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`500

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APPLIED MATHEMATICS

APPLIED MATHEMATICS

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APPLIED MATHEMATICS CONTENTS: UNIT – 1:

01-42

Characteristic equation – Eigen Values and Eigen Vectors – Properities – Problems – Rank of Matrix – Problems – Solutions of simultaneous equations using matrices – consistency condition. Polynomial equations – relation between roots and coefficients imaginary roots and irrational roots – solving equations under given conditions – transformation of equations. UNIT – II:

43-68

Definition of a derivative , different types of differentiation – standard formulae – successive differentiation – nth derivative – Leibnitz formula – problems. Partial differentiation – Euler’s theorem – Curvature – Radius of curvature in Cartesian co-ordinates. UNIT - III :

69-99 

 2

Integration by Parts :

 sin 0

2

n

xdx

 cos 0

a

2

n

xdx ,

 tan 0

n

xdx ,

x e

n ax

0

a

dx ,

x e

n x

dx

0

Definite integrals – properties – reduction formulae – Problems. Second order differential equations with constant coefficients – Particular Integrals of type eaxV - where V is x or x 2 or cos ax or sin ax

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UNIT – IV :

100-117

Definition Complete, - Singular and General integrals solutions of Standard types f (p, q) = 0, f ( x, p, q) = 0, f ( y, p, q) = 0, f ( z, p, q ) = 0, f (x, p )= f

2

1

( x, p) – Clariant’s form – Lagrange’s equation Pp + Qq = R –

Problems.

UNIT - V:

118-142

Definition – Laplace transform of standard function simple theorems – problems inverse Laplace transform – Fourier coefficients – periodic functions with period 2 p – half range series – cosine series – sine series – problems.

UNIT QUESTIONS:

143-147

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APPLIED MATHEMATICS

1

UNIT-I

1.0

Introduction

1.1

Objectives

1.2

Characteristic Equation

1.3

1.2.1

Eigen values and Eigen vectors

1.2.2

Characteristic roots

1.2.4

Cayley â&#x20AC;&#x201C; Hamiltonâ&#x20AC;&#x2122;s Theorem

1.2.5

Properties of Characteristics roots

1.2.6

Problems

Rank of a Matrix 1.3.1

Problems

1.4

Simultaneous equations using matrices

1.5

1.4.1 Elementary Transformation 1.4.2 Equivalent matrices Relation between the roots and coefficients of an Equation

1.6

Imaginary and irrational roots

1.7

Transformation of equation

1.0

Introduction The Source material of Mathematics is considered to be a source of study

organized according to a set plan and a learning guide rather than a source book of information.

The content is given as simple, brief, exact, accessible and definite. The

source of mathematics is satifying the demand of the examinations, especially new type tests in mathematics. This syllabus is more suitable for the application of mathematics in the field of technology.

It is completely deal with applied oriented mathematics with

equation, derivatives, radius of curvature in Cartesian co-ordinates.

Integration,

Singularities and Laplace transforms and Fourier series and all problems presented in a very simple language with suitable examples.

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APPLIED MATHEMATICS 1.1

2

Objectives

To develop certain mathematical abilities and skills among students.

To develop the knowledge of given topics which covered by syllabus.

The pupils may have proper view of the subject covered all the syllabus mater to be studied.

To crease new knowledge to solve the problem given in the exercise by understanding the solved examples.

It helps the pupils to acquire the required information with speed and accuracy.

To develop the skills of measuring, thinking, analyzing logically.

To crease the knowledge of application oriented mathematics and solving problems in allied mathematics related in all other service subjects.

1.2

Characteristic equation of a matrix

Let A be a n × n square matrix over a field F and I be the unit matrix of the same order. Let I be an unknown. Then determinant A – I  is called the characteristic Polynomial of a matrix A. The equation det.A – I  = 0 is called the characteristic equation of the matrix A. The roots of this equation are called the characteristic roots of the matrix A. Characteristic roots are also called latent roots or eigen values. The characteristic vectors of a Matrix Let A be a n × n matrix. Let X be any non-zero column vector, x1 . . i.e X = . . xn then any solution of the equation AX = X other than X = 0 corresponding to some particular value of  is called a characteristic vector or Eigen vector or latent vector. The equation AX = I X may be written as ( A = I) X = 0. This equation gives a set of linear homogenous equations. If it is to possess a non-trivial solution, then.A – I = 0. Then  is a characteristic root of the matrix A. Hence to find all the characteristic vectors, we adopt the following procedure . Step I : Solve A – I = 0 for  Let 1, 2, 3 …………..n be the characteristic root of A. Step II : Find the non-trivial solution of A – I = 0 . This solution is the characteristic vector corresponding to be characteristic value or characteristic roots of  Step III : Similarly find the characteristic vector corresponding to the characteristic roots.

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APPLIED MATHEMATICS

3

Step IV : Note that the number of linearly independent solution of (A – I ) X = 0 is equal to n-Rank of A – I .

Problems: Example : 1 1.

1

2

2

1

Find the characteristic equation of A =

Solution: The characteristic equation is A – I  = 0 1

2

2

1

1-

2

2

1-

(i.e)

(i.e)

(i.e)

-

1

0

0

1

2 6 4

0 0 0

1

0

1 0 0

0 1 0

0 0 1

= 0

 0 0

0  0

0 0 

= 0

= 0

= 0

(1- ) ( 1- ) – 4 = 0  –2-3=0 2

Example : 2

Find the characteristic equation of A =

2

Solution: The characteristic equation is A – I  = 0

(i.e)

2 0 4

0 6 0

4 0 2

-

(i.e)

2 0 4

0 6 0

4 0 2

-

(i.e)

2- 0 4

0 6- 0

4 0 2-

=

( 2-  ) [ (6-) (2 -  ) – 0 ] – 0 + 4 [ 0 – 4 ( 6- ) ] = 0

= 0

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APPLIED MATHEMATICS

4

=

( 2-  ) [ 12- 6 - 2  + 

=

( 2-  ) [ 12- 8 + 

=

24 - 16  + 2  – 12  + 8  -  - 96 + 16  = 0

2

2

) + 4 [ -24 + 4  ] = 0

) + 96 + 16  = 0

2

2

3

 - 10  + 12  + 72 = 0 3

1.2.3

2

Characteristic Roots:

Find the eigen value of A

0 - 4 4

=

1 4 - 3

1 2 -1

1 0 0

0 1 0

Solution: The characteristic equation is A – I  = 0

(i.e)

0 -4 4

1 4 -3

(i.e)

- -4 4

1 4- - 3

1 2 - 1

1 2 -1- 

-

0 0 1

= 0

= 0

-  [ (4-) ( -1 -  0 + 6 ] – 1 [ -4 (-1- ) – 8 ] + 1[ 12-4(4-)]

=0

-  ( 4 - 4 +  +  + 6 ) - ( - 4 + 4  – 8 ) + 1 ( 12- 16 + 4 ) = 0 2

-  +3 -2 =0 3

2

 -3 +2 =0 3

2

 (  - 3  +2 ) = 0 2

 = 0 and

(  - 3  +2 ) = 0

 = 0 and

(-1) ( -2)=0

2

 = 0,  = 1,  = 2

SELF – ASSESSMENT QUESTIONS : I 1.

Find the eigen values and eigen vectors of the matrices

a)

1 2 3  0 4 2    0 0 7 

……………………………………………………………………………………..

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APPLIED MATHEMATICS

5

…………………………………………………………………………………….. …………………………………………………………………………………….. Ans: 1, -4, 7 ; [k,0,0]; k [ 1, ½ 0 ]

1.2.4

Cayley Hamilton’s Theorem

Theorem : Every square matrix satisfies its own characteristic equation (i.e. If the characteristic polynomials is n n-1 n-2  () =  + P1  + P2  + … + P n-1  + Pn then,  (A) = 0  n n-1 n-2 A + P1 A + P2 A + … + P n-1 A + P n I = 0 Proof:  () = | A - I | a11 -  a21 a31

a12 … a22 -  … a32 A33 -  …

a1n a2n a3n

. . .

. . .

. . .

. . .

an1

an2

an3

= P0  + P1  n

n-1

+ P2 

n-2

. . .

ann - 

- … + P n ( Say)

Then we have to prove that  () = P0 A + P1 A n

n-1

+ P2 A

n-2

+ … + Pn I = 0

Consider ( A - I )* which is the adjacent of A -  I. Let us call it B. The elements of B are minors of order (n-1) of A -  I. The element of B are of degree at most n-1 in  and hence can be written as B0 

n-1

+ B1 

n-2

+ …. + B n-1

Where B0, B1, B2, …. B n-1 are matrices of order n whose elements are polynomials in the elements of A We know that (A-I) Adj. (A-I)

= | A- I| I =  () I

(A-I) (B0  + B1  + …. + B n-1 ) I n n-1 n-2 = ( P0  + P1  + P2  - … + P n ) I n-1

n-2

This is an identity in  . Therefore equating the coefficient of like powers of , we obtain  B0 AB0 – B1 AB1 – B2

= P0 I = P1 I = P2 I

. . .

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APPLIED MATHEMATICS ABr-1 – Br

6

= Pr I

. . .

Abn-1 – B

= Pn I n

By pre-multiplying these equations with A , A 0

n

=

P0 A + P1 A

n-1

+ P2 A

n-2

n-1

… A, I respectively

-…+Pn I

Which prove Cayley – Hamilton’s Theorem Note : : Cayley-Hamilton’s Theorem can be used to find the inverse of a Matrix. By Cayley- Hamilton’s theorem, 0

n

=

P0 A + P1 A

n-1

+ P2 A

n-2

- … + Pn I

-1

Multiplying both side by A , we have 0=

P0 A

n-1

+ P1 A

n-2

+ P2 A

n-2

-1

+… + Pn-2 + Pn A

1 -1

A =

---- P0 A n P

n-1

+ P1 A

n-2

+ P2 A

n-2

-1

+… + Pn-2 + Pn A

Note : Cayley – Hamilton’s theorem can also be used in computation of arbitrary integral power of A.

Example : 1

0 1 2    Determine the characteristic roots of the matrix 1 0 1    2 1 0  Let

A =

A - I =

=

0 1 2

1 0 -1

2 -1 0

0 1 2

1 0 -1

2 -1 0

- 1 2

1 - -1

2 -1 -

- 

1 0 0

0 1 0

0 0 1

The characteristic equation of A is | A -  I | = 0 - 1 2 

1 - -1

2 -1 = 0 -

 ( – 1 ) – 1 ( -  + 2 ) + 2 ( - 1 + 2  ) = 0 2

 +  +  - 2 -2 + 4  = 0 3

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APPLIED MATHEMATICS

7

 –6+4=0 3

 ( – 4 ) – 2 (  - 2 ) = 0 2

( – 2 ) [  (  + 2 ) – 2 ] = 0 ( – 2 ) (  + 2 – 2 ) = 0 2

 = 2,

-2±4+8 2 - 2 ± 23 2

= 2, - 1 ± 3

= 2,

 The characteristic roots are

2,

- 1 ± 3

Example : 2

Find the eigen values of

Let A

a 0 0

=

a 0 0

h b 0

h b 0

g 0 c

g 0 c

The characteristic equation is | A -  I | = 0 a- h 0 b- 0 0 i.e. (a - ) (b - ) (c - ) = 0  = a, b, c the eigen values are a, b, c (i.e )

g 0 c- 

=

Example : 3 Prove that the matrices A, B, C given below have the same characteristic values.

A=

0 a b

a 0 c

b c 0

0 b a

B=

b 0 c

a c 0

C=

0 c b

=

0

c 0 c

c a 0

Solution : The characteristic equation of A is A -  I = 0 |A-I|=0

   

- a B

a - c

b c -

-  ( – c ) – a ( - a - bc ) + b (ac + b ) = 0 3 2 2 2 -  + c + a  + abc + abc + b  = 0 3 2 2 2  -  ( a + b + c ) – 2 abc = 0 2

2

The characteristic equation of B is | B - I | = 0 |B-I|=0 ;

0

- -

b c

c = 0

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APPLIED MATHEMATICS

8 a

c

-

Equating the determinant, we get  -  ( a + b + c ) – 2 abc = 0 3

2

2

2

The characteristic equation of C is |C- I | = 0 |C-I|=0 ;

c

- - d

c a a

b = 0 -

Expanding we get, 2  -  ( a + b + c ) – 2 abc = 0 From I, II, III , we conclude that the matrices A, B, C have the same characteristic equation and hence the same characteristic values.

1.2.5

Properties of characteristic Equations

Property : 1 Show that if A is a square matrix, the characteristic equation of A and A’ are identical Solution: Let A = ( aij) i, j = 1, 2 , …. n Then A - I =

A - I =

a11 -  a21 a31

a12 … a22 -  … a32 a33 -  …

. . .

A′ - I =

an3

an1

an2

a11 -  a21 a31

a12 … a22 -  … a32 a33 -  …

. . .

an1

an2

an3

a1n a2n a3n

and

ann -  a1n a2n a3n

ann - 

the value of a determinant is unaffected by interchanging of row into columns and columns into rows. |A-I|

=

| A′ -  I |

i.e., the scalar equation | A -  I | = 0, is the same as the scalar equation | A′ -  I | = 0 Hence the characteristic equations of a A and A’ are identical.

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9

Property : 2 -1

Show that the two matrices A and P A P have the same characteristic roots. Solution Let

-1

B = B-I = =

P AP -1 P AP-I -1 P [A -  I] P

B-I

= = = =

| P | |A -  I | | P | -1 |A - I | | P | | P | -1 |A - I | | P | I |A - I |

| A -  I |

= 0  |B-  I | = 0

-1

The characteristic roots of A and B =

-1

P A P are the same.

Property : 3 If and B are two non-singular matrices, Prove that AB and BA have the same characteristic roots. AB

=

I ( AB)

= =

(B B ) AB -1 B (BA) B

-1

From example same 5, it follows that AB and BA have the same characteristic roots.

Property : 4 2

of A

If the characteristic roots of A are 1, 2, 3, ….. n show that the characteristic roots 2 2 2 are 1 , 2 , ….. n ,

Proof: Since, 1 , 2, ….. n, are the characteristic roots of A, we have |A - r I | = 0 , r = 1, 2, ….. n, Then

(A - r I ) (A + r I ) A - r I 2

2

| A - r I | = | A - r I | | A + r I | = 0

r ( r = 1, 2, … n) are the characteristic roots of A

2

2

2

2

Property : 5 If 1 , 2, ….. n, are the characteristic roots of A, the 1 are the characteristic roots of A

-1

1

1 1 , 2, ….. n

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Solution Since r ( r = 1, 2, … n) is a characteristic roots of A, we have |A - r I | = 0 |A - r A A | = 0 -1

- r A A – _1_ I r -1

=0

( - r ) | A | A – _1_ I r n

-1

=0

Since A is non-singular, | A |  0 and r  0 

the

-1

A

_1_ I r

=0

1 1 1 -1 1 , 2, ….. n are the characteristic roots of A

Property : 6

Find the characteristic roots of the orthogonal matrix

cos   sin  

 sin   and verify that they cos  

are of unit modulus. Solution : A=

| A - I | =

=

cos   sin  

 sin   cos  

cos     sin    sin  cos     

 cos   

2

+ sin

2

The characteristic equation is | A - I | = 0 i.e.

 cos   

2

+ sin

2

=0

2 – 2  cos  + 1 = 0

2cos   4cos 2   4 = 2

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11

= cos   i sin 

 the characteristic roots are cos  i sin  =||=

cos2   sin 2  = 1

| cos   i sin  | =

Example 7 If A is a skew Symmetric Matrix of order n, show that | A -  I | = ( - 1) | A +  I| . Hence show that if  is a characteristic roots of A, then -  is also a characteristic root. n

Solution :

If a is skew symmetric matrix, then |A-I|

= = = =

If

|A-I|

' | A ' ( I ) ' | |  A  ( I ) | since I '  I n ( - 1 ) |  A  ( I ) |

 is a characteristic roots of A, then |A -  I| = 0

Then (1) gives

-

A'   A

| A  ( I ) | = 0

is also a characteristic root of A

c b  0  Show that the Matrix c 0 a  satisfies Cayley – Hamilton’s theorem   b a 0  Solution By Cayley- Hamilton’s theorem

A2  A  7 I  0 Divide

f ( )by 2  5  7

i.e.

|A -  I | X = 0 |A - I | X = 0

i.e.

2 2 x1 0 2 2 x2 2 x1  2 x2  0 2 x1  2 x2  0

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APPLIED MATHEMATICS Since

12

x1 = k, x2 = - k satisfies the equation for all values of k. Then the characteristic

vector corresponding to

 c 2  b 2    ac  ac 

ab

= 1, is

1 X1  k    1

  bc  b 2  a 2  ac

c 2  a 2 bc

 c 2  b 2  A3  A2 A   ac  ac 

ab c  a 2

 0 c b   bc   c 0 a  b 2  a 2   b a 0  ac

2

bc

 0 c ( a 2  b 2  c 2 ) b ( a 2  b 2  c 2 )    2 2 2 =  c(a  b  c ) 0 a (a 2  b 2  c 2 )   b(a 2  b 2  c 2 ) c(a 2  b 2  c 2 )  0    0 c b   (a 2  b 2  c 2 )  c 0 a   b a 0  =

(a 2  b2  c2 ) A

=> Example 8

A3  (a 2  b2  c2 ) A  0

1 1 3  2 6  and show that the matrix A satisfies Find the characteristic equation of A = 5   2 1 1 the equation.

Solution: The characteristic equation is

1    5 i.e.   2

1 2 1

A  I  0

 6  = 0 3    3

(1   )  (2   )(3   )  6   5(3   )  12   5  4  2   0 i.e.

(1   )( 2   )  5  3  6  3  0

 2     3   2  5  3  6  3  0  3  0or 3  0

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13

We will show that the matrix satisfies the characteristic equation. We have to verify that

A3  0

 1 1 3  1 1 3  A   5 2 6   5 2 6   2 1 3  2 1 3 2

0 0 0  = 3 3 9    1 1 3  0 0 0  1 1 3  A   3 3 9   5 2 6   1 1 3  2 1 3 3

=

0 0 0 0 0 0   0 0 0 

 A3  0 Example 9 Use Cayley-Hamilton theorem to express A when A =

2 A5  3 A4  A2  4I as a linear polynomial in

 3 1  1 2   

Solution : The characteristic equation of A is |A-  I| = 0

i.e.

1  3    1 2    = 0   (3   )(2   )  1  0

 2  5  7 I  0 By Cayley-Hamilton Theorem 2

λ –5λ+7I Consider the polynomial

f ( )  2 2  3 4   2  4 f ( )by 2  5  7

Divide

3

2

(2λ – 7λ + 21λ +57) 2

( λ +5λ + 7)

2

4

2

2λ – 3λ + λ – 4

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7 4 14 3   2  4 7 4  35 3  49 2 ------------------------

21 3  48 2  4 21 3 105 2 147 ---------------------------

57 2  147  4 57 2  285  399 -----------------------

138  403

 2 2  3 4  4)  (2 2  5  7)(2 3  7 2  21  57)  138  403  A5  3 A4  A2  4I A2  5 A  7 I 2 A3  7 A2  21A  571)  138 A  403I  0  138 A  403I  138 A  403I Which is linear Polynomial is A. Example 10

Find the characteristic vectors of the matrix

3 2 2 3  0  

Solution: The characteristic Equation is

2  3   0  2 3    

(3   )2  4  0 or  2  6  5  0 ( 1)(  5)  0    1,5 The characteristic roots are 1, 5. The characteristic vector of A corresponding to the characteristic root  =1,

(A  I )X  0 (A  I)X  0

 2 2 x1 2 2 x  0   2 2 X1  2 X 2  0

i . e.

2 X1  2 X 2  0 Since X1  k , X 2  k satisfies the equation to all the values of k.

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The characteristic vector corresponding to

When

1  = 1, is k    1 

 2 2  x1 0  = 5, we have    2 2 x2 2 X 1  2 X 2  0 2 X1  2 X 2  0

X1  k , X 2  k satisfies the equation for all the vectors of k, the characteristic vector corresponding to  = 5 is Here

1 X2  k   1

SELF ASSESSMENT - II

1 0 2    1. Find the characteristic Equation of A = 0 2 1    2 0 3  ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… Ans:

 3  6 2  7  2  0

1 0 3    2. Verify Cayley – Hamilton theorem for the matrix 2 1 1   1 1 1  ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… 1.4

Rank of Matrix The number r is said to be the rank of Matrix A if (i) (ii )

A possesses atleast one minor of order r which does not vanish Every minor of A of order ( r + 1) and higher orders vanish

In other words the rank of a matrix is the order of any highest order non vanishing minor of the matrix. The rank of a matrix of A is denoted by  ( A) Note 1 : If I is the unit matrix of order n, its rank is n Note 2 : If A is any matrix of order m  n its rank is  m and

n

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Note 3 : The rank of A is the same as the rank of the transpose of A Note 4 : The rank of a null matrix is taken as zero. Example : 1 Consider the matrix

1 2 3    A= 2 4 6    3 1 2  Here the highest minor is of order 3. But |A| = 0

  ( A)  3 But the 2

nd

order

2 4 3 1  0  

The rank of A = 2

1 2 3   Consider A = 2 4 0    3 6 9  Hence |A| = 0 Also every minor of order 2 vanish. But there exists non-zero minors of order 1

  ( A)  1 1.4.1

Elementary Transformation

In some cases the evaluation of determinant of matrices of highest order may not be simple. In such cases we can obtain easily the rank of a matrix using elementary transformations. An elementary transformation on a matrix is an operation of any one of the following three types. (i)

The interchange of any two rows ( or column )

(ii) The multiplication of the elements of any row ( or column) by a non-zero scalar (iii) The addition to the elements of any row ( or column) the scalar multiples of the corresponding elements of any other row ( or column )

1.4.2

Equivalent matrices

Definition : Two matrices A and B of the same order are said to be equivalent if one can be obtained from the other by the application of a finite chain of elementary transformations. Symbolically we write A – B and read it as A equivalent B

Example 1 Find the rank of the matrix

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 3 1 2    A = 6 2 4    3 1 2  Solution :

 3 1 2  0 0 0    0 0 0 

R1,

R2  2 R1 , R3  R1

Since all minors of order 3 and order 2 vanish There exists non-zero minors of order 1

 ( A)  3 ,  ( A)  2

  ( A)  1

Example 2 Find the rank of

1 2 3 4  A= 2 4 6 8    1 2 3 4  Solution:

A

=

  ( A)  1

1 2 3 4 2 4 6 8    1 2 3 4 

R2  2R1 , R3  R1

Example 3 Find the rank of the Matrix

A=

1 0  1  0

a b 0 c d 1  If a, b, c, d are all different a b 0  c d 1

Solution:

1 0 A=  0  0

R1 a b 0  R2 c d 1 0 0 0  R3  R1  0 0 0  R4  R2

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Example 4 Find the rank of the Matrix

1 a A=   a 2

1 c  c 2 

1 b b2

If a, b, c, d are all different

Solution: |A| = (a-b) (b – c ) ( c – a ) Case 1 Suppose a, b,c are all different Then |A|

 0 ,  ( A)  3

Case 2

rd

Suppose two of a, b,c are equal and the 3 is different i.e. Say a = b,

 c

then |A| = 0, since 2 columns are identical But

b

c

2

a2

b

= bc

1

1

b

a

= bc ( c – b )

 0

  ( A)  2 Case 3: Suppose all a, b, c are equal i.e. a = b = c

then

1  a  a2 

1 a a2

1  a a 2 

In this case | A| = 0

Example 5

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Find the Rank of

1 7 3 3   A = 7 20 2 25    5 2 4 7  Solution:

R1 1 7 3 3   A ~ 7 20 2 25 R2    0 33 11 22  R3  5 R1 R1 1 7 3 3   A ~ 0 69 23 46 R2  7 R1   0 33 11 22  R3

 1 2 0 0   - 0 0 23 46 C1, C2 + 3 C3, C3 + C4, C4 + 3C1   0 0 11 22   1 0 0 0   - 0 0 23 0 C1, C2 – 2C1, C3, C4 - 2C3   0 0 11 0  rd

All 3 order minors vanish

  ( A)  3 But,

1 0    = 23  0,  0 23 

  ( A)  2

Example 6 : Find the Rank of

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1 1 A=  2  3

2 1 2 3 2 2  4 3 4  7 4 6

1 0 A~  1  0

R1 2  R2  R1 1 1 0 R3  R2 0 0 2  0 1 0  R4  ( R2  R4 )

1 0 A~  1  0

2

1

2  R1 1 1 0  R2 0 0 2  R3  R2  0 1 0  R4

1 0 A~  1  0 1 0 A~  1  0

2

2  R1  R3 1 0  R2 R3 0 2  1 0  R4 1 2 1 0  C1 , C2 , C3 , C4  2C1 0 0  1 0 

Solution:

2

1 0 0 2 1 0 0

1

1

0 0 ~  1  1

1 0 0  R1  R2 1 0 0  R2  R4 R3 0 0 0  R4 0 0 0

0 0 ~  1  0

0 0 1 0 0  0 0 0  0 1 0  0

 ( A)  4

R1  R2 R2 R3 R4

since A = 0

0 1 1   Consider the Minor 1 0 0 = 1  0   0 0 1

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  ( A)  3 Self Assessment - III Find the Rank of the following Matrices

1 3 4 3  3 9 12 9  (1)    1 3 4 1 ……………………………………………………………………………………….. ……………………………………………………………………………………….. ……………………………………………………………………………………….. Answer: 1

1 3 4 3  3 9 12 4  (2 )   1 3 4 1  ……………………………………………………………………………………….. ……………………………………………………………………………………….. ……………………………………………………………………………………….. Answer 2

1.4

Test of Consistency of linear equation

Working rule for finding the solution of the equation AX = B. Suppose we are given m equations and n unknowns. Step (i)

Write down the coefficient Matrix A

Step ( ii)

Write down the augmented matrix [ A, B]

Step ( iii)

Apply elementary transformation to find the ranks of A and [A,B]

The following situations may arise Case I Rank A < Rank [A,B] In this case equations are inconsistent i.e. They have no solution Case II Rank A = Rank [A,B] = r (say ) In this case the equations are consistent

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Ie. They possess a solution If r = n then the solution is unique. If r < n then there are infinite number of solution. Example :1 Show that the system of equation 3x – 4y = 2 5x + 2y = 12 - x + 3y = 1 are consistent Solution:

 3 4    A= 5 2 ; B =   1 3    3 5 (A,B) =   1

2 12     1  4 2 3

2 12  1 

-

 0 5 5  R1  3R3  0 17 17  R  5R 3   2  1 3 1  R3

-

0 0 0  0 17 17     1 3 1 

We note that P (A) = 2, and

5 R2 17 R2

R 

R3

 (A,B) = 2

Hence the given system of equation is consistent and therefore possesses a solution.

Example : 2

x  3 y  8 z  10 Show that the system of equations

3x  y  4 z  0

are consistent and solve them

2 x  5 y  6 z  13 Solution

1 3 8  4  A= 3 1   2 5 6 

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23

R1 1 3 8 10    = 3 1 4 0  R2   2 5 6 13  11 R3  R2 10 = Here

 (A) = 2,  (A,B) = 2

 The given system of equations is consistent. Since,

 (A), 2 < 3 there are infinite number of solutions.

x  3 y  8 z  10  10 y  20 z  30

y  3  2z Where z is a variable. i.e. x = -1 +2z Take z = k, the solutions are x = - 1 + 2z y = 3 – 2k z = k where k is any arbitrary value.

Example 3 Show that the equation

x yz 6 x  2 y  3z  14 are consistent and solve them x  4 y  7 z  30 Solution:

1 1 1    A= 1 2 3   1 4 7    The Augmented matrix is

1 1 1 6    (A, B) = 1 2 3 14   1 4 7 30 

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1 1 1 6  R1   - 0 1 2 8 R2  R1   0 3 6 24  R3  R1 1 1 1 6  R1   R - 0 1 2 8 2   0 0 0 0  R3  3R2 We note that

 ( A)  2

and also

 ( A, B)  2   ( A)  2 =  ( A, B)  2 Hence the given equations are consistent, But,

  ( A)  2 =  ( A, B)  2 <3

 there are infinite number of solutions for the given equations. Also, x yz 6

y  2z  8 y  9  2z x  6  (8  2 z )  z x  z2 Taking z = k where k is arbitary the solutions are

x  k 2 y  8  2k zk Example : 4

Show that the equation

x  2y  z  3 3x  y  2 z  1

are consistent and solve them.

x  y  z  1 Solution

1 2 1   A = 3 1 2   1 1 1 

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1 2 1 3  3 1 2 1   (A,B) =  1 1 1 2    1 1 1 1

-

R1 1 2 1 3  0 7 5 8 R  R 3   2 0 6 5 4  R3  2 R1   0 3 2 4  R4  R1

-

1 2 1 3  R1 0 1 0 4  R  R 3   2 0 6 5 4  R3   0 3 2 4  R4

-

R1 1 2 1 3  0 1 0 4  R2   0 0 5 20  R3  6 R2   0 0 2 8  R4  3R2

-

1 2 1 3  R1 0 1 0 4  R   2 0 0 1 4  R3/ 5   0 0 1 4  R4 / 2

-

1 2 1 3  R1 0 1 0 4  R 2   0 0 1 4  R3   0 0 0 0  R4  R3

   2 =  ( A, B)  3  The given system of equations is consistent and has a unique solution  the solution is z = 4 y=4 x + 2y – z = 6 x = -1 x = -1, y = 4, z = 4 SELF – ASSESSMENT IV Test of Consistency and hence solve the equation:

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x  2 y  3z  2 (i)

2 x  3z  3 x yz

……………………………………………………………………………………….. ……………………………………………………………………………………….. ………………….…………………………………………………………………….. Answer : k = - ½ , K = -4, Y = ½ where K -6 where z = k 1.5

Relation between the roots and coefficients of an Equation Let 1, 2, … n be the roots of the equation

x n  p1 x n1  p2 x n2  ...  pn  0

… (1)

The equation corresponding to the roots 1, 2, … n is

(x  α1 )(x  α 2 ) . . . (x  α n )  0

x n  1 x n1  1 2 x n2 1 2 3  x n3 . . .  (-1)n 1 2 ... n  0

…(2)

Comparing equations (1) & (2) we get,

1   p1

1 ,  2  p2 1 ,  2 3   p 3 . .

1 2 ... n  (1) n p n If Sr is defined as the sum of the products of the roots of the equation taken r at a time we have the following relations between the roots and coefficients.

S1   p1 S 2  (1) 2 p2 S 3  (1) 3 p3 S n  (1) n pn Note : If the given equation is

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a0 x n  a1 x n1  a2 x n2  ...an1 x  an  0 then divide the equation by a0

xn 

Then

S1  

a a1 n1 a2 n2 a n1 x  x  ...  x n 0 a0 a0 a0 a0 a1 a0

S2  

a2 a0

S3  

a3 a0

S n  (1) n

an a0

Note 1: (1) If , ,  are the roots of the equation ax +bx +cx+d =0 then 3

2

b S1         . a c S 2        . a

d S3     . a Note 2: (2) If , , ,  are the roots of the equation x +px +qx +rx+s=0 then. 4

3

2

S1     p S 2     q S3     r S4    s Example 1 3

2

3

Show that if the roots of the equation x +px +qx+r=0 are in A.P. then 2p - 9pq + 27r = 0. Solution

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Let the roots of the equation x +px +qx+r=0 be -, ,+ . 3

2

Sum of the roots is  -  +  +  +  = -p i.e.

3   p or   

p 3

Since  is a root of the given equation it has to satisfy the equation x +px +qx+r=0. 3

3

2

2

 p  p  p     p    q     r  0  3  3  3

p 3 p 3 pq   r 0 27 9 3

 p3  3 p3  9 pq  27r  0 2 p 3  9 pq  27r  0 Example 2 3

2

Solve the equation x -12x +39x-28= 0 whose roots are in A.P. Solution Let  - ,,  + be the roots Then  -  +  +  +  = 12 i.e. 3 = 12 or = 4 4 is a root of the given equation. Let us remove this root by synthetic division.

1

-12

39

-28

0

4

-32

28

1

-8

7

0

2

The other roots are given by x – 8x + 7 = 0 (x – 1) (x - 7) = 0 x = 1 or x = 7  The roots of the given equation are 1, 4, 7 Example 3 4

3

2

Solve the equation x -2x -21x +22x+40=0 given that the roots are in A.P. Solution

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29

2

x -2x -21x +22x+40=0 Let the roots be , , ,  Since they are in A.P. +=+

…(1)

 x4-2x3-21x2+22x+40=0 = (x-) (x-) (x-) (x-) = [x -x) ( + ) +  ] [x -x (+)x+] 2

2

Equating the coefficients of x

3

+++=2 2(+)=2

[From (1)]

+=1

…(2)

Equating the coefficient of x

2

++(+)(+)= -21 ++1=-21 +=-22 Equating the coefficient of x

…(3)

(+)+(+)=-22

+=-22 Equating the constant term =40

…(4)

By inspection from (4) & (5) =-20 =-2 +=1, =-2 =-1, =2 +=1, =-20 =-4, =5 The roots are -4, -1, 2, 5 Example 4 4

3

2

If the sum of the two roots of the equation x +px +qx +x+ s=0 equals the sum of 3

the other two, prove that p +8r= 4pq Solution

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Let ,, ,  be the roots such that +=+ Then

……(I)

x 4  px3  qx 2  rx  s  ( x   )( x   )( x   )( x   ) = x – ( - )x + ) ( x –(+ )x +  2

Equating the coefficient of x

2

3

 + ++ = - p Equating the of efficient of x

… (1)

2

(+) ( + ) + +  = q

… (2)

Equating the coefficient of x -  ( +) -  (+) = r  (+)+  (+) = - r

… (3)

Equating the constant term,  = s … (4)

Using (I) in (1) we get 2 (+) = - p or

p 2

         

p 2

    q 

From (2),

p2 4

… (5)

Using I in (3)

αβ  γδ    p   r  2

   

2r p

From (5) and (6)

q

p 2 2r  4 p

4 pq  p 3  8r or

p 3  8r  4 pq

Example 5 Find the condition that the roots of the equation

ax3  3bx 2  3cx  d  0 are in

G.P.

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Solution

Let the roots be

Then

 ,  ,  

 d . .    a i.e.

3  

d a

…(1)

Since  is a root of the equations

ax3  3bx 2  3cx  d  0, a 3  3b 2  3c  d  0

d  2 a   3b.  3c  d  0  a   d  3b 2  3c  d  0

3 b  c   0 Since  0, b+c = 0 or b = -c b  = -c 3 3

Cubing,

3

d  3 b3    c  a 

d  3 3 3 b3    c or b d  c a a   This is the required condition. Example 7 3

2

Solve the equation x -4x -3x+18=0 given that two of its roots are equation. Solution Let the roots of the equation

x3  4 x 2  3x  18  0 be  ,  ,  . Then 2+=4  +2=-3 2

 =-18 2

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From (1), = 4- 2 Substituting this in (2)  + 2 (4-2)= - 3 2

 2  8  4 2  3 3 2  8  3  0 (3  1)(  3)  0

1 3

   or 3 1 14 When    ,   3 3 When

But

  3,   2

 2   -18 is not satisfied by 1 3

  , 

14 3

 The roots are 3, 3, -2. Example 8 Solve the equation

x 4  2 x3  4 x 2  22 x  40  0 given that its roots are in A.P.

Solution Let the roots be

  3 ,    ,    ,   3

Sum of the roots is

  3            3  2 4  2 or   

Also

1 2

(  3 ) (   ) (   ) (  3 )  40

( 2   2 ) ( 2   2 )  40

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1  2  1 2      9  40  4  4 

1  4 1  36   640 2

2

144 2  40 2  639  0 

40  368064 288

40  608 288

648 568  288' 288 =

648

- 568

288 ,

288

9

- 71 4

36

9 3   4' 2

When

2 

When

 

Where

,

1 3   the roots are -5, -2, -1, 4. 2' 2

 

1 3    the roots are 4, -1, -2, -5. 2' 2

These roots satisfy the given equation and hence, -5,-2, -1, 4 are the only roots of the given equation.

Example 9 Solve the equation

x 4  2 x3  25x 2  26 x  120  0 given that the product of two

of its roots is 8. Solution Let the roots be , , , ; given that =8

x 4  2 x3  25x 2  26 x  120  ( x   )( x   )( x   )( x   )

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 [x 2  (   ) x   ][ x 2  (   ) x  y ] ...(1) 2

Equating the co-efficient of x ,  +  +  +  = -2

…(1) 2

Equation the co-efficient of x , ++(+)(+)= -25

…(2)

Equating the co-efficient of x …(3)

- [(+)+(+)] = -26 Equating the constant term, =120

y 

From (4),

From (3),

…(4)

120  15 8

15(   )  8(   )  26

(1)x 8 : 8(   )  8(   )  16 Subtracting

7(   )  42 (   )  6

    8

From (1), From I,

( x 2  6 x  8)( x 2  8x  15)  0 i.e.

( x  2)( x  4)( x  3)( x  5)  0

the roots are 2, 4, -3, -5. Example 12 4

3

2

Solve the equation 15x -8x – 14x + 8x -1 =0 given that the roots are in H.P.

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Solution 4

3

2

15x -8x -14x +8x-1 = 0 If the roots of the equation are in H.P. then the roots of the equation 4

3

2

x -8x +14x +8x-15=0 are in A.P. Let the roots be -3, -, +, +3 Sum of the roots is 4 = 8  =2 Product of the roots is 2

2

2

2

( - ) ( -9 ) = -15 2

2

(4- ) (4-9 ) = -15 4

2

9 - 40 + 31 = 0 4

2

2

9 -9 -31 + 31 = 0 2

2

2

9 ( -1) -31 ( -1) = 0 2

2

( -1) (9 -31) = 0

β  1 or 

31 3

When =2, =1, roots are -1,1,3,5. When =2, = -1 roots are 5, 3, 1, -1 Which are the same as the roots given by =2, = 1. These roots satisfy the given equation. Hence these are only possible roots of the given equation and hence

β

SELF – ASSESSMENT – 5

1)

Solve the value of k so that the roots of the equation

2 x2  6 x2  5x  k  0 are in A.P ……………………………………………………………………………………….. ……………………………………………………………………………………….. ……………………………………………………………………………………….. Answer

3 2 1 , ,10, 2 3 10

2. Solve the equation x  x  16 x  20  0 given that the difference between two ratio is 7 ……………………………………………………………………………………….. 3

2

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……………………………………………………………………………………….. ……………………………………………………………………………………….. Answer ( 2, -2, 5) 1.6

Imaginary and irrational roots

Theorem In a polynomial equation with real co-efficient imaginary roots occur in conjugate pairs. Proof Let f(x) = 0 be a polynomial equation with real co-efficient. Let  + i where  and  are real be an imaginary root of the equation

f(x)=0.

We will now prove that  - i is also a root of the equation. Now

x  α  iβx  α - iβ  x  α  iβx  α  iβ = (x-) +  2

2

which is a quadratic polynomial with real co-efficient. Let us divide the polynomial f(x) by (x-) +  . Let the quotient be  (x) and the 2

2

remainder be Ax + B. Then f (x) = [(x-) +  ]  (x) + AX + B 2

2

By hypothesis  + i is a root of f(x) = 0  f (+i) = 0 f (+ i) = 0 + A(+ i) + B = 0 Equating real and imaginary parts we get … (1)

A + B=0 A = 0

… (2)

Since  + i is an imaginary root   0 and therefore from (2), A =0 Consequently from (1), B = 0 Hence the remainder =Ax +B =0 This means (x-) +  is a factor of f (x). 2

2

In other words,  + i and  - i are the roots of f (x)=0.

Hence in a polynomial equation with real co-efficient imaginary occur in conjugate pairs.

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Example 2 4

2

Solve the equation x – 11x +2x +12 = 0 given that

5  1 is a root.

Solution In an equation with irrational roots occur in conjugate airs. Since

5  1 is a root.

 5  1 is also a root. The real factor corresponding to these roots is ( x  1  5 )( x  1  5 )  ( x  1) 2  5 =

( x 2  2 x  4)

( x 2  2 x  4)

x4 -11x 2 + 2x + 12 x4 +2x 3 - 4x2

x2  2x  3

- 2x3 -7x 2 + 2x - 2x3 -4x 2 + 8x - 3x2 - 6x + 12 - 3x2 - 6x + 12 0  The other roots are given by 2

x -2x-3 = 0 (x-3) (x+1)

=0

x=3, -1  The roots of the equation are – 1 -

5 , - 1 + 5 , - 1, 3.

Example 3 Given that

 2   7 is a root of the equation x4+2x2-16x+77=0 solve it completely

Solution

 2   7  2  i 7 Since

 2  i  7 is a root,  2  i 7 is also a root.

The real factor corresponding to these roots is

( x  2  7i )( x  2  7i )

 ( x  2) 2  7  x 2  4 x  11

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38

2 x4 +2x 2 - 16x + 77 x  2 x  3 x4 +4x 3 +11x2

- 4x3 -9x 2 - 16x - 4x3 -16x 2 - 44x 7x2 +28x + 77 7x2 +28x + 77  The other roots are given by

x

x 02  4 x  7  0

4  16  28 4  3i   2i 3 2 2

The roots are

 2  i 7 , 2  3i

Example 4 Solve the equation

x 4  5x3  4 x 2  8x  8  0 given that 1 5 is a root.

Solution

1 5 is a root. 1 5 is also a root. Let the other roots be , . Sum of the roots =

1 5 1 5      5

+= 3 Product of the roots = (1 

5 )(1  5 )  8

 4  8

  2  &  are the roots of

x 2  x(   )    0

x 2  3x  2  0 ( x  1)( x  2)  0

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x  1,2  roots of the given equation are

1  5, 1  5 , 1, 2

Self – Assessment - 6 1.

Solve the equation

x4  4 x3  8x  35  0 given that 2  3i is roots

……………………………………………………………………………………….. ……………………………………………………………………………………….. ……………………………………………………………………………………….. Answer : - 1, 1.7

 5 , 3, -1

Transformation of equation If 1, 2,….n are the roots of the equation f(x)=0, then forming the equation whose

roots are (1), (2),…. (n) is called transformation of the equation. The relation between a root of f(x)=0 and a root y of the required equation is y=(x). If x is eliminated between the equation f(x)=0 and y=(x), an equation in y is obtained and this is the required equation whose roots are (1), (2),… (n). Example 1 If , ,  are the roots of the equation x + px +qx + r = 0 form the equation whose 3

roots are

2

α 2  βγ β 2  γα γ 2  αβ , , α β γ

Solution , ,  are the roots of 3

2

X + px + qx + r ++

=0 = -p

…. (2)

+  +  = q  = -1

… (3)

We have to form the equation whose roots are

α 2  βγ β 2  γα γ 2  αβ , , . α β γ

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α 3  αβγ α 3  r  α2 α2

Let y =  – = 2

x3  r (i.e) y = x2 3

2

X - yx + r 3

=0

2

….(1)

But x + px + qx + r = 0 2

Subtracting we get x (p+ y) + qx = 0 Since x  0, x (p+y) = - q

q py

or x =

Substituting in the given equation (1) we get, 3

2

 q   q   q     p    q    r  0 p  y p  y p  y       3

2

2

2

3

r (p+y) – q (p+y) + pq (p+y) – q = 0 3

2

2

2

2

3

3

(i.e.) ry +y (3pr- q ) + y (3p r- q p) + rp - q = 0 is the required equation. Example 2 If , ,  are the roots of the equation x +qx+r=0 find the equation whose roots are 3

       ,  ,       

.

Solution If , ,  are the roots of

x3  qx  r  0       0

      q

  r Let y 

   2  2      (    )  2y 2



   / 

 2  2

 3  2  3  2r    r

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i.e. y 

41

 3  2r r

i.e.

x3  ry  2r  0

But

x3  qx  r  0

Subtracting

ry  qx  r  0

x 

r ( y  1) q

Substituting in the given equation we get,

r 3 ( y  1)3 qr ( y  1)  r 0 q3 q r 3 ( y  1)3  q3r ( y  1)  rq3  0 r 2 ( y 3  3 y 2  3 y  1)  q3 ( y  1)  q3  0 r 2 y 3  3r 2 y 2  y(3r 2  q3 )  r 3  2q3  0

Example 3 , ,  are roots of are

x3  2 x 2  3x  3  0 . First let us form the equation whose roots

2 2 2    13 (  1) 2 (   1) 2 (  1) 2

Solution , ,  are the roots of roots are

Let

   . , ,  1  1  1

x3  2 x 2  3x  3  0 . First let us form the equation whose

 x   1 x 1 yx  y  x or x 

y 1 y

Substituting this in the given equation we get,

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APPLIED MATHEMATICS 3

42 2

 y   y   y     2   3   3  0 1  y 1  y 1  y       y 3  2 y 2 (1  y)  3 y(1  y) 2  3(1  y 3 )  0 y e  2 y 2 (1  y)  3 y(1  2 y  y 2 )  3(1  3 y  3 y 2  y 3 )  0 y3  5 y 2  6 y  3  0 Let us now form the equation whose roots are

2 2 2 , , (1   ) 2 (1   ) 2 (1   ) 2 Let

2 z  y2 2 (1   ) 2

Substitute z=y in (2)

zy  5z  6 y  3  0 y( z  6)  5z  3

Squaring

y 2 ( z  6) 2  (5z  3) 2 z( z 2  12 z  36)  25z 2  30 z  9 z 3  13z 2  6 z  9  0  Sum of the roots

2 2 2     13 (1   ) 2 (1   ) 2 (1   ) 2

Self – Assessment : 8 1.

If

roots are

 ,  ,  are the roots of the equation x3  qx  r  0  2,  2, 2

form the equation whose

……………………………………………………………………………………….. ……………………………………………………………………………………….. ……………………………………………………………………………………….. Ans:

x3  2qx 2  q 2 x  r 2  0

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43

UNIT – II

2.0

Introduction

2.1

Objectives

2.2

Successive Differentiation 2.2.1

2.3

n-th Derivative 2.3.1

2.4

SELF ASSESSMENT QUESTION - III

Euler’s Theorem on homogeneous functions 2.5.1

2.6

SELF ASSESSMENT QUESTIN – II

LEIBNITZ THEOREM 2.4.1

2.5

SELF-ASSESSMENT QUESTION - I

SELF ASSESSMENT QUESTION - IV

Curvature and Radius of Curvature 2.6.1

Cartesian Formula for Radius of curvature

2.6.2

Parametric formula for Radius of Curvature

2.6.3

SELF ASSESSMENT – V

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2.0

44

Introduction

In this chapter, we have to deal with successive differentiation and understanding the definition and applied in to the problem. It will lead up to the nth derivative. 2.1

Objective To develop the skill in the successive differentiation To create new idea to application oriented sum. To understanding the way of approach for the nth derivative

2.2

Successive Differentiation If y is a function of x, its derivative

dy will be some other function of x and the dx

differentiation of this function with respect to x is called second derivative and is denoted by

d2y dx 2 i.e.

d  dy  d 2 y   dx  dx  dx 2

Similarly the third derivative is denoted by

d3y dx 3

d  d2y  d3y i.e.   dx  dx 2  dx3 Thus if we differentiate y twice with respect to x, we get the second derivative. If y is differentiated thrice with respect to x we get the third derivative. Example 1

If y =

ax+b d2 y find cx+d dx 2

Solution:

5 1 5 1 13 1 .  .  32 2 x  1 16 2 x  1 32 2 x  3 5 (1) n (n !2n ) 5 ( 1) n ( n!2 n ) 13 ( 1) n ( n!2 n )  yn    32 (2 x  1) n 1 16 (2 x  1) n 1 32 (2 x  3) n 1 5 1 5 1 13 1 = (-1) n n!2n [ .  .  . ] n 1 n 1 32 (2 x  1) 16 (2 x  1) 32 (2 x  3) n 1 (i.e.) y 

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d 2 y 0  (ad  bc)(2)(cx  d )c  dx 2 (cx  d )4 =

acx  ad  acx  bc (cx  d )2

=

ad  bc (cx  d ) 2

d 2 y 0  (ad  bc)(2)(cx  d )c  dx 2 (cx  d )4 =

2c(ad  bc) (cx  d )3

Example 2 If x = a(Cost + tSint) y =a(Sint-tCost) find

d2y dx 2

Solution y =a(Sint-tCost)

dx  a(S int  tCost  S int) dt = atSint x = a(Cost + tSint)

dx  a(S int  tCost  S int) dt = atcost

dy dy dt atS int  .   tan t dx dt dx atCost d2y d dy d dy dt  ( ) ( ) 2 dx dx dx dt dx dx =

d dt (tan t ). dt dx 2

= sec t.

=

1 at cos t

sec3 t at

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Example 3

If y = acos5x+bsin5x show that

d2y  25 y  0 dx 2

Solution y = acos5x+bsin5x Differentiating with respect to x

dy  5a sin 5 x  5b cos 5 x dx d2y  25a cos 5 x  25b sin 5 x dx 2 = -25(acos5x+bsin5x) = - 25y

d2y  25 y  0 dx 2

Example 5 If y =

( x  1  x 2 )m show that (1+x2)y2+xy1-m2y=0

Solution y = (x  1 x ) Differentiating with respect to x 2 m

dy 2x  m( x  1  x 2 ) m 1.[1  ] dx 2 1  x2 = = =

m( x  1  x 2 ) m 1[ 1  x 2  x] 1  x2 m( x  1  x 2 ) m 1  x2 my 1  x2

Cross multiplying and squaring we get,

dy 2 )  m2 y 2 dx d2y dy (1  x 2 ) 2  x( )  m 2 y  0 dx dx 2  (1  x ) y2  xy1  m 2 y  0 (1  x 2 )(

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Canceling,

2

47

dy we get, dx

d2y dy  x( )  m 2 y  0 2 dx dx 2  (1  x ) y2  xy1  m2 y  0 (1  x 2 )

Example 5 -1

2

2

2

If y = (tan x) show that (1+x )y2+2x(1+x )y1=2 Solution -1 2 y = (tan x) Differentiating with respect to x,

y1  2

2 tan 1 x 1  x2 -1

(1+x )y1 = 2tan x 2

Again differentiating (1+x )y2+y12x =

2 1  x2

2 2

2

= (1+x ) y2+2x(1+x ) y1=2 Example 6

d2y 2 2 If y = sin x prove that sin x = (m cos x-m)y 2 dx m

2

Solution m y = sin x Differentiating with respect to x

dy  m sin m1 x cos x dx d2y  m(m  1)sin m 2 x cos 2 x  m sin m x 2 dx 2

Multiplying both sides by sin x,

 sin 2 x

d2y dx 2

m

2

m

2

= m(m-1)sin xcos x-msin x.sin x 2 2 = m(m-1)ycos x-mysin x 2 2 2 2 = m ycos x-mycos x-mysin x 2 2 2 2 = m ycos x-my(cos x+sin x)

d2y  sin x 2  m2 y cos 2 x  my dx 2

2

= (mcos x-m)y 2.2.1

SELF-ASSESSMENT QUESTION - I

1.

Find

d2y dx 2

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APPLIED MATHEMATICS y

48

x 1 x 1

…………………………………………………………………………………….. …………………………………………………………………………………….. …………………………………………………………………………………….. Ans:

4 ( x  1)3

2. If xy  x  1 show that x y  2 xy  2 y  0 …………………………………………………………………………………….. 2

2

…………………………………………………………………………………….. 2.3

n-th Derivative y

yx

ax

n ax

e

ae

1 (ax  b) 2

b (a  b ) eax cos(bx  c  n tan 1 ) a

1 (ax  b) 2

(1) n (n  1)!a n (ax  b)n  2 (1)n 1 (n  1)!a n (ax  b)n n n a sin(ax+b+ ) 2 n n a cosx(ax+b+ ) 2

2

log(ax+b)

sin(ax+b) coa(ax+b) ax

e sin(bx+c) ax

e cos(bx+c)

Find the nth derivative of

n 2 2

b (a  b ) eax sin(bx  c  n tan 1 ) a n b (a 2  b2 ) 2 eax cos(bx  c  n tan 1 ) a 2

Example 1

n 2 2

2x 1 (2 x  1)(2 x  3)

Solution

2x 1 (2 x  1)(2 x  3)

Let

y

Let

2x 1 A B =  (2 x  1)(2 x  3) (2 x  1) (2 x  3)

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2x+1=A(2x+3)+B(2x-1) Put

x

1 2

 2  A(4) or A =

3 2

2  B(4) or B =

Put x 

1 2

1 2

1/ 2 1/ 2  (2 x  1) (2 x  3) 1 1 (1) n 2n n ! (1) n 2n n ! 2 2  yn =  (2 x  1) n 1 (2 x  3) n 1 1 = (-1)n 2n 1 n ![  ] n 1 (2 x  1) (2 x  3) n 1 y

Example 2

x2  1 Find the nth derivative of (2 x  1)(2 x  1)(2 x  3) Solution

let y 

x2  1 A B C    (2 x  1)(2 x  1)(2 x  3) 2 x  1 2 x  1 2 x  3

2

x +1 = A(2x+1)(2x+3)+B(2x-1)(2x+3)+C(2x-1)(2x+1)

1 ; 2 1 put x = - ; 2 -3 put x = ; 2 put x =

5 5  A(2)(4)  A= 4 32 5 -5  B(2)(2)  B = 4 16 13 13  C (4)(2)  C = 4 32

5 1 5 1 13 1 .  .  32 2 x  1 16 2 x  1 32 2 x  3 5 (1) n (n !2n ) 5 ( 1) n ( n!2 n ) 13 ( 1) n ( n!2 n )  yn    32 (2 x  1) n 1 16 (2 x  1) n 1 32 (2 x  3) n 1 5 1 5 1 13 1 = (-1) n n!2n [ .  .  . ] n 1 n 1 32 (2 x  1) 16 (2 x  1) 32 (2 x  3) n 1 (i.e.) y 

Example 3 If

x a tan 1 ( ) show that yn = (-1)n-1(n-1)!a-n sinn  sinn  where   tan 1 ( ) a x

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Solution

x let y = tan -1 ( ) a 2 a y1  2 x  a2 1 1 1 = [  ] 2i x  ai x  ai 1 1 1 y n  (1) n 1 (n  1)![  ] n 2i ( x  ai ) ( x  ai ) n Put x = rcos  ; a = rsin  1  y n  (1) n 1 (n  1)![r  n (cos   i sin  )  n  r  n (cos   i sin  )  n ] 2i 1 = ( 1) n 1 ( n  1)! r  n 2i sin n 2i = ( 1) n 1 ( n  1)! a  n sin n  sin n ( a = rsin ) Example 4 3

2

Find the nth derivative of sin xcos x

Solution 3 2 y = sin xcos x

(3sin x  sin 3x) 1  cos 2 x [ ] 4 2 1 = [3sin x  sin 3 x  3cos 2 x sin x  sin 3 x cos 2 x] 8 1 3 1 = [3sin x  sin 3 x  (sin 3 x  sin x)  (sin 5 x  sin x)] 8 2 2 1 = [6sin x  2sin 3 x  3sin 3 x  3sin x  sin 5 x  sin x] 16 1 = [2sin x  sin 3 x  sin 5 x] 16 1 n n n y n  [2sin( x  )  3n.sin(3x  )  5n sin(5 x  )] 16 2 2 2 =

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51

Example 5 5

4

Find the nth derivative of sin xcos x Solution 5

4

Let y = sin xcos x

let z = cosx+isinx

1 5 1 4 Consider (z- ) (z+ ) z z

1 1 1 1 (z- )5 (z+ ) 4 = (z 2 - 2 ) 4 (z- ) z z z z 4 1 1  8 )(z- ) 4 z z z 4 1 6 4 1 = z9  4 z5  6 z  3  7  z 7  4 z3   5  9 z z z z z 1 1 1 1 1 = (z 9  9 )  (z 7  7 )  4(z 5  5 )  4(z 3  3 )  6(z  ) z z z z z = (z 8  4z 4  6 

5

4

(2isinx) (2cosx) = 2isin9x-2isin7x-4(2isin5x)+4(2isin3x)+6(2isinx) Canceling 2isinx, 5 4 Let y = sin xcos x

1 [sin 9 x  sin 7 x  4sin 5 x  4sin 3 x  6sin x] 28 1 n n n y n  8 [9n sin(9 x  )  7 n sin(7 x  )  4.5n sin(5 x  ) 2 2 2 2 n  n +4.3n sin(3 x  )  6sin( x  )] 2 2

=

2.3.1 1. If

SELF ASSESSMENT QUESTIN – II

y  sin 1 x Prove that (1  x2 ) y2  xy1  0

…………………………………………………………………………………….. …………………………………………………………………………………….. ……………………………………………………………………………………..

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2.4

52

LEIBNITZ THEOREM

If u and v are functions of x and n is a positive integer then n

D (uv) = unv+nc1un-1v1+ nc2un-2v2+…+ ncrun-rvr+…+uvn n

Where D (uv) standards for the nth derivative of uv. Proof: Let us prove this result by induction. When n = 1, D(uv) = u1v+uv1 which is the product rule for the differentiating of the product function uv. Therefore the result in true for n = 1. Let us now assume the result to be true for n = m where m is a fixed integer. m (i.e.) D (uv) = umv+mc1um-1v1+ mc2um-2v2+…+ mcrum-rvr+…+uvm …..(1) let us prove that the result in true for n = m+1 i.e. to prove, D

m+1

(uv) = um+1v+

m+1

C1umv1+

m+1

C2um-1v2+…+

m+1

Crum-r+1vr+…+uvm+1

To prove this let us differentiate (1) with respect to x. D

m+1

(uv) = (um+1v+umv1)+mc1(umv1+um-1v2)+mc2(um-1v2+um-2v3)+….+ m Cr(um-r+1vr+um-rvr+1)+….+ (u1vm+uvm+1) …..(2) m

m

m

m

m

m

= um+1v+( C0+ C1)umv1+( C1+ C2)um-1v2+( C2+ C3)um-3v2+….+ m m ( Cr-1+ Cr)um-r+1vr+…+uvm+1 m+1

= um+1v+

C1umv1+

m+2

C1um-1v2+…+

m+1

(

Crum-r+1vr+…+uvm+1 m

C r-1 +

m

C r=

m+1

Cr )

Which is (2). Therefore the result is true for n = m+1 Since the result is true for n=1, it is true for n=2, n=3 and so on. Therefore the result is true for all positive integer values of n.

Example 1 2 5x

Find the nth derivative of x e

Solution 5x Let us choose u = e

and v = x

2

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53

In 1 1 1 1  I1     ...  n! 2 3 4 n I 1 1 1 1  n  I1     ...  n! 2 3 4 n 1 1 1 1 Then I n  n ![I1     ...  ] 2 3 4 n d But I1  ( x log x)  log x  1 dx 1 1 1  I n  n ![log x  1    ...  ] 2 3 n n 5x 2

= 5 e x +nc15

n-15x

n-2 5x

.2x+nc25 e .2

 e5 x .5n2 [52 x 2  n  5  2 x  5x

n-2

n(n  1) .2] 2

2

= e .5 [25x +10nx+n(n-1)] Example 2 x

Find the nth derivative of e logx Solution Let us choose u = logx and v = e

x

dny  un v  nc1un1v1  nc2un2v2  ....  uvn dx n x

If u = e then un = e

x

If v = logx then

vn 

(1)n 1 (n  1)! xn

1 1 2! (1) n1 (n  1)!  yn  ex .log x  nc1e x .  nc2e x .( 2 )  nc3e x . 3  ...  e x x x x xn 1 1 2! (1) n1 (n  1)!  ex [log x  nc1.  nc2 2  nc3 . 3  ...  ] x x x xn Example 3

If In 

dn n 1 1 1 [ x log x] prove that I n  n![log x  1    ...  ] n dx 2 3 n

Solution

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54

dn n I n  n ( x log x) dx d n 1 d = n 1 . ( x n log x) dx dx n 1 d 1 = n 1 [ x n .  n.x n 1 log x] dx x n 1 d d n 1 = n 1 ( x n 1 )  n n 1 ( x n 1 log x) dx dx = (n-1)!+nIn-1 Dividing by n!.

In In-1 1 = n ! (n  1)! n

Changing n into n-1, we get

I n-1 I n-2 1 = (n  1)! (n  2)! n-1 I n-3 I n-2 1 similarly, = (n  2)! (n  3)! n-2 . . I2 I 1  1 2! 1! 2 Adding all these we get,

In 1 1 1 1  I1     ...  n! 2 3 4 n In 1 1 1 1   I1     ...  n! 2 3 4 n 1 1 1 1 I n  n ![I1     ...  ] 2 3 4 n d But I1  ( x log x)  log x  1 dx 1 1 1  I n  n ![log x  1    ...  ] 2 3 n Example 4 -1

If cos (

x y 2 2 ) = nlog( ), prove that x yn+2+(2n+1)xyn+1+2n yn=0 b n

Solution -1

cos (

y )=n(logx-logn) n

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55

Differentiating with respect to x,

1

.

y2 b2

1

1 dy n  b dx x

1

dy n  b  y dx x

2

2

Squaring and cross multiplying

x2 (

dy 2 )  n 2 (b 2  a 2 ) dx

Differentiating with respect to x,

dy d 2 y dy dy )  ( )2 .2 x  n 2 (2 y ) 2 dx dx dx dx dy 2 2 Canceling 2 ,x y2+xy1+ny =0 dx x 2 (2

Applying leibnitz theorem to differentiate this further n times, 2 n n n 2 i.e.,[yn+2x + C1yn+1.2x+ C2yn.2]+[yn+1.x+ C1yn.1]+n yn=0 2 2 2 i.e., x yn+2+2nxyn+1+n yn-nyn+xyn+1+nyn+n yn=0 2 2 i.e., x yn+2+(2n+1)xyn+1+2n yn=0

Example 5

If

1 m

y y

1 m

 2 x prove that (x2+1)yn+2+(2n+1)xyn+1+(n2-m2)yn=0

Solution 1

ym  y

1 m

 2x

Differentiating with respect to x, 1

1

1 m 1 1  1 y y1  y m y1  2 m m 1 1  y1 m m i.e., (y  y )  2 my Squaring and cross multiplying 1

y12 ( y m  y

1

y12 [( y m  y 2

1 m 2

)  4m 2 y 2

1 m 2

)  4]m 2 y 2

2

2 2

Y1 (4x -4) = 4m y 2

2

2 2

(x -1)y1 = m y

Differentiating with respect to x, 2

2

2

(x -1)2y1y2+y1 (2x)=m .2yy1

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Canceling 2y1, 2

2

(x -1)y2+xy1-m y=0 Applying Leibnitz theorem to differentiate this further n times, we get

[( x 2  1) y2 ]n  [ xy1 ]n  [m2 y]n  0 2

2

Yn+2(x -1)+nc1yn+1.2x+nc2yn.2+(yn+1.x+nc1yn.1)-m yn=0 2

2

2

(x -1)yn+2+2nxyn+1+n yn-nyn+xyn+1+nyn-m yn=0 2

2

2

i.e., (x +1)yn+2+(2n+1)xyn+1+(n -m )yn=0 2.4.1

SELF ASSESSMENT QUESTION - III

1.

If

y  (1  x)2e

 ax

P.T .

(1  x) yn1  (n  2 x) yn  n yn1  0

……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… 2.5

Euler’s Theorem on homogeneous functions If ‘f’ is a homogeneous function of degree ‘n’ in x and y then

x

f f y  nf x y

Proof Since f is a homogeneous function of degree ‘n’ in x and y we can write ‘f’ as

y f  xn F ( ) x

….(1)

Differentiating partially w.r.t x,

f y y y  x n F 1 ( )( 2 )  nx n 1F ( ) x x x x f y y x   x n 1F 1 ( )  nx n F ( ) x x x

….(2)

Differentiating (1) partially w.r.t y

f y 1  x n F ( )( ) y x x f y y.  yx n 1F 1 ( ) y x

.....(3)

f f y n +y =nx F( )  x y x f f i.e., x +y =nf  x y x

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57

Note: The generalized form of Euler’s theorem for homogeneous function of degree ‘n’ in k variables x1,x2,…,xk is

x1

f f f f  x2  x3  ....  xk  nf x1 x2 x3 xk

Example 1 If u = (x-y)(y-z)(z-x) show that

u u u   0 x y z

Solution u = (x-y)(y-z)(z-x)

u =(y-z)(z-x)-(x-y)(y-z) x u =(z-x)(x-y)-(z-x)(y-z) y u =(x-y)(y-z)-(x-y)(z-x) z Adding

u u u   0 x y z

Example 2 x

If z = e (x cosy – y siny), show that

2 z 2 z  0 x 2 y 2

Solution x

z = e (x cosy – y siny)

z x x =e (x cosy – y siny)+e .cosy x 2z x x = e (x cosy – y siny)+2e cosy x 2 x

= e (x cosy – y siny +2 cosy)

z x = e (-x siny –y cosy –siny) y

….(1)

2z x =e (-x cosy + y siny –cosy – cosy) y 2 x

= e (-x cosy -2 cosy + y siny) Adding (1) and (2)

2 z 2 z  0 x 2 y 2 Example 3

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APPLIED MATHEMATICS

If

1 x2  y 2  z 2

58

2 f 2 f 2 f   0 x 2 y 2 z 2

show that

Solution

f  ( x2  y 2  z 2 )

1 2

 f 1   ( x 2  y 2  z 2 ) 2 .2 x x 2 3

 - x( x2  y 2  z 2 )

3 2

  2 f 3  x. ( x 2  y 2  z 2 ) 2 .2 x  ( x 2  y 2  z 2 ) 2 2 x 2 5

 3x ( x  y  z ) 2

2

2

2

5 2

3

 (x  y  z ) 2

 f  3y 2 ( x 2  y 2  z ) 2 y

5  2 2

2

Similarly

2

2

3 2

 ( x2  y 2  z 2 )

3 2

  2 f  3z 2 ( x 2  y 2  z 2 ) 2  ( x 2  y 2  z 2 ) 2 2 z 5 3 2    f 2 f 2 f 2 2 2 2 2 2 2 2 2 2 Adding = 3(x  y  z )( x  y  z )  3( x  y  z ) 2   x 2 y 2 z 2 5

= 3( x =0

2

3

y z ) 2

2

3 2

 3( x  y  z ) 2

2

2

3 2

Example 4

If f =

x 2  y 2  z 2 prove that

2 f 2 f 2 f 2 =   x 2 y 2 z 2 f

Solution 1 2 2

f  (x  y  z ) 2

2

 f 1 2  ( x  y 2  z 2 ) 2 .2 x x 2 1

 x( x2  y 2  z 2 )

1 2

  2 f 1   x( x 2  y 2  z 2 ) 2 .2 x  ( x 2  y 2  z 2 ) 2 2 x 2 3 1   2 f 2 2 2 2 2 2 2 2  - x (x  y  z )  (x  y  z ) 2 x 2 3 1   2 f 2 2 2 2 2 2 2 2 2 Similarly  y ( x  y  z )  ( x  y  z ) y 2 3

2 f z 2

 - z2 ( x2  y 2  z 2 )

1

3 2

 ( x2  y 2  z 2 )

1 2

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59

  2 f 2 f 2 f  2  2 = - (x 2  y 2  z 2 )( x 2  y 2  z 2 ) 2  3( x 2  y 2  z 2 ) 2 2 x y z 3

=

 ( x2  y 2  z 2 )

= 2 (x

2

1 2

 3( x 2  y 2  z 2 )

1 2

y z ) = 2

2

1

1 2

2 f

Example 5 2

2

2

If u = x (y-z) + y (z-x) +z (x-y) show that

u u u   0 x y z

Solution 2

2

2

u = x (y-z) + y (z-x) +z (x-y)

u 2 2 = 2x(y-z) – y +z x u 2 2 = x –z +2y(z-x) y u 2 2 = 2x(x-y) –x +y z

u u u = 2[x(y-z)+y(z-x)+z(x-y)]=0   x y z

Example 6 3/2

If z = tan(y+ax) + (y-ax) , show that

2 2 z 2  z  a x 2 y 2

Solution 3/2 z = tan(y+ax) + (y-ax)

z 3  sec2 ( y  ax).a  ( y  ax) 2 (a) x 2 1 2   z 3 2 2 2 2  2 sec ( y  ax ) tan( y  ax ). a  ( y  ax ) a 2 x 4 1  3  a 2 [2 sec2 ( y  ax) tan( y  ax)  ( y  ax) 2 4 1 z 3  sec2 ( y  ax).a  ( y  ax) 2 y 2 1

 2z 3  2 sec2 ( y  ax) tan( y  ax)  ( y  ax) 2 2 4 y

.....(1)]

1

From (1) and (2),

.....(2)

2 2 z 2  z  a x 2 y 2

Example 7 If u = log(tanx + tany + tanz) prove that

sin 2 x

u u u  sin 2 y  sin 2 z 2 x y z

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Solution u = log(tanx + tany + tanz)

u 1  .sec2 x x tan x  tan y  tan z

u sin 2 x.sec 2 x sin 2 x  x tan x  tan y  tan z u 2 tan x sin 2 x  x tan x  tan y  tan z u 2 tan y Similarly, sin 2 y  y tan x  tan y  tan z u 2 tan z sin 2 z  z tan x  tan y  tan z u u u 2(tan x  tan y  tan z ) Adding sin 2 x = =2  sin 2 y  sin 2 z tan x  tan y  tan z x y z Example 8 Verify Euler’s theorem for the following functions (i)

y x

1

(ii)

3

x y 2

3

2

2

(iii) u = x +y +3x y-3xy

2

Solution (i)

let f =

y x

This is a homogeneous function of degree 0. We have to verify that x

f f +y =0  x y

f y  2 x x f 1  y x f f Adding x +y =0  x y

f y  x x f y y  y x x

Therefore Euler’s theorem is verified. (ii)

let f =

1 x y 2

2

2

2 -1/2

= (x +y )

This is a homogeneous function of degree -1. We have to verify that x

f f +y = -f.  x y

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 f 1   ( x 2  y 2 ) 2 .2 x x 2 3  f x   x2 ( x2  y 2 ) 2 x 3  f 1   ( x 2  y 2 ) 2 .2 y y 2 3

 f   y2 ( x2  y2 ) 2 y 3

y

 f f Adding , x  y  ( x 2  y 2 )( x 2  y 2 ) 2 x y 3

 - (x  y ) 2

2

1 2

f

Therefore Euler’s theorem is verified. 3

3

2

2

(iii) u = x +y +3x y-3xy

u 2 2 =3x +6xy+3y x

u 2 2 =3y +3x +6xy y x

u u y  x(3x 2  6 xy  3 y 2 )  y(3 y 2  3x 2  6 xy ) x y 3

2

3

3

2

3

2

= 3x +6x y+3xy +3y +6xy +3x 2

2

2

= 3x +3y +9x y+9xy 3

3

2

2

= 3(x +y +3x y+3xy ) = 3u Therefore Euler’s theorem is verified.

Example 9 Verify Euler’s theorem in the following cases:

(i) u = sin

x2  y 2 xy

3

(ii) u = x cos(

y ) x

(iii) u =

x y x y

Solution (i)

Let u = sin -1

Sin u =

x2  y 2 xy

x2  y 2 =f xy

This is a homogeneous function of degree 0.

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62

f f +y =0.  x y

We have to verify that x

Differentiating with respect to x,

f xy.2 x  ( x 2  y 2 ). y x 2 y  y y ( x 2  y 2 )   2 2  dx ( xy ) 2 x y x2 y2  x

( x2  y 2 ) x2 y

f x 2  y 2  dx x y

f y 2  x2  dy xy f f Therefore x +y =0  x y Similarly

y

Therefore Euler’s theorem is verified. 3

(ii)

u = x cos(

y ) x

u y y y  3x 2 cos( )  x 3 sin( ).( 2 ) x x x x u y y x  3x 3 cos( )  x 2 y sin( ) x x x u y 1   x 3 sin( ). y x x u y y   x 2 y sin( ) y x

....(1)

....(2)

u u y 3 +y =3 x cos( ) =3u. x y x

Therefore Euler’s theorem is verified. 2.5.1

SELF ASSESSMENT QUESTION – 5

1.

u=

u u x y Verify Euler’s theorem x +y =0 x y x y

……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………

2. If

x 2 f 2 f f  a tan   verify that  xy yx  y 1

……………………………………………………………………………………………… ………………………………………………………………………………………………

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………………………………………………………………………………………………

2.6

The direction of the curve at a given point on the curve is determined by the tangent at that point. The tangent line rotates as the point moves along the curve. The curvature of the curve at a given point is defined as the rate of change of bending of the curve at the point. 2.6.1

Cartesian Formula for Radius of curvature Let P be any point on the curve y = f(x) and PT be the tangent making an angle

 with x-axis. Then we know that

dy  tan  dx

Differentiating with respect to x, y

x

d2y d  sec 2  2 dx dx d  ds = sec 2  ds dx 2 ds sec  ds   2 . d y dx d dx 2 (1+tan 2 Ψ)secΨ = d2 y dx 2

[Since

dx  cos Ψ] ds

(1+tan 2 Ψ) 1+tan 2Ψ = d2 y dx 2

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=

(1+tan Ψ) d2 y dx 2 2

3

64

[1+(

2

=

(1+y12 ) i.e.,  = y2

3

2

dy 2 3 2 ) ] dx d2 y dx 2

Where y1 

dy d2 y and y2  2 dx dx

Note: This formula does not hold good where the tangent at the point(x,y) is parallel to y axis. In that case

dy is not defined. Since the value of  is independent of the choice of dx

axis of coordinates, in this case we take the formula for

=

2.6.2

[1+(

as

dy 2 3 2 ) ] dx d2 y dx 2

Parametric formula for Radius of Curvature Let the parametric equations of the curve be x = f(t) and y = (t).

dy dy y1 Then = dt  1 dx dx x dt d 2 y d dy Also  ( ) dx 2 dx dx d y1 = ( ) dx x1 d y1 dt = ( ) dt x1 dx x1 y11  y1 x11 1 = . 1 2 x x1 1 11 1 11 x y yx = 3 x1

(1+y12 ) = y2

3

2

2

y1

3

x1  [1  1 2 ] . 1 11 1 11 x y yx x 3

2

3

( x1  y1 ) 2 = 1 11 1 11 x y yx 2

2

Example 1 Find the radius of curvature at x = y =

3a 3 3 to the curve x +y = 3axy. 2

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APPLIED MATHEMATICS Solution

3

65

3

…(1)

x +y = 3axy Differentiating with respect to x 2

2

3x +3y

dy dy dx = 3a( x dx  y )

dy 2 ( y  ax)  ay  x 2 dx dy ay  x 2  dx y 2  ax

...(2)

3a 2 9a 2  dy 3a 3a 4  1 ( , )  22 9a 3a 2 dx 2 2  4 2 Also 2

d y  dx 2

( y 2  ax)(a

dy dy  2 x)  (ay  x 2 )(2 y  a) dx dx 2 2 ( y  ax)

9a 2 3a 2 3a 2 9a 2  )(a  3a)  (  )(3a  a) d y 3a 3a 4 2 2 4 ( , )  9a 2 3a 2 2 dx 2 2 2 (  ) 4 2 (-4a - 4a) 4 32   8a  2   2 3a 3a 3a ( ) 4 (

2

2 32

(1  y1 )  y2  

3

(1  1) 2 6 2a 3 2a  3a     32 32 16

3 2a 16

Example 2 2

3

3

Find the radius of curvature of the curve xy =a -x at the point (a,0). Solution 2 3 3 The equation of the curve is xy = a -x . Differentiating with respect to x,

dy 2 2 +y = -3x dx dy 2 2 2xy = -y -3x dx 2xy

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dy  y 2  3 x 2  dx 2 xy dy (a,0)   dx dx  2 xy let us consider  2 dy y  3x 2 dx (a,0)  0 dy dy dx ( y 2  3x 2 )(2 x  2 y )  2 xy (2 y  6 x ) 2 d x dx dy  2 2 2 2 dy ( y  3x ) d 2x 3a 2 (2a  0)  0 2a 2 (a,0)    2  2 2 2 dy (3a ) 3a 3a dx 3 3 [1  ( ) 2 ] 2 (1  0) 2 3a dy    2 2 d x 2 2 3a dy

 

3a 2

Example 3 3

3

Prove that the radius of a curvature at the point (acos θ, asin θ) on the curve x 2/3 a is 3asin θcos θ

2/3

2/3

+y

=

Solution The parametric equations of the curve are 3 x = acos θ,

3

y = asin θ

dx dy 2 2 =-3acos θsin θ; =3asin θ cos θ d d dy 3a sin 2  cos     tan dx 3a cos 2  sin  d2y d  sec 2  1 2   sec    2 2 4 dx dx  3a cos  sin  3a cos  sin  2 32

(1  y1 )  y2

3

 (1  tan 2  ) 2 .3a cos 4  sin 

 sec3  .3a cos 4  sin   3a sin  cos  Example 4 2

Show that the radius of the curvature at(a,0) on the curve y =

a 2 (a  x) a is x 2

Solution 2

y =

a 2 (a  x) x

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Differentiating with respect to x,

dy x(a)  a 2 (a  x)1  a 2 x  a 3  a 2 x a3     dx x2 x2 x2 dy a3  dx 2 yx 2 dy ( )( a ,0)   dx 2y

Consider

dx 2 yx 2  3 ; dy a

(

dy )( a , 0)  0 dx

d 2x 2 dx   3 [ x 2  2 xy ] 2 dy a dy (

d 2x 2 2 )   3 (a 2  0)   2 ( a ,0) dy a a

dx 2 3 2 3 ) ] (1  0) 2 a dy   2  2 d x 2 2 a dy

[1  (



 

a 2

Example 5 Find the radius of the curvature at the point ‘θ’ on the curve x = a(cos θ+ θsin θ),y = a(sin θθcos θ). Solution x = a(cos θ+ θsin θ)

dx = a(-sin θ+sin θ+ θcos θ) = a θcos θ d y = a(sin θ- θcos θ)

dy = a(cos θ-cos θ+ θsin θ)=a θsin θ d dy dy d a sin     tan dx d dx a cos  d2y d sec2  1 2  sec    2 dx dx a cos  a cos 3  2 32

(1  y1 )  y2

3

 (1  tan 2  ) 2 .a cos 3   sec3  .a cos 3 

 a

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APPLIED MATHEMATICS 2.6.4

1. If

68

SELF ASSESSMENT – VI

 x2  y 2  u u 1 U  tan 1   sin 2u  show that x  y x y 2  x y 

…………………………………………………………………………………………….. …………………………………………………………………………………………….. ……………………………………………………………………………………………..

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69

UNIT – III 3.0

Introduction

3.1

Objectives

3.2

Integration by Parts 3.2.1

3.3

Definite Integral 3.3.1 3.3.2 3.3.3

3.4

Definition of Definite Integral Properties of Definite Integral Self Assessment Question – I

Bernoulli’s Formula 3.4.1 3.4.1

3.5

SELF-ASSESSMENT QUESTION - I

Reduction Formula Self Assessment Question – II

Particular Integral 3.5.1

Self Assessment Question – III

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3.0

70

Introduction In this chapter, we have to deal with Integration for Higher grade level and

understanding the formulae of the integration, Leibnitz Formulae and applied in to the problem. Student study for the particular integrals and all the types of the integration. 3.1

Objectives To develop the skill in the Integration To create new idea to application oriented sum in the particular integration. To understanding the way of approach for the integration in applied sciences.

3.2

Integration by Parts

If u and v are functions of x we know that

dv du d (uv) = u v dx dx dx

Integrating both sides with respect to x, we get

d

du

du

 dx (uv) dx   u dx dx   v dx dx i.e.

 d(uv)   u dv   v du

uv = 

 u dv   v du

 u dv  uv   v du

This is called the by parts rule.

Note: By – parts is applied when the integral is a product of two functions. The evaluation of this integral depends upon the proper choice of u and v. u is chosen such that  v du is easily integrated. Example 1 Integrate the following with respect to x. (i) x e

x

n

(ii) x log x

(iii) x sin 2x

2

(iv) x sin x.

Solution

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APPLIED MATHEMATICS (i)  x e dx x

71 x

Take u = x and dv = e dx. Then du= dx; V=e

x

  x e dx = uv -  v du x

= x e -  e dx x

x

x

x

= x e –e + c x

= e (x-1) + c (ii)  x log x dx n

n

Take u = log x and dv = x dx du =

1 x n 1 dx; v  x n 1

  x log x dx = uv -  v du n

x n 1 x n 1 1  dx n 1 n 1 x x n 1 1 = logx  x n dx  n 1 n 1 n 1 x x n 1 = logx  c n 1 (n  1) 2 x n 1  1  =  log x  c n 1  n 1 = log x

(iii)  x sin 2x dx = Take u = x; dv = sin 2x dx du = dx ;

v=

 cos 2x x

  x sin 2x dx = uv -  v du

 cos2x   cos2x  dx   2  2 

=x

=

(iv)  x sin x dx Take u = x; 2

x cos 2x sin 2x  c 2 4 2

dv = sin x dx

du = dx ; v =  sin x dx 2

=

=

1  cos 2 x dx 2

1 sin 2 x  x  2 2 

  x sin x dx = uv -  v du 2

=

1 sin 2 x  1  sin 2 x  x x    x  dx  2 2  2  2 

=

x 2 x sin 2 x x 2 cos 2 x    c 2 4 4 8

x 2 x sin 2 x cos 2 x =   c 4 4 8

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APPLIED MATHEMATICS

72

Example 2 Integrate the following -1

-1

(i) sin x

-1

(ii) tan x

(iii) x tan x

(iv)

x sin -1 x 1- x 2

Solution (i)  sin x dx -1

-1

Take u = sin x dv = dx

1

du =  sin x dx

dx, v  x

1- x 2

= uv -  v du

-1

x

= x sin x-  -1

dx

1 x2

 -2x dx = dt

2

Put 1 - x = t

  sin x dx = x sin x + -1

-1

d t/2 t

1

1 t 2 = x sin x + c 2 1/2 -1

1 x2  c

-1

= x sin x + (ii)

 tan x dx -1

-1

Take u =tan x; dv = dx

du    tan x dx

1 dx; 1 x2

= uv - v du

-1

-1

= (tan x) x-

= (iii)

 x tan

vx

-1

x dx 1 x 2

1 x tan 1 x  log(1  x 2 )  c 2

x dx

Take u  tan-1 x; dv  x dx du

1 x2 dx ; v  1 x2 2

  x tan -1 x dx  uv -  v du

x2 2 = tan 1 x   x2 2

1 1 x 2

dx

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APPLIED MATHEMATICS

73

=

x2 1 x2 tan 1x   dx 2 2 1 x2

=

x2 1 x 2  1 1 tan 1x   dx, 2 2 1 x2

=

x 2 tan 1 x 1  1    1  dx 2 2  1 x2 

x 2 tan 1 x 1 =   [x  tan 1x]  c 2 2

(iv)

x sin 1 x dx

1 x2 -1

Take u = sin x ; dv =

du =

1 1 x2

; v

x

dx

1 x2 x 1 x2

dx put u = 1-x

2

du = -2x dx v

3.2.1

=



du / 2 u

x sin 1x dx

=-

1 x2

=

 1  x 2 sin 1 x   1  x 2

=

 1  x 2 sin -1 x  x  c

= uv-  v du

1 x2

1 1 x2

dx

Self-Assessment Question - I  4

1.

Evaluate

 2sin

2

x sin 2 xdx

0

……………………………………………………………………………………………………….… …….. ……………………………………………………………………………………………………….… …….. Ans: ¼

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APPLIED MATHEMATICS

74

3.3

Definite Integral

3.3.1

Definition of a Definite Integral Let f be a function defined on a closed integral [a,b]. Let P be a partition of [a, b]

with partition points x0, x1,…xn where ax0,<x1,<x2 <……<xn=b. Choose points xi* in [xi-1, xi]

xi  xi  xi 1 and || p || max{ xi } . Then the definite integral of the form

and let b

f ( x)dx  lim

|| p|| 0

a

n

 f (x

*

i

i 1

)xi , if this limit exists. If this limit does exist then it is called

integrable on the interval [a, b].

The definite integral of f (x) between the limits x=a and x= b is defined by b

 f (x)

dx and its value is F (b) – F (a).

a

Here ‘a’ is called the lower limit and ‘b’ is called the upper limit of the integral, and F (x) is the integral of f (x). The value of the definite integral is obtained by finding out the indefinite integral first and then substituting the upper limit and lower limit for the variable in the indefinite integral.

3.3.2

Properties of Definite Integral

 f (x) dx  F (x)  C.

Let

b

Then

 f (x) dx  F (b)  F (a)  [F (x)]

b a

a

Property 1 b

b

a

a

 f (x) dx   f (t) dt . Proof b

 f (x) dx F(x) 

b a

 F (b)  F (a) .

a b

 f (t) dt F(t)

b a

 F (b)  F (a) .

a

b

b

a

a

 f (x) dx   f (t) dt .

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APPLIED MATHEMATICS

75

Property 2: b

b

a

a

 f (x) dx   f (x) dx . Proof : a

RHS

=

 f (x) dx  -F(x) 

a b

b

 [F(a)  F(b)] = [F(b)  F(a)] =

b

=

 f (x) dx a

Property 3 : b

c

b

a

a

c

c

b

 f(x)dx   f(x) dx   f(x) dx Proof

RHS   f(x) dx   f(x) dx a

c

 f(x)   [F(x)]cb c a

= F(c) - F(a) + F(b) – F(c). = F(b) – F(a). b

  f(x)dx a

Property 4 : a

a

0

0

 f (x) dx  f (a - x) dx . Proof: Put a - x =t dx = -dt When x = 0, t= a; When x= a, t= 0. 0

a

 f (t) dt   f (t) dt

R.H.S = -

a

0 a

=

 f (x) dx by property(1) 0

a

a

0

0

 f (x) dx   f (a - x) dx

Property 5:

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APPLIED MATHEMATICS 2a

a

a

0

0

0

a

2a

76

 f (x) dx   f (x) dx   f (2a - x) dx

Proof 2a

 f (x) dx   f (x) dx   f (x)dx . 0

0

…… (1)

a nd

Put x = 2a – t in the 2 Integral. dx = -dt When x= a, t=a; x=2a, t=0. 

2a

0

0

a

 f (x) dx    f (2a  t) dt a

=

 f (2a - t) dt 0 a

=

 f (2a - x) dx 0

Using this in equation (1) we get 2a

a

a

0

0

0

 f (x) dx   f (x) dx   f (2a - x) dx

Property 6:

i)

2a

2a

0

0

If f (2a-x) = f (x) then

 f (x) dx  2  f (x) dx 2a

ii)

If f (2a-x) = -f (x) then

 f (x) dx =0. 0

i)

by property 5

2a

a

0

0

a

 f (x) dx   f (x) dx   f (2a - x) dx . 0 a

=

a

 f (x) dx   f (x) dx . 0

0 a

=

2 f (x) dx . 0

ii)

If f (2a-x) = -f (x) then 2a

a

a

 f (x) dx   f (x) dx   f (2a - x) dx 0

0

=

0

a

a

0

0

 f (x) dx   f (x) dx.  0 .

Property 7

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APPLIED MATHEMATICS a

a

a

0

77

 f (x) dx  2 f (x) dx if f (x) is an even function.

(i)

a

 f (x) dx =0 if f (x) is odd function.

(ii)

a

Proof: a

a

a

 f (x) dx   f (x) dx   f (x) dx

a

a

0

0

a

 f (-x) dx   f (x) dx

=

a

=

0

a

a

0

0

 f (t) dt   f (x) dx

st

(Putting –x = t in the 1 integral) a

=

a

 f (x) dx   f (x) dx 0

0 a

=

2 f (x) dx 0

Suppose f (x) is an odd function, Then f (x) = - f (-x). a

0

a

 f (x) dx   f (x) dx   f (x) dx

a

a

0

0

a

a

0

 - f (-x) dx   f (x) dx

=

(Putting –x = t in the integral) 0

=

a

 - f (t) dt   f (x) dx . a

=

0 a

a

0

0

  f (x) dx   f (x) dx .

=0 Property 8 b

b

a

a

 f (x) dx   f (a  b - x) dx . Proof: b

RHS

=

 f (a  b - x) dx a

Put a + b – x = t.  dx = -dt. When x = a, t=b, x=b, t=a.

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APPLIED MATHEMATICS a

78

b

 f (t) (-dt)   f (t) dt

RHS=

b

a b

=

 f (x) dx  L.H.S . a

Example : 1 π

i)

2

 sin

Evaluate

2

x dx.

0

Solution π

Let I=

2

 sin

2

x dx.

0 π

=

π   2  x  dx  

2

 sin

2

0 π

=

(by property 4)

2

 cos

2

x dx

(2)

0

2I =

π

2

2

 (sin x  cos ) dx   dx 2

2

0

0

=

x  0  π

I=

π

2

 4

2

Example 2 

 x sin x dx 2

Evaluate

0

Solution 

Let I

=

 x sin

2

x dx

(1)

0 π

=

 (π - x) sin

2

(π - x) dx

2

x dx

0

π

=

 (π - x) sin

(2)

0

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APPLIED MATHEMATICS

79

π

2I

=

 [x sin

x  (π - x)sin 2 x] dx

2

0

π

=

 π sin

2

x dx .

0 π

=

 1 - cos 2x   dx 2 

 π  0

π

π sin2x  = x  2 2  0 2 π π   π 2 2 2 π I= 4

=

Example 3 π

2

Evaluate

0

sinx dx sinx  cosx

Solution π

Let I

=

2

 0

π

=

0

=

π  sin x  x  2 

2

 π

sinx dx . sinx  cosx

2

 0

π  π  sinx   x   cosx  x  2  2  cos x dx sin x  cos x

……

(1)

…..

(2)

dx .

2I

=

2

 0 π

=

sin x  cos x dx sin x  cos x

 dx  x  2

π

2

0

0

I=

π 2

 4

Example 4 π

Evaluate

=

4

 log(1  tan  ) d 0

Solution π

Let I

=

4

 log (1  tan ) d 0

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APPLIED MATHEMATICS π



4

80



 log 1  tan 4    d

=

0

   tan  tan    4 =  log 1   d  0  1  tan tan   4   π

4

π

4

1  tan  

 log 1  1  tan   d

=

0

π

4

1  tan   1  tan    d 1  tan  

 log 

=

0

π

4

2

 log 1  tan   d

=

0

π

 log 2  log(1  tan  )d

=

4

0 π

4

4

 log 2d   log (1  tan  ) d

=

0

0 

4

= log 2 [] 0 -I 2I

log 2 π

=

I 

4

π log 2 8

Example 5 π

Evaluate

x sin x dx 2 x

 1  cos 0

Solution π

Let I

=

x sin x 2 x

 1  cos 0

....

(1)

....

(2)

π

=

(π - x) sin (π - x) dx 1  cos 2 (x - π) 0

 π

=

(π - x) sin dx 2 x 0

 1  cos

2I

=

 0 π

=

 sin x

 1  cos 0

Put cos x = t, When x= 0, When x=,

x sin x  (π - x)sin x dx 1  cos 2 x 2

x

dx

- sin xdx = dt t= cos 0 = 1 t= cos  = -1

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APPLIED MATHEMATICS



-1

(ie) 2I

=

 1 t

81

dt

2

1

1

I

1

dt dt  2π  2 1 t 1 t2 -1 0

=

π

=

2π tan 1t 0  2π tan 11

=

  2π  4

=

2 4

1

Example 6 π

x tan x

 sec x  tan x dx

Evaluate

0

Solution π

Let I

=

x tan x

 sec x  tan x dx

. . . .(1)

0

π

=

(π  x) tan (π  x)

 sec (x - π)  tan (π - x) dx 0

π

=

- (π  x) tan x

 - sec x  tan x dx 0

π

2I

=

(π  x) tan x

 sec x  tan x dx 0

=

π tan x

 sec x  tan π dx 0

x sin x cos x =  dx 1 sin x 0  cos x cos x π π sin x dx =  1  sin x 0 π

π

=

 0

π sin x(1 - sinx) dx 1 - sin 2 x

π

=

π (sin - sin 2 x) 0 cos 2 x dx π

=

π  (secx tan x - tan 2 x) dx 0

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APPLIED MATHEMATICS

82

π

=

π  (sec x tan x - sec 2 x  1) dx 0

πsec x - tan x  x 0 = π sec π - tan x  π)  sec 0  tan 0  0 = π - 1 - 0  π   1  0 

=

=  (-2)

I

ππ  2 = 2

Example 7 1

Evaluate

 x (1 - x)

10

dx

0

Solution 1

 x (1 - x)

10

dx

0 1

=

 (1 - x) 1 - (1 - x) 

10

dx

0 1

=

 (1 - x)x

1

10

dx   (x10  x11) dx

0

0 12 1

 x11 x  1 1       11 12  0 11 12 12  11 1 =  132 132 =

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APPLIED MATHEMATICS

83

Example 8

log (1  x) dx 1 x2 0

1

Evaluate

=

log (1  x) dx 1 x2 0

1

Let I

=

Put x = tan   dx When x = 0,  =0 When x = 1,  =

= sec  d 2

π 4 π/4

I

=

log (1  tan θ) 2 sec θ d θ 1  tan 2θ 0

π/4

=

log (1  tan θ) 2 sec θ d θ 2 s ec 0

π/4

=

 log (1  tan θ) d θ 0

=

π log (2) see Example 5 2

Example 9 π -β

Show that

θdθ

β

π

 sin θ  π log cot 2, 0  β  2 β

Solution : π -β

I

=

θdθ

 sin θ

.....

(1)

β

π -β

=

β  π -β -θ

 sin (β  π  β  θ) d θ

(using property (6))

β

π -β

=

 β

π -β

=

 β

π-θ sin (π - θ) π-θ dθ sin θ π -β

Adding (1) and (2), 2I =

 β

. . . . . . (2)

π dθ sin θ π β

=

θ  π log tan  2 β 

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APPLIED MATHEMATICS

84

 β π β π log tan     log tan  2  2 2  β β  = π log cot  log cot  2 2  β β = 2π log cot  I  π log cot 2 2 =

2I Example 10

b

b

a

a

 f (x) dx   f (a  b  x) dx

Using property 3

 (x

Evaluate

4

- 20 x 3  150 x 2  500x  625) dx

2

Solution 3

 (x

4

- 20 x 3  150 x 2  500x  625) dx

2 3

=

 (x

4

- 4c1x 3 5  4c 2 x 2 52  4c3 x 3 53  54 ) dx

2 3

=

3

4 4  (x - 5) dx   (5  x) dx 2 3

=

2

 (2  3 - x)

4

dx

2

By the given property 3

 x5  =  x dx     5 2 2 35 25 211 =   5 5 5 3

4

Example 11  /2

Show that

 log(tan x  cot x)dx   log 2 0

Solution  /2

 log(tan x  cot x)dx

LHS 

0

 /2

 sin x

cos x 

 log  cos x  sin x  dx

0

 /2

1

 log  sin x cos x  dx 0

 /2

 /2

0

0

   log sin x dx    log cos x dx.

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APPLIED MATHEMATICS 

 2

log 2 

 2

85

log 2   log 2.

Example 12  /4

 (log sin x  cos x)dx   4 log 2

Show that

 4

Solution  /4

 log(sin x  cos x)dx

 / 4

 /4

1  1  log 2  sin x  cos x  dx 2  2  /4

 

 /4

  log 2 sin  x  dx 4  /4

 

 /4

 log 2 dx 

 / 4

 /4

 /4

  log sin   x dx 4  /4

 

 /4

1    log 2  dx   log sin  x   dx 2 4   / 4  / 4

Put

4

xt

dx= dt When

x

When

x

 4

4

, t  0. ,t 

 2

1     RHS  log 2.    2 4 4 

 /2

 log sin t dt 0

log 2.  log 2. 4 2   log 2. 4

Example 13 a

Prove that

 x f ( x)dx  0

a

a x f ( x)dx. If f(a-x)  f(x) 2 0

Solution a

a

LHS   xf ( x)dx   (a  x) f (a  x)dx 0

0 a

a

 af (a  x)dx  xf (a  x)dx 0

(ie)

0

a

a

a

0

0

0

 xf ( x)dx  af ( x)dx   xf ( x)dx [ f (a  x)  f ( x)]

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APPLIED MATHEMATICS a

86

a

 2 xf ( x)dx  a  f ( x)dx 0

0

a

(ie)

  xf ( x)dx  0

a

a f ( x)dx 2 0

Example 14 b

b

a

a

If f (a  b  x)  f ( x) prove that 2 xf ( x)dx  f (a  b)  f ( x)dx Solution Put a + b – x = t dx = - dt When x= a, t=b. When x=b, t=a. b

b

  xf ( x)dx    (a  b  t ) f (a  b  t )dt a

a b

  (a  b  t ) f (a  b  t )dt a b

b

a

a

 (a  b)  (a  b  t )dt   tf (a  b  t )dt b

b

a

a

 (a  b)  (a  b  x)dx   xf (a  b  x)dx b

b

a

a

 (a  b)  f ( x)dx   xf ( x)dx

 f (a  b  x)  f ( x) b

b

a

a

 2 xf ( x)dx  (a  b)  f ( x)dx Example 15 2a

If f (a  x)   f (a  x)prove that  f ( x)dx  0 0

Solution 2a

a

2a

0

0

0

 f ( x)dx  f ( x)dx   f ( x)dx

....(1)

nd

Consider the 2 integral Put t=x-a. dt=dx. When x=a, t=0. x=2a, t= a.

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APPLIED MATHEMATICS 2a

a

0

0

87

  f ( x)dx   f (a  t )dt a

   f (a  t )dt 0

 f (a  x)   f (a  x) (Property 4) a

   f (t )dt 0 a

   f ( x)dx 0 2a

a

a

0

0

0

 From (1)  f ( x)dx   f ( x)dx   f ( x)dx 0 Example 16 1

Evaluate

 cot

1

(1  x  x 2 )dx

0

Solution 1

1

1 2 1  cot (1  x  x )dx   tan 0

0 1

  tan 1 0 1

  tan 1 0

1 dx 1 x  x2

1 dx 1  x(1  x) x 1 x dx 1  x(1  x)

(property 4) 1

1

  tan xdx   tan 1 xdx 1

0

0

(using property 5) 1

 2 tan 1 xdx 0

x  2x tan x   1 x 1

1

1

0

2

dx

0

1

1    2 x tan 1 x  log(1  x 2 ) 2  0  1   2  log 2 4 2   3.3.3

2

 log 2

Self Assessment Question – I

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APPLIED MATHEMATICS

1. Evaluate

88

dx 4  1)

 x( x

……………………………………………………………………………………………………….… …….. ……………………………………………………………………………………………………….… ……..

Ans:

 x4  1 log  4   c 4  x 1 

2. Evaluate

1 x dx 1 x

……………………………………………………………………………………………………….… …….. ……………………………………………………………………………………………………….… …….. Ans:

3.4

Sin1 x  1  x 2  c Bernoulli’s formula Consider the integral a

x

3

sin 2 xdx . We can evaluate this integral by

repeatedly applying ‘by parts’ rule. Let

u=x

3

dv = sin 2x dx

du  3x dx 3

v

cos 2 x 2

 I   x3 sin 2 xdx =

uv   vdu

cos 2 x cos 2 x 2  .3x dx 2 2 cos 2 x 2 Again take u = 3x dv   dx 2 sin 2 x du = 6xdx v 4   x3

Again applying ‘by parts’ rule we get

 cos 2 x   2  sin 2 x   sin 2 x  I  x3    3x      6 xdx 4    4   2    sin 2 x   sin 2 x   cos 2 x  2   x3    3x      6 xdx 4    4   2  

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APPLIED MATHEMATICS

89 dv  

Again take u = 6x

v

du = 6dx

sin 2 x dx 4

cos 2 x 8

sin2x  cos2x  cos2x   cos2x  2 I  x3   .6dx   3x     6 x  2  4  8    8  sin 2 x   cos 2 x   cos 2 x   sin 2 x  2  x3     3x     6 x   6  2  4     8   16 

..........(1)

If u’, u”, u’“, etc denote first, second, third derivatives of the function u and v 1, v2, v3 etc are the successive integrals of functions of the function v then form (1), we arrive at a formula for integration of u dv. (i.e.)

 udv  uv  u' v  u"v 1

2

 u' " v3  .......

This is called Bernoulli’s formula for integration.

Example 1 ax

Evaluate

3  e .x dx

Solution Here

u  x3 u '  3x 2 u" 6 x u'"  6

dv  eaxdx e ax v a e ax v1  2 a e ax v2  3 a e ax v3  4 a

  e ax.x3dx  uv  u' v1  u" v2  u' " v3

 x3

 e ax   e ax   e ax  e ax  3x 2  2   6 x 3   6 4   4 a a  a  a 

 x 2 3x 2 6 x 6  e   2  3  4 c a a  a a e ax  4 a 3 x 3  3a 2 x 2  6ax  6  c a ax

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APPLIED MATHEMATICS

90

Example 2 3

Evaluate

x  x e dx

Solution 3

x  x e dx

u  x3 u '  3x 2 u" 6 x

dv  e x v  ex v1  e x

u'"  6

v2  e x v3  e x

  x3e x dx  x3e x  3x 2e x  6 xe x  6e x  c

 e3 x 3  3x 2  6 x  6  c Example 3 2

Evaluate

2 x  x e dx

Solution 2

2 x  x e dx

 e 2 x   e 2 x   e 2 x   x2   2 x     2 c  2   4   8   x 1 2 x  x  e      c  2 2 4 

e 2 x 2x  2x 1  c 4

3.4.1 Reduction formula Example 1 1. Reduction formula for Let

x

m

(log x) n dx

I m,n   x m (log x) n dx

u  (log x) n and dv  x m dx

Let

du  n

(log x) n 1 x m 1 dx and v  x m 1

Then applying by parts Rule,

I m,n

m 1   x m1 (log x) n1 n x  (log x) .n dx  m  1  m  1 x 

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APPLIED MATHEMATICS

91

x m1 n (logx) n  x m (logx) n 1 dx  m 1 m 1 m 1 x n (i.e) I m,n  (log x) n  I m,n 1 m 1 m 1 

Example 2 Reduction formula for

I n   x n e axdx

Solution Let

u  xn

dv  eaxdx e ax v a

du  nx n1dx

x n e ax e ax n1  .nx dx a a x n e ax n n 1 ax    n e dx a a n ax xe n  In   I n 1 a a

In 

Example 3 Reduction formula for

 sin

n

xdx

Solution Let

I n   sin n xdx

  sin n1 x sin xdx

u  sin n1 x du = (n-1) sin

n-2

x cosx dx

dv = sinx v=-cos x

 I n  sin n 1x( cosx)   (n  1)sin n 2 xcosx.( cosx)dx =

 sin n 1xcosx  (n  1)  sin n 2 xcos 2 x.dx

 sin n 1xcosx  (n  1) sin n 2 x(1  sin 2 x)dx  sin n 1xcosx  (n  1) sin n 2 xdx  (n  1) sin n x dx

 sin n 1xcosx  (n  1)In 2  (n  1)In  (1  n  1)In  sin n 1xcosx  (n  1)In 2

sin n 1xcosx n  1 In    I n 2 n n

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APPLIED MATHEMATICS

92

Example 5 π 2

 sin

Reduction formula for

n

x dx

0

Solution π2

Let

I n   sin n x dx 0 π 2

=

 sin

n -1

x (sin x) dx

0

Let du In

n-1

= u=sin x n-2 = (n-1)sin x cos x dx =

v=sinx dx dv= -cos x

π

π

2

I n   cosxsin n 1x 0 2   cosx(n  1)sin n 2 x cosx dx 0 π

2

 0  (n  1)  sin n 2 xcos 2 x dx 0 π

π2

=

2

(n  1)  sin n 2 xdx  (n  1)  sin n x dx 0

0

 I n  (n  1)In 2  (n  1)In (1  n  1)In  (n  1)In 2 (n  1) In  I n 2 n Evaluation of In Case 1: let n be an even integer

n -1 I n2 n (n  1) n - 3  . I n 4 n n-2 In 

Proceeding in this way

In 

(n  1) n - 3 n  5 1 . . ........ .I 0 n n-2 n4 2 

But

2

I 0   dx 

2 (n  1) n  3 n  5 1 In  . . ..... . n n2 n4 22 0

Case 2 : let n be an odd integer. Then

In 

(n  1) n  3 2 .  I1 n n2 3

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APPLIED MATHEMATICS 

93

I1   sinxdx   cos 0 2  1

But

1

0

(n  1) n  3 n  5 2  In  .  n n2 n4 3 Example 6 π

2

 cos

Reduction formula for

n

x dx

0

Solution π

Let

2

I n   cos n x dx 0 π

=

2

 cos

n -1

x cos x dx

0 n-1

Let u = cos x dv = cos x dx n-2 du = (n-1) cos x(-sin x dx) v= sin x

π

 I n  cos

n 1

xsinx

0

π 2

2

  sinx(n  1)cos n 2 xsin x dx 0 π

2

 0  (n  1)  cos n 2 x(1  cos 2 x)dx 0

= (n-1) In-2 – (n-1) In In +(n-1) In = (n-1)In-2 (1+n-1) In = (n-1)In-2

 In 

n 1 I n 2 n

Proceeding as in the last example

n 1 n  3 n  5 1 π . . ...... . n n2 n4 2 2 n 1 n  3 n  5 2 When n is odd I n  . . ...... n n2 n4 3 When n is even

In 

Example 8 

If

4

I n   tan n xdx prove that I n  I n 2  0

1 and hence evaluate I5 n 1

Solution

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APPLIED MATHEMATICS 

94

4

I n   tan n xdx 0

4

  tan n 2 x(sec 2 x  1)dx 0

4

4

  tan n 2 xd (tan x)   tan n 2 xdx 0

0

 tan n 1 x  4 In     I n2  n 1 0  I n  I n2 

1 n 1

 In 

1  I n2 n 1

1  I3 4 1 I 3   I1 2 I5 

11   I 5    I1  42  1 1    I1 4 2 

1 4     tanxdx 4 0  1   log sec x0 4 4 1   log 2 4 1  log 2 4 Example 9 

Reduction formula for

e

x

x n dx

0

Solution 

Let

I n   e  x x n dx 0

  e x x n

   nx  0

n 1  x

e dx

0

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APPLIED MATHEMATICS

95

 0  n  x n 1e  x dx 0

 I n  nI n1 Also  I n 1  (n  1) I n 2 I n2  (n  2) I n3 Multiplying all these results

I n  n(n  1)(n  2).......2.1.I 0 

 

I 0   e  x dx  e  x

 0

1

0

 I n  n! 3.4.1

Self Assessment Question – II

1.

Reduction formula for

 cos

n

xdx

…………………………………………………………………………………………….. …………………………………………………………………………………………….. Ans:

In 

cos n1 x sin x n  1  I n2 n n  2

2.

Evaluate :

 sin

7

xcox5 xdx

0

…………………………………………………………………………………………….. …………………………………………………………………………………………….. Ans:

1 120

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APPLIED MATHEMATICS

3.5

96

Particular Integral

Example 1 2

3

2x

Solve (D -4D+3)y=x e Solution 2 The auxiliary equation is m -4m+3=0 (m-1) (m-3)=0 m=1, 3 x 3x The complementary function is y=Ae + Be

1 x 3e 2 x D  4D  3 1  e2 x x3 2 ( D  2)  4( D  2)  3 1  e2 x 2 x3 D  4D  4  4D  8  3 1  e2 x 2 x3 D 1 1  e2 x x3 2 1 D 2x  e (1  D 2 ) 1 x3

PI 

2

 e2 x (1  D 2 ) x3  e2 x ( x3  6 x) x 3x 2x 3 The complete solution is y  Ae  Be  e ( x  6 x)

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APPLIED MATHEMATICS

97

Example 2

d2y dy Solve  5  6 y  e x cos 2 x dx dx 2 Solution 2 The auxiliary equation is m -5m+6=0 (m-2) (m-3)=0 m=2, 3 x 3x The complementary function is y=Ae + Be

1 .e x cos 2 x D  5D  6 1  ex cos 2 x 2 ( D  1)  5( D  1)  6 1  ex 2 cos 2 x D  3D  2 1  ex cos 2 x  4  3D  2 1  e x cos 2 x 3D  2 3D  2  e x cos 2 x  40 ex  (6 sin 2 x  2 cos 2 x) 40  e x (3 sin 2 x  cos 2 x)  20

PI 

2

 The complete solution is

y  Ae 2 x  Be 3 x 

ex (3 sin 2 x  cos 2 x) 20

Example 3 Solve ( D  2D  1) y  x sin x Solution 2 The auxiliary equation is m -2m+1=0 2

ex

m=1, 1  The complementary function is y=Ae + Be x

3x

1 xe x sin x D  2D  1 1  xe x sin x 2 ( D  1) 1  ex x sin x ( D  1  1) 2 1  e x 2 x sin x D  2D  1  e x  x  2  2 sin x D D  2   e x  x  ( sin x) D 

PI 

2

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APPLIED MATHEMATICS

98

 e x ( x sin x  2 cos x)  The complete solution is

y  e x ( Ax  B)  e x ( x sin x  2 cos x) Example 4 Solve

( D 2  D  2) y  e2 x  e x

Solution 2

The auxiliary equation is m – m – 2 = 0  m=2, -1 x 3x  The complementary function is y=Ae + Be

1 1 e2 x  2 ex D D2 D D2 1 1  e2 x  ex ( D  2)( D  1) 1 1  2 1 1 1  . e2 x  e x 3 D2 2 1 2x 1 x  xe  e 3 2

PI 

2

 The complete solution is

1 1 y  Ae 2 x  Be x  xe 2 x  e x 3 2 Example 5 Solve Solution

( D 2  6D  9) y  e3 x 2

The auxiliary equation is m –6 m+9 = 0 2 (m-3) =0  m=3, 3.  The complementary function is

y  e3 x ( Ax  B) 1 PI  2 e3 x ( D  6 D  9) 1  e3 x 2 ( D  3)

(failure case)

x2 3x e 2

x2 3x  the complete solution is y  e ( Ax  B)  e 2 3x

3.5.1

Self Assessment Question - III

1.

Solve :

D2  2D  3) y  e2 x (1  x2 )

………………………………………………………………………………………….

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APPLIED MATHEMATICS

99

…………………………………………………………………………………………. Ans:

y  Ae

3 x

e2 x 2 12 x 87  Be  (x   ) 5 5 25 x

2. Solve : D  4D  4) y  8x e Sin2 x …………………………………………………………………………………………. …………………………………………………………………………………………. 2

Ans:

2 2x

 3  y  ( A  Bx)e2 x  2e2 x  x 2   sin 2 x  2 x cos 2 x  2  

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APPLIED MATHEMATICS

100

UNIT – IV

4.0

Introduction

4.1

Objectives

4.2

Non-linear differential equation of first order standard type 4.2.1

Complete solution

4.1.2

Singular Solution

4.1.3

General solution

4.1.4

Charpits Method

4.1.5

Standard type I

4.1.6

Standard type II

4.1.7

Standard type III

4.1.8

Standard type I

4.1.9 Self Assessment Question - I 4.2

Lagrange’s Linear Partial Differential Equations 4.2.1 Self Assessment Question - II

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APPLIED MATHEMATICS

4.0

101

Introduction In this chapter, we have to deal with Partial Differential Equation which involves

higher powers in non-linear partial differential equation, as understanding the concept of differential equation. In the differential equation with standard type are deal with all features in the high level and low level magnitude of the equations. Students will study for the second order differential equation with all standard types. 4.1

Objectives To develop the skill in the differential equation with standard types.

To create new idea to application oriented in the all applied sciences, like tsunami tides which deal with high and low attitude of the waves. To understanding the way of approach for the non-linear differential equation in all applied sciences. 4.2

Non-linear differential equation of first order standard type

A partial differential equation involving the first order partial derivatives

and

z ( p) x

z ( q) is called a Linear partial differential equations. Partial differential equations y

which involve the higher powers of

p, q, pq are called non-linear partial differential

equations. Linear partial differential equations are obtained by eliminating arbitrary constants or arbitrary functions.

4.2.1

Complete solution

A solution of a partial differential equation which contains as many arbitrary constants as the number of independent variables is called a complete solution.

The

complete solution is also referred to as complete integral. For example the partial differential equation

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

may be obtained from the equation

g(x, y, a, b) = 0.

(1)

(2)

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102

In this case (2) is called the complete solution of the partial differential equation (1). There are two independent variables. In the solution (2). There are two arbitrary constants.

3.1.2

Singular Solution

The complete solution represents two parameters family of surfaces which may or may not have an envelope. The envelope if it exists, is obtained by eliminating ‘a’ and ‘b’ from

g ( x, y, z, a, b)  0

(1).

 g ( x, y, z, a, b)  0, a

(2).

 g ( x, y, z, a, b)  0. b

(3).

If the eliminent  (x, y, z)= 0 exists it is called the singular solution of equation (1). Hence if f(x, y, z, p, q) = 0 is the partial differential equation whose complete solution is g(x, y, z, a, b)=0 then the eliminent (if it exits) of

g ( x, y, z, a, b)  0 .

(1).

 g ( x, y, z, a, b)  0. a

(2).

 g ( x, y, z, a, b)  0. b

(3).

Is called the singular solution or singular integral. The singular solution represents the envelope of family of surface given by the general solution

g x, y z, a, b  0. The

envelope in general may or may not exist.

4.1.3

General solution

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APPLIED MATHEMATICS

103

In the non-linear partial different equation f(x,y,z,a,b)=0, if g(x,y,z,a,b)=0 is the complete

solution it may be possible for the existence of a relationship between the

arbitrary constants a and b. Suppose this relationship given by b=(a). Then the complete

g x, y, z, a,  , (a)  0. This is a one parameter family of

solution takes the form

surfaces of the partial differential equation f(x,y,z,p,q) = 0. If the family has an envelope it is obtained by eliminating a from

and

g ( x, y, z, a,  (b))  0.

(1)

 g ( x, y, z, a,  (b))  0. a

(2)

If this eliminent if it exists if called the general solution of equation (1). This solution is also referred to as general integral. Standard type I: f(p,q) = 0

Here the partial differential equation involves only p and q and the variables x, y, z are absent. Standard type II: z = px + qy + f (p, q) This called the clairaut’s form Standard type III: F (z, p, q) = 0. Here the partial differential equation will contain p, q and the dependent variable z only. The independent variables x and y are absent. Standard type IV: Partial differential equation of the form f1(x, p) = f2(y, q). Here z is absent and also it is in a separable form (i.e) the terms containing x and p only on one side and terms containing y and q only on the other side.

4.1.4

Charpits Method

This is a general method of solving linear partial differential equation where the given partial differential equation cannot be reduce to any one of the four general forms. 4.1.5

Standard type I;

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104

F (p, q) = 0.

(1)

Consider the equation z = ax + by + c

………(2)

where a and b are connected by the relation f(a, b) = 0.

Differentiating the above equation partially with respect to x and y we get p = a, q =b. Substituting these in the given partial differential equation w get

f(a, b) = 0.

From the relationship f(a, b) =0 we can find b in terms of a, say b =  (a) Substituting this in (2) we get, z = ax+  (a) y +c

…….(3)

We see that (3) is the complete solution of equation (1). Hence we note that for the linear partial differential equation of type (1), the compete solution is of the form z=ax + by + c where a and b are connected by relation  (a, b) = 0 or

b=

g (a). Here a and b are arbitrary constants. To find the singular solution we have to determine the eliminent of a, and c from z = ax + g (a) y + c 0 = x + g’ (a) y (differentiating partially with respect to a) Differentiating partially with respect to c, 0 =1. As the last equation in inconsistent no singular solutions will exist for the partial differential equation of the type F (p, q) = 0. Take c =  (a). The complete solution becomes one parameter family z = ax + g (a) y +  (a) Differentiating partially with respect to a, 0 = a + g’ (a) + ’ (a) Eliminating ‘a’ from the last two equations we get the general solution which is the envelop of the single parameter family given by 1.

Example : 1 2

2

Solve p + q = 1

Solution This is of the type f (p, q) = 0. Therefore the complete solution of this partial differential equation is in the form z= ax + by + c

…….(1)

Differentiating (1) partially with respect to x and y we get p = a, q = b. Substituting in the given partial differential equation we have,

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APPLIED MATHEMATICS

a 2  b 2  1,

b  1  a 2

The complete solution is

105

z  ax  1  a 2 y  c

….(1)

Where a and c are arbitrary constants There is no singular integral to the partial differential equation of this type. To obtain the general solution take c = (a).

 z  ax  1  a 2 y   (a)

…… (2)

Differentiating partially with respect to a we get

 a 0  x   2  1 a

  y   ' (a)  

……..

(3)

The eliminent of a between (2) and (3) gives the general solution.

Example 2

Solve

p  q  1.

Solution The equation is of the type f (p, q) = 0.  The complete solution is of the form z = ax + by + c

……..(1)

Differentiating (1) partially with respect to x and y we get p = a and q = b. Therefore the given equation becomes

a  b  1.

The complete solution is

2

z  ax  1  a y  c Where a and c are arbitrary constants. This type of equation has no singular solution. Take c = (a).

2

 z  ax  1  a y   (a)

……(2)

Differentiating partially with respect to a we get

0  x  (1  a)

1 y   ' (a) a

………(3)

The eliminent of ‘a’ between (2) and (3) gives the general solution.

4.1.6

Standard Type II

Equation of the type z = px + qy + f (p, q).

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106

This type of equation may be considered as analogous to clairaut’s form y = px + f (p) in the ordinary differential equation. Here

p

dy dx

Now consider the equation

z  ax  by  f (a, b)

…..(1)

Where a and b are arbitrary constants. Differentiating (1) partially with respect to x and y we get p = a …………(2)

q=b

……….(3)

Therefore by eliminating the arbitrary constants from (1) (2), and (3), we get,

z  px  qy  f ( p, q) …..(4). Therefore (1) can be considered as the complete solution of the given partial differential equation. This type of partial differential equation is known as extended Clairaut’s form. The complete solution consists of a two parameter family of planes. The singular solution if it exists is a surface having the complete solution as tangent planes.

Example 1 Solve z = px + qy – qb

Solution This equation is a partial differential equation of Clairauts type. Therefore the complete solution is got by replacing p by a and q by b where a and b are arbitrary contests. (i e) the complete solution is z = ax + by + ab ………(1). Differentiating (1) partially with respect to a and b we get

0  xb

……(2)

0 ya

……(3)

Eliminating a and b from (1), (2) and (3) we get

z   xy  xy  xy (i.e)

z  xy  0.

This gives the singular solution of the given partial differential equation and to get the general solution, Put

b   (a)

in (1)

z  ax   (a) y  a (a) …..(4) Differentiating this partially with respect to ‘a’ we get

0  x   ' (a) y  a ' (a)   (a) ……. (5) Eliminating

‘a’ from (4) and (5) we get the general solution.

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107

Example 2 2 2

Solve z = px + qv + p q Solution

This equation is a partial differential equation of Clairauts type. Therefore the complete solution is

z  ax  by  a 2b 2

……..(1) where a and b arbitrary constants. Differentiating (1) partially with respect to a and b we get

0  x  2ab 2

……(2)

0  y  2a 2b

……(3)

x  2ab 2

 2ay 2  y 2   4a 4 2a 3

z  x3

y2 x2  y3 2x 2y 2

2

2

z  2 3 x 3 y 3 This

is

the

singular

solution

where

b

=

f(a).

The

complete

z  ax  f (a) y  a  f (a) (2) Differentiating partially with respect to ‘a’.

solution

is

2

2

0  a  f ' (a) y  2a 2 f (a) f ' (a)  2a f (a) Eliminant of ‘a’ from (2) and (3) gives the 2

general solution.

Example 3

Obtain the complete solution and singular solution of

z  px  qy  p 2  pq  q 2 .

Solution This equation is of clairaut’s form. Therefore

the complete solution is

z  ax  by  a  ab  b (1) where a and b are arbitrary constant. 2

2

Differentiating (1) partially with respect to a and b we get

0  x  2a  b(2) 0  y  2b  a(3) 2x-y =3a and 2y- x = 3b.

a

2x  y 2y  x ,b  3 3

Substituting this in equation (1) we get

 2 x  y   2 x  x   2 x  y  (2 x  y)(2 y  x)  2 x  x  z  x   y      9  3   3   3   3  2

2

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APPLIED MATHEMATICS Simplifying we get

108

3z  xy  x 2  y 2

This is the singular solution,

4.1.7

Standard type III

Partial differential equation of the form F (z,p,q) = 0. For this partial differential equation we assume z = F (x + ay) as solution. Let u  z  ay, then Then

q

p

z  F (u)

z z u dz   x u x du

z z u dz  a y u y du

Then the given differential equation becomes

 dz dz  F  z, , a   0 which is an ordinary differential equation of first order.  du du  Procedure for solving partial differential equation of the type

F(z, p, q) = 0. Step I :

Assume u = x + ay

Step II:

Replace p and q by

dz dz and a respectively in the given partial du du

differential equation. Step III:

Solve the resulting ordinary differential equation of first order.

Example 2 2

2

Solve z= p +q for complete and singular solution.

Solution

Assume z = F (x+ay)= F (u). then

p

dz dz ,q  a du dy

The given equation becomes 2

 dz   dz  z     a2    du   du 

2

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109

2

=

 dz  2   (1  a )  du 

dz z  du 1 a2 (i.e)

dz du  z 1 a2 u

2 z 

1 a2

1 1 a2

 c.

(u  b)

 4(1  a 2 ) z  ( x  ay  b) 2 (1) This is the complete solution. Differentiating equation (1) partially with respect a and b we get 8az = 2(x + ay + b) y (2) 0 =2 (x + ay + b) 1 (3) Substuting (3) in (2) we get z = 0 This is the singular solution.

Example 3

Find the complete solution of p (1+q)=qz.

Solution Assume that z = f(x+ay) then

p

The given equation becomes

dz  dz  dz 1  a   a z du  du  du (ie)

1

dz az  1  du a adz  du az  1 log (az-1) =u+c. log (az-1) =x+ay+c.

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110

This is the complete solution.

Example 4 2

2

2

Solve p z + q = 1

Solution

Assume z=F (x + ay) = F (u). then

p

dz dz ,q  a . du du

The given equation becomes. 2

2

 dz   dz  z    a2    1  du   du  2

z

2

2

 dz   a2    1  du 

z

2

 a 2 dz  du.

Integration we get

 z  a  a2 logz  z  a   u  c z  z  a  a log z  z  a   2( x  ay)  c 2

z 2

2

2

4.1.8

2

2

2

2

2

2

2

Standard type IV

Partial differential equation of the type a

f1 ( x, p)  f 2 ( y, q)

In this type of equation z is absent. Also the terms containing p and x can be separated from those containing q and y Let f1(x, p) = f2 (y, q) = k (1) (i.e.) f1 (x, p) = k. Solving for p we get, p = F1 (x). f2(y, q) = k. Solving for q we get q = F2 (y). Also

dz 

z z dx  dy. x x

= pdx + qdy = F1 (x) dx + F2 (y) dy.

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APPLIED MATHEMATICS Integrating

111

 dz   F ( x)dx   F ( y)dy 1

2

z   F1 ( x)dx   F2 ( y)dy  c. This is the complete solution

Example 1

Solve

p2  q2  x  y

Solution

p2  q2  x  y .

p 2  x  q 2  y  k.  p 2  x  k ; q 2  y  k.

p   x  k.q   y  k . dz  pdx  qdy

 

  x  k dx 

yk

Integrating we get the complete solution is

z

3 3 2 x  k  2  2  y  k  2  c. 3 3

3 3 2   ( x  k ) 2  ( y  k ) 2   c.  3 

Example 2

p 2  q 2  z 2 ( x  y). Solution Dividing by

z2,

2

2

 p q       x  y.  z z Take Z= log z

dZ 

dz z

The given equation becomes 2

 Z   Z        x  y.  x   y  2

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112 2

 Z   Z  (i.e.)    x   y     k .  x   y  2

2

 Z   Z     x  k ; y     k  x   y  2

Z Z   x  k;  yk x x herefore

dZ 

ntegrating

Z Z dx  dy x y

3 3 2 Z   ( x  k ) 2  ( y  k ) 2   c  3 

3 3 2 log z   ( x  k ) 2  ( y  k ) 2   c  3 

Example 3

p  q  sin x  sin y. Solution

p  sin x  sin y  q  k.

 p  k  sin x; q  sin y  k. dz  pdx  qdy  (k  sin x)dx  (sin y  k )dy. Integrating we get

z  (kx  cos x)  (ky  cos y)  c. z  k ( x  y)  (cos x  cos y)  c. This is the complete solution.

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4.1.9

1.

113

Self Assessment Question – I

Solve

p3  q3  0

………………………………………………………………………………………… ………………………………………………………………………………………… Ans: 2.

z  a( x  y )  c Solve

q2  p

………………………………………………………………………………………… ………………………………………………………………………………………… Ans: 3.

z  a 2  ay  c Solve pq = z

………………………………………………………………………………………… ………………………………………………………………………………………… Ans: 4.

4az  ( x  ay  b)2

x y  1 p q

………………………………………………………………………………………… …………………………………………………………………………………………

x2 y2 Ans:  b a 1 a 5.

Solve

z  p( x  q)  qy

………………………………………………………………………………………… ………………………………………………………………………………………… Ans:

4.2

z  ax  by  ab Lagrange’s Linear Partial Differential Equations

he partial differential Pp + Qq = R where P,Q,R are functions of x, y, z and

p

z z , q  is called Langrange’s linear differential equation. x y

We shall now consider the method of solving such an equation. It has already been shown that a partial differential equation of this type is obtained by eliminating the arbitrary function from the equation F(x, y, z) =0. In other words F (u,v)=0 is the solution of the partial differential equation Pp + Qq=R. We are now required to find the values of u and v and have the solution F(u,v)=0. Let us assume that u(x, y, z) = c1 and v (x, y, z)= c2 be two solutions of the given linear partial differential equation. Taking the differentials of the above two solutions,

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114

du 

u u u du  dy  dz  0. x y z

(1)

dv 

v v v dx  dy  dz  0. x y z

(2)

dx u v u v  y z z y

dy u v u v  z x x z

dz u v u v   x y y x

(i.e)

dx dy dz   P Q R

(3).

It is now easily seen that u = c1 and v=c2 are solutions of the equation (3). Then,

 (u,v) = 0. (i.e) u =  (v) is the solution of the linear partial differential equation Pp + Qq = R. Hence Procedure for solving Pp +Qq = R.

dx dy dz   P Q R

Step 1 :

From the auxiliary equations

Step 2:

Find two independent solutions of the auxiliary equations say u = c1 & v = c2.

Step 3:

Then the solution of the given partial differential equation is

 (u, v)=0, or u =  (v). Example 1 2

2

2

Solve x p + y q= z Solution

This is lagrange’s linear equation The auxiliary equations are

dx dy dz   x2 y2 z 2 Consider

dx dy  x2 y

Integrating

Consider

1 1   c. x y

dx dz  y2 z2

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Integrating

115

1 1   c. y z

The solution is

1

1

1 1

   ,    0. x y y z 

Example 2

Solve

( y  z) p  ( z  x)q  x  y.

Solution This is a lagrange’s linear equation

The auxiliary equation are

dx dy dz   y z z x x y (i.e)

dx  dy dy  dz dz  dx dx  dy  dz    x y yz zx 2( x  y  z )

Considering first two members and integrating we get

x y c yz

(1)

Considering

first

x  y  ( x  y  z )  c

and

last

members

and

integrating

we

get

2

 The Solution is

x y  , ( x  y ) 2 ( x  y  z )  0 yz 



Example 3 Solve

y 2 p  x2q  x2 y 2 z 2

Solution The auxiliary equations are

dx dy dz  2  2 2 2 2 y x x y z st

nd

Considering 1 and 2 members,

x 2 dx  y 2 dy. Integrating we get

x3 y 3  c 3 3

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x3  y 3  c Considering 2

y 2 dy 

nd

rd

and 3 members,

dz z2

y 1 Integrating  c 3 z 3

3

y 1   c 3 z 3  3 1 3 y  The solution is   x  y ,    0. 3 z 

Example 5 Solve

(mz  ny) p  (nx  lz )q  ly  mx.

Solution

The auxiliary equations are

dx dy dz   mz  ny nx  lz ly  mx Using multipliers x, y, z we get each ratio

xdx  ydy  zdx x(mz  ny)  y(nx  lz )  z (lz  mx)

xdx  ydy  zdx 0

 x 2  y 2  z   c is one solution. Also using multipliers l, m, n, we get. each ratio

ldx  mdy  ndz l (mz  ny)  m(nx  lz )  n(ly  mx)

ldx  mdy  ndz 0

lx  my  nz  c is the 2nd solution. The general solution

 ( x 2  y 2  z 2 , lx  my  nz)  0.

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4.2.1 Self Assessment Question - II

1.

Solve

z  p tan x  q tan y  tan z

………………………………………………………………………………………… ………………………………………………………………………………………… Ans:

2.

 sin y sin z  Q , 0  sin x sin x 

Solve

p( x  y )  q ( y  x  z )  z

………………………………………………………………………………………… ………………………………………………………………………………………… Ans:

 yxz Q  x  y  z, 0 z2  

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UNIT â&#x20AC;&#x201C; V 5.0

Introduction

5.1

Objectives

5.2

Laplace Transforms of some elementary function 5.2.1

First Shifting Theorem

5.2.2

Second Shifting Theorem

5.2.3

Change of Scale property

5.2.4 Self Assessment Question - I 5.3

5.4

Inverse Laplace Transform 5.3.1

Table of Inverse Laplace Transforms

5.3.2

Inverse Laplace Transforms by first shifting theorem

5.3.2

Inverse Laplace Transforms by second shifting theorem

5.3.3

Self Assessment Questions - I

Laplace Transform of Periodic Function 5.4.1

Self Assessment Questions - II

5.5

Fourier Series 5.5.1 Fourier Coefficient 5.5.2 Self Assessment Questions - III

5.6

Half range Fourier series 5.6.1 Self Assessment Question - IV

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119

Introduction In this chapter, we have to deal with Laplace transform in a linear transformation. as

understanding the concept of Laplace transformation and inverse Laplace transformation. In the Laplace Transformation with Shifting theorem deal with all features in the application of mathematics in the field of applied sciences.

Students will study for the Laplace

transformation with periodic functions. 5.1

Objectives To develop the skill in the Laplace transformation with standard types.

To create new idea to application oriented in the all applied sciences in the engineering systems To understanding the way of approach for the Laplace transformation in all applied sciences. 5.2

Laplace Transform: Let f(t) be defined for 0< t <  .The transform of f(t) is defined by 

L  f (t )   e st f (t )dt 0

Note: In the above definition, the R.H.S. integral is a function of s only ,we can Write

L  f (t ) =F(s).

Notations : Laplace transform of f(t) is denoted by F(s),

(s) or

L  f (t ) ,

f (s)

Instead of s, the parameter p can also be used . Linearity property of the Laplace Transformation

The Laplace transformation is a linear transformation i.e.,

L a1 f1 (t )  a2 f 2 (t )  a1L  f1 (t )  a2 L  f 2 (t ) Where a1 , a2 are constants.

Piecewise (or) sectionally Continuous Function A function f(t) is said to be piecewise (or) Sectionally continuous on a closed interval

a  t  b , if is defined on that interval and is such that the interval can be divided into a finite number of subintervals, in each of which f(t) is continuous Functions of exponential order A function f(t) is said to be of exponential order if

lim e st f (t )  a finite quantity. t 

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Condition for existence of Laplace transform (sufficient condition) If f(t) is a function which is piecewise continuous on every finite interval in the range t  0 and is of exponential order, then the Laplace transform of f(t) exists. This is the sufficient condition but not necessary for the existence of Laplace transform. 5.2.1

Laplace transforms of some elementary functions

Sl.No.

L  f (t )

f(t)

1 2

K,a constant

3

1 t

4

e at

5

e at

6

Cos at

7

Sin at

8

Sinh at

9

Cosh at

t

k/s ,s>0

(n  1) n! or n 1 n 1 s s

n

 s 1 sa 1 sa s 2 s  a2 a 2 s  a2 a 2 s  a2 s 2 s  a2

Example 1

L t n  ,n is a positive integer and n  -1

Find Solution:

L t n  =  e st t n dt 0 

n

 x  dx L t  =  e   s s 0 n

=

=

1 s

x

put st

1  x  dt  dx s

e x x ( n 1)1dx n 1  0

(n  1) n !  n 1 s n 1 s

Since

  n    e x x n 1dx 0

Where n > -1

Example 2 Find

L eat  and L e at 

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Solution

L e

at

  e

 st at

e dt

0

=

e

 ( s a )t

0

III

ly

L e

 at

  e

 1  e ( s a )t   dt   , s >a   s  a  0 s  a 

 ( s  a ) t at

e dt 

0

1 , s >-a sa

Example 3

Find (i)

L cos at and (ii) L sin at 

Solution (i)

L cos at   e st cos atdt 0 

 e st  =  2 ( s cos at  a sin at )  2 s  a 0 s = 2 s  a2

(ii)

L sin at   e st sin atdt 0 

 e st  ( s sin at  a cos at )  =  2 2 s  a 0 a = 2 s  a2 Example 4 Find Solution

L sin t cos t .

L sin t cos t = L sin 2t / 2 =

1 1 2 1 L sin 2t  . 2  2 2 2 s 4 s 4

Example 5 Find

Solution

L sin 2 2t

L sin 2 2t 

1 L 1  cos 4t 2

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1 1 L 1  L cos 4t 2 2 1 s =  2 2s 2( s  16) =

Example 6 Find

L 7e2t  9e2t  5t 3

Solution

L 7e2t  9e2t  5t 3 = 7 L e2t   9L e2t   5L t 3

7 9 6 16 s  4 30   5. 4  2  s2 s2 s s  4 s4

=

5.2.1

If

First shifting theorem

L  f (t )  F (s), then L eat f (t )  F (s  a)

Proof 

F ( s)  L  f (t )   e st f (t )dt. 0 

F ( s  a)   e ( s a )t f (t )dt. 0 

=

 e e  st

at

f (t ) dt.

0

 F (s  a)  L eat f (t ) 5.2.2

If

Second Shifting theorem

 f (t  a), t  a  sa L  f (t )  F (s), and g (t )   then L  g (t )  e F (s) .  0, t  a

Proof By definition, we have 

L  g (t )   e st g (t )dt 0

=

0

0

 st  st  e 0dt   e f (t  a)dt

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123 

=

e

 st

f (t  a)dt

0 

 L  g (t )   e (u  a ) s . f (u )du

Put t-a = u. then dt = du

0

=e

=e

 sa

 sa

e

0 

e

 su

 st

. f (u )du

. f (t )dt by property of definite integrals

0

 L  g (t )  e sa F (s) 5.2.3 If

Change of scale Property

L  f (t )  F (s), then L  f (at ) 

1 f ( s / a) a

Proof by the definition, we have 

L  f (at )   e st f (at )dt 0 

1 ( s / a) x e f ( x)dx a 0

1  ( s / a )t e f (t )dt a 0

Put at = x

 dt = 1/a dx

1  F ( s / a) a Note: If

Since

F ( s)   e st f (t )dt 0

L  f (t )  F (s), then L  f  t / a   aF (as)

Example 1 Find

L t 3e2t 

Solution

L t 3e2t   L t 3

s s  2

6  3!   4 4  s  s s  2 (6  2)

= Example 2 Find L

e

2t

(3cos 6t  5sin 6t ) .

Solution We have

L (3cos 6t  5sin 6t )  3.

s 6  5. 2 s  36 s  36 2

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3s  30  F ( s) s 2  36 L e2t (3cos 6t  5sin 6t )  F ( s  2) =

=

3( s  2)  30 3s  24  2 2 2( s  2)  36 s  4s  40

Example 3 Find

L et (3sinh 2t  5cosh 2t .

Solution

2 s  5. 2 s 4 s 4 6  5s = 2  F ( s) s 4  L et (3sinh t  5cos 2t )  F ( s 1)

We have

L (3sinh 2t  5cosh 2t  3

=

2

6  5( s  1) 11  5s  2 2 ( s  1)  4 s  2s  3

Applying the change of scale property show that

L  f (2t )  We have

s 2  2s  4 s2  s  1 If L f ( t )    4( s  1) 2 ( s  2) (2s  1) 2 ( s  1)

s2  s  1 (2s  1)2 ( s  1) 1 L  f (2t )  F ( s / 2) 2 2 s / 2   s / 2  1  1 = 2  2  s / 2   1 2  s / 2   1     2 s  2s  4 =  4  s  12 ( s  2)   

L  f (t )  F ( s) 

Example 4

e t  a , t  0 Find L  g (t ) where g (t )    0, t  a Solution

 f (t  a), t  0 g (t )   where F(s) = L  f (t )  0, t  a 1 t Here F(s) = L e   s 1 L g (t )  e sa .F (s)

e sa L  g (t )  , s 1 s 1

(Using second shifting theorem)

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Example 5 If

1 L  f (t )  e s

1 s

, prove that

L et f (3t ) 

3 ( s 1)

e . ( s  1)

Solution

1 1 L  f (t )  e s  F (s) s t  L e f (3t )  L  f (3t )ss 1 Given

By change of scale property,

L  f (at )  3

1 F ( s / a) a

3

1 3e s e s   L  f (3t )  3 s s 3

e ( s 1)  L et f (3t )  s 1 Example 6 If

 sin t   sin at  1 1 L   tan (1/ s) then prove that L    tan (a / s). t t    

We know by the change of scale property If

L  f (t )  F (s)thenL  f (at ) 

Given

1 F (s / a) a

 sin t  1 L   tan (1/ s)  t  1 1 a  sin at  1 1 L )  tan 1 ( )   tan ( s/a a s  at  a  sin at  1 a L    tan ( ) s  t 

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5.2.4

Self Assessment Questions - I

1.

Find he Laplace transform of

sin 3 2t

…………………………………………………………………………………………… …………………………………………………………………………………………… Ans :

2.

48 ( s  4)( s 2  36) 2

Find he Laplace transform of

e3t  e 2t t

…………………………………………………………………………………………… …………………………………………………………………………………………… Ans :

 s2  log    ( s  3) 

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127

Inverse Laplace Transform Definition: Inverse Laplace Transform

L  f (t )  F (s), then f(t) is called

If F(s) is the Laplace transform of a function f(t),

the inverse Laplace transform of the function F(s) and is written as

f (t )  L1 F (s)

L1 is called the Inverse Transformation Operator Linear Property Theorem Let

F1 ( s) and F2 ( s) be the inverse Laplace transforms of function f1 (t ) and f 2 (t )

respectively and a, b be two constants Then

L1 aF1 (s)  bF2 (s)  aL1 F1 (s)  bL1 F2 (s) =

5.3.1

af1 (t )  bf 2 (t )

Tables of Inverse Laplace Transforms S.No.

F(s)

1

1 s

2

1 1

s

f (t )  L1 F (s)

n 1

, n is a

t n!

positive integer 3

1 s

4 5 6 7 8 9

Example 1 Find the inverse transform of

, n  1 n 1

1 sa 1 sa 1 2 s  a2 s 2 s  a2 a 2 s  a2 s 2 s  a2

tn (n  1) e at e at

sin at a Cosat Sinhat coshat

3 2s  1 .  2 s 2 s 4 2

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2s  1  s  1  1   3 1  1  1  L1  2  2   3L    2L  2  L  2  s  2 s  4 s  2 s  4 s  4 sinh 2t 2t = 3e  2cosh 2t  2

Solution

Example 2

1

Find L

 s 3 1   .    2 7  s  2 s  3   s2  

Solution

 s     3 1 s   1  1  1  1  L1  2   7  = L1  2  3 L  L     7  s  2 s 3 s  ( 2) 2   s  3  2    s   s2   7

1

t2  cos 2t  3e3t  7   2 5

t2  cos 2t  3e3t  5 3 1 1 . .   2 2 2 2  cos 2t  3e3t 

8t 2 15

t

Example 3

Find (i)

 3s 2  4s  5  1  2 s  3  L1   and (ii) L  2  4 s  s   

Solution

 3s 2  4s  5  1  1  1  1  1  1  L1    3L  2   4 L  3   5 L  4  4 s s  s  s    = 3t  4

t2 t3  5. 2! 3!

5t 3  3t  2t  6 2

(ii)

 2s  3  1  1 L1  2   2 L1    3L1  2   2  3t  s  s s 

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129

Inverse Laplace Transform by First Shifting Theorem

L1 F (s)  f (t ) ,then L1 F (s  a)  eat L1 F (s)

Note:

L1 F (s  a)  e at L1 F (s) Example 4

Find

 3s  2  L1  2   s  6s  13  3s  2 3( s  3)  7 = s  6s  13 ( s  3) 2  4 2

 3( s  3)  7  L1   2  ( s  3)  4 

 3s  2  L1  2 =  s  6s  13 

 3s  7  e3t L1  2  s  4

=

=e

3t

=e

 1  s  1  1   L  s 2  4   7 L  s 2  4      

3t

7   3cos 2 t  sin 2t   2 

Example 5

Find

 3s  7  L1  2   s  2s  3 

3s  7 3s  7 = s  2s  3 ( s  1) 2  4 2

=

3( s  1)  10 ( s  1) 2  4

 3s  7  1  3( s  1)  10  L1  2  =L  2  s  2s  3   ( s  1)  4   ( s  1)  1 1  1  = 3L    10 L   2 2  ( s  1)  4   ( s  1)  4  1   s  t 1  3et L1  2   10e L  2  s  4 s  4 t t = e cosh 2t  5e sinh 2t

=

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= 3e Example 6

t

130

2t 2t  e2t  e2t   t e  e 3t t  5 e      4e  e 2 2    

14s  10 49s 2  28s  13  5 14  s   14s  10  7 = 2 4 13  49s  28s  13  49  s 2  s   7 49   5 2 3 s (s  )  2 2 7 7 7 = = 2 2 2 7 2 9 7 2 3 s   s    7  49 7 7  

Find the inverse Laplace transform of

 14s  10  2  L1  = e 2  49s  28s  13  7

=

2 72t e 7

 3   s   1  7 L  2  s2   3       7 

     3      s  1   7  L1  L  2 2   2 3   s 2   3    s          7   7   =

5.3.3

2 t 7

2 72t e 7

3t 3t   cos 7  sin 7 

Inverse Laplace Transform by second shifting theorem

If

 f (t  a), t  a L1 F (s)  f (t ), thenL1 e as F (s)  G(t ) where G (t )    0, t  a

Or G(t) = f(t-a) H(t-a) Or Equivalently if

L1 F (s)  f (t ) , then L1 e as F (s) = f(t-a) H(t-a) Where H(t-a) is Heaviside ‘s unit step function

Example 7

Find

 e3s  L1  . 4  ( s  2) 

Solution Let

F(s) =

1 ( s  2) 4

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131

 1  L1  4  ( s  2)  e2t t 3 2t 1  1  2t t =e L  4   e  3! 6 s   e3s  1 = f (t  3) H (t  3) L1  4  ( s  2)  6 1 2(t 3) = e (t  3)3 H (t  3) 6 f (t) =

Example 8

Find

 se s  L1  2  s  9

Solution

s  s 1  ; f(t)= L  2   cos 3t s 9 s  9  s  1  s 1  se L  2  = L e F (s) s  9 = cos3(t   ) H (t   )  cos(3  t ) H (t   )   cos3tH (t   )

Taking F(s) =

2

Example 9

1  e s  Show that L  2  = sin tH (t   )  sin t  s 1  1

Solution

1 s 1 1 f (t )  L F (s)  sin t

Take

F ( s) 

2

1  e s  1 1  s L  2  = L F (s)  L e F (s)  s 1  = sin t  f (t   ) H (t   ) = sin t  sin(t   ) H (t   ) = sin t  sin(  t ) H (t   ) = sin t  sin tH (t   ) . 1

5.3.4

Self Assessment Questions - I

1.

Find

 10  L1  6   ( S  2) 

……………………………………………………………………………………………

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132

……………………………………………………………………………………………

Ans :

5.4

e

2 t

t5 12

Laplace Transform of Periodic Function Definition Periodic Function A function f(t) is said to be a periodic function with period P if f(t+P) = f(t) for all values of t.

Note: If P is a period of the function f(t) then 2P,3P,…. Are also periods

Theorem P

1 If f(t) is a periodic function with period P , then L  f (t )  e st f (t )dt  sP  1 e 0 Proof By definition 

L  f (t )   e st f (t )dt 0 P

 L  f (t )   e

 st

f (t )dt   e  st f (t )dt ………(1)

0

P

Consider the second integral on R.H.S and put t =u+P . Then dt=du and u varies from 0 to  

 e

 st

f (t )dt   e s (u  P ) f (u  P)du

P

0

 e sP  e su f (u )du

since f(u+P)=f(u)

P

=e

 sP

L  f (t )

From (1) P

L  f (t )   e st f (t )dt  e  sP L  f (t ) 0 P

1  e sP  L  f (t )   est f (t )dt 0 P

1  L  f (t )  e st f (t )dt  sP  1  e  0 Example 1 If

f (t )  t 2 ,0  t  2 and f (t  2)  f (t ) . find L  f (t ) .

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133

Solution Here f(t) is a periodic function with period P=2

 L  f (t ) 

2

1 e st f (t )dt 2 s  1  e  0 2

1 t 2e st dt 2 s  1  e  0

 2  e st 1  t  1  e2s    s

 L  f (t ) 

= 5.4.1

1.

  e st  2 t   2   s

  e2 s 1 4  1  e2s    s

2

  e st     2   3    s 0

  e2 s   4 2   s

  e2 s  2    2 3   3    s  s 

2  (4s 2  4s  2)e2 s s3 (1  e2 s )

Self Assessment Questions - III

sin t 0  t    f (t )    extended periodically with period 2 and find its laplace 0   t  2 

transform …………………………………………………………………………………………… …………………………………………………………………………………………… 5.5

Fourier Series

It has been shown by Fourier that a function f(x) which has only a finite number of discontinuities can be expressed as a trigonometric series in a given range of x in the form

f ( x) 

a0  (a1 cos x  a2 cos 2 x  a3 cos 3x  ...  an cos nx  ...) 2 +(b1 sin x  b 2 sin 2 x  b3 sin 3 x  ...  b n sin nx  ...)

Fourier has shown that the expansion of f(x) in the above form is possible only if it satisfies certain conditions. These conditions called Dirichlet conditions are stated below. Let f(x) be defined in the interval c<x<c+2 with period 2 and satisfy the following conditions. i) f(x) is single valued. ii) It has a finite number of discontinuities in a period of 2. iii) It has a finite number of maxima and minima in a given period. c  2

iv)

f(x) dx is convergent

c

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134

If f(x) satisfies the above conditions then it is possible to express f(x) as

bn 

1

2

 1

=

f ( x) sin nxdx

0 2

1

 2 (  x) sin nxdx 0

1 cos nx sin nx = [(  x).(  )  ( 1)( 2 )]02 2 n n 1 1 1 1 = [  ]  2 n n n 2   1 f(x) =   2 co s nx 12 n 1 n sin2x sin3x = sinx+   .... 2 3  x 2 ( ) 2 This representation of f(x) is called a Fourier expansion or a Fourier series. Here the coefficients a0,an,bn are called Fourier coefficients. 5.5.1

Fourier Coefficients

Let us now to find the coefficients a0,an,bn on the assumption that the series given in (1) Is uniformly convergent in the interval c<x<c+2 and that the term by term integration is permissible in that interval. Integrate (1) with respect to x between the limits c and c+2 c  2

a0 2

f ( x)dx 

c

c  2

c  2

dx 

c

c  2

1

c

( an cos nx)dx 

c

= a0  0  0  a0 

1

( bn cos nx)dx 1

(Using the properties)

c  2

f ( x) dx

c

Multiply both sides of (1) by cosnx and integrate with respect to x between the limits c and c+2 c  2

c  2

f ( x) cos nxdx 

c

a0 cos nxdx  2

 c

c  2

c  2

1

c

( an cos nx) cos nxdx 

c

( bn cos nx) cos nxdx 1

c  2

=0 

an cos 2 nxdx  0

c

= a n

an 

1

c  2

f ( x) cos nxdx

c

Now multiply both sides of (1) by sinnx and integrate with respect to x between the limits c and c+2

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c  2

f ( x) sin nxdx 

c

 c

135

a0 sin nxdx  2

c  2

 c

c  2

1

c

( an cos nx) sin nxdx 

( bn sin nx) sin nxdx 1

c  2

= 00

bn sin 2 nxdx

c

= b n  bn 

1

c  2

f ( x) sin nxdx

c

Thus to summarise, the fourier expansion of a function f(x) satisfying Dirichlet conditions is given by

1 f ( x)  (  x), 0  x  2 2 Note 1: If c = 0, then the interval of definition for f(x) is (0,2) and in this case.

a0 

1

2

an  bn 

f ( x)dx

0 2

1

f ( x) cos nxdx

0

1

2

f ( x) sin nxdx

0

Note 2: If c = -, then the interval of definition for f(x) is (-,) and in this interval

a0 

1

an  bn 

1

 1

f ( x)dx



 f ( x) cos nxdx

f ( x) sin nxdx



Value of f(x) at a point: Note: (1) The Fourier series converges to f(x) at all points where f(x) is continuous. i.e. If x = x0 is a point of continuity, the sum of the Fourier series at x = x 0 is f(x0). (2) If x = x0 is a point of discontinuity then the value of f(x) at x = x 0 is

f ( x0  0)  f ( x0  0) 2

Example 1: Obtain a Fourier expansion for the function

1 f ( x)  (  x), 0  x  2 2

Solution:

1 f ( x)  (  x), 0  x  2 2 Let us Fourier series for the function f(x) be

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f ( x) 

136

 a0    an cos nx   bn sin nx 2 n 1 n 1

a0 

Then

1

2

=

1

f ( x )dx

0 2

1

 2 (  x)dx 0

1 x2 [ x  ]02 2 2 1 = [2 2  2 2 ]  0 2 =

 f ( x)  (i.e.)(

 a0    an cos nx   bn sin nx 2 n 1 n 1

(  x) 2  2  1 )    co s nx 2 12 n 1 n 2

bn 

1

2

f ( x) sin nxdx

0

1 = 4

2

( 0

 x 2

) 2 sin nxdx

1 co s nx sin nx cos nx [(  x) 2 .(  )  2(  x)( 1)(  2 )  2( 3 )]02 4 n n n 2 2 1  2  2 = [  3  3]0 4 n n n n   -x 1  f(x) =   sin nx 2 n 1 n sin2x sin3x = sinx+   .... 2 3 =

Example 2:

If f(x) =

(

 x 2

) 2 in (0,2) show that f(x) =

Solution: f(x) =

(

 x 2

2 12

1 co s nx 2 n 1 n



)2

let

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a0  = = =

1

 1

 1

 1

137

2

f ( x) dx

0 2

  dx  0

1

[[  x]0n  [

2

 ( x   )dx 0

2

x   x]2 ] 2

[[  2 ]  [2 2  2 2 ]  [

= - +

2



2 2

  2 ]]

 2

1  3 2 = [  ] 4 3 3 6 3

an 

1

2

f ( x) cos nxdx

0

=

1

2

(

 x

0

2

) 2 cos nxdx

1 sin nx  cos nx sin nx [(  x) 2 .  2(  x)( 1)( )  2( 1) 2 ( 3 )]02 2 4 n n n 1 sin nx 2(   x ) cos nx 2sin nx = [(  x) 2   ]02 2 3 4 n n n 1 2 2 1 = [ 3  3 ] 2 4 n n n =

bn 

1

2

f ( x) sin nxdx

0

1 = 4

2

( 0

 x 2

) 2 sin nxdx

1 co s nx sin nx cos nx [(  x) 2 .(  )  2(  x)( 1)(  2 )  2( 3 )]02 4 n n n 2 2 1  2  2 = [  3  3]0 4 n n n n

=

 f ( x)  (i.e.)(

 a0    an cos nx   bn sin nx 2 n 1 n 1

(  x) 2  2  1 )    co s nx 2 12 n 1 n 2

Example 3: Determine Fourier expansion for f(x) = - 0<x< = x -  <x<2 and show that

1 2   2 8 r 1 (2r  1) 

Solution:

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a0 

138

2

1

f ( x) dx

0 2

1

=

  dx 

0

1

=

 1

=

an 

 1

=

2

 ( x   )dx

[[  x]0n  [

0 2

x   x]2 ] 2

[[  2 ]  [2 2  2 2 ]  [

= - + 1

1

 2



2

  2 ]]

 2

2

0

[   cos nxdx  [ (

2

 sin nx

 ( x   ) cos nxdx]

)0  (( x   )

n 1 cos 2nx cos nx = [  ]  n2 n2 1 1  (1) n = [ ]  n2

sin nx 2 cos nx )  1(  2 )2 ] n n

an = 0 when n is even

1 2 . when n is odd  n2  2 1 bn  [   sin nxdx   ( x   ) sin nxdx] =

=

0

1

{[  ( 

cos nx  cos nx 2 sin nx )]0  [( x   )( )  (1)( 2 )]2 } n n n

1    [ (1) n   ]  n n n n (-1) 2 =  n n =

The Fourier series for f(x) is

f(x)=-

 4

2

cos(2r  1) x  2  (1) n  sin nx (2r  1)2 n r 0 n 1 

 

f(-0) = -

f(+0) = 0

f(  0)  f (  0) -  0  f(x)= = =2 2 2 2 1         2  (2r  1) 2 4 4

1 2   2 8 r 1 (2r  1) 

Therefore

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139

5.5.2

Self Assessment Question – IV

1.

Show that for all values of x on (-  ,  ),

x sin 2 x sin 3x sin 4 x  sin x     ....... 2 2 2 2 …………………………………………………………………………………………… …………………………………………………………………………………………… 5.6

Half Range Fourier series

If a function is defined over the range 0 to , instead of 0 to 2 or - to , it may be expanded in a series of sine terms only or of cosine terms only. This is particularly useful in the solution of some partial differential equations whose boundary conditions may restrict us to a series which contains sine terms only or cosine terms only. The series produced is then called a half range fourier series. In such cases, we may have to construct a function in the full range 0 to 2, or - to  in such a way that the fourier expansion for this series is valid in the half range. Suppose we are given a function f(x) defined in the interval (0, ). Here we extend the function in the interval (-,0) and then find the fourier coefficients in the interval (-,). Such an extended function should converge to given function f(x) between 0 to  also. For such an extension here we consider two types of function.

I.

Suppose f(x) is defined in the interval (0, ). We define a extension of f(x) as follows and call the new function as F(x).

Let F(x) = f(x), 0<x< And F(x) = F(-x), -<x< Clearly F(x) is an even function in the interval -<x< and is identified with f(x) in 0<x<. The Fourier series for F(x) is 

F ( x)  an 

a0  2   an cos nx Where a 0   f ( x)dx 2 n 1  0 2

 f ( x) cos nxdx 0

Since F(x) =f(x) in 0<x<, the function f(x) in 0<x< is given by the expansion

f( )  k 2 2 2 k 4k 1 1 1   [ 2  2  2  .....] 2  1 3 5 1 1 1 2  2  2  2  .....  1 3 5 8

Put x =

II.

,

Suppose f(x) is defined in the interval 0<x<. Then we define as extension of f(x) as follows:

F(x) = f(x), in 0<x< =- F(-x), in -<x< Clearly F(x) is an odd function in the interval -<x< and is identical with f(x) in 0<x<. The Fourier series for the odd function F(x) is

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APPLIED MATHEMATICS 

F(x)= bn sin nx

140

Where b n 

1

2

 f ( x)sin nxdx

0

Since F(x) =f(x), in 0<x<, the Fourier series for f(x) in 0<x< is 

f(x)= bn sin nx

Where bn 

1

2

 f ( x)sin nxdx 0

i) A Fourier cosine series of a function f(x) in 0<x<is 

a0  2 f ( x)    an cos nx Where a 0   f ( x)dx 2  0 1 an 

2

 f ( x) cos nxdx 0

ii) A Fourier series of f(x) on 0<x< is given by 

f ( x)   bn sin nx Where b n  1

2

 f ( x)sin nxdx 0

Example 1:

4k 1 . ;  12 4k 1 a3  ( 2 );  3 4k 1 a1  . 2 ;  5 a1 

a2  0 a4  0 a6  0

etc

Solution: We note that the given series for f(x) is a cosine series for f9x) in the half range(0, ). Therefore let us now find the half range cosine series for f(x).

a0 

2

 f ( x)dx 0

=

2

2

xdx 

0

2

 (  x)dx

2

2

2 x 2 2 x2 [ ]0  [ x  ]  2  2 2 2 2 2  2  2 2 = .  [( 2  )  (  )]  8  2 2 8  3 =   4 4 =

=

2

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APPLIED MATHEMATICS 

an 

2

2

141

2

 x cos nxdx    (  x) cos nxdx 0

2

2 x sin nx 1.( cos nx)  2 2 sin nx [  ]0  [(  x) 2  n n  n -cosnx  -(-1)( )] 2 n2 =

 2 2 cos n 2 1  cos n = [  ]  n2 n2

Also, a1 = 0, a3 = 0, a5 = 0…..

2 4 2 1 [ 2] ( 2 )  2  1 2 2 a4  2  2  0 4 4 2 2 2 2 4 2 1 a6  [ 2  2 ]  [ 2 2 ]   . 2  6 6  2 .3  3 2 2 2 2 1 a10  [ 2  2 ]   [ 2 ]  10 10  5  2 cos 2 x cos 6 x cos10 x  f(x) =  ( 2    .....) 4  1 32 52 a2 

Example 2:

If f(x) = kx = k( -x)

0 x 

2

  x  , 2

Show that  f(x) =

4k sin x sin 3x sin 5 x 1 1 1 2 [ 2  2  2  .....] and deduce that 2  2  2  .....   1 3 5 1 3 5 8

Solution: We have to find the half range sine series for f(x)

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APPLIED MATHEMATICS

142

f ( x)   bn sin nx 1

bn = =

2

2

2

 kx sin nxdx    k (  x) sin nxdx 0

2

2k x( cos nx) 1(  sin nx)  2 [  ]0  n n2 2k  cos nx  sin nx  + [(  x)( )  ( 1)( )] 2  n n2

=

2k sin n 2 sin n 2 [  ] n n2 n2

=

2k 2sin n 2 [ ] n n2

4k sin n 2 = [ ] n n2 4k 1 a1  . 2 ; a2  0  1 4k 1 a3  ( 2 ); a4  0  3 4k 1 a1  . 2 ; a 6  0 etc  5 4k sin x sin 3x sin 5 x  f(x) = [  2  2  .....]  12 3 5

f( )  k 2 2 2 k 4k 1 1 1   [ 2  2  2  .....] 2  1 3 5 1 1 1 2  2  2  2  .....  1 3 5 8

Put x =

,

5.6.1 Self Assessment Question - IV 1.

If f(x) =

x 2 . 0  x   , find the cosine series for f(x)

…………………………………………………………………………………………

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APPLIED MATHEMATICS

143

UNIT QUESTIONS

UNIT-I QUESTIONS

2 2 1   1 1. Find the characteristic equation of the matrix A = 2 1 1 and then find A    7 2 3 2. Find the eigen values and eigen vectors of the following matrices

3 2 (i)   2 3

 2 0 1 1 2 1     2 2 3 

(ii)

2 2 0 2 2 0    0 0 1 

(iii)

 1 1 2   3. Verify Cayley – Hamilton Theorem for the matrix A = 2 1 3    3 2 3 x  2y  2  3 4. Show that the equations :

3x  y  2 z  1 2 x  2 y  3z  2

are consistent and solve them

x  y  z  1 5. Solve the equation 6. Solve x  8x the other two. 4

7. Solve

3

4 x2  24 x2  23x  18  0, given that the roots are in A.P.

 14 x2  18x  15  0, given that sum of two roots is equal to the sum of

x4  4 x3  2 x2 12 x  9  0 given it has two pairs of equal roots

8. Determine a and b so that the equation equal roots, find the roots. 9. If

is a root of the equation ,

x4  4 x3  ax2  4 x  b  0 has two pair of

x3  x2  2 x  1  0 show that  2  2 is also a roots.

 ,  ,  ,  are the roots of x4  7 x3  8x2  5x  10  0 2 2 2 2 (   2 ) (   2) (  2) (  2) 10. If

, find the value of

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APPLIED MATHEMATICS

144 1  4 3  2  2 1

1

 2  5  2 =0, A1

2.

i)

  1,5; x1  K1   ; X 2  K 2   1 1

ii)

  3,3,5; x1  [3K2  2K1; k1k2 ]; X 2  K[1, 2, 1]

=

1

1



(iii) 1, 2, 2 ; K[0,0,1]; K K1[1, 3.

1 , 0] 2

---

4. -1, 4, 4 5.

1 9  , 2, 2 2

6. 5, -1, 3, 1 7. ---

1 2 ; 1 3 9. x  x  2 x  1  0 8. a = 2, b = 1, 3

2

10. Product of the roots – 166

UNIT-II QUESTIONS th

Cos 4 x

th

x 2 sin 5x

1.

Find the n derivative of

2.

Find the n derivative of

3.

If

4.

 2u u  x  y  z  3xuz, find 2 x

5.

if

u  sin 1

2

y  2u  2u s.t.  x xy xy

2

2

 x4  y 4  u u u  log  3  show that x  y x y  x y 

Ans:

1.

2.

1 n   n n    yn   4.2n cos  2 x     4 cos  4 x   8 2  2   

yn  5n2

 n  2   25 x sin  5 x  2  

      10nx sin 5 x  n  2   n(n  1)sin(5 x  n  2) 2 2 

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APPLIED MATHEMATICS 3

proof

4.

6x

5.

3

145

UNIT-III QUESTIONS

1. Evaluate

e

x

sin xdx

 2

2. Prove that

dx

 1  tan 0

3

x

 4

 4

3. Evaluate :

 Cos x sin 5

5

xdx

0

4. Solve :

( D2  4) y  x 2 sin 2 x

5. Solve :

( D2  D) y  3xe x

1.

I

3.

1/60

4.

5.

ex (sin x  cos x)  c 2

1  x 2 cos 2 x   x3 x  y  A cos 2 x  B sin 2 x        sin 2 x 8  32   3 8   3e x  x 2 x y  A  Be    x 2  2  UNIT-IV QUESTIONS z  p2  q2

1.

Solve :

2.

Solve : Sin x – p = - Siny + q

3.

Solve :

x2 p  y 2q  z 2  0

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APPLIED MATHEMATICS 4.

Solve :

5.

Solve

146

p  3q  5z  tan(q  3x)

z( x  y)  x 2 p  y 2q

Z=0

2.

Z = K ( X +Y) – ( Cos x + Cos y) +c

3.

1 1 1 1 Q  ,    0 x y y z

4.

 1  Q  y  3x, log 5 z  tan( y  3x)   0 5   1 1 x y 5. Q  , 0 x y 2  

UNIT-V QUESTIONS 1.

Find the Laplace transform of (i)

sinh 5t

(ii )

Sinht t

  1 L1  2 2   S (S  a ) 

2.

Find

3.

Find the Inverse Laplace Transform of

4.

Graph the following function and find its lapalce transform

 1 S  log    S 

0t   t f (t )   ...  where f(t) = f(t + 2  )    t   t  2  f ( x)  x sin xin0  x  2

5.

Obtain the fourier series for

6.

Find the sine Series for the function

f (t )  1 in 0  t   ANSWERS: 1. 2. 3.

5 S  25 1  cos at a2 1  et f (t )  t

(i )

2

( ii)

1  S 1  log   2  S 1 

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APPLIED MATHEMATICS

4.

147

1  (1  S )e s s 2 (1  e s

5.

 1 cos nx f ( x)  1   sin x  cos x   2 2 n2 n  1

6.

f ( x) 

4  sin x sin 3x sin 5 x     .....   1 3 5 

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APPLI EDMATHEMATI CS

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IMTS Mechanical Eng. (Applied mathematics)

IMTS Mechanical Eng. (Applied mathematics)