1

2 4 2 41 . 1., 2

2 4 2 41 1

., log 2log

1̀ 2consider 2 tan ( )

2

Using (2) in (1) we have,

1y = 2 / . tan ( )

12 tan ,

p dp pie p x

dxx p x p

x p p dp p x pie p x

dxx p x p

ie x p k

y px xp

and xp c

c x x c

Thus y cx c is the general solution.



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That is, 1=2cx or c=1/2x.

The general solution now becomes,

2

1 1

2 4xy x

x x

Thus 24 1 0,x y is the singular solution.

3) Solve y=p sin p + cos p

>> y = p sin p + cos p

Differentiating w.r.t. x,

cos sin sin

., 1 cos cos cos

.,sin sin

Thus we can say that sin cos and sin - constitutes

the general solution of the given d.e

:si

dp dp dpp p p p p

dx dx dx

dpie p or p dp dx p dp dx c

dx

ie p x c or x p c

y p p p x p c

Note 1n sin ( ).

We can as well substitute for in (1) and present the solution in the form,

1 1( )sin ( ) cossin ( )

p x c p x c

p

y x c x c x c

4) Obtain the general solution and singular solution of the equation22y px p y .

Solution: The given equation is solvable for x and it can be written as

2 ........(1)y

x pyp

Differentiating w.r.t y we get

2

2 1

11 0

y dp dpp y

p p p dy dy

y dpp

p p dy



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2 2

1,

1 0 0

........(2)

2 (1)

2

dpIgnoring p which does not contain this gives

p dy

y dp dy dpor

p dy y p

Integrating we get

yp c

substituting for p from in

y cx c

5) Solve 2 22 cot p py x y .

Solution: Dividing throughout by p2, the equation can be written as

22

2

22 2

2

2cot 1 cot .

2cot cot 1 cot

y yx adding x tob s

p p

y yx x x

p p2

2 cot cos

cot cos

yor x ec x

p

yx ecx

p

1 2

cot cos/

sin sin

cos 1 cos 1

Integrating these two equations we get

(cos 1) (cos 1)

(cos 1) (cos 1) 0

yx ecx

dy dx

dy x dy xdx and

y x y x

y x c and y x c

general solutionis

y x c y x c

6) Solve:2 5 44 12 0 , obtain the singular solution also.p x p x y .

Solution: The given equation is solvable for y only.

2 5 4

2 5

4

4 12 0 ...........(1)_

4( , )

12

(1) . . . ,

p x p x y

p x py f x p

x

Differentiating w r t x



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5 4 4 3

2 55 3

4

5 5

2 4 20 12 48 0

4(2 4 ) 8 ( ) 0

2

2( 2 ) ( 2 )

20

log log

dp dpp x x p x p x y

dx dx

dp p x pp x x xp

dx x

dp pp x p x

dx x

dp p

dx x

Integrating p x k2 2

4 2 3

(1)

4 12

p c x equation becomes

c c x y

2

2 3

3

3

6

4 12

. .

2 4 0

Using 2 in general solution we get

3 0 as the singular solution

Setting c k the general solutionbecomes

k kx y

Differentiating w r t k partially we get

k x

k x

x y

7) Solve3 24 8 0 by solving for x.p xyp y

Solution: The given equation is solvable for x only.

y

ypyp

dy

dp

p

y

yp

p

yp

dy

dp

yy

yp

p

ypp

dy

dp

ypxxypdy

dp

ypxp

ypdy

dpxy

dy

dpp

ytrwatingDifferenti

pyfyp

ypx

yxypp

2323

2323

23232

2

2

23

23

4)4(

2

482

1288

3

124)43(

01641

.443

,...)1(

),(4

8

084



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yxciscsolutiongeneraltheThus

yxcc

ycxcc

haveweyyybythroughoutDividing

ycyxycycy

haveweincyPgU

cyp

ydy

dp

p

64)4(

8)4(

084

,

084

,)1(sin

logloglog2

12

2

23

2

Clairaut’s Equation

The equation of the form ( )y px f p is known as Clairaut’s equation.

This being in the form y = F (x , p), that is solvable for y, we differentiate (1) w.r.t.x

( )

This implies that 0 and hence p=c

Using in (1) we obtain the genertal solution of clairaut's equation in the form

( )

dy dp dpp p x f p

dx dx dx

dp

dx

p c

y cx f c

1. Solve:a

y pxp

>> The given equation is Clairaut’s equation of the form ( )y px f p , whose general

solution is ( )y cx f c

Thus the general solution is a

y cxc

Singular solution

Differentiating partially w.r.t c the above equation we have,



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02

( / ) ,

/ . /

2Thus 4 is the singular solution.

ax

c

ac

x

Hence y cx a c becomes

y a x x a x a

y ax

2. Modify the following equation into Clairaut’s form. Hence obtain the associated general

and singular solutions.

2 0

2 0, by data

2ie.,

( ).,

.,

xp py kp a

xp py kp a

xp kp a py

p xp k aie y

p

aie y px k

p

Here (1) is in the Clairaut’s form y=px+f(p) whose general solution is y = cx + f (c)

Thus the general solution is a

y cx kc

Now differentiating partially w.r.t c we have,

2

2

0

/

Hence the general solution becomes,

y -k = 2

Thus the singular solution is (y - k) 4 .

ax

c

c a x

ax

ax

Remark: We can also obtain the solution in the method: solvable for y.



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3. Solve the equation (px – y) (py + x) = 2p by reducing into Clairaut’s form, taking the

substitutions X = x2 , Y = y

2

2 2

2 2

,

1., . .2

2

.,

( )( ) 2

dXX x x

dx

dYY y y

dy

dy dy dY dX dYNow p and let p

dx dY dX dx dx

ie p P xy

Xie p P

Y

Consider px y py x p

., 2

., ( ) ( 1) 2

2., is in the Clairaut's form and hence the associated genertal solution is

1

2

1

Thus the required general solution of the given equat

X X Xie P X Y P Y X P

Y Y Y

ie PX Y P P

Pie Y PX

P

cY cX

c

22 2ion is y1

ccx

c

4) Solve 2 2 2, use the substitution , .px y py x a p X x Y y

Solution: Let 2 2dX

X x xdx

2 2

,

dXY x y

dy

dy dy dY dX dYNow p and let P

dx dY dX dx dx

1. .2

2

xP p x or p P

y y

Xp P

Y



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( ) ( ) 2

2

( )( 1) 2

2

1

Consider px y py x p

X X XP X Y P Y X P

Y Y Y

PX Y P P

PY PX

P Is in the Clairaut’s form and hence the associated general solution is

1

2

c

ccXY

Thus the required general solution of the given equation is 1

222

c

ccxy

5) Obtain the general solution and singular solution of the Clairaut’s equation3 2 1 0xp yp .( Dec

2011) Solution: The given equation can be written as

3

2 2

2

1 1sin ' ( )

( )

1

xpy y px i theClairaut s form y px f p

p p

whose general solutionis y cx f c

Thus general solutionis y cxc

. . .Differnetiating partially w r t c we get1/3

3

1/3 2/3

2/3 2/3

3 2

2 20

22 3

2

4 27

x cc x

Thus general solutionbecomes

xy x y x

x

or y x

APPLICATIONS OF DIFFERENTIAL EQUATIONS OF FIRST

ORDERWe are familiar with an application of differential equation of first order in the form of

Orthogonal Trajectories.

We present a few illustrative problems on Application to Electric Circuits.

A governing first order differential equation by Kirchhoff’s law is given by

diL Ri E

dt

Where L is the inductance, R is the resistance and E is the electromotive force.



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1. A constant electromotive force E volts is applied to a circuit containing a

constant resistance R ohms in series and a constant inductance L henries. If

the initial current is Zero, find the current in the circuit at any time t.

>> We have the governing differential equation,

diL Ri E

dt

This is a linear d.e of the form dy

py Qdx

whose solution is

P dx P dxye Qe dx c

Hence the solution of (1) is given by,

/ /

//.,

( / )

/ /.,

ERt L Rt Lie e dt cL

Rt LE eRt Lie ie cL R L

ERt L Rt Lie ie e cR

Using the initial condition, i=0 when t=0, (2) becomes,

0E

cR

Using this value of c in (2), we obtain the current in the circuit at any time t,

/1 Rt LEi e

R

2. The differential equation for the current in in an electric circuit containing an inductance L

and a resistance R in series and acted on by an electromotive force E sin wt satisfies the

equation di

L Ri Edt

sin wt. Find the value of current at any time t, if initially there is

no current in the circuit.

>> The given equation is put in the form,

sindi R E

i wtdt L L

The solution of this linear equation is



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/ sin

/ /., sin

ERt L Rt Lie wt e dt cL

ERt L Rt Lie ie e wt dt cL

Note: We are proceed as in the previous problem and obtain the solution. We can also obtain

the solution using the following alternative formula for the integral of eat sin bt.

1sin sin ( tan / )

2 2

// 1sin tan /

2 2 2/

// 1., sin tan /

2 2 2

1tan / ,

/sin2 2 2

ateatWe have e bt dt bt b a

a b

Rt LE eRt Lie wt wL R cL

R L w

Rt LEeRt Lie ie wt wL R c

R w L

Putting wL R we have

E Rt Li wt ce

R w L

Using the initial condition, I = 0 when t = 0, we have

sin( )0

2 2 2

Ec

R w L

Using this value of c in (2) we have,

/ 1sin sin , tan2 2 2

E Rt Li wt e where wt R

R w L



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UNIT II

DIFFERENTIAL EQUATIONS – II

INTRODUCTION:

LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND

HIGHER ORDER WITH CONSTANT COEFFICIENTS



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SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR

DIFFERENTIAL EQUATION



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INVERSE DIFFERENTIAL OPERATOR AND PARTICULAR INTEGRAL



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SPECIAL FORMS OF THE PARTICULAR INTEGRAL



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Type2:

1.



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SOLUTION OF SIMULTANEOUS DIFFERENTIAL

EQUATIONS



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UNIT III

DIFFERENTIAL EQUATIONS – III

METHOD OF VARIATION OF PARAMETERS



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SOLUTION OF CAUCHY’S HOMOGENEOUS LINEAR EQUATION AND LEGENDRE’S

LINEAR EQUATION



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PROBLEMS:



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Series solution of differential equations:

Consider the second order ODE 0)()()(2

2

yxhdx

dyxg

dx

ydxf , where f(x), g(x) and h(x) are

functions in x and f(x) 0. The series solution of above type of DE is explained as follows:

1. Assume the solution of (1) in the form0r

r

r xay

2. Find the derivatives y/ and y

// from the assumed solution and substitute in to the given DE

which results in an infinite series with various powers of x equal to zero.

3. Now equate the coefficients of various powers of x to zero and try to obtain recurrence

relation from which the constants a0, a

1, a

2,…. can be determined.

4. When substituted the values of a0, a

1, a

2,…. in to the assumed solution, we get the power

series solution of the given DE in the form y = Ay1(x) + By2(x), where A and B are

arbitrary constants.

In general, The above type of DE can be solved by the following two methods:

Type – I (Frobenius Method)

Suppose f(x) = 0 at x= 0 in the above differential equation, we assume solution in the

form 0r

rk

r xay where k, a0, a

1, a

2,…. are all constants to be determined and a

0 ≠ 0.

Type – II (Power Series Method)

Suppose f(x) ≠ 0 at x= 0 in the above differential equation, we assume solution in the

form 0r

r

r xay where a0, a

1, a

2,…. are all constants to be determined. Here all the constants

ar’s will be expressed in terms of a0 and a

1 only.

Problems :

1. Obtain the series solution of the equation 02

2

ydx

yd----- (1)

Let 0r

r

r xay ----- (2) be the series solution of (1).



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Hence, 0

2//

0

1/ ,)1(,r

r

r

r

r

r xrrayxray

Now (1) becomes

12

2 0)1(r

r

r

r

r

r xaxrra

Equating the coefficients of various powers of x to zero, we get

Coefficient of x–2

: a0 (0) (-1) = 0 and a0 ≠ 0.

Coefficient of x–1

: a1 (1) (0) = 0 and a1 ≠ 0.

Equating the coefficient of xr (r >=0)

0)1)(2(2 rr arra or )0()1)(2(

2 rrr

aa r

r -------(3)

Putting r = 0,1, 2, 3, …… in (3) we obtain,

;12020

;2412

;6

;2

135

024

13

02

aaa

aaa

aa

aa

;504042

;72030

157

046

aaa

aaa and so on.

Substituting these values in the expanded form of (1), we get,

4

4

3

3

2

210 xaxaxaxaay

i.e., 50401206720242

1753

1

642

0

xxxxa

xxxay

Hence !7!5!3!6!4!2

1753

1

642

0

xxxxa

xxxay is the required solution of

the given DE.



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POWER SERIES METHOD (Frobenious method)

1. Solve by Frobenious method: 0)( 22

2

22 ynx

dx

dyx

dx

ydx , where n is a non negative real

constant or parameter.

>> We assume the series solution of (i) in the form

0r

rk

r xay where a0 0 (ii)

Hence, 0

1)(r

rk

r xrkadx

dy

0

2

2

2

)1)((r

rk

r xrkrkadx

yd

Substituting these in (i) we get,

0

22

0

1

0

22 0)()1)((r

rk

r

r

rk

r

r

rk

r xanxxrkaxxrkrkax

i.e.,0

2

0

2

00

0)()1)((r

rk

r

r

rk

r

r

rk

r

r

rk

r xanxaxrkaxrkrka

Grouping the like powers, we get

0)()1)((0

2

0

2

r

rk

r

r

rk

r xaxnrkrkrka

0)(0

2

0

22

r

rk

r

r

rk

r xaxnrka (iii)

Now we shall equate the coefficient of various powers of x to zero

Equating the coefficient of xk from the first term and equating it to zero, we get

nknkanka ,0get we,0 Since .0 22

0

22

0

Coefficient of xk+1

is got by putting r = 1 in the first term and equating it to zero, we get



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i.e., nknkanka 1,gives0)1( since ,0 gives This.0)1( 22

1

22

1

which is a contradiction to k = n.

Let us consider the coefficient of xk+r

from (iii) and equate it to zero.

i.e, .0)( 2

22

rr anrka

)( 22

2

nrk

aa r

r (iv)

If k = +n, (iv) becomes

nrr

a

nrn

aa rr

r2)( 2

2

22

2

Now putting r = 1,3,5, ….., (odd vales of n) we obtain,

0a,09n6

aa 1

13

Similarly a5, a7… are equal to zero.

i.e., a1 = a5 = a7 = …… = 0

Now, putting r = 2, 4, 6 … (even values of n) we get,

;)1(444

002

n

a

n

aa ;

)2)(1(32168

024

nn

a

n

aa

Similarly we can obtain a6, a8.......

We shall substitute the values of ,,,, 4321 aaaa in the assumed series solution, we get

)xaxaxaxaa(xxay 44

33

2210

k

0r

rkr

Let y1 be the solution for k = +n

402001

)2)(1(32)1(4x

nn

ax

n

aaxy n



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)2)(1(2)1(2

1.,.5

4

2

2

01nn

x

n

xxayei n (v)

This is a solution of the Bessel’s equation.

Let y2 be the solution corresponding to k = - n. Replacing n be – n in (v) we get

)2)(1(2)1(2

15

4

2

2

02nn

x

n

xxay n

(vi)

The complete or general solution of the Bessel’s differential equation is y = c1y1 + c2y2, where c1,

c2 are arbitrary constants.

Now we will proceed to find the solution in terms of Bessel’s function by choosing

)1(2

10

na

n and let us denote it as Y1.

2)2)(1(

1

2)1(

1

21

)1(2.,.

42

1nn

x

n

x

n

xYei

n

n

2)1()2)(1(

1

2)1()1(

1

2)1(

1

2

42

nnn

x

nn

x

n

xn

We have the result (n) = (n – 1) (n – 1) from Gamma function

Hence, (n + 2) = (n + 1) (n + 1) and

(n + 3) = (n + 2) (n + 2) = (n + 2) (n + 1) (n + 1)

Using the above results in Y1, we get

2)3(

1

2)2(

1

2)1(

1

2

42

1n

x

n

x

n

xY

n

which can be further put in the following form

422100

12!2)3(

)1(

2!1)2(

)1(

2!0)1(

)1(

2

x

n

x

n

x

n

xY

n



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0

2

2!)1(

)1(

2 r

rrnx

rrn

x

0

2

!)1(

1

2)1(

r

rn

r

rrn

x

This function is called the Bessel function of the first kind of order n and is denoted by Jn(x).

Thus 0

2

!)1(

1

2)1()(

r

rn

r

nrrn

xxJ

Further the particular solution for k = -n ( replacing n by –n ) be denoted as J-n(x). Hence the

general solution of the Bessel’s equation is given by y = AJn(x) + BJ-n(x), where A and B are


2. Solve '' '2 3 0 by Frobenius method.xy y y

>> For the given equation, we seek a series solution in the from

0

0

21 2

20 0

2

2

1

, 0

,

( ) , ( )( 1)

Substituting for y, and from the above expressions in the given equation, we get

2 ( )( 1)

m r

r

r

m r m r

r r

r r

m r

r

y a x a

From this we find

dy d ym r a x m r m r a x

dx dx

dy d y

dx dx

m r m r a x 1

0 0 0

3 ( ) 0m r m r

r r

r r r

m r a x a x

We observe that in this expression the lowest degree term in x is xm-1

and this occurs in the first

and second series only. Let us rewrite this expression as shown below in which the coefficients

of xm-1

and 1m r

x for 1r are shown explicitly.

4. Solve by Frobenius method, the equation

2

24 2 0

d y dyx y

dx dxsoln: For the given equation we seek series solution in the form



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0

0

21 2

20 0

2

2

1 1

0 0 0

, 0.........(1)

( ) , ( )( 1)

, ,

4 ( )( 1) 2 ( ) 0

Re

m r

r

r

m r m r

r r

r r

m r m r m r

r r r

r r r

y a x a

dy d ym r a x m r m r a x

dx dx

dy d ySubstituting for y in the givenequation

dx dx

m r m r a x m r a x a x

write t

1 1

0

1

1 1 1

0 1

1 1

-1 -1

exp

4 ( 1) 4 ( )( 1)

2 2 ( ) 0..............(2)

1

m m r

r

r

m m r m r

r r

r r

m m r

he ression as follows

m m a x m r m r a x

ma x m r a x a x

Equating the coefficients of x and x for r we get2

0 0

1

1

1 2

1

4 ( 1) 2 0 2 0.........(3)

4 1 2 0

11.........(4)

2 2 2 1

(3) 1/ 2, 0...........(5)

1/ 2 (4)

1 1

2 1/ 2 2 2 2 1

r r r

r r

r r

m m a ma or m m

and m r m r a m r a a

or a a for rm r m r

equation gives m m

For m m equation becomes

a ar r r r

1 1.....(6)for r

Fromthis we get

1 0 2 1 0 0

3 2 0 0

1

2 3 4

0 1 2 3 4

2 31/2

0

1 1 1 1 1, ,

2.3 4.5 4.5 2.3 5!

1 1 1,

6.7 6.7.5! 7!

1/ 2 (1)

................

1 ....... ............(73! 5! 7!

m

a a a a a a

a a a a and so on

Putting these and m m in we have

y x a a x a x a x a x

x x xa x )



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2

1

1 0 2 1 0 0

3 2 0 0

2

2 3 4

0 1 2 3 4

0, (4)

11..........(8)

2 2 1

1 1 1 1, ,

2 4.3 4.3.2 4!

1 1 1,

6.5 (6.5)4! 6!

0 (1)

.........

r r

m

For m m equation becomes

a a for rr r

a a a a a a

a a a a and so on

Putting these and m m in we have

y x a a x a x a x a x

2 3

0

.......

1 ....... ...........(9)2 4! 6!

x x xa

The solution (7 ) and (9) are two linearly independent Frobenius type series solution of the given

equation, a0 is arbitrary constant

A general solution is y=Ay1+By2 where y1 is the solution (7) and y2 is the solution (9) and A & B are


1 1 1 1 1

0 0 1

1 1 1

m-1m-1

2 ( 1) 2 ( )( 1) 3 3 ( ) 0

Equating the coefficients of x x 1 present in the LHS of this expression to zero,

we get

m m r m m r m r

r r r

r r r

r

m m a x m r m r a x ma x m r a x a x

and for r

0 0

1

1 2

the following equaions:

2 ( 1) 3 0,

2( )( 1) 3( ) 0

We note that (iii) is the Indicial equation. Its roots are

1m 0,

2

r r r

m m a ma

m r m r a m r a a

m

1

1

which are real and distinct and their difference is not an integer.

Setting m = m 0 ( )

11

(2 1)r r

in iv we get

a a for rr r



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1 0 2 1 0

,

1 1 1, ,

1.3 2.5 (1.2)(3.5)

From this we get

a a a a a

3 2 0 0

4 3 0

1 1 1,

3.7 (1.2.3)(3.5.7) 3!(3.5.7)

1 1 1 1.

4.9 (4.9) 3!(3.5.7) 4!(3.5.7.9)

a a a a

a a a and so on

The solutions (vii) and (ix) are two linearly independent solutions of the given equation in series from.

In these solutions, a0 is an arbitrary constant.

A general solution of the given equation is written down by taking a linear combination of the

solutions (vii) and (ix). This means that a general solution of the given equation is

1 2y Ay By

Where 1y is the solution (vii) and 2y is the solution (ix) and A and B are arbitrary contents.



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UNIT IV

PARTIAL DIFFERENTIAL EQUATIONS (PDE)

Introduction:

Many problems in vibration of strings, heat conduction, electrostatics involve two or more variables.

Analysis of these problems leads to partial derivatives and equations involving them. In this unit we first

discuss the formation of PDE analogous to that of formation of ODE. Later we discuss some methods of

solving PDE.

Definitions:

An equation involving one or more derivatives of a function of two or more variables is called a partial

differential equation.

The order of a PDE is the order of the highest derivative and the degree of the PDE is the degree of highest

order derivative after clearing the equation of fractional powers.

A PDE is said to be linear if it is of first degree in the dependent variable and its partial derivative.

In each term of the PDE contains either the dependent variable or one of its partial derivatives, the PDE is

said to be homogeneous. Otherwise it is said to be a nonhomogeneous PDE.

Formation of pde by eliminating the arbitrary constants

Formation of pde by eliminating the arbitrary functions

Solutions to first order first degree pde of the type

P p + Q q =R

Formation of pde by eliminating the arbitrary constants:

(1) Solve:

Differentiating (i) partially with respect to x and y,

x

p

x

z

xaor

a

x

x

z 1122

22

2

2

2

2

b

y

a

xz2



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y

q

x

z

ybor

b

y

y

z 112222

Substituting these values of 1/a2 and 1/b

2 in (i), we get

(2) 2 z = x p + y q

(3) z = (x2 + a) (y

2 + b)

Differentiating the given relation partially

(x-a) 2 + (y-b)

2 + z

2 = k

2 …

Differentiating (i) partially w. r. t. x and y,

0y

zz)by(,0

x

zz)ax(

Substituting for (x- a) and (y- b) from these in (i), we get

2

22

2 ky

z

x

z1z This is the required partial differential equation.

(4) z = ax + by + cxy ...(i)

Differentiating (i) partially w.r.t. x y, we get

)ii..(cyax

z

)iii..(cxby

z

It is not possible to eliminate a,b,c from relations (i)-(iii).

Partially differentiating (ii),

cyx

z2

Using this in (ii) and (iii)



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yx

zy

x

za

2

yx

zx

y

zb

2

Substituting for a, b, c in (i), we get

yx

zxy

yx

zx

y

zy

yx

zy

x

zxz

222

yx

zxy

y

zy

x

zxz

2

(5) 1c

z

b

y

a

x2

2

2

2

2

2

Differentiating partially w.r.t. x,

x

z

c

z

a

xor,0

x

z

c

z2

a

x22222

Differentiating this partially w.r.t. x, we get

2

22

22 x

zz

x

z

c

1

a

1 or 2

22

2

2

x

zz

x

z

a

c

: Differentiating the given equation partially w.r.t. y twice we get

2

22

y

zz

y

z

y

z

y

z2

22

x

zz

x

z

x

z

x

z

Is the required p. d. e..

Note:

As another required partial differential equation.

P.D.E. obtained by elimination of arbitrary constants need not be not unique



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Formation of p d e by eliminating the arbitrary functions:

1) z = f(x2 + y

2)

Differentiating z partially w.r.t. x and y,

yyxfy

zqxyxf

x

zp 2).(',2).(' 2222

p /q = x / y or y p –x q=0 is the required pde

(2) z = f ( x +ct ) + g (x -ct)

Differentiating z partially with respect to x and t,

)(")("),(')('2

2

ctxgctxfx

zctxgctxf

x

z

Thus the pde is

(3) x + y + z = f(x2 + y

2 + z

2)

Differentiating partially w.r.t. x and y

x

zzxzyxf

x

z22)('1 222

)/(

)/(1

)/(

)/(1)('2 222

yzzy

yz

xzzx

xzzyxf

yxy

zxz

x

zzy )()( is the required pde

02

2

2

2

x

z

t

z

y

zzyzyxf

y

z22)('1 222



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(4) z = f ( xy / z ).

Differentiating partially w.r.t. x and y

x

z

z

xy

z

y

z

xyf

x

z2

'

x

z

z

xy

z

x

z

xyf

y

z2

'

}/)(/(1){/(

/

}/)(/(1){/(

/'

yzzyzx

yz

xzzxzy

xz

z

xyf

y

zy

x

zx

or xp = yq is the required pde.

(5) z = y2 + 2 f(1/x + logy)

yyxfy

y

z 1)log/1('22

2

1)log/1('2

xyxf

x

z

yy

zy

x

zxyxf 2)log/1('2 2

Hence

22 2yy

zy

x

zx

(6) Z = xΦ(y) + y (x)

)()(');(')( xyxy

zxyy

x

z



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Substituting )(')(' xandy

)]()([2

xyyxy

zy

x

zx

yx

zxy

is the required pde.

7) Form the partial differential equation by eliminating the arbitrary functions from

z = f(y-2x)+g(2y-x) (Dec 2011) Solution: By data, z = f(y-2x)+g(2y-x)

2

2

2

2

2

2 ( 2 ) (2 )

( 2 ) 2 (2 )

4 ( 2 ) (2 )...............(1)

2 ( 2 ) 2 (2 ).........(2)

( 2 ) 4 (2 )................(3)

(1) 2 (2)

zp f y x g y x

x

zq f y x g y x

y

zr f y x g y x

x

zs f y x g y x

x y

zt f y x g y x

y

2 2 2

2 2

2 6 ( 2 )..............(4)

(2) 2 (3) 2 3 ( 2 )............(5)

(4) (5)

22 2 5 2 0

2

2 5 2 0

r s f y x

s t f y x

Nowdividing by we get

r sor r s t

s t

z z zThus is the required PDE

x x y y

zy

zy

x

zx

yx

zxy

2



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LAGRANGE’S FIRST ORDER FIRST DEGREE PDE: Pp+Qq=R

(1) Solve : yzp + zxq = xy.

xy

dz

zx

dy

yz

dx

Subsidiary equations are

From the first two and the last two terms, we get, respectively

0 ydyxdxorx

dy

y

dx and .0 zdzydyor

y

dz

z

dy

Integrating we get x2 - y

2 = a, y

2 – z

2 = b.

Hence, a general solution is

Φ(x2-y

2, y

2 –z

2) = 0

(2) Solve : y2p - xyq = x(z-2y)

)2(2 yzx

dz

xy

dy

y

dx

From the first two ratios we get

x2 + y

2 = a from the last ratios two we get

2y

z

dy

dz

from the last ratios two we get

2y

z

dy

dzordinary linear differential equation hence

yz – y2= b

solution is Φ( x2 + y

2, yz – y

2) = 0



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(3) Solve : z(xp – yq) = y2 –x

2

22 xy

dz

zy

dy

zx

dx

0,d(xy)or 0or , ydxxdyy

dy

x

dx

on integration, yields xy = a

xdx +ydy + zdz = 0 x2 + y

2 + z

2 = b

Hence, a general solution of the given equation

Φ(xy,x2+y2+z2)=0

(4) Solve: xy

yxq

zx

xzp

yz

zy

dzyx

xydy

xz

zxdx

zy

yz

x dx + y dy + z dz = 0 …(i)

Integrating (i) we get

x2 + y

2 + z

2 = a

yz dx + zx dy + xy dz = 0 …(ii)

Dividing (ii) throughout by xyz and then integrating,

we get xyz = b

Φ( x2 + y

2 + z

2, xyz ) = 0

(5) (x+2z)p + (4zx – y)q = 2x2 + y

)..(242 2

iyx

dz

yzx

dy

zx

dx

Using multipliers 2x, -1, -1 we obtain 2x dx – dy – dz = 0



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Using multipliers y, x, -2z in (i), we obtain

y dx + x dy – 2z dz = 0 which on integration yields

xy – z2 = b ….(iii)

5) Solve sin sin 2sin 0 0xy yz x y for which z y when x and z

when y is an odd multiple of2

.

Solution: Here we first find z by integration and apply the given conditions to determine the

arbitrary functions occurring as constants of integration.

The given PDF can be written as yxy

z

xsinsin

Integrating w.r.t x treating y as constant,

sin sin ( ) sin cos ( )z

y x dx f y y x f yy

Integrating w.r.t y treating x as constant

cos sin ( ) ( )

cos ( cos ) ( ) ( ),

( ) ( ) .

cos cos ( ) ( )

, 2sin 0. sin (1)

2sin ( sin ).1 ( ) (cos0 1)

( ) ( ) sin cos

z x y dy f y dy g x

z x y F y g x

where F y f y dy

Thus z x y F y g x

zAlsoby data y when x U g this in

y

y y f y

Hence F y f y dy y dy y

Wi , (2) cos cos cos ( )

sin 0 (2 1) (3)2

0 cos cos(2 1) cos c(2 1) ( )2 2

cos (2 1) 0. 0 0 0 ( )2

Thus the solution of the PDE is given by

z=cos

th this becomes z x y y g x

U g theconditionthat z if y n in we have

x n x n g x

But n and hence g x

x cosy + cosy



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Solution of PDE by the method of Separation of Variables

1) Solve by the method of variables 3 2 0, ( ,0) 4 x

x yu u giventhat u x e

Solution: 3 2 0u u

Givenx x

Assume solution of (1) as

U=XY where X=X(x); ( )

3 ( ) 2 ( ) 0

23 2 0

3 3

Y Y y

u uxy xy

x x

dX dY dX dYY

dx dy X dx Y dy

dX dXLet K kdx

X dx X

1

1 1

3

3log log3

kxc

KxX kx c X c

X e

2

1 2

22

3 2

1

2

6 3

33 2

2

2

log2

(2)&(3) (1)

( ) 4

. ., 4 4

4 & 3

4 is required solution.

kyc

x yK c c

x

x kxkx

x y

dY dY KdyLet k

Y dy Y

KdyY c Y e

Substituting in

U e

Also u x o e

i e e Ae e Ae

Comparing we get A K

U e

2) Solve by the method of variables 54 3 , (0, ) 2 yu u

u giventhat u y edx y

(Dec 2011,

June2012)

Solution: 4 3u u

Given ux y

Assume solution of (1) as

( ); ( )u XY where X X x Y Y y



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4 ( ) ( ) 3

4 14 3 3

XY XY XYx y

dX dY dX dYY X XY

dx dy X dx Y dy

4 1, 3

dX dYLet k k

X dx Y dy

12

1 2 1 2

1 2

34

3 34 4

5

35

var int

log , log 34

0 2

2 2 2

kxc k y c

kx kxk y k yc c c c

y

k yy

Separating iables and egrating we get

kxX c Y k y c

X e and Y e

Hence u XY e e Ae where A e

put x and u e

The general solutionbecomes

e Ae A and k

Pa

522x

y

rticular solutionis

u e



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UNIT V

INTEGRAL CACULUS

MULTIPLE INTERGRALS

DOUBLE INTEGRAL



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PROBLEMS:



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Evaluation of a Double Integral by Changing the Order of Integration

Evaluation of a Double Integral by Change of Variables

Applications to Area and Volume



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BETA AND GAMMA FUNCTIONS

Definitions

Properties of Beta and Gamma Functions



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Relationship between Beta and Gamma Functions



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1.



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2.



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UNIT VI

VECTOR CALCULUS

Vector Line Integral

PROBLEMS:



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INTEGRAL THEOREM

Green’s Theorem in a Plane

Surface integral and Volume integral



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Stoke’s Theorem

Gauss Divergence Theorem



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= 2/3



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UNIT VII

LAPLACE TRANSFORMS - I

INTRODUCTION

Laplace transform is an integral transform employed in solving physical

problems.

Many physical problems when analysed assumes the form of a differential

equation subjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable

are specified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different

values of the independent variable, the conditions are called boundary

conditions.

The problem with initial conditions is referred to as the Initial value

problem.

The problem with boundary conditions is referred to as the Boundary value

problem.

Example 1: The problem of solving the equation xydx

dy

dx

yd2

2

with conditions

y(0) = y (0) = 1 is an initial value problem.

Example 2: The problem of solving the equation xydx

dy

dx

ydcos23

2

2

with

y(1)=1, y(2)=3 is called Boundary value problem.

Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827).

The subject is divided into the following sub topics.



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Definition:

Let f(t) be a real-valued function defined for all t 0 and s be a parameter,

real or complex. Suppose the integral 0

)( dttfe stexists (converges). Then this

integral is called the Laplace transform of f(t) and is denoted by L[f(t)].

Thus, L[f(t)] = 0

)( dttfe st

(1)

We note that the value of the integral on the right hand side of (1) depends

on s. Hence L[f(t)] is a function of s denoted by F(s) or )(sf .

Thus, L[f(t)] = F(s) (2)

Consider relation (2). Here f(t) is called the Inverse Laplace transform of

F(s) and is denoted by L-1

[F(s)].

Thus, L-1

[F(s)] = f(t) (3)

Suppose f(t) is defined as follows :

f1(t), 0 < t < a

f(t) = f2(t), a < t < b

f3(t), t > b

Note that f(t) is piecewise continuous. The Laplace transform of f(t) is defined as

LAPLACE TRANSFORMS

Definition and

Properties

Transforms of

some functions

Convolution

theorem

Inverse

transforms

Solution of

differential

equations



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L[f(t)] = 0

)(tfe st

=

a b

a b

ststst dttfedttfedttfe0

321 )()()(

NOTE: In a practical situation, the variable t represents the time and s represents

frequency.

Hence the Laplace transform converts the time domain into the frequency

domain.

Basic properties

The following are some basic properties of Laplace transforms:

1. Linearity property: For any two functions f(t) and (t) (whose Laplace

transforms exist)

and any two constants a and b, we have

L [a f(t) + b (t)] = a L[f(t)] + b L[ (t)]

Proof :- By definition, we have

L [af (t) + b (t)] = dttbtafe st

0

)()( = 0 0

)()( dttebdttfea stst

= a L[f(t)] + b L[ (t)]

This is the desired property.

In particular, for a=b=1, we have

L [ f(t) + (t)] = L [f(t)] + L[ (t)]

and for a = -b = 1, we have L [ f(t) - (t)] = L [f(t) ]- L[ (t)]



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2. Change of scale property: If L L[f(t)] = F(s), then L[f(at)] =a

sF

a

1, where a is

a positive constant.

Proof: - By definition, we have

L[f(at)] = 0

)( dtatfe st(1)

Let us set at = x. Then expression (1) becomes,

L f(at) = 0

)(1

dxxfea

xa

s

a

sF

a

1

This is the desired property.

3. Shifting property: - Let a be any real constant. Then

L [eatf (t)] = F(s-a)

Proof :- By definition, we have

L [eatf (t)] =

0

)( dttfee atst

= 0

)( )( dttfe as

= F(s-a)

This is the desired property. Here we note that the Laplace transform of eat

f(t)

can be written down directly by changing s to s-a in the Laplace transform of f(t).



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LAPLACE TRANSFORMS OF STANDARD FUNCTIONS

1. Let a be a constant. Then

L[(eat)] =

0 0

)( dtedtee tasatst

= asas

e tas 1

)(0

)(

, s > a

Thus,

L[(eat)] =

as

1

In particular, when a=0, we get

L(1) = s

1, s > 0

By inversion formula, we have

atat es

Leas

L11 11

2. L(cosh at) =2

atat eeL =

02

1dteee atatst

= 0

)()(

2

1dtee tastas

Let s > |a|. Then,

0

)()(

)()(2

1)(cosh

as

e

as

eatL

tastas

= 22 as

s



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Thus, L (cosh at) = 22 as

s, s > |a|

and so

atas

sL cosh

22

1

3. L (sinh at) = 222 as

aeeL

atat

, s > |a|

Thus,

L (sinh at) = 22 as

a, s > |a|

and so,

a

at

asL

sinh122

1

4. L (sin at) =0

sin ate stdt

Here we suppose that s > 0 and then integrate by using the formula

bxbbxaba

ebxdxe

axax cossinsin

22

Thus,

L (sinh at) = 22 as

a, s > 0

and so

a

at

asL

sinh122

1



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5. L (cos at) = atdte st cos0

Here we suppose that s>0 and integrate by using the formula

bxbbxaba

ebxdxe

axax sincoscos

22

Thus, L (cos at) = 22 as

s, s > 0

and so atas

sL cos

22

1

6. Let n be a constant, which is a non-negative real number or a negative non-

integer. Then

L(tn) =

0

dtte nst

Let s > 0 and set st = x, then

0 0

1

1)( dxxe

ss

dx

s

xetL nx

n

n

xn

The integral dxxe nx

0

is called gamma function of (n+1) denoted by )1(n .

Thus 1

)1()(

n

n

s

ntL

In particular, if n is a non-negative integer then )1(n =n!. Hence



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1

!)(

n

n

s

ntL

and so

)1(

11

1

n

t

sL

n

n or !n

t n

as the case may be

Application of shifting property:-

The shifting property is

If L f(t) = F(s), then L [eatf(t)] = F(s-a)

Application of this property leads to the following results :

1. ass

ass

at

bs

sbtLbteL

22)(cosh)cosh( = 22)( bas

as

Thus,

L(eatcoshbt) = 22)( bas

as

and

btebas

asL at cosh

)( 22

1

2. 22)()sinh(

bas

abteL at

and

btebas

L at sinh)(

122

1



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3. 22)()cos(

bas

asbteL at

and

btebas

asL at cos

)( 22

1

4. 22)()sin(

bas

bbteL at

and

b

bte

basL

at sin

)(

122

1

5. 1)(

)1()(

n

nat

as

nteL or 1)(

!nas

n as the case may be

Hence

)1()(

11

1

n

te

asL

nat

n or 1)(

!nas

n as the case may be

Examples :-

1. Find L[f(t)] given f(t) = t, 0 < t < 3

4, t > 3

Here

L[f(t)]= 0

3

0 3

4)( dtetdtedttfe ststst

Integrating the terms on the RHS, we get



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L[f(t)] = )1(11 3

2

3 ss es

es

This is the desired result.

2. Find L[f(t)] given L[f(t)] = sin2t, 0 < t

0, t >

Here

L[f(t)] = dttfedttfe stst )()(0

= 0

2sin tdte st

= 0

22cos22sin

4tts

s

e st

= se

s1

4

22


3. Evaluate: (i) L(sin3t sin4t)

(ii) L(cos2 4t)

(iii) L(sin32t)

(i) Here L(sin3t sin4t) = L [ )]7cos(cos2

1tt

= )7(cos)(cos2

1tLtL , by using linearity property

= )49)(1(

24

4912

12222 ss

s

s

s

s

s

(ii) Here

L(cos24t) =

64

1

2

1)8cos1(

2

12s

s

stL



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(iii) We have

3sinsin34

1sin 3

For =2t, we get

ttt 6sin2sin34

12sin 3

so that

)36)(4(

48

36

6

4

6

4

1)2(sin

2222

3

sssstL


4. Find L(cost cos2t cos3t)

Here cos2t cos3t = ]cos5[cos2

1tt

so that

cost cos2t cos3t = ]coscos5[cos2

1 2 ttt

= ]2cos14cos6[cos4

1ttt

Thus L(cost cos2t cos3t) = 4

1

16364

1222 s

s

ss

s

s

s

5. Find L(cosh22t)

We have

2

2cosh1cosh2

For = 2t, we get



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2

4cosh12cosh2 tt

Thus,

16

1

2

1)2(cosh

2

2

s

s

stL

6. Evaluate (i) L( t ) (ii) t

L1

(iii) L(t-3/2

)

We have L(tn) =

1

)1(ns

n

(i) For n= 2

1, we get

L(t1/2

) = 2/3

)12

1(

s

Since )()1( nnn , we have 22

1

2

11

2

1

Thus, 2

3

2)(

stL

(ii) For n = -2

1, we get

sstL

21

21 2

1

)(

(iii) For n = -2

3, we get

sss

tL 222

1

)(2

12

12

3



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7. Evaluate: (i) L(t2) (ii) L(t

3)

We have,

L (tn) =

1

!ns

n

(i) For n = 2, we get

L (t2) =

33

2!2

ss

(ii) For n=3, we get

L (t3) =

44

6!3

ss

8. Find L [e-3t

(2cos5t – 3sin5t)]

Given =

2L (e-3t

cos5t) – 3L(e-3t

sin5t)

=25)3(

15

25)3(

32

22 ss

s, by using shifting property

= 346

922 ss

s, on simplification

9. Find L [coshat sinhat]

Here L [coshat sinat] = atee

Latat

sin2

=2222 )()(2

1

aas

a

aas

a

, on simplification ])][()[(

)2(2222

22

aasaas

asa



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10. Find L (cosht sin3 2t)

Given

4

6sin2sin3

2

tteeL

tt

= )6sin()2sin(3)6sin(2sin38

1teLteLteLteL tttt

= 36)1(

6

4)1(

6

36)1(

6

4)1(

6

8

12222 ssss

= 36)1(

1

24)1(

1

36)1(

1

4)1(

1

4

32222 ssss

11. Find )( 25

4 teL t

We have

L(tn) =

1

)1(ns

n Put n= -5/2. Hence

L(t-5/2

) = 2/32/3 3

4)2/3(

ss Change s to s+4.

Therefore, 2/3

2/54

)4(3

4)(

steL t

Transform of tn f(t)

Here we suppose that n is a positive integer. By definition, we have

F(s) = 0

)( dttfe st



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Differentiating ‘n’ times on both sides w.r.t. s, we get

0

)()( dttfes

sFds

d st

n

n

n

n

Performing differentiation under the integral sign, we get

0

)()()( dttfetsFds

d stn

n

n

Multiplying on both sides by (-1)n , we get

0

)]([)(()()1( tftLdtetftsFds

d nstn

n

nn

, by definition

Thus,

L [tnf(t)]= )()1( sF

ds

dn

nn

This is the transform of tn f (t).

Also,

)()1()(1 tftsFds

dL nn

n

n

In particular, we have

L[t f(t)] = )(sFds

d, for n=1

L [t2 f(t)]= )(

2

2

sFds

d, for n=2, etc.

Also, )()(1 ttfsFds

dL and

)()( 2

2

21 tftsF

ds

dL



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Transform of t

f(t)

We have, F(s) = 0

)( dttfe st

Therefore,

s sdsdttfstedssF

0)()(

= dtdsetfs

st

0

)(

= 0

)( dtt

etf

s

st

= 0

)()(

t

tfLdt

t

tfe st

Thus, s

dssFt

tfL )(

)(

This is the transform of t

tf )(

Also, s

t

tfdssFL

)()(1

Examples :

1. Find L [te-t sin4t]

We have, 16)1(

4]4sin[

2steL t

So that,



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L [te-t sin4t] =

172

14

2 ssds

d

= 22 )172(

)1(8

ss

s

2. Find L (t2 sin3t)

We have L (sin3t) = 9

32s

So that,

L (t2 sin3t) =

9

322

2

sds

d

= 22 )9(

6s

s

ds

d

=32

2

)9(

)3(18

s

s

3. Findt

teL

t sin

We have

1)1(

1)sin(

2steL t

Hence t

teL

t sin=

0

1

2)1(tan

1)1(ss

s

ds

= )1(tan2

1 s = cot –1

(s+1)

4. Findt

tL

sin. Using this, evaluate L

t

atsin

We have L (sint) = 1

12s

So that L [f (t)] = t

tL

sin= S

s

ss

ds 1

2tan

1



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= )(cottan2

11 sFss

Consider

Lt

atsin= a L )(

sinataLf

at

at

=a

sF

aa

1, in view of the change of scale property

= a

s1cot

5. Find Lt

btat coscos

We have L [cosat – cosbt] = 2222 bs

s

as

s

So that Lt

btat coscos= ds

bs

s

as

s

s

2222

= s

bs

as22

22

log2

1

= 22

22

22

22

loglog2

1

bs

as

bs

asLt

s

= 22

22

log02

1

as

bs

=22

22

log2

1

as

bs



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6. Prove that0

3

50

3sin tdtte t

We have

0

)sin(sin ttLtdtte st = )(sin tL

ds

d =

1

12sds

d

= 22 )1(

2

s

s

Putting s = 3 in this result, we get

0

3

50

3sin tdtte t

This is the result as required.

Consider

L )(tf =0

)( dttfe st

= 0

0 )()()( dttfestfe stst, by using integration by parts

= )()0()(( tsLfftfeLt st

t

= 0 - f (0) + s L[f(t)]

Thus

L )(tf = s L[f(t)] – f(0)

Similarly,

L )(tf = s2 L[f(t)] – s f(0) - )0(f



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In general, we have

)0(.......)0()0()()( 121 nnnnn ffsfstLfstLf

Transform of

t

0

f(t)dt

Let (t) =

t

dttf0

)( . Then (0) = 0 and (t) = f(t)

Now, L (t) = 0

)( dtte st

= 00

)()( dts

et

s

et

stst

= 0

)(1

)00( dtetfs

st

Thus, t

tfLs

dttfL0

)]([1

)(

Also,

t

dttftfLs

L0

1 )()]([1

Examples:

1. By using the Laplace transform of sinat, find the Laplace transforms of cosat.

Let f(t) = sin at, then Lf(t) = 22 as

a

We note that

atatf cos)(

Taking Laplace transforms, we get

)(cos)cos()( ataLataLtfL



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or L(cosat) = )0()(1

)(1

ftsLfa

tfLa

= 01

22 as

sa

a

Thus

L(cosat) = 22 as

s


2. Given2/3

12

s

tL , show that

stL

11

Let f(t) = t

2 , given L[f(t)] = 2/3

1

s

We note that, tt

tf1

2

12)(

Taking Laplace transforms, we get

tLtfL

1)(

Hence

)0()()(1

ftsLftfLt

L

= 01

2/3ss

Thus st

L11

This is the result as required.



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3. Findt

dtt

btatL

0

coscos

Here

L[f(t)] = 22

22

log2

1coscos

as

bs

t

btatL

Using the result

L

t

tLfs

dttf0

)(1

)(

We get,

t

dtt

btatL

0

coscos=

22

22

log2

1

as

bs

s

4. Find

t

t tdtteL0

4sin

Here

22 )172(

)1(84sin

ss

stteL t

Thus

t

t tdtteL0

4sin = 22 )172(

)1(8

sss

s



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Laplace Transform of a periodic function

A function f(t) is said to be a periodic function of period T > 0 if f(t) = f(t +

nT) where n=1,2,3,….. The graph of the periodic function repeats itself in equal

intervals.

For example, sint, cost are periodic functions of period 2 since

sin (t + 2n ) = sin t, cos(t + 2n ) = cos t.

The graph of f(t) = sin t is shown below :

Note that the graph of the function between 0 and 2 is the same as that between

2 and 4 and so on.

The graph of f(t) = cos t is shown below :



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Note that the graph of the function between 0 and 2 is the same as that between

2 and 4 and so on.

Formula: Let f (t) be a periodic function of period T. Then

T

st

STdttfe

etLf

0

)(1

1)(

Proof :By definition, we have

L f (t) = 0 0

)()( duufedttfe sust

= ....)(.......)()(

)1(2

0

Tn

nT

su

T

T

su

T

su duufeduufeduufe

= 0

)1(

)(n

Tn

nT

su duufe

Let us set u = t + nT, then

L f(t) = 0 0

)( )(n

T

t

nTts dtnTtfe

Here

f(t+nT) = f(t), by periodic property

Hence

T

st

n

nsT dttfeetLf00

)()()(

=

T

st

STdttfe

e0

)(1

1, identifying the above series as a geometric

series.

Thus



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L[ f(t)] =

T

st

sTdttfe

e0

)(1

1


Examples:-

1. For the periodic function f(t) of period 4, defined by f(t) = 3t, 0 < t < 2

6, 2 < t < 4

find L [f(t)]

Here, period of f(t) = T = 4

We have,

L f(t) =

T

st

sTdttfe

e0

)(1

1

=

4

0

4)(

1

1dttfe

e

st

s

=

4

2

2

0

463

1

1dtedtte

e

stst

s

=

4

2

2

0

2

0

46.13

1

1

s

edt

s

e

s

et

e

ststst

s

= 2

42

4

213

1

1

s

see

e

ss

s

Thus,

L[f(t)] = )1(

)21(342

42

s

ss

es

see



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3. A periodic function of period2

is defined by

f (t)= Esin t, 0 t <

0, t 2

where E and are positive constants. Show that

L f(t) = )1)(( /22 wsews

E

Here T = 2

. Therefore

L f(t) =

/2

0

)/2()(

1

1dttfe

e

st

s

=

/

0

)/2(sin

1

1tdtEe

e

st

s

=

/

0

22)/2(cossin

1tts

s

e

e

E st

s

= 22

/

)/2(

)1(

1 s

e

e

E s

s

= ))(1)(1(

)1(22//

/

see

eEss

s

= ))(1( 22/ se

Es




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3. A periodic function f(t) of period 2a, a>0 is defined by

f (t)= E , 0 t a

-E, a < t 2a

show that

L [f (t)] = 2

tanhas

s

E

Here T = 2a. Therefore

L [f (t)] =

a

st

asdttfe

e

2

0

2)(

1

1

=

a a

a

stst

asdtEedtEe

e0

2

21

1

= )(1)1(

2

2

asassa

aseee

es

E

2

21

)1(

as

ase

es

E

)1)(1(

)1( 2

asas

as

ees

eE

= 2/2/

2/2/

asas

asas

ee

ee

s

E

2tanh

as

s

E

This is the result as desired.



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Step Function:

In many Engineering applications, we deal with an important discontinuous

function H (t-a) defined as follows:

0, t a

H (t-a) = 1, t > a

where a is a non-negative constant.

This function is known as the unit step function or the Heaviside function. The

function is named after the British electrical engineer Oliver Heaviside.The

function is also denoted by u (t-a). The graph of the function is shown below:

H (t-1)

Note that the value of the function suddenly jumps from value zero to the

value 1 as at from the left and retains the value 1 for all t>a. Hence the

function H (t-a) is called the unit step function.

In particular, when a=0, the function H(t-a) become H(t), where

0 , t 0

H(t) = 1 , t > 0

Transform of step function

By definition, we have L [H(t-a)] = 0

)( dtatHe st

=

a

a

stst dtedte0

)1(0

= s

e as



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In particular, we have L H(t) = s

1

Also, )(1 atHs

eL

as

and )(11 tHs

L

Unit step function (Heaviside function)

Statement:- L [f (t-a) H (t-a)] = e-as

Lf(t)

Proof: - We have

L [f(t-a) H(t-a)] = 0

)()( dteatHatf st

= a

st dtatfe )(

Setting t-a = u, we get

L[f(t-a) H(t-a)] = duufe uas )(0

)(

= e-as

L [f(t)]

This is the desired shift theorem.

Also, L-1

[e-as

L f(t)] = f(t-a) H(t-a)

Examples:

1. Find L [et-2

+ sin(t-2)] H(t-2)

Let

f (t-2) = [et-2

+ sin (t-2)]

Then f (t) = [et + sint]

so that L f(t) = 1

1

1

12ss



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By Heaviside shift theorem, we have

L[f(t-2) H(t-2)] = e-2s

Lf(t)

Thus,

1

1

1

1)2()]2sin([

2

2)2(

ssetHteL st

2. Find L (3t2 +2t +3) H(t-1)

Let

f(t-1) = 3t2 +2t +3

so that

f (t) = 3(t+1)2 +2(t+1) +3 = 3t

2 +8t +8

Hence

ssstfL

886)]([

23

Thus

L [3t2 +2t +3] H(t-1) = L[f(t-1) H(t-1)]

= e-s L [f(t)]

= sss

e s 88623

3. Find Le-t H (t-2)

Let f (t-2) = e-t , so that, f(t) = e

-(t+2)



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Thus,

L [f(t)] = 1

2

s

e

By shift theorem, we have

1)()]2()2([

)1(22

s

etLfetHtfL

ss

Thus

1)2(

)1(2

s

etHeL

st

f1 (t), t a

4. Let f (t) = f2 (t) , t > a

Verify that f(t) = f1(t) + [f2(t) – f1(t)]H(t-a)

Consider

f1(t) + [f2(t) – f1(t)]H(t-a) = f1(t) + f2 (t) – f1(t), t > a

0 , t a

= f2 (t), t > a

f1(t), t a = f(t), given

Thus the required result is verified.

5. Express the following functions in terms of unit step function and hence

find their Laplace transforms.

1. f(t) = t2 , 1 < t 2

4t , t > 2



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Here, f(t) = t2 + (4t-t

2) H(t-2)

Hence, L f(t) = )2()4(2 2

3tHttL

s(i)

Let (t-2) = 4t – t2

so that (t) = 4(t+2) – (t+2)2 = -t

2 + 4

Now,ss

tL42

)]([3

Expression (i) reads as

L f(t) = )2()2(23

tHtLs

= )(2 2

3tLe

s

s

= 3

2

3

242

sse

s

s


2. cost, 0 < t <

f (t) = sint, t >

>> Here f(t) = cost + (sint-cost)H(t- )

Hence,

L[ f(t)] = )()cos(sin12

tHttLs

s(ii)

Let

(t- ) = sint – cost

Then



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(t) = sin(t + ) – cos(t + ) = -sint + cost

so that

L[ (t)] = 11

122 s

s

s

Expression (ii) reads as

L [f(t)] = )()(12

tHtLs

s

= )(12

tLes

s s

= 1

1

1 22 s

se

s

s s

UNIT IMPULSE FUNCTION



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Sol:



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3.

4.



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UNIT VIIILAPLACE TRANSFORMS – 2

Introduction:

Let L [f (t)]= F(s). Then f(t) is defined as the inverse Laplace transform of

F(s) and is denoted by L-1

F(s). Thus L-1

[F(s)] = f (t).

Linearity Property

Let L-1

[F(s)] = f(t) and L-1

[G(s) = g(t)] and a and b be any two constants. Then

L-1

[a F(s) + b G(s)] = a L-1

[F(s)] + b L-1

[G(s)] Table of Inverse Laplace Transforms

F(s) )()( 1 sFLtf

0,1

ss

1

asas

,1 ate

0,22

sas

s Cos at

0,1

22s

as a

atSin

asas

,1

22 a

ath Sin

asas

s,

22

ath Cos

0,1

1s

sn

n = 0, 1, 2, 3, . . .

!n

t n

0,1

1s

sn

n > -1

1n

tn



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Examples

1. Find the inverse Laplace transforms of the following:

22)(

as

bsii

22 9

94

254

52)(

s

s

s

siii

Here

2

5

11

2

1

25

1

2

1

52

1)(

t

es

Ls

Li

ata

bat

asLb

as

sL

as

bsLii sincos

1)(

22

1

22

1

22

1

9

29

4

4

25

25

4

2

9

84

254

52)(

2

1

2

1

22

1

s

sL

s

sL

s

s

s

sLiii

ththtt

3sin2

33cos4

2

5sin

2

5cos

2

1

Evaluation of L-1

F(s – a)

We have, if L [f(t)] = F(s), then L[eat f(t)] = F(s – a), and so

L-1

[F(s – a) ]= eat f (t) = e

at L

-1 [F(s)]

Examples

4

1

1

13: Evaluate 1.

s

sL

4

1

3

1

4

1-

1

12

1

13

1

11-1s3L Given

sL

sL

s

52

1)(

si



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4

1

3

1 12

13

sLe

sLe tt

Using the formula

getweandntakingandn

t

sL

n

n,32

!

11

1

32

3 32 teteGiven

tt

52s-s

2sL : Evaluate 2.

2

1-

41

13

41

1

41

31

41

2L Given

2

1

2

1

2

1

2

1-

sL

s

sL

s

sL

s

s

4

13

4 2

1

2

1

sLe

s

s Le t-t

tet e tt 2sin2

32cos

13

12: Evaluate

2

1

ss

sL

45

1

45

2

45

12L Given

2

2

3

1

2

2

3

2

3

1

2

2

3

2

3

1-

sL

s

sL

s

s

45

1

45

22

12

3

2

12

3

sLe

s

sLe

tt

ththe

t

2

5sin

5

2

2

5cos2 2

3



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2

452L : Evaluate 4.

23

21-

sss

ss

havewe

12

12

452

2

452

2

452

2

2

2

23

2

s

C

s

B

s

A

sss

ss

sss

ss

sss

ss

Then 2s2+5s-4 = A(s+2) (s-1) + Bs (s-1) + Cs (s+2)

For s = 0, we get A = 2, for s = 1, we get C = 1 and for s = -2, we get B = -1. Using these values

in (1), we get

1

1

2

12

2

45223

2

ssssss

ss

Hence

tt eess

ssL 2

22

21 2

25

452

21

54: Evaluate 5.

2

1

ss

sL

Let us take

21121

5422 s

C

s

B

s

A

ss

s

Then 4s + 5 =A(s + 2) + B(s + 1) (s + 2) + C (s + 1)2

For s = -1, we get A = 1, for s = -2, we get C = -3



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Comparing the coefficients of s2, we get B + C = 0, so that B = 3. Using these values in (1),

we get2

3

1

3

1

1

21

5422 sssss

s

Hences

Les

Les

Less

sL ttt 1

31

31

21

54 121

2

1

2

1

ttt eete 233

44

31: Evaluate 5.

as

sL

Let)1(

2244

3

as

DCs

as

B

as

A

as

s

Hence s3 = A(s + a) (s

2 + a

2) + B (s-a)(s

2+a

2)+(Cs + D) (s

2 – a

2)

For s = a, we get A = ¼; for s = -a, we get B = ¼; comparing the constant terms, we get

D = a(A-B) = 0; comparing the coefficients of s3, we get

1 = A + B + C and so C = ½. Using these values in (1), we get

2244

3

2

111

4

1

as

s

asasas

s

Taking inverse transforms, we get

ateeas

sL atat cos

2

1

4

144

31

athat coscos2

1

1: Evaluate 6.

24

1

ss

sL

Consider 11

2

2

1

111 222224 ssss

s

ssss

s

ss

s



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4

3

1

4

3

1

2

1

1

4

3

1

4

3

1

2

1

1

1

1

1

2

1

11

11

2

1

2

12

1

2

12

1

24

1

2

2

12

2

1

22

22

22

s

Le

s

Less

sL

Therefore

ss

ssss

ssss

ssss

tt

2

3

2

3sin

2

3

2

3sin

2

12

1

2

1 te

te

tt

2sin

2

3sin

3

2 tht

Evaluation of L-1

[e-as

F (s)]

We have, if L [f (t)] = F(s), then L[f(t-a) H(t-a) = e-as

F(s), and so

L-1

[e-as

F(s)] = f(t-a) H(t-a)



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Examples

4

1

2:)1(

s

eLEvaluate

ss

Here

56

5

)()(2

6

1

2

1)()(

2

1)(,5

352

4

51

32

4

12

4

11

4

tHte

atHatfs

eL

Thus

te

sLe

sLsFLtfTherefore

ssFa

t

s

tt

41:)2(

2

2

2

1

s

se

s

eLEvaluate

ss

22coscos

222cossin

)1(

2cos4

)(

sin1

1)(

)1(22

2

12

2

11

21

tHttHt

tHttHtGiven

asreadsrelationNow

ts

sLtf

ts

LtfHere

tHtftHtfGiven



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Inverse transform of logarithmic functions

sFds

dttfL then F(s), f(t) L if have, We

Hence

)(1 ttfsFds

dL

Examples:

bs

asLEvaluate log:)1( 1

b

eetfThus

eetftor

eesFds

dLthatSo

bsassF

ds

dThen

bsasbs

assFLet

atbt

atbt

btat1

11

logloglog)(

(2) Evaluate 1L

s

a1tan

22

1tan)(

as

asF

ds

dThen

s

asFLet



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F s

Inverse transform ofs

22

1 1:)1(

assLEvaluate

Convolution Theorem:

2

0

1

22

1

1

22

cos1

sin1

sin)(

1

a

at

dta

at

s

sFL

assLThen

a

atsFLtf

thatsoas

sFdenoteusLet

t

a

attf

attftor

thatsoatsFds

dLor

sin

sin

sin1



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Using Convolution theorem find the inverse laplace transforms of the following



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t

atat

t

at-

at

t

at

at

duugutftgtfsGsFL

soandsGsFtLgtLf

eeata

dtateaass

L

atea

dtteass

LHence

teas

Lhavewe

assLEvaluate

0

1

3

0

222

1

2

0

2

1

2

1

22

1

)()()()(

)()()()(g(t) f(t)L

thenG(s), Lg(t) and F(s) L(t) if have, We

1211

1111

get we this,Using

parts.by n integratioon,111

1

1

1:)2(

theorem nconvolutio using by F(s) of transformInverse

transformLaplace inversefor n theoremconvolutio thecalled is expression This



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Examples

Employ convolution theorem to evaluate the following:

bsasL

1)1( 1

ba

ee

ba

ee

dueedueebsas

L

e, g(t) ef(t)

bssG

as

atbt

tbaat

t

ubaat

t

buuta-

-bt-at

1

1

n theorem,convolutioby Therefore,

get weinverse, theTaking

1)(,

1 F(s) denote usLet

00

1

222

1)2(as

sL

a

att

a

auatatu

duauatat

a

duauutaaas

sL

at , g(t) at

f(t)

as

ssG

asF(s)

t

t

t-

2

sin

2

2cossin

2a

1

formula angle compound usingby ,2

2sinsin1

cossin1

n theorem,convolutioby Hence

cosa

sin

Then)(,1

denote usLet

0

0

0

222

1

2222



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11)3(

2

1

ss

sL

Here

1)(,

1

1

2s

ssG

sF(s)

Therefore

ttettee

uue

eduuess

L

t , g(t)ef(t)

ttt

tu

tut-

t

cossin2

11cossin

2

cossin2

sin11

1

have wen theorem,convolutioBy

sin

0

2

1

LAPLACE TRANSFORM METHOD FOR DIFFERENTIAL EQUATIONS

As noted earlier, Laplace transform technique is employed to solve initial-value

problems. The solution of such a problem is obtained by using the Laplace Transform of the

derivatives of function and then the inverse Laplace Transform.

The following are the expressions for the derivatives derived earlier.

(0)f - (0)f s - f(0) s - f(t) L s (t)f[ L

(0)f - f(o) s - f(t) L s (t)fL[

f(0) - f(t) L s (t)]fL[

23

2



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1. Solve by using Laplace transform method

2 y(o) t y y ,te

Taking the Laplace transform of the given equation, we get

42

1

1

1

1

2

1

3114112

1

342

get we, transformLaplace inverse theTaking

1

342

1

121s

1

1

2

3

1

3

2

1

3

21

3

2

2

2

te

ssL

s

ssL

s

ssLtY

s

sstyL

thatso

styL

styLoytysL

t

This is the solution of the given equation.

2. Solve by using Laplace transform method:

0)()(,sin32 oyoytyyy

Taking the Laplace transform of the given equation, we get

1

1)(3)0()(2)0()0()(

2

2

styLytLysysytLys

Using the given conditions, we get



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equation.given theofsolution required theis This

sin2cos10

1

40

1

8

1

sums, partial of method theusingby

1

5

1

10

3

1

40

1

1

1

8

1

131

131

1)(

131

1)(

1

132)(

3

2

1

2

1

2

1

2

2

2

ttee

s

s

ssL

s

DCs

s

B

s

AL

sssLty

or

ssstyL

or

ssstyL

tt

equation. integral thesolve tomethod Transform LaplaceEmploy3)

t

0

sinlf(t) duutuf

get wehere, n theorem,convolutio usingBy

sin1

get weequation,given theof transformLaplace Taking

t

0

duutufLs

L f(t)



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1

)(1sin)(

12s

tfL

stLtLf

sL f(t)

equation. integralgiven theofsolution theis This

21

1)(

1)(

2

3

21

3

2 t

s

sLtfor

s

stfL

Thus

s. transformLaplace using t any timeat particle theofnt displaceme thefind 10, is speed initial theand 20at x is particle theofposition initial theIf t.at time particle theofnt displaceme thedenotesx

where025dt

dx6

dt

xdequation the,satisfyingpath a along moving is particleA 4)(

2

2

x

get weequation, theof transformLaplace theTaking

10. (o) x'20, x(o)are conditions initial theHere

025x(t)(t)6x'(t)'x'

asrewritten bemay equation Given

163

13020x(t)

256

13020Lx(t)

013020256ssLx(t)

2

1

2

2

s

sL

thatso

ss

s

ors

163

170

163

320

163

70320

2

1

2

1

2

1

sL

s

sL

s

sL



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problem.given theofsolution desired theis This

2

4sin354cos20

3

3 tete

t

t

L

Rt

at eeaL-R

E is t any timeat current

that theShow R. resistance and L inductance ofcircuit a to0at t applied is Ee A voltage (5) -at

The circuit is an LR circuit. The differential equation with respect to the circuit is

)(tERidt

diL

Here L denotes the inductance, i denotes current at any time t and E(t) denotes the E.M.F.

It is given that E(t) = E e-at

. With this, we have

Thus, we have

at

at

EetiRtLi

orEeRidt

diL

)()('

orER ateL(t)i'L(t)i'LL TTT

get wesides,both on )( transformLaplace Taking TL

asEtiLRitiLsL TT

1)()0()(

oras

ERsLtiLT )(get weo, i(o) Since



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))(()(get we,L transforminverse Taking

)(

1

RsLas

ELti

RsLas

EtiL

T

T

desired. asresult theis This

)(

11 11

L

Rt

at

TT

eeaLR

Eti

Thus

RsLLL

asL

aLR

E

1. y(o) 2, with x(o)09dt

dy

,04dt

dxgiven t of in termsy andfor x equations ussimultaneo theSolve )6(

x

y

Taking Laplace transforms of the given equations, we get

0)()()(9

0)(4 x(o)- Lx(t) s

oytLystxL

tLy

get we,conditions initialgiven theUsing

1)(5)(9

2)(4)(

tyLtxL

tyLtxLs

get weLy(t),for equations theseSolving

36

182s

sL y(t)

thatso

tt

ss

sLy(t)

6sin36cos

36

18

36 22

1

(1)



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get we,09dt

dyin thisUsing x

tt 6cos186sin69

1x(t)

or

tt 6sin6cos33

2x(t)

(2)

(1) and (2) together represents the solution of the given equations.



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Engineering Mathematics - 2 · PDF fileEquations of first order and higher degree (p-y-x equations), Equations solvable for p, y, x. General and singular solutions, Clarauit’s equation