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1

1
MATRICES
UNIT STRUCTURE
1
...
1 Introduction
1
...
3

Illustrative examples

1
...
5 Canonical form or Normal form
1
...
7
Let Us Sum Up
1
...
0 OBJECTIVES
In this chapter a student has to learn the
Concept of adjoint of a matrix
...

Rank of a matrix and methods finding these
...
1 INTRODUCTION
At higher secondary level, we have studied the definition of a matrix,
operations on the matrices, types of matrices inverse of a matrix etc
...


1
...

The matrix of order m n is written as

2

a11 a12 a13 a1 j a 1n
a 21 a 22 a 23 a 2 j a 2n

...

The elements are generally denoted by corresponding small letters
...

For e
...


x

1
2
4

3 1

3) Row Matrix :
It is a matrix in which there is only one row
...

i
...


3

e
...

A

2 3
4 6

2 2

5) Diagonal Matrix:
It is a square matrix in which all non-diagonal elements are zero
...
g
...

e
...


A

5 0 0
0 5 0
0 0 5

3 3

7) Unit Matrix:
It is a scalar matrix with diagonal elements as unity
...
g
...


4

e
...


A

1 3 0
0 0 1
0 0 5

3 3

9) Lower Triangular Matrix:
It is a square matrix in which all the elements above
diagonal are zero
...
g
...

A

1 3 5
3 7 9

2 3

1 3
T

A

Transpose of A

3 7
5 9

3 2

11) Symmetric Matrix:
If for a square matrix A, A

A

AT then A is symmetric

1 3 5
3 4 1
5 1 9

12) Skew Symmeric Matrix :
If for a square matrix A, A

AT then it is skew -symmetric matrix
...

Determinant of a Matrix:
Let A be a square matrix then
A = determinant of A i
...


Note : for non-singular matrix A-1 exists
...

E
...
Consider,
a11 a12
A = a 21 a 22
a 31 a 32

a13
a 23
a 33

M 11 = Minor of an element a 11

6

A=

a 22

a 23

a 32

a 33

II y
M12 =

a 21
a 31

a 22
a 33

E
...

(ii) Let,
2 5 8
A= 1 3 2
0 4 6

M11 =

M 21 =

3 2
1 2
1 3
, M12 =
, M13 =
4 6
0 6
0 4
5 8

, M12 =

4 6

2 8

, M 23 =

0 6

2 5
0 4

(b) Cofactor of an element :If A = a ij

is a square matrix of order n and a ij denotes cofactor of the

element a ij
...
M ij Where M ij is minor of a ij
...
g
...


2 1
7 6

=

1


...


Of a matrix A of order nxn, is called a cofactor matrix
...


8

Thus in the notations used,
Adjoint of A CT
A1
Adj A = A2
A3

B1
B2
B3

C1
C2
C3

Adjoint of a matrix A is denoted as Adj
...
A =
3 7 6

10

3

3
6

6
9

1
2

Note :
If A =

a b
c d

than Adj
...

Inverse of A is denoted as A-1 and read as A inverse
...

A-1 =

1
Adj
...

A

Note:- From this relation it is clear that A-1 exist if and only if A
A is non singular matrix
...
3 ILLUSTRATIVE EXAMPLES

0 i
...
F
...
F
...
F
...
F
...
F
...
F
...
F
...
F
...
F
...


1 2

1 3


...


2 1

3 1

3 2

3 3

3 1
1 2


...



...


...


2 2

2 3

1 3

1

1 2
1
1 1
0 2
3 1
0 1
3 1

6

3

1 2
2 3
0 2
1 3
0 1
1 2

1
2
1

Thus,
Cofactor of matrix C =

And Adjoint of A= C1

1
1
1

8
6
2

5
3
1

11

1
=

1

8
5

6
3

1
2
1

A

1
1
8
2
5

1

1
6
3

1
2
1

Note:- A Rectangular matrix does not process inverse
...
e
ii)
The inverse of the transpose of a matrix is the transpose of inverse
T 1
(A 1 )T
i
...
(A )
iii)
If A & B are two non-singular matrices of the same order
1
(AB)
B 1A 1
This property is called reversal law
...
:If a square matrix it satisfies the relation AAT
is called an orthogonal matrix
...


Sin

Solution:
To show that A is orthogonal i
...


Check Your Progress:
Q
...

i)

1

2

2

2

1
iv)

,

3
3 0

2
1
vii) 1
2

2
5,

cos
v) sin
0

5 0
1
2
1

Q
...
C-1
4 3
Q
...
A= A
3

1 2 1
Q
...
A)1= (Adj
...
6) Find the inverse of A = 0 1
2 2
3 6
A= 0 3
6 6

3
3
9

1
1 , hence find inverse of
3

13

1
...
The determinant of any
submatrix of a square order is called minor of the matrix A
...

e
...
Let
1 3
A= 4 0
8 5

1
1
4

4
7
3

The determinants
1 3
4 0
8 5

1
3
1 , 0
4
5

1
1
4

4
1
7 , 4
3
8

1 3 0 1 3 4
,
,
, 1, 0 ,
4 0 5 4 0 7

1
1
4

4
7 ,
3

3,

Are some examples of minors of A
...
g
...
g
...


2

19 0

Rank of A= 3
1
(ii) A = 1
0

1
2

2
3

1

1

2
3

11

Here,
1
A1 = 1
0
A2 =

1 1
1 2

1
2
1

1 1

2

1

0

1
1 0

Thus minor of order 3 is zero and atleast one minor of order 2 is non-zero
Rank of A = 2
...

(ii)
Rank of any non-zero matrix is always greater than or equal to 1
...

(iv)
Rank of transpose of matrix A is always equal to rank of A
...

(vi)
Rank of a matrix remains unleasted by elementary
transformations
...

(i)
Interchanging any two rows (or columns)
...


15

(iii) Adding non-zero scalar multitudes of all the elements of any row
(or columns) into the corresponding elements of any another row (or
column)
...
Two
equivalent matrices have the same order & the same rank
...

1 2 3
A= 1 4 2
2 6 5
Solution:
1 2 3
Given A = 1 4 2
2 6 5
R2

R2 - R1 & R3

1 2
0 2

3
1

0 2

R3 - 2R1

1

Here two column are Identical
...


1
...
g
...
A if
2
A= 3
1

1
3
1

3
1
1

6
2
2

1
3
1

3
1
1

6
2
2

Solution:
2
A= 3
1
R1

R3
1
3
2

1
3
1

1 2
1 2
3 6

R 2 3R1 , R3 2R1
1

1

1

2

3
0

6
1

2
5

4
10

1 1
1 33
1 5

2
66
10

R 2 7R 3
1
0
0

17

R1 R 2 , R3 R 2
1 0
0 1

32
33

64
66

0 0

28

56

32
33
1

64
66
2

1
28

R3

1 0
0 1
0 0
R1

32 R 3 , R 2

33 R 3

1 0 0 0
0 1 0 0
0 0 1 0
I3

o

Rank of A=3
Example 6: Determine the rank of matrix
1 2

7

A= 2 4 7
3 6 10
Solution:
1 2
A= 2 4

3
7

3 6 10
R 2 2R1 , R3 3R1
1 2

3

2 4 7
3 6 10
R3 R 2
1 2 3
0 0 1
0 0 0
R1 3R 2

18

1 2 0
0 0 1
0 0 0
C2 2C1
1 0 0
0 0 1
0 0 0
C1

C3
1 0 0
0 1 0
0 0 0
I2

0

Rank of A= 2
Example 7: Determine the rank of matrix A if
1
2
A=
3
6

1
3
1
3

2
1
3
0

4
1
2
7

1
3
1
3

2
1
3
0

4
1
2
7

Solution:
1
2
A=
3
6

R 2 2R1 , R3 3R1 , R 4 6R1 ,
1
0
0
0

1 2 4
5 3 7
4 9 10
9 12 17

R 2 R3

19

1
0
0
0

1 2 4
1
6 3
4 9 10
9 12 17

R1 + R 2 , R3 4R 2 , R 4 9R 2
1
0
0
0

0 8 7
1 6 3
0 33 22
0 66 44

R 4 2R 3
1
0
0
0
R3

0 8 7
1 6 3
0 33 22
0 0 0

1
11
1 0

0 1
0 0
0 0

8
6
3
0

7
2
2
0

1
3
1
0

7
3
2
0

C3 - C 4
1
0
0
0

0
1
0
0

R1 + R3 , R 2 3R 3
1
0
0
0

0
1
0
0

0
0
1
0

5
3
2
0

C4 - 5C1 3C2 2C2
1
0
0
0

0
1
0
0

0
0
1
0

0
0
0
0

20

I3 0
0 0
Rank of A= 3
Check Your Progress:Reduce the following to normal form and hence find the ranks of the
matrices
...
6 NORMAL FORM PAQ
such that,
Ir
0

0
0

PAQ

We observe that, the matrix A can be expressed as
A = Im
Where Im In are the identity matrices of order m and n respectively
...
A in L
...
S
...
The equation can be transformal into the
equations
...
H
...

and prefactor (i
...
Im in equation (i)) and column operations can be
performed simultaneously on L
...
S
...
H
...
i
...
[(In in
eqn (i)]

Examples 8: Find the non-singular matrices P and Q such that PAQ is in
normal and hence find the rank of A
...

Also,
Rank of A = 3
...


i)

1 0
2 3
3 3

2
4
6

ii)

1 2 3 2
2 3 5 1
1 3 4 5

3
iii) 1
1

1 1
1 1
1 1

25

2
iv)

3 4

7

3 4 7
5 4 6

9
5

(v)

1 3 5 7
4 6 8 10
15 27 39 51
6 12 18 24

1
...

Using Adjoint method to find the A
1
using formula A 1
adjA
A

1

by

Rank of the matrix using row & column transformation
Using canonical & normal form to find Rank of matrix
...
8 UNIT END EXERCISE

1)

1 2 3
4 5 6
7 8 9

Find the inverse of matrix A

1
ii)

Find Adjoint of Matrix A

1

0
2

2
1

1
1
1

if exists
...


vi)

Reduce the matrix A

0
1
3
1

1
0
1
1

3
1
0
2

1
1
to the normal form
2
0

and

find its rank
...
such that

1

Also find the rank of matrix B
1

1

X= 2
3

viii)

3
2

1
4 &Y
3

1
6
5

2
12
10

1
6
5

Under what condition the rank of the matrix will be 3!
2 4 2
A= 2 1 2
1 0

is the normal

27

1
ix)

If X = 2
3

1
3
2

Then show that

x)

1
4 &Y
3
xy

1
6
5

2
12
10

yx where

1
6
5
denotes Rank
...
1
Objectives
2
...
3
Canonical or echelon form of matrix
2
...
5
Let Us Sum Up
2
...
1 OBJECTIVES
After going through this chapter you will be able to
Find the rank of Matrix
...

Type of linear equations
...

Find solution of non-Homogeneous equations
...
2 INTRODUCTION
In XIIth we have solved linear equations by using method of
reduction also by rule
...
Using matrix we can
discuss consistency of system of equation
...
3 CACONICAL OR ECHOLON FORM OF MATRIX
Let A be a given matrix
...

(ii)

The elements in the remaining rows are zero
...

2)
To reduce the matrix to Echelon form only row transformations are
to be applied
...


A

2
1
3
6

3 -1
-1 -2
1 3
3 0

1
4
2
7

Solution:
2
1
3
6

A

R1

R2 2 R1

R3

R3 3 R1

R4

R4 6 R1

1
4
2
7

1
2
3
6

-1 -2
3 -1
1 3
3 0

4
1
2
7

R2

A

R2

3 -1
-1 -2
1 3
3 0

29

1
0
0
0

A

1 2 4
5 3 7
4 9 10
9 12 17

4
5
9
R4
5
1 1
0 5
0 0
0 0

R3

R3

R4

A

R4

R4
1
0
0
0

A

R2
R2
2
4
3
7
33 5 22 5
33 5 22 5

R3
1
2
4
5
3
7
0 33 5 22 5
0
0
0

Rank of A

e A

No
...

1 2 3
i)
A 2 4 7
Ans : 2
3 6 10

ii)

A

iii)

A

1
3
1
1
3
2
-1
1

2
4
0
2
4
4
-2
-1

1
0
2
3
1
3
6
2

3
1
7
1
1
6
Ans : 4
4
-3

30

2
...
e
...
D
...
e
...


( AD)
( AD)

( A) then the system is consistent and if
( A) Number of unknowns then the system is

consistent and has unique solution
...


( A)

Number of unknowns and has infinitely many

Non- Homogeneous equation:System of simultaneous equation in the matrix form is
Pre-multiplying both sides of I by A 1 we set

31

A 1 AX A 1 D
IX A 1B
X A 1B
which is required solution of the given non-homogeneous equation
...

AX D
If all elements of D are zero
i
...

In this case coefficient matrix A and the augmented matrix [A,O]
are the same
...
It follow that the system has solution
x1 , x2 , x3
...


Example 2: Solve the following system of equations
2 x1 3x2

x3

0

x1 2x2 3x3 0
4x1 x2 2x3 0
Solution: The system is written as

AX 0
2 3 1
x1
0
1 2
3
x2
0
4 1 2
x3
0
Hence the coefficient and augmented matrix are the same
We consider

A

2
1
4

3
2
1

1
3
2

A

2
1
4

3
2
1

1
3
2

R1

R1

R2

32

1
2

R2

1

2

R2 -2 R1 & R3
1
0
0

R2

2
7
9
R2

1
0
0

R3

3
1

4

A

2
3

3
7
10
1

7

2
1
9

3
1
10

R3 +9R2 & R1
1 0
0 1
0 0

R2

R1 2R2

1
1
19

R3

1

1 0
0 1
0 0

R3

R3 4R1

1
1
1

7

R2 +R3 & R1

R1 R3

1 0 0
0 1 0
0 0 1
Hence Rank of A is 3

( A) 3,
The coefficient matrix is non-singular
Therefore there exist a trivial solution
x1 x2 x3
Example 3:

0

Solve the following system of equations
x1 3x2 2 x3 0

2x1 x2 4x3 0
x1 11x2 14x3 0
Solution: The given equations can be written as

AX

0

33

1
2

3
1

2
4

x1
x2

0
0

1 11 14
x3
0
Here the coefficient & augmented matrix are the same
1
A

R2

3

2

2
1

1 4
11 14

R2 -2 R1 & R3

1
0
0

3
2
7 8
14 16

R3

R3 R1

R3 2R2

1 3
2
0 7 8
0 0
0
Here rank of A is 2 i
...

1
0
0

3
7
0

2
8
0

x1 3x2 2x3

x1
x2
x3

0

7 x2 8x3 0
7 x2 8x3
8
x2
x3
7
Let x3 8 x3
8
x2
7
8
x1 3
2
0
7
24
x1
2
0
7
24
x1 2
7
10
x1
7

0
0
0

34

Hence x1

x1
x2
x3

10
7

x2

8
7

and

x3

10
7
8
7

Hence infinite solution as deferred upon value of

Example 4: Discuss the consistency of

2x 3 y 4z
2
x y 3z 4
3x 2 y z
5
Solution: In the matrix form
2
1
3

3
1
2

4
3
1

x
y
z

-2
4
-5

Consider an Agumental matrix
2
1
3

A: D

R2

R2

R3

R3

3
1
2

0

R3

3
5

0

R2

5

4 :

2

5

:

5

5

:

2

4 :

2

2

2

5

:

5

5

:

7

R2
2

A: D

2
4
5

1
R1
2
3
R1
2
2

A: D

4 :
3 :
1 :

0

3
5

0

0

2

35

AD
A

3

2
AD

A

The system is inconsistent and it has no solution
...
(1)
8
7

AD
A
AD

3

3

A

Number of unknowns

system is consist and has unique solution
...
At the end of the
row transformation the value of z is calculated then values of y and the
value of x in the last
...
g
...

Step (2) :- To find the solution we proceed as follows:
Let

38

z

k
...

x1 2x2 3x3

v)

2 x1 4 x2 7 x3

0

3x1 6x2 10x3

0

0

Ans : Definitely many solution
x1
x2

1
2
0

x3

2
...

Representing linear equation m x n in to argumented matrix
...

Solution of Homogeneous equations
...


2
...


i)

ii)

A

A

1 3 6
1 4 5
1 5 4
1
4
3
1

2
1
1
2

1
1
3
1
2
1
0

3
1
2
1

Ans : Rank = 2

Ans : Rank = 3

40

iii)

A

1 2 3
2 1 0

Ans : Rank = 2

0 1 2

iv)

A

1
1
3

1
1
1

1
1
1

Ans : Rank = 2

2) Solve the following system of equations
...

ii) 2x1 - x2 - x3 = 0, x1 - x3 = 0, 2x1 + x2 - 3x3 = 0
Ans:iii)

x1

x2

x3


...


iv)

x1 - 4x2 - x3 = 3
3x1 + x2 - 2x3 =7
2x1 - 3x2 + x3 = 10
...

1


...

Conjugate of a Matrix:
The matrix of order mxn is obtained by replacing the elements by their
corresponding conjugate elements, is called conjugate of a matrix
...
g
...
B

Conjugate Transpose:
Transpose of the conjugate matrix A is called conjugate transpose
...

For e
...
A

A

1 i
3

1 i
i
1
3 i 2 3i 2
i

1
then A
3i 2

i 2

1 i

3

i
1

i 2
3i 2

Properties of Transpose of Conjugate of a matrix:
(1)

A

(2)

A B

(3)

AB

A
A
B
...
e
...


Example 10:
1

Show that the matrix A

2 i
3 i

2 i 3 i
i is Hermitian
3

3
i

Solution:
1

2 i 3 i
2 i
3
i

Here A

3 i
1
A

i

3

2 i 3 i

2 i
3 i

3
i

i
3

1

2 i 3 i
2 i
3
i

A

3 i

A

i

3

A

Hence by definition A is Hermitian matrix
...

i
...
if A aij m n is Skew Hermitian if aij
a ji i and j
...

Example 11:
2i

Show that the matrix A

5 i
6 i

Matrix
...

Note:
Let A be a square matrix expressed as B+iC where B and C are Hermitian
and Skew Hermitian Matrices respectively
...

3 i

73

1 15 0
15 0 15
AA

1 0
0 1

I

I

Hence A is Unitary Matrix
...

Solution:
2 i

Let

A

1

i

1 i
1 i
3

2 i

1
1 i

2 i

1 i

3

2 i
...
( II )
2 i

5

Adding I and II we get

A A

4
i 1
4 4i

B

2 i 1 3 3i
i
1 i 2 i
1 i
3
5
1 i
2
i 1

1
A A
2

2 i
i 1 i
1
1 i
3
3 3i 2 i 5

4 4i
i 1
10
4
1
i 1
2
4 4i

1 i

4 4i

2
i 1

i 1
...
( IV )

5 i

0

Now, A=B+iC
4
1
i 1
2
4 4i

A

1 i
2
i 1

10

1 i
2
5 i

2 2i
5 i
0

Example 14:
Prove that the matrix, A
Solution:
1 1
Let A
2 i

A A

1
i
1

1 1
2 i

i
1

i i
i2 1

1 2 0
2 0 2
AA

1

i

1 1
2 i

1 1 i2
2 i i

i

i
1

1 1
2 i

A

1 1
2 i

1 0
0 1

I

I

Hence A is Unitary
...

2i

(i) A
(2)

2

3

2 4i

6

4i

(ii) A

i 1

1 i 2 2i
i

5i

3 6 0
2 2i
5i
3i
Show that the following matrices are Unitary matrices
...


4
...

Minimal
...

Complex matrix
...
i
...
i
...


4
...

i)

Show that the given matrix A satisfies its characteristics equation
...


Using Cayley Hermitian theorem find inverse of the matrix A
...


Calculate A5 by Cayley Hamilton Theorem if A
2
2

Let A

3

1

6

1

4
...
Find a similarity transformation that

diagonalises matrix A
...


2

Let A

2
3

2

6
...


1 0
Diagonalise the matrix 1 2
2 2

7
...

0 1 4

Determine a matrix P such that is diagonal matrix
...


If show that is Hermitian matrix
...


Show that the following matrix are skew Hermitian matrix
...


i)

1 i
2 3i

1 i 2 3i
0
6i

6i
0

Show that the following matrix are unitary matrix

A

1 i
2
1 i
2

1 i
2
1 i
2

MEAN VALUE THEOREMS
Unit Structure
12
...
1 Introduction
12
...
1
Theorem:
12
...
2
12
...
3
12
...
4
12
...
5 Some Important Deductions from the Mean Value Theorem:
12
...
2
...
2
...
3 Summary
12
...
0 OBJECTIVES:
After going through this chapter you will be able to:
State and prove

12
...
Let us consider the following real life event to understand the concept
of this theorem: If a train travels 120 km in one hour, then its average speed
during is 120 km/hr
...
Thus, the Mean Value Theorem tells us
that at some point during the journey, the train must
have been traveling at exactly 120 km/hr
...
Geometrically we can say that MVT states
that given a continuous and differentiable curve in an
interval [a, b], there exists a point c [a, b] such that
the tangent at c is parallel to the secant joining (a, f(a)) and (b, f(b))
...
1
...
1
We know that
is the slope of the tangent to the graph of f at x = c
...
1
...

Algebraic Interpretation of Rolle's Theorem:
We have seen that the third condition of the hypothesis of Rolle's theorem is that
f(a) = f(b)
...

Thus Rolle's theorem implies that between two roots a and b of f(x) = 0 there
always exists at least one root c of
= 0 where a < c < b
...

Example 1:
2
2
(1) x in
(2) x in

x2 , x

Solution: (1) Let f x

As f x is a polynomial in x, it is continuous and differentiable everywhere on
its domain
...
f(x) is continuous and differentiable everywhere on
its domain
...
e
...

But we have
and f 3 9 which are not equal
...
e
...


(ii)

=

=

exists in (-3, 0)

(iii)
There exists

such

that
=0
c = 3, -2
c = - 2 is the required value
...

Also
There exists

such that
(i
...
)

Example 4:

, which lies in (a, b)
...

Solution: Since e x , sinx, cosx are continuous and differentiable functions, the
given functions is also continuous in
Also,

and differentiable in

=

=0

Now,

=

c
Example 5:

f(x) =

Solution: We have f(x) =
Since

/ 2 , which lies in
,


...
Now

],

is also continuous

=

=0

The derivative of f x should vanish for at least one point
that

Since c =

Now,


...

Example 6:

If

,

,

are differentiable in

exists a value c in (a, b) such that

Solution: Consider the function

Since f x ,

such

x ,

=0

defined by, F(x) =

x are differentiable in

a, b
...


c

i
...


=

then show that

has

Solution

in three intervals

,

,
We observe that
(i)
is continuous in all the intervals since it is a polynomial in x
...


(iii)

has three real roots
...


Solution: Let a and b be two real roots of the equation
(i
...
) of sin x

e

x

(i
...
) of

Let

, which is continuous and differentiable
...
Since a and b are roots of


...


Example 9:
has a root between 0 and 1
...

Here
is continuous in

and differentiable in (0, 1) and

such that
Now,
equation,

and this is zero at x = c which means the
has a root between 0 and 1
...


Show that the equation

where

has exactly

Solution: Let
and
Since

is a polynomial, it is continuous
...

Now, let if possible
have two roots, say a and b
...


represents a polynomial, it is differentiable on (a, b) and continuous

on
c between a and b such that

But
,
Hence

for any x, which is a contradiction
...


The equation

has exactly one root
...

Theorem for the function f defined on the intervals as given below:
a)

on

b)
c)

on
on

d)
e)

on
on

f)

in

2
...

3
...

5
...


1,1

Ans:(1)

12
...
2
Theorem 6
...
1
...
1
...
The slope m of the line AB is given by, m

=
Also,

is the slope of the tangent at the point C

there exists at least one point C c, f c on the graph
where the slope of the tangent line is same as the slope of
line AB
...
e
...

Physical Significance:
We note that
a to b, so that
over


...


Thus the theorem states that the average rate of change of a function over an
interval is also the actual rate of change of the function at some point of the
interval
...
1
...
Let

such that

is said to be a monotonically increasing function
...
Let
then

Note:
(i)
If


...
If

is said to be a monotonically decreasing function
...


is its minimum value and

its maximum value
...


is

then we can

is its maximum value and

is its minimum value
...
Let

and

x2

, we get

or
(1)

(*)
Let

for every value of x in
for

then from equation (*)

and

both are positive i
...


We have thus proved: A function whose derivative is positive for every value of
x in an interval is a monotonically increasing function of x in that interval
...


is a decreasing function of x
...

Example 11: Verify mean value theorem for
Solution: The given function is
We know that

is continuous on

such that

on
on
and differentiable on (1, e)
...

Example

12:

Separate the interval in which
is increasing or decreasing
...
e
...
e
...
e
...
e
...
e
...
e
...
e
...
e
...
e
...

is decreasing in (2, 3)

is increasing in

Example 13:

and

Find the interval in which

decreasing
...

x2
i
...
if x 2 1 0
...
e
...
e
...

is increasing or

Hence

is increasing in the interval

(ii)

is a decreasing function if
i
...


i
...

i
...
if x

1

Hence

1 x 1
is decreasing in (-1, 1)
...


Solution: Let us assume,

for all x

0 except at

and

is an increasing function in
increasing from 0 and hence


...


is an increasing function in
increasing from 0 and hence
for

From (i) and (ii),

for

Example 15:

Show that

Solution:

Let

f b f a
b a

f' c
for

But,

Example 16:

0

(

Show that

c 2 is positive)

where

and

1

Solution:
For

Let

=

we have,

putting, a = 0 and h = x
...
Hence deduce that,
Solution:

Let

in terms of a

Now,

But,
Now by substituting h = x in the above equation, we get,

Example 18:

Show that,

Hence show that
Solution:

Let

in

where
(1)
Since

(2)
From (1) and (2)

b a
1 b2

tan 1 b tan 1 a

b a
1 a2

(3)

For the second part;
Since

we put a = 1 and

in (3)

Example 19:

Prove that,

for

Hence deduce that
Solution:

Let

in

Since

is (i) continuous in

and (ii) differentiable in (a, b)
such that

But

(1)
But a

c b,

1
a

1
c

1
b

(2)

From (1) and (2) we get,

For the second part a = 3, b = 4
...

Examine the validity of the conditions and the conclusions of LMVT for
the functions given below:
(i)

e x on

(ii)

on

(iii)
(iv)

in
on

2
...
Hence deduce that:

in


...
Applying LMVT show that:
for x

(i)

0

(ii)

for

x 0
...
Prove that,
Hence deduce that,
(i)

(ii)

5
...

[

(ii) x 3 6x 2 36x + 7

(i)

]
for 1 x
...
Show that,
7
...
2
If functions f and g are (i) continuous in a closed interval [a, b], (ii) differentiable
in the open interval (a, b) and (iii)
for any point of the open interval
(a, b) then for some

,

i
...

12
...
1
If two function

and

are derivable in a closed interval [a, a + h] and

for any x in (a, a + h) then there exists at least one number
such that,
Th
mean value theorem
...

(ii)
Usefulness of this theorem depends on the fact that
f
g
for some

and

are

12
...
2
:
Geometrically, we consider a curve whose paramedic equations are

...
If
that the tangent to the curve at

As

such

is parallel to the chord AB
...

Solution:

and

Let
and

and let


...

Example 21: Using CMVT show that

Solution:
Here,

Let
and

and
are continuous on [a, b] and differentiable on (a, b) and

for any c in (a, b), thus CMVT can be applied
...

Solution: Now

and

If can be proved that function

and

are continuous on any closed

interval [a, b] and differentiable in (a, b)
...


Now

and

such that,

and

e 2c

ea

where

b

Thus, c is the arithmetic mean between a and b
...
By putting

deduce that

n b

1

n

1

log b
...
Hence

If f x

x

1

n

and

satisfy the condition of continuity and

c

differentiability

a, b such that,

then by putting a = 1 we get in the interval (1,

b)
where 1 c

b

[
If 1 a

Example 24:
that

as

]

b , show that there exists c satisfying a c b such

Solution: We have to prove that,
This suggests us to take

and

Now,

continuous on [a, b] and differentiable on (a, b) and

and

are

for any c in (a,

b)
...


such that,

Check Your Progress
1
...
(iv)

f x

ex ,

on [0, 1]

x2

ex , g x

f x

[Ans :- (i)
12
...
]

Summary
In this chapter we have learnt about the mean value theorems
...


based on these theorems have been done in order to understand the Mean Value
theorems
...

12
...
Verify LMVT for the following functions
...
Gamma Function
Definition
Γ(n) =

∞
0

x n-1 e -x dx ; n > 0

& Γ(n) = Γ(n+1) / n

; n Є R - Z0

Results:
(1) Γ(n+1) = n Γ(n) ; n > 0 , where Γ(1) = 1

(2) Γ(n+1) = n!

; nЄ N

(3) Γ(n) Γ(1- n) = π /sin(nπ)

( convention: 0! = 1)

;0
In Particular;
Γ(1/2) = π

1

Examples:

Example(1)
Evaluate

∞
0

x 4 e -x dx

Solution
∞
0

x 4 e -x dx =

∞
0

x 5-1 e -x dx = Γ(5)

Γ(5) = Γ(4+1) = 4! = 4(3)(2)(1) = 24

Exercise
Evaluate

∞
0

x 5 e -x dx

Example(2)
Evaluate
∞
0

∞
0

x 1/2 e -x dx

x 1/2 e -x dx =

∞
0

x 3/2-1 e -x dx = Γ(3/2)

3/2 = ½ + 1
Γ(3/2) = Γ(½+ 1) = ½ Γ(½ ) = ½ π

Exercise
Evaluate

∞
0

x 3/2 e -x dx

2

Example(3)
Evaluate
∞
0

∞
0

x 3/2 e -x dx

x 3/2 e -x dx =

∞
0

x 5/2-1 e -x dx = Γ(5/2)

5/2 = 3/2 + 1
Γ(5/2) = Γ(3/2+ 1) = 3/2 Γ(3/2 ) = 3/2
...
π =

Exercise
Evaluate

∞
0

x 5/2 e -x dx

Example(4)
Find Γ(-½)
(-½) + 1 = ½
Γ(-1/2) = Γ(-½ + 1) / (-½) = - 2 Γ(1/2 ) = - 2 π

3

¾ π

Example(5)
Find Γ(-3/2)
(-3/2) + 1 = - ½
Γ(-3/2) = Γ(-3/2 + 1) / (-3/2) = Γ(-1/2 ) / (-2/3) = ( - 2 π ) / (-2/3) = 4 π /3

II
...
cos 2n-1x dx = Γ(m) Γ(n) / 2 Γ(m+ n) ; m>0 & n>0
x q-1 / (1+x)
...
3! / 8! = 3!/(8
...
6
...
7
...
(1 – x ) dx

Solution
I =

1
0

x 1/2 (1 – x ) dx =

1
0

x 3/2 - 1 (1 – x ) 2 - 1 dx

= B(3/2 , 2) = Γ(3/2) Γ(2) / Γ(7/2)
Γ(3/2) = ½ π
Γ(5/2) = Γ(3/2+ 1) = (3/2) Γ(3/2 ) = (3/2)
...
(3π / 4) = 15 π / 8

Thus,

I =

(½ π )
...
Using Gamma Function to Evaluate Integrals

Example(1)
Evaluate: I =

∞
0

x 6 e -2x dx

Solution:
Letting y = 2x, we get
∞

I = (1/128) 0

y 6 e -y dy = (1/128) Γ(7) = (1/128) 6! = 45/8

Example(2)

6

Evaluate: I =

∞
0

x e –x^3 dx

Solution:
Letting y = x3 , we get
∞

I = (1/3) 0

y -1/2 e -y dy = (1/3) Γ(1/2) = π / 3

Example(3)
Evaluate: I =

∞
0

xm e – k x^n dx

Solution:
Letting y = k xn , we get
∞

I = [ 1 / ( n
...
k (m+1)/n) ] Γ[(m+1)/n ]

II
...
cos 2n-1x dx = (1/2) B(m,n)

; m>0 & n>0

; m>0 & n>0

x q-1 / (1+x)
...
dx

Solution:
Letting x = 2y, we get
1

I = (8/2) 0  y

2

(1 – y ) -1/2 dy = (8/2)
...
dx

Solution:
2

2

Letting x = a y , we get
6

1

I = (a / 2) 0  y

3/2

(1 – y )1/2 dy = (a6 / 2)
...
dx
Hint
3

Lett x = 8y
Answer

8

1

-1/3

(1 – y ) 1/3
...
Γ (1/4)
...
[ π / sin ( ¼
...
Evaluate: I = 0 

sin 3
...
Evaluate: I = 0π/2 sin 4
...
Notice that: 2m - 1 = 3 → m = 2

& 2n - 1 = 2 → m = 3/ 2

I = (1 / 2) B( 2 , 3/2 ) = 8/15

b
...
Evaluate: I = 0 

sin6 dx

b
...
Notice that: 2m - 1 = 6 → m = 7/2 & 2n - 1 = 0 → m = 1/ 2
I = (1 / 2) B( 7/2 , 1/2 ) = 5π /32

b
...
Evaluate: I = 0π cos4x dx
2π

b
...
I = 0  cos x = 2 0 

π

8

cos4x = 2 (1/2) B (1/2 , 5/2 ) = 3π / 8

π/2

b
...

Example(1)
Evaluate: I =
x = y/2
x 6 = y 6 /64

∞
0

x 6 e -2x dx

dx = (1/2)dy
x 6 e -2x dx = y 6 /64 e –y
...
(1/3)y-2/3 dy

Example(3)
∞

m

– k x^n

Evaluate: I = 0 x e
dx
y = k xn
x = y1/n / k1/n
xm = ym/n / km/n
dx = (1/n) y(1/n-1) / k1/n dy
xm e – k x^n dx = ( ym/n / km/n )
...
(1/n) y(1/n-1) / k1/n dy
m/n + 1/n – 1 = (m+1)/n - 1
-m/n – 1/n = - (m+1)/n
∞
[(m+1)/n – 1]
I = [ 1 / ( n
...

Example(1)

11

Example(1)
2

2

I = 0  x / (2 – x )
...
dx = 4 y2 / 2

(1 – y )

2dy

y=0 when x=0
y=1 when x=2

Example(2)
a

4

2

Evaluate: I = 0  x
2

2

 (a – x )
...
dx = a4 y2 a (1 – y )1/2 (1/2)a y-1/2 dy
y=0 when x=0
y=1 when x=a

Example(3)
∞

4

I = 0  dx / ( 1+x )
4
x = y

12

1/4

x=y
-3/4
dy= (1/4) y
dy

dx / ( 1+x4 ) = (1 / 4) y -3/4 dy / (1 + y )

Proofs of formulas (2) & (3)

Formula (2)
We have,

B(m,n) =

1
0

x m-1 (1 – x ) n-1 dx

Let x = sin2y
Then dy = 2 sinx cox dx
&
x m-1 (1 – x ) n-1 dx = (sin2y) m-1 ( cos2y ) n-1 ( dy / 2 sinx cox )
= 2 sin 2m-1y
...
cos 2n-1y dy

I = 0π/2 sin 2m-1y
...
dy= 1 / (1-y)2
...
1 / (1-y)2
...
1
Objective
6
...
3
Differential Equation
6
...
5
Let Us Sum Up
6
...
1 OBJECTIVE
After going through this chapter you will able to
i
...

Order & degree of differential
equation
iii
...
2 INTRODUCTION
We have already learned differential equation in XIIth
...
In this chapter we
discuss only formulation of differential equation
...
3 DIFFERENTIAL EQUATION
Definition:An equation involving independent and dependent variables and the
differential coefficients or differentials is called a differential equation
...
g
...

The differential equation is said to be ordinary if it contains only one
independent variable
...

Order and Degree of a Differential Equations:(i) Order:The order of the differential equations is the order of the highest orderal
derivatives present in the function or equation
...

dx 2 dx dx
d2 y
dy
e
...

e
...

1

2

d2 y
k2 y 0
dx 2
order = 2, degree = 1
d2 y
2
dx 2

dy
dx

2

Y

0

Order =2, degree =1
3

4

dy
x
dx

1
dy
dx
Order=1, degree=2
y

3

dy 2
dx 2
d2 y
dx 2

dy
dx
1

Cubing both sides

3

dy
dx

1

2

105
3 2

d2y
dy
2
dx
dx
Squaring both sides
2

3

d 2y
dy
2
dx
dx
Order=2, degree=2
Solved examples:
Example 1: Find the order and degree of the following
dy
dx

1
i)

e

3

2

2

d2y
dx 2

Solution:
d2y
e
dx 2

2

dy
dx

1

3

2

Squaring both sides
e

d2y
dx 2

2

2

1

dy
dx

2 3

order = 2, degree = 2
ii)

d
d2y
x
dx
dx3

3

sin( xy ) e x

Solution:
3

d 3y
dx 3

d3 y
x 3
dx 3

Order = 4, degree=1
dy
5
y x
dy
dx
dx

iii)

Solution:
dy
y
dx

x

dy
dx

2

5

Order =1, degree=2

2

d4y
sin( xy ) e x
4
dx

106

iv)

dy
x
dx

y

2

5 1

dy
dx

dy
dx

5 1

Solution:
y x

dy
dx

2

Squaring on both sides
y-x

y

dy
dx

2

dy
2 xy
dx

2

dy
dx

25 1

x

2

dy
dx

2

2

25 1

dy
dx

2

Order =1, degree=2
Check your progress:
2

u
x2

1)

u
y

u

Ans : order =2, degree=1
2)

d3y
dx 3

4

7

d 3y
5
dx 3

7

d2y
dx 2

11

dy
dx

y

ex

Ans : order=3, degree=7
3)

y11

3

y1

4

ex

Ans : order=2, degree=3
x
4)
y11
1
y11
Ans : order=2, degree=2
y11

5)

1 y12

Ans : order=2, degree=2
y1

x

y xy1

2

Ans : order =1, degree=3

6
...

Consider

107

y ax2

1

Where y= independent variable
x = dependent variable
Differentiating equation (1) with respect to x
dy
we have
2ax
(2)
dx
From equation (1) we have
y
a
x2
Put value of a in equation (2), we have
dy
y
2 2 x
dx
x
dy 2 y
dx
x
dy
x
2y
dx
dy
x
2y 0
dx
This is the required differential equation
Note:To eliminate two arbitrary constants, three equations are required
...

In general to eliminate n arbitrary constants
...

In other words elimination of n arbitrary consonants will bring us to
differential equation of nth order
...

Differentiating equation (1) with respect to x
dy
= -c1 cos x c2 cos x
...

Example 3: Form the differential equation from
x= a sin (wt+c) where a and c are arbitrary constants
...
r
...
't'
d2x
dt 2

-a wsin cot c w

d2x
dt 2

-w 2 a sin cot c

d2x
dt 2

-w2 x
...

dy 1
A
dx Ax
dy 1
dx x
dy
x
1
dx
This is the required differential equation
...

Example 6: Obtain the differential equation for the relation
y= a e2x b e3 xWhere a,b are constants
...

Here the number of arbitrary constants is two
Hence we shall require three equations to
Eliminate and b
...

dy
= 2a e2x 3b e3 x -------------(2)
...

Example 7: Find the differential equation of all circles
touching y axis at the origin and centers on x-axis
Solution:

The equation of such a circle is
2

y2

a2

x2 2ax a2

y2

x a

i
...


x2

y2

a2

2ax 0

Where a is the only arbitrary contents
Differentiate equation (1) with respect to x We have
dy
2x 2 y
2a 0
dx
dy
x 2 y 2 2x x y
0
dx
x2

y 2 2 x 2 2 xy

-x 2

y 2 2 xy

dy
dx

dy
dx
0

0

(1)

111

dy
x2 y2 0
dx
Which is the require differential equation
...


Form the differential equation of all circles of radius a
...
5 LET US SUM UP
In this chapter we have learn
dy
Equation in term
of is called differential equation
...

Formation of differential equation while removing arbitrary
constant likes A&B,&C
...
6

UNIT END EXERCISE

1)

Find the order 7 degree of Differential equation given below
2

d3 y
dx 3

i
...


i

iii
...


d2y
2 2
dx

3

3

2

k

5

1

3

1

dy
dx

3

y

d2y
dx2

dy
dx
dy
dx

2

y

2)
i
...


1
dy
dx
Formulate the differential equation
Y A Blogx
X = asin(w++c)

iii
...


iv
...

vi
...

viii
...


y

x

dy
dx

0

1

Y = em cos x
Y = ax2 bx
Y = cx+2c 2 c3
X 2 +Y 2 = 2ax
Y 2 = 4ax
e x +Ce y = 1

113

x
...
1
Objectives
7
...
3
Solution of Differential equation
7
...
5
Let Us Sum Up
7
...
1 OBJECTIVES
After going through this chapter you will able to
Find general & particular solution of differential equations
...

Apply particular method first find the solution of differential
equation
...
2 INTRODUCTION
We have already formed differential equation in previous chapter
...
It is very useful in different field
...
3 SOLUTION OF DIFFERENTIAL EQUATION
General Solutions:The general Solution of a differential equation is the most general
relation between the dependent and the independent variable occurring in
the equation which satisfies the given differential equation
...
g
...

This equation can also be written as
Mdx Ndy 0

7
...
Important one among those are listed below
...

2)
Equations reducible to variable seperable form
...

4)
Exact differential equations
...

6)
differential equation)

115

7)
Methods of substitution
...

7
...
1 Variable Separable form:Working Rule
1)
Consider the differential equation Mdx+ Ndy=0
2)
If possible rearrange the terms and get f(x) dx +g(y)
...

4)
Simplify if possible
...
dy

1 e x Sec 2 y
...

y dy
1 x 2 y 2 x2 y 2
Example 2: Solve
x dx

c
log c

116

y dy
x dx

Solution:

1 x

2

y 2 1 x2

y dy
× = 1+ x 2 × 1+ y 2
x dx
y dy
× = 1+ x 2 × 1+ y 2
x dx
y
× dy = x 1+ x 2 × dx - - - - - - - - - -1
2
1+ y
This is in variable separable form
Integrate equation (1)
1
2y
1
dy
2 x 1 x 2 dx c
2
2
2
1 y
f1 x

\

dx = 2 f x +c

f x
f x
=

f x

n

× f 1 x dx
n+1

n+1

1
1 2
× 2 1+ y 2 = × × 1+ x 2
2
2 3
3
1
1+ x2 2 +c
3
This is in required general solution
...

Example 4: Solve 3e x tan y dx

1 e x sec2 y dy

0

given y

when x=0
4
Solution: The given equation is
3e x tan y dx

1 e x sec2 y dy

through out by 1+e x

0

tan y

3e x
sec 2 y
dx
dy
1+ex
tan y

0

(1)

This is in variable separable form,
Integrate equation (1) we get
3e x
dx
1+ex
3

sec 2 y
dy
tan y

ex
dx
1+ex

3log 1 e x
3

log 1+ex
3

sec2 y
dy
tan y

log tany=log c
3

tan y

4

log c

tan y =c---------------(2)

This is the required general solution
To final particular solution:put y=

log c

log tany=log c

log 1+e x

1+e x

log c

at x=0 in equation --------(2)

118

1+1

3

tan

c

4

c=8

Put value of c in equation (2)
1 ex

3

tan y 8

This is a particular solution
dy x 2log x 1
Example 5: Solve
dx sin y y cos y
Solution: The given equation is
x 2 log x 1
sin y y cos y

dy
dx

sin y y cos y dy

x 2log x 1 dx

(1)

This is in variable separable form
Integrate equation (1), we get
sin y

y cos y dy

x 2log x 1 dx cons tan t

siny dy
cos y

y cos y dy

y sin y cos y

y sin y

x2 log x x2

y sin y

2 x log x dx

2 log x

x2
2

x2
2

x2 c

x2 log x c

This is required general solution
Check Your Progress:
1)
solve:
dy
ex y x 2 e y
dx
x3
ex
ey c
3
2)
ans

solve:

y x

1 ay x a

3)
ans

eax
a

4)

solve:

ans

e

a

y2

dy
dx

cy
log

solve:

dy
dx

dy
dx

ax by

by

b

c
x cos x cos y sin y

x sin x cos x logcos y c

dy
dx

0

xdx c
x2
2

c

119

5)
solve:
Sec2 x tan y dx sec2 y tan x dy 0
ans
tan x tan y c
dy
6)
solve:
ex 2 y
dx
1 2y x
e
e c
ans
2
7
...
2 Equations Reducible to variable separable forms:
Sometimes we come across differential equations which cannot be
converted into variable separable form by mere rearrangement of its terms
...
4
...


127

x3

y 3 dx 3xy 2 dy

0

This is a homogenous equation
x3

y3 dx 3xy 2 dy

x3 y3
3xy 2

dy
dx
Put

(1)

y vx

dy
dv
v x
dx
dx
Using equation 1 we have
dv
dx

x3 v3 x3
3x v 2 x 2

dv
v x
dx

x3 1 v3

v x

3v 2 x3

x

dv
dx

1 v3
3v2

x

dv
dx

1 2v 3
3v2

v

3v 2
1
dv
dx
3
1 2v
x
This is in variable separable form
Integrating we have
6v 2
dv
2v 3 1
1
log 2v3 1
2

1
2

log 2v3 1

Put

v=

log cx

1
c x2
2

y
x

y3
1
1 2 2
3
x
c x
x
2y3 x3
c2
2

log x log c
2 log cx

log 2v 3 1

2v3 1

1
dx cons tan t
x

2

128

2y3 x3

kx where k=

1
c2

This is the required general solution
y
x

x tan

Example 14: solve

y sec2

y
y
dx x sec2 dy
x
x

Solution:
The given equation is
x tan

y
x

y
y
dx x sec2 dy
x
x

y sec2
y sec 2

dy
dx

y
x

x tan

x sec2

y
x

y
x

y
dy y
x
dx x sec 2 y
x
This is a homogeneous equation
Put y vx
tan

(1)

dy
dv
v x
dx
dx
Using equation 1 we have

v+x
x

dv
dx

dv
dx

v

tan v
sec 2 v

tan v
sec2 v

sec 2 v
dv
tan v

1
dx
x

sec2 v
1
dv
dx
tan v
x
This is in variable separable form
Integrating we get,
sec2 v
dv
tan 2 v
log tanv + log x =log c
log tanv x
x tanv = c

=log c

1
dx
x

0

cons tan t

0

0

129

y
x
y
x tan = c
x
This is the required general solution
Check Your Progress:
1) solve the following
Put v =

i)
ans

y

x2

xdy ydx
x2

x

ii)

y2

y cot

ans

y

c sec

y2

x2

cy

e

x
dy
y

dy
dx

ans

cx2

ydx

0

x
y

iii)

y 2 dx

xy

dy
dx

y

iv)

x2

x

y 2 dx

ans

x x2 3y 2

v)

x

2 xydy

dy
dx

ans
vi)
ans

y
x

x2 a2

y
x2

c e
y

c

3 y2

dy
dx

y 2 2xy x2

x y
c

7
...
4 Exact Differential Equation
Definition:The equation Mdx+Ndy=0 is said to be an exact differential equation if
and only it
...
g
...

M
N
y
x
Rules for the General solution:If the equation Mdx+Ndy=0 is exact then its general solution is given by
M treat y as constant dx

N terms free from x

dy

c

Where
(1)
In first integral with respect to x, treat y as constant
(ii)
In second integral do not take the terms containing x i
...
take only
those terms of N which are free from x
...

(iii) c is arbitrary constant of Integration
...
(1)

Comparing with Mdx+Ndy=0; we have
M

4 x3 y 2

N

4

y cos xy

2x y x cos xy
M
y

y

M
y

4 x3 y 2

y cos xy

8 x3 y 2 cos xy xy sin xy

N
2 x 4 y y cos xy
x x
N
8 x3 y cos xy xy sin xy
x
M
N
y
x
Hence differential equation (1) is exact
Its solution is given by
Min treat y constant dx
4 x3 y 2
4 y 2 x3 dx

4y

2

x4
4

y cos xy dx

y cos xy

y

sin xy
y

N terms free from x dy
ody

c

c

c

x 4 y 2 sin xy c
This is the required general solution
Example 17: Solve

x 2e y dy

y x sin x dx

0


...


y 2 dx

2

a x x y xy
x

(2)

1 e

dx e
x

Ans
...


M
y

sin xy xy cos( xy) 2 xy cos( xy) x 2 y 2 sin( xy )

N
x

General solution is given by
xy sin xy

dy

c

2 xy e y dx

Ans
...
y x log xy
(6)

0

x
dy 0
y

dx

sin x tan y sin x y

Ans
...


4 x 3 dx

2

c

1

cos x tan y cos x y

2

y

c

x 2 y cos xy

dy

0

o

135

7
...
E:- general solution, particular solution
variable separable form:- dx
f x dx

f y dy c

Equations reducible to variable separable form
...
e
dx g ( xy )
With substituting Y=Yx
...
6 UNIT END EXERCISE
Solve the following differential equation
...

dx Y(1+2logu)
dy
x 2 x 2 e3 y
ii
...

2x Cosy dx -(1+ x2 )siny dy = 0
dy
iv
...

ax by c
dx
dy
vi
...

=e x +y
x
dx
dy
= (4x+ y +1)2
viii
...

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

x
x
dx
dy
xii
...

4 - 2 dx +
dy = 0
x
x

136

xiv
...


dy x2 +3y 2
+
=0
dx 3x 2 + y 2
dy
4( x y ) = 3x - 4y
dx

*****

EQUATION REDUCIBLE TO EXACT
EQUATIONS
UNIT STRUCTURE
8
...
2
Introduction
8
...
4
Linear Equation And Equations Reducible To Linear Form
8
...
6
Let Us Sum Up
8
...
8
Unit End Exercise

8
...
differential equation
...
differential equation
...

Find the solution of non-linear equation
...
2 INTRODUCTION
In previous chapter we have learn about exact differential equation & its
solution
...
To find the solution of non-exact differential equation we use
integrating factor which convert non-exact differential equation to exact
differential equation
...


137

In some cases equations which are not exact can be converted to exact
differential equation by multiplying by some suitable factor called as
Integrating factor
...
3 DEFINITION
Integrating Factor
If the equation leMdx +leNdy=0 is exact
then le is said to be an integrating factor of the equation Mdx+ Ndy = 0
8
...
1 Rules of finding Integrating factor :Rule (1)
If the equation Mdx+Ndy=0 is homogeneous then

1
Mx Ny

integrating factor
Solved Example:
Example 1: x 2 y 2 xy 2 dx- x 3 3x 2 y dy 0
Solution: The given equation is
x 2 y 2 xy 2 dx- x 3 3x 2 y dy

0
...

Comparing with Mdx +Ndy=0 ; we have
M

x2 y 2 xy 2

N

- x 3 3x 2 y

1

I
...


Mx Ny
1

3

2

x y 2 x y x3 y 3x 2 y 2
x 2 y 2 xy 2
x2 y 2

i
...


1
y

dx-

x 3 3x 2 y

2
x
dx- 2
x
y

x2 y2

dy

0 is exact

3
dy 0 is exact
y

Its general solution is given by
M

treat y const dx

N terms free from x dy

c

is

138

1
y

2
dx
x

1
y

dx-2

3
dy c
y
1
dx
x

3

1
dy c
y

x
2 log x 3log y c
2
This is required general solution
Check your progress:
Solve
3xy 2

i)

y3 dx

xy 2 2 x 2 y dy

0

1
x y2
General solution is given by
Hint : I
...


cy 2
x3

2

y

c

x

x 2 3xy 2 y 2 dx

ii)

x 3x-2y dy

0

1
x3
General solution is given by
Hint: I
...

x2 log x

3xy y2 cx2

8
...
2 Rule (II) :
If the equation Mdx+Ndy=0 can be written as
M

y f1 xy dx, N xf2 xy dy 0

i
...
M
then

y f1 xy , N x f2 xy

1
is an integration factor
...

Solved Examples :Example 2: Solve x 2 y 2 2 ydx

2 2 x 2 y 2 xdy

Solution: The equation is given by
x 2 y 2 2 ydx

2 2 x 2 y 2 xdy

0

0

139

Comparing with Mdx+Ndy=0; we have
x 2 y2

2 y

2-2x 2 y2

x

M
N

1

I
...


Mx Ny

I
...


2

xy x y

2

1
2 2 2x 2 y2

1
3x3 y 3

I
...


x 2 y2

2 y

3x3 y 3

1
3x

i
...
S
...

solve :
2 2
x y xy 1 y dx x 2

c

y2

2c

xy 1 xdy

0

Hint: I
...


2
...
3
...
Say f(x) then e f x dx
N
integrated
...

Solved Examples :Example 3: Solve ( y4 2 y) dx ( xy3 2 y4 4 x) dy 0
Solution: The given equation is
( y 4 2 y)dx ( xy3 2 y 4 4x)dy 0
Comparing with Mdx+Ndy=0; we get
M

y4 2 y

N

xy3 2 y 4 4 x
M
y
M
y
N
x

N
x

y

4y3 2

x

xy3 2 y 4 4 x

y3 4
N
x

M
y
M

-3 y3 2
y y3 2
3
y

function of y alone

I
...

-3

e

e
1
dy
y

e-3 log y

f y dy

y4 2 y

is

141
log

e

1
y3

1
y3

I
...

y4

2y
y3

xy3 2 y 4 4 x

dx

y3

dy

0

This is exact differential equation
Comparing with Mdx+ Ndy=0; we get
2
M y
y2
N

x 2y 4

x
y3

General solution is given by
M treat y constant dx

y

2
y2

y
y

2
dx 2
y2

2
y2

y dy

dx 2

x y2

N terms free from x dy

y2
2

c

c

c

c

This is required general solution
...

y4
General solution is given by
Hint : I
...


x2e y

x2
y

ii)

x 2 y 3dx

Ans

3x3 y 2 y 6 cy e

x
y3

c
x3 y 2 dy
3

0

y

8
...
and p& Q
are functions of x only
...

Working Rule:
1)
Consider linear differential equation
...
F
...
F
...
F
...

2)
For linear differential equation
dx
p1 x Q 1
dy
Where p1 and Q1 are functions of y or constants only
Its integrating factor is given by
p1dy

I
...

e
Its solution is given by
x IF
Q ( IF ) dy

c

Where c is arbitrary constant
...
(1)

This is of the type
dy
py Q
dx
Hence equation (1) is linear differential equation
...
F
...
F
...
F
...
F
...

Example 5: Solve 1 y 2 dx

tan y -1 x dy

Solution: The given equation is
1 y 2 dx

tan y

dx tan y 1 x
dy
1 y2
dx tan 1 y
dy 1 y 2 1
dx
1
x
dy 1 y 2

x dy

1
x
y2
tan 1 y

...
f
e

e

pdy

1
dy
1 y2

I
...


etan

1

y

144

The solution of differential equation (i) is
x I
...

Q I
...
dy c
tan 1 y tan 1 y
e
dy c
1 y2
consider the integral
x e tan 1 y

tan -1 y tan 1 y
e
dy
1 y2
put z tan-1 y
Differentiating with respect to z
1 dy
1
1 Y 2 dz
1
dy dz
1 Y2
z ez dz
d
z ez dz dz
dz

e z dz

z

z ez

l e z dz

z ez e z
ez z 1
put z tan-1 y
1

e tan

y

tan 1 y 1

solution is given by
x e tan

-1

y

e tan

1

y

tan 1 y 1

x tan-1 y 1 c e
This is the required solution
...
(1)
x 1-x 2

Hence equation (1) is linear in dependent variable y
This is of the type

145

dy
dx

py

Q
2x2 1

where P

x 1 x2

I
...
e

x3
x 1 x2

,Q

pdx

2x2 1
x 1 x 1 x

Let P

1
1
1
x 21 x 21 x
( By partial fraction)
P

IF
e
e

e
IF

1
1
x 21 x

e

-logx

1
log 1 x
2

1
21 x

dx

1
log 1 x
2

- logx 1-x 2

log

x 1-x 2

1

1

x 1 x2
Hence solution of differential equation (i) is
y IF
Q IF dx c
x2
1 x2

1

y

x 1-x 2
x
1 x

2

1
2

3

1
x 1 x2

dx c

dx c

2

2 x 1 x2

2
1 1 x
1
2

1

3

2

dx c

2

c
2
fn f1

y

1

x 1 x2

1 x2

y

x cx 1 x 2

Which is the required solution
...
(1)
dx
x
x
Which is of the type
dy
py Q
dx
1
e-2 x
where P
,Q
x
x
The equation (1) is linear in y
I
...

1

pdx

e
dx

e x
I
...
e 2 x
Hence the solution of differential equation (1) is
y IF
Q IF dx c
y e2

e

x

2 x

e2

x

x

dx c

1
dx c
x
y e2 x 2 x c
This is the required general solution
...
(1)
dy 1 y 2 1 y 2
Which is of the type
dx
px Q
dy

1 y2

dy
dx
1 y2
dx
dy

147
1

1
etan y
where p
,Q
1 y2
1 y2
The equation (1) is linear differential equation
Hence
IF
1

e

1 y2

pdy

e

dy
1

IF etan y
Hence solution of differential equation (1) is given by
x IF
Q (IF) dy c
1

x e

e tan y tan 1 y
e
dy c
1 y2

tan 1 y

1

x e

e2 tan y
dy c
...

tan

1

y

Check your progress:
1)
i)

Solve
2 y x 2 dx

xdy

Ans: y x 2 log cx
dy
y
x2 x
dx 1 x
Ans : 2y 1-x c 2 x 2

ii)

iii)

dy
dx
x4
1 y
4

x2 1

Ans : x 2

x 3 -2xy x
x2
2

c

148

iv)

dy
dx

y

1 x2

2

2

1
1 x2

e

1

1

y

3

1 x2
-

H int I
...


1 x2

x 1

x

1 x2

1 x2

c c

dx xdy e y sec2 dy

v)

H int : I
...
ey
x e y tan y c
vi

x cos x

dy
dx

cos x x sin x y 1

x
sec x
xy cos x x c
3 dy
x2 1
4x x2 1
dx
H int : I
...


vii

x2 1

H int : If
x2 1

viii

ix

2

2

y 1

2

tan 1 x c

y

dy
1
dx
H int : I
...
e-y
x y 2 c ey
x

y 1

x 2 y 3 dy

H int I
...

x

ydx

1
y

y3 cy

8
...
(1)
dx

149

Let y1-n

u
1-n

y

dy
dx

n

du
dx

u sin g equation (1) we get
1 du
Pu Q
1-n dx
du
1-n pu 1 n Q
dx
Note:
We divide by xn and substitute u= x1-n and proceed
...

Solution:
dy y
xe x y 2
...

dx
1
Where p
, Q xe x , n 2
x
Equation (1) is Bernoulli's differential equation
throughtout by y2 , we get
dy 1 1
y-2
y
x e x
...

1
where p
, Q -x e x
x
I
...
e Pdx

150
-

1

dx

e x
e log x

log

I
...
e

1
x

1
x
Hence, General solution is given by
u IF
Q IF dx c
I
...


u
Put u
y-1

y-1
1
x

1
x

-x e x

1
dx c
x

e x dx c

1
ex c
xy
This is the required solution
...
(1)
dy
which is of the type,
dx
px Q x n
dy
where p -y, Q y3 , n 2

through out by x2 , we get
dx
x -2
x 1 y y3
...

p y, Q
y3
I
...
e
e y dy

pdy

y2

I
...
e 2
Hence general solution is given by
u IF
Q IF dy c
y2
2

u e

y

y2

3

2

e

dy c

2

y
t
2
y dy dt

Let

y2

2t et dt c

2

u e
y2

u e

2 t e et

2

y2

x 1, t

Put u
1 y
e
x

c

2
2

2

2

2

1 y2 2
e
x

y2
y
e 2
2

y2 e

y2

e

y2

2

c

y2
2

2 e

2

c

y2
y2
1 y2 2
e
y2 e 2 2 e 2 c
x
This is the required general solution
...
F
...
F
...
5
...
(1)
dx
which is of the form
dy
f1 y
pf y Q
dx
secy, p -1
...

Where p -1, Q -x
pdx

I
...
e
e x
I
...
6 LET US SUM UP
In this chapter we have learned
Integrating factor for non-exact equation
...

Using integrating factor find the solution of linear differential
equation
...
7 UNIT END EXERCISE
Solve the following D
...

y
2
2
dx ( x 1)
( x 1)3
dy
ii
...

y 1
dx
x2
iv
...

(x2 + y2 +1)dx - 2xy dy = 0
vi
...

viii
...


(x2 + y 2 )dx -(x2 + xy )dy = 0
y(1+ xy)dx+(1- xy ) dy = 0
(2y 2 +4x2 y)dx+(4xy+3x3 )dy = 0
dy
+(cotx)y = Cosx
dx
dy
+ y secx = tanx
dx
dy
(1 x 2 ) + 2xy - 4x 2 = 0
dx
dy
(1 x 2 ) + y = etan 1 x
dx
dy
y
+
=1 x
dx (1 x) x
Sec x dy =(y+Sin x)dx
(y log x - 1)y dx = x dy

x
...

xii
...

xiv
...

xvi
...

xviii
...

xx
...
1
Objective
9
...
3
Geometrical
9
...
5
Simple Electric Circuits
9
...
7
Let Us Sum Up
9
...
1 OBJECTIVE
After going through this chapter you will able to
Use differential equation to find the equation of any curve
...


9
...

We differ type
...


156

9
...
e
...


The tangent and normal at p meet X axis in T and respectively
...
[Geometrical Construction]
Then,
dy
,y
Slope of Tangent at p = tan
=
dx x 1 1
Equation of tangent at p is

y

X

x1

intercept of tangent =

y

1

y1

dy
dx
dx
p
dy

x

x

1

157

x
y

intercept of tangent =

y x
1

1

y1
dy
dx

p

dy
P
dx

1

Equation of the normal at P is given by
dx
y- y
(x x 1)
1
dy

6) Length of tangent = PT =

7) Length of Normal at

y

1

1

dx
dy

P

PN

y1

dy
1+
dx

2

y1
dy
dx

8) Length of Sub tangent

9) Length of Sub normal

2

y1

dy
dx

10) If e is a radius of curvature at p then

1
e

dy
dx

2

3
2

d2 y
d x2

Solved Examples:
Example 1:
Find the curve which passes through the points [ 2, 1 ] and [ 8, 2 ]
for which sub tangent at any point varies as the abscissa of that point
...
[k constant]
dy
dx
dy
y kx
dx
1
k
dx
dy
x
y
dy 1
k
dx
y x
which is in variable seperate form
integrate both side
1
k
...
log y =log x + log c
subtangent

log

y

k

y =log

k

cx

cx
...

Solution: Let p [x , y ] be a point on the curve
We Known that,

Radius of curvature =

dy
1+
dx

2

3/2

d2 y
dx 2

Lenght of normal = y 1+

dy
dx

2

From given condition 2 3/2

dy
1+
dx

dy
25 1+
dx

d2 y
dx 2
2 1/2

dy
1+
dx

dy
1+
dx
d2 y
dx 2

dy
1+
dx
Let

dy
=z
dx

dy2
d dy
=
2
dx
dx dx
=
=

d
z
dx

dz
dy

dy
dx

2

= 25

d2 y
dx 2

2

2

dy
= 25 1+
dx

[1]

2 1/2

160

=

dz
dy

z

d2 y
dz
=z
2
dx
dy
From eq n 1
1 + z 2 = 25 z
2zy

dz
dy

dz
= 1 + z2
dy

2z
1
dz =
dy
2
1+z
y

which is in variable separable form
Integrate both side
2z
dz
1 + z2

1
dy constant
y

log 1+z2

log y

log 1+z 2

log cy

log c

1+z2 = cy
z2 = cy - 1
z = cy - 1
Again put z =
dy
dx

dy
dx

= cy - 1

1
dy = dx
cy - 1
This is in variable separable form

Integrate both sides
1
c

c1
dy
cy 1

1
2 cy 1 = x + c1
c

dx constant

161

2 cy 1 = cx + cc1
2 cy 1 = cx + c2
Where c2

cc1

which is the eqn of the curve
...

Ans : x + 2y

x+y

2

=c

9
...
Let p be the position
of the body at any instant
Where OP = X , then
1) velocity v

dx
dt
dv
dt
2
dx
= 2
dt
dv
=v
dx

2) The acceleration =

By chain rule dv
dt

dv dx
dx dt
dv
v
dx
v

dv
dx

162

f = ma

dv
dt

=m

d2x
=m 2
dt
f = mv

dv
dx

where f = effective force

Algebraic sum of the forces acting on a body along the given direction is
equal to the product of mass and acceleration in that direction
...
Show that the velocity of the
particle
...

c
1 e2bx
2
2b

v2

cx
b

Solution: Consider the motion
Step 1) :
Let m be the mass of the particle moving to right
...

mcx

mv

dv
dx

mbv 2
ie

mcx and

mbv2 are forces to the right

mv

dv
dx

mcx mbv 2

step[2]
v

dv
bv 2
dx

cx

[1]

163

Let v2

z

dv
dx

2v

dz
dx

eqn [1] becomes
1 dz
2 dx
dz
dx

bz

cx

2bz

2cx

which is a linear equation in z
p = 2b
...

pdx

I
...
= e

2bdx

=e

e

I
...
=

2bx

Its general solution is given by
z [ IF ] =
ze

Q

2bx

IF dx constant

-2cx

e

2bx

-2c

-2c x

ze

2bx

=

x

2c

2

2

e

2bx

2b

1 e

=

e

c
2b 2

d
x
dn

dx-

e

c1
2bx

2b

2bx

c
2b 2

b
cx
b

dx

e

1
2b

-cx

dx c1

2bx

2b

2bx

2bx

2bx

2bx

2b

v e

v

e

e

-2c x

e

x

dx c1

c e

-2bx

1

+ c1

e

2bx

+ c1
[3]

[III] to find c1 , we impose initial conditions
ie for x = o , v = o in eqn [3]

2bx

dx

c1

164

o=0+

g

1

=-

c
+ c1
2b 2

c
2b 2

put values of

g

1

in eq n [3]

v2 =

cx
c
+
b
2b 2

c
e
2b 2

v2 =

c
1 e
2b 2

2 bx

cx
b

2 bx

Example 4: A body of mass m
...
If it falls through a distance x and possesses a velocity v at that
instant show that
2k x
m

log

a2
a

2

v

, where mg = ka 2

2

Solution:

Step
The forces acting on the body are
1)
Its weight mg acting vertically downwards
...

The net forces acting on the body vertically downwards
= mg kv2
...
[ 1 ]
Step [ 2 ]
mv

dv
dx

k a2

v
a

2

v

2

dv

v2
k
dx
m

This is in variable separable form
Integrating both sides
1
2

2v
dv
a v2
2

1
log a 2 v 2
2

k
m

dx c1
...

From 2
1
log a 2 c1
2
put value of c1 in eqn [ 2 ]
1
log a 2 v 2
2
1
log a 2 v 2
2
log a 2 v2
log a 2
2kx
m

log a 2 v2

k
x
m
k
x
m
2kx
m

1
log a 2
2
1
log a 2
2
log a 2

2kx
m

a2

log

a 2 v2

Check your progress:
1)
A particle of Unit mass is projected upward with velocity u and the
resistance of air produces a retardation kv2 and v is the velocity at any
instant show that the velocity v with which the particle will return to the
point of projection is given by
1
1 k
= 2
2
v
u
g
2)
Determine the least velocity with which a particle must be
projected vertically upwards so that it does not return to the Earth
...

Ans : Least Velocity vo
2 gR
R
Radius of earth
3)
A paratrooper and his parachute weigh 50 kg
...
He is Travelling vertically downward at the speed of 20
m/s
...
When the velocity is 10 m/s Find the limiting velocity, the
v

e

5 s

x =5 st+
x =25t

-gt/25

25
g

e


...
5 SIMPLE ELECTRIC CIRCUITS
The following Notations are frequently used
...

t seconds
q coulombs
i ampere

Ch arg e on capacitor
Current

e volts

voltage

R ohms

Re sis tan ce

L Hentries
C Farads

Time

Indua tan ce

capaci tan ce

Current is the rate of electricity
i=

dq
dt

[ II ]
1) The algebraic sum of the currents into any point is zero
...

3)
Voltage drops as current i flows through a resistance R is Ri ;
di
q
through an induction L is L
and through a capacitor C is
...
containing a
constant resistance
...
It the initial current is zero,
show that the current builds upto half its theoretical maximum
L log 2
in
seconds
...

R
E
P=
Q
L
L
I
...
= e
=e

pdt
R
dt
L
R

t

I
...
= e L
The general solution is given by
i
...
( IF ) dt + constant
R

E Rt
e L dt c
L
E L Rt
eL c
L R
R
t
E Rt
i
...
e
...
e L

E
R
E
i=
R

i=

E
e
R
1 e

t

(2)

R
t
L

R
t
L

(3)

This is the expression for i at any time t
...


168

From eqn ( 3 )

1 E
E
=
2 R
R

1

e

1
=1
2
e

R
t
L

R
t
L

e

R
t
L

1
2
R
1
t = log
L
2
1
= log
2
= log 1 log 2
= O log 2
R
t = log 2
L
= 1

L log 2
R
Check Your Progress:
1)
The equation of the eml in terms of current i for an electrical
circuit having resistance R and a condenser of capacity C, in series is
...
Find current if
it is zero when
t = o ; at t = 0
...
2) 100 cos (0
...
e
Ans :
10144
125

9

OOLING

The law states that the rate at which the temp of a body changes is
proportional to the difference between the instantaneous temp of the body
and the temp of the surrounding medium
...


169

dQ
dt
dQ
dt

Q Qo
k Q Qo

Where k is a constant and Q decreases as t increases i
...


dQ
is negative
dt

hence negative sign is added
...
and the substance cools
from 1000C to 700C in 15 minutes, find when the temperature will be
400C
...


The study of a differential equation in applied mathematics consists of three phases
...

Solutions of this differential equation, evaluating the arbitrary constants
from the given conditions, and
Physical interpretation of the solution
...

General form of a linear differential equation of the nth order with constant
coefficients is
dny
d n −1 y
d n−2 y
+ K 1 n −1 + K 2 n − 2 +
...
K n are constants
...
(1)

The symbol D stands for the operation of differential
dy
d 2y
d3y
2
3
, similarly D y = 2 , D y = 3 , etc
...
e
...
+ K n ) y = X i
...
, f(D)y = X
Where f (D) = D n + K 1 D n −1 +
...
X = ∫ Xdx
D
1
X = e ax ∫ Xe − ax dx
2
...

X = e −ax ∫ Xe ax dx
D+a
(i) The general form of the differential equation of second order is

2

d2y
dy
+ P + Qy = R …………………(1)
2
dx
dx

Where P and Q are constants and R is a function of x or constant
...
e
...

D2
(1) can be written in the operator form
D 2 y + PDy + Qy = R (Or) ( D 2 + PD + Q) y = R
(iv)

Complete solution = Complementary function + Particular Integral

PROBLEMS
1
...
e
...
F = Ae 2 x + Be 3 x
The general solution is given by
y = Ae 2 x + Be 3 x

dy
d2y
− 6 + 3y = 0
2
dx
dx
2
Solution: Given (D − 6 D + 3 y ) = 0
The auxiliary equation is m 2 − 6m + 13 = 0
6 ± 36 − 52
i
...
, m =
2
= 3 ± 2i
Hence the solution is y = e 3 x ( A cos 2 x + B sin 2 x)
2
...
Solve (D 2 +1) = 0 given y(0) =0, y’’(0) = 1

3
Solution: Given (D 2 +1) = 0
A
...
e
...
Solve ( D 2 − 4 D + 13) y = e 2 x
Solution: Given ( D 2 − 4 D + 13) y = e 2 x
The auxiliary equation is m 2 − 4m + 13 = 0
4 ± 16 − 52 4 ± − 36
=
= 2 ± 3i
m=
2
2
∴ C
...
I
...
F +P
...

1
y = e 2 x ( A cos 3x + B sin 3 x ) + e 2 x
9
5
...
I 1 = 2
ex
D − 3D + 2
1
ex
=
1− 3 + 4
1
= ex
0
1
=x
ex
2D − 3
1 x
e
=x
2−3
= − xe x

4

(

1
− e2x
D − 3D + 2
1
=e2x
4−6+2
1
e2x
=-x
2D − 3
1 2x
e
=-x
4−3
= - xe 2 x

P
...
I
...
I 1 + P
...
Solve
2
dx
dx
d2y
dy
Solution: Given 2 − 4 + 5 y = −2 cosh x
dx
dx
2
The A
...
F = e −2 x ( A cos x + B sin x )
e x + e−x 
1
(− 2 cosh x ) = −2 2 1


D 2 + 4D + 5
D + 4D + 5  2 
−1
−1
ex + 2
e −x
= 2
D + 4D + 5
D + 4D + 5
x
x
−e
e
−
=
1+ 4 + 5 1− 4 + 5
− e x e−x
−
=
10
2
∴ y = C
...
I

P
...
I =

7
...
E is m +3m + 2 = 0
(m+1)(m+2) = 0
M = -1, m = -2
C
...
I = 2
D + 3D + 2
1
sin 3 x (Replace D 2 by − a 2 )
=
2
− 3 + 3D + 2
1
1 (3D + 7)
sin 3x
=
sin 3x =
3D − 7
3D − 7 (3D + 7)
3D + 7
sin 3x
=
(3D) 2 − (7) 2
3D + 7
sin 3x
=
9 D 2 − 49
3D + 7
sin 3x
=
9(−3 2 ) − 49
3D + 7
=
sin 3x
− 130
1
(3D(sin 3x) + 7 sin 3x )
= −
130
1
(9 cos 3x + 7 sin 3x )
= −
130

Solution: Given

∴ y = C
...
I
Y = Ae − x + Be −2 x −

1
(9 cos 3x + 7 sin 3x )
130

8
...
I of (D 2 +1) = sin x
Solution: Given (D 2 +1) = sin x
1
sin x
P
...
= 2
D +1
1
=
sin x
−1+1
1
sin x
=x
2D
x 1
sin x
=
2D
x
= ∫ sin xdx
2
x cos x
P
...
Find the particular integral of (D 2 +1) y = sin 2 x sin x
Solution: Given (D 2 +1) y = sin 2 x sin x
1
=- (cos 3x − cos x )
2
1
1
= - cos 3x + cos x
2
2
1  1

P
...
I =
cos 3x + sin x
4
16
Problems based on R
...
S = e ax + cos ax(or )e ax + cos ax
P
...
Solve (D 2 −4 D + 4) y = e 2 x + cos 2 x
Solution: Given (D 2 −4 D + 4) y = e 2 x + cos 2 x
The Auxiliary equation is m 2 −4m + 4 = 0
(m – 2 ) 2 = 0
m = 2 ,2
2x
C
...
I 1 = 2
D − 4D + 4
1
=
e2x
4−8+ 4
1
= e2x
0
1
e2x
=x
2D − 4
1 2x
=x e
0

7
1
= x 2 e2x
2
1
cos 2 x
P
...
F + P
...
H
...

• (1 − x) −1 = 1 + x + x 2 + x 3 +
...

• (1 − x) −2 = 1 + 2 x + 3x 2 + 4 x 3 +
...
Find the Particular Integral of (D 2 +1) y = x
Solution: Given (D 2 +1) y = x
A
...
F = Ae + Be x
1
x
P
...
x
= − [x + 0 + 0 + 0
...
Solve: D 4 − 2 D 3 + D 2 y = x 3

8

(

)

Solution: Given D 4 − 2 D 3 + D 2 y = x 3
The A
...
F = (A + Bx)e 0 x + (C + Dx)e x

1
x3
3
2
D − 2D + D
1
x3
= 2
D 1 + (D 2 − 2 D )
−1
1
= 2 1 + (D 2 − 2 D ) x 3
D
2
3
1
= 2 1 − D 2 − 2 D + D 2 − 2 D − D 2 − 2 D +
...
I =

4

[
[

[

]
]

(

) (

) (

[

)

]

[

]

x 5 6 x 4 18 x 3 24 x 2
+
+
+
20 12
6
2
5
4
x
x
+
+ 3x 3 + 12 x 2
=
20 2
=

∴ y = C
...
I
y = (A + Bx)e 0 x + (C + Dx)e x +

x5 x4
+
+ 3x 3 + 12 x 2
20 2

Problems based on R
...
S = e ax x
1
1
e ax x = e ax
x
P
...
Obtain the particular integral of ( D 2 − 2 D + 5) y = e x cos 2 x
Solution: Given ( D 2 − 2 D + 5) y = e x cos 2 x

1
e x cos 2 x
D − 2D + 5


1
= ex 
 cos 2 x(Re placeDbyD = 1)
2
 (D + 1) − 2(D + 1) + 5 



P
...
I =
4
14
...
E is (m 2 +1) = 0
m = -2, -2
C
...
= (Ax + B)e −2 x
1
e − 2 x sin x
P
...
F + P
...
I when f(x) = x n sin ax(or ) x n cos ax
1
x n sin ax(or ) x n cos ax
P
...
e
...
Solve ( D 2 + 4 D + 3) y = e − x sin x + xe 3 x
Solution: The auxiliary equation is m 2 +4m + 3 = 0
m= -1,-3
−x
C
...
I 1 =
2
( D + 4 D + 3)
1
sin x
=
2
(D − 1) + 4(D − 1) + 3
1
= e −x 2
sin x
D + 2D
(1 + 2 D ) sin x
= e −x
− 1 + 4D 2
e−x
[2 cos x + sin x]
=
−5
1
P
...
F + P
...
E : m 2 +2m + 1 = 0
m = -1,-1
C
...
I =
( D + 1) 2

16 Solve


1
2( D + 1) 
(cos x )
= x −
2 
(D + 1)  (D + 1)2


11

2 
1
= x −
 D 2 + 2 D + 1 (cos x )
(D + 1)

2 
1

= x −
 (− 1 + 2 D + 1) (cos x )
D + 1

2  sin x

= x −
D + 1 2


x sin x
=
2
x sin x sin x
=
−
2
D +1
x sin x (D − 1)sin x
=
−
2
D2 −1
x sin x cos x − sin x
+
=
2
2

(

The solution is y = ( A + Bx)e − x +

(

)

x sin x cos x − sin x
+
2
2

)

17
...
E : m 2 +1 = 0
m = ±i
C
...
I = 2
D +1
1  1 − cos 2 x 
= 2


2
D +1

1 1
1

e0x − 2
cos 2 x 
=  2
2 D +1
D +1

1 1

= 1 + cos 2 x 
2 3

1 1
= + cos 2 x
2 6
1 1
∴ y = A cosx +B sinx + + cos 2 x
2 6

d2y
− y = x sin x + (1 + x )e x
dx 2
Solution: A
...
F = A e − x + Be x
1
(xV ) =  x − f ' ( D)  1 (V )
P
...
Solve

12
2D 
1

(sin x )
= x − 2  2
D − 1 (D − 1)

2 D  sin x

= x − 2 
D − 1 − 2

 x sin x
2 cos x 
= −
+

2
2(D 2 − 1)

x sin x cos x
−
=−
2
2
1
1+ x2 ex
P
...
E : m 2 −1 = 0
m =±1
C
...
I =
2
D −1
1
(x sin x )
= ex
(D + 1)2 − 1
1
(x sin x )
= ex 2
D + 2D

19
...
I = e x 

25
 5

Complete Solution is
y = A e − x + Be x +
− x
(sin x + 2 cos x ) + (14 sin x − 2 cos x )
ex

25
 5


METHOD OF VARIATION OF PARAMETERS
This method is very useful in finding the general solution of the second order
equation
...
(1)
2
dx
dx
The complementary function of (1)
C
...
I = Pf 1 +Qf 2
f2 X
dx
f 1 f − f1' f 2
f1 X
dx
'
f1 f 2 − f1' f 2

P= −∫
Q=

∫

1
2

∴ y = c1 f 1 + c 2 f 2 + P
...
Solve D 2 + 4 y = sec 2 x
Solution: The A
...
F = C 1 cos 2 x + C 2 sin 2 x
f 1 = cos 2 x
f 2 = sin 2 x
'
f 1 = −2 sin 2 x
f '2 = 2 cos 2 x
f 1 f 2' − f 1' f 2 = 2 cos 2 2 x + 2 sin 2 2 x
= 2 cos 2 2 x + sin 2 2 x
= 2 [1]
=2
f2 X
dx
P= −∫
1
f 1 f 2 − f1' f 2

[

]

14
sin 2 x sec 2 x
dx
2
1
1
= − ∫ sin 2 x
dx
cos 2 x
2
1 − 2 sin 2 x
dx
= ∫
4 cos 2 x
1
= log(cos 2 x)
4
f1 X
dx
Q= ∫
'
f1 f 2 − f1' f 2
= −∫

cos 2 x sec 2 x
dx
2
1
1
dx
= ∫ cos 2 x
2
cos 2 x
1
= ∫ dx
2
1
= x
2
P
...
Solve by the method of variation of parameters
Solution: The A
...
F = C 1 cos x + C 2 sin x
Here f1 = cos x
f 2 = sin x
'
'
f1 = − sin x
f 2 = cos x
'
'
2
f1 f 2 − f 1 f 2 = cos x + sin 2 x = 1
f2 X
dx
f 1 f − f1' f 2
sin x( x sin x)
dx
= −∫
1
= − ∫ x sin 2 xdx

P= −∫

= −∫x

1
2

(1 − cos 2 x ) dx
2

1
(x − x cos 2 x )dx
2∫
1
1
= − ∫ xdx + ∫ x cos 2 xdx
2
2

=−

d2y
+ y = x sin x
dx 2

15

=−
=−

∫

Q =

1  x 2  1   sin x 
 − cos 2 x 
 − (1)

  +  x
4
2 2  2  2 



1
x2 x
+ sin 2 x + cos 2 x
4 4
8
f1 X
dx
f1 f 2' − f1' f 2

(cos x) x(sin x)

∫
1
= ∫ x sin x cos xdx
sin 2 x
dx
= ∫x
2
=

dx

1
x sin 2 xdx
2∫
1   − cos 2 x 
 − sin 2 x 
=  x
 − (1)

2 
2

 4 
x
1
= − cos 2 x + sin 2 x
4
8
=

P
...
Solve (D 2 −4 D + 4) y = e 2 x by the method of variation of parameters
...
E is m 2 − 4m + 4 = 0
(m − 2)2 = 0
m = 2,2
C
...
I = x 2 e 2 x −

x 2 2x x 2 2x
e =
e
2
2

y = C
...
I
x 2 2x
e
= (Ax +B) e +
2
5
...
E is m 2 + 1 = 0
m = ±i
C
...
I = Pf 1 + Qf 2
= log(cos x) cos x + x sin x
y = c1 cos x + c 2 sin x + log(cos x) cos x + x sin x

17
6
...

Solution: Given (D 2 + a 2 )y = tan ax
The A
...
F = c1 cos ax + c 2 sin ax
f1 = cos ax, f 2 = sin ax
f1' = − a sin x, f 2' = a cos ax
f1 f 2' − f 2 f1' = a cos ax cos ax − sin ax(− a sin ax)
= a cos 2 ax + a sin 2 ax
= a(cos 2 ax + sin 2 ax)
=a
P
...
I = Pf1 + Qf 2
1
1
= 2 cos ax[sin ax − log(sec ax + tan ax )] − 2 sin ax[cos ax ]
a
a
=

18
1
[cos ax log(sec ax + tan ax )]
a2
y = C
...
I
1
= c1 cos ax + c 2 sin ax − 2 [cos ax log(sec ax + tan ax )]
a
=−

d2y
+ y = tan x by the method of variation of parameters
...
Solve
dx 2
Solution: The A
...
F = c1 cos x + c 2 sin x
Here f1 = cos x, f 2 = sin x
f1' = − sin x, f 2' = cos x
f1 f 2' − f 2 f1' = cos 2 x + sin 2 x = 1
f2 X
P= − ∫
dx
1
f 1 f 2 − f1' f 2
sin x tan x
= −∫
dx
1
sin 2 x
dx
= −∫
cos x
1 − cos 2 x
dx
= −∫
cos x
= − ∫ (sec x − cos x )dx

= − log(sec x + tan x) + sin x
f1 X
dx
Q= ∫
'
f1 f 2 − f1' f 2
cos x tan x

∫ 1
= ∫ sin dx
=

dx

= − cos x
∴ P
...
F + P
...
Solve by method of variation of parameters y '' −

4 ' 4
y + 2 y = x2 +1
x
x

4 ' 4
y + 2 y = x2 +1
x
x
2 ''
i
...
, x y − 4 xy ' + 4 y = x 4 + x 2
i
...
, x 2 D 2 − 4 xD + 4 y = x 4 + x 2 …………
...
E is m 2 − 5m + 4 = 0
(m − 4)(m − 1) = 0
m = 1,4
∴ C
...
I = Pf1 + Qf 2
f2 X
dx
P= − ∫
1
f 1 f 2 − f1' f 2

[

]

e z e4z + e2z
dx
− 3e 5 z
e5z + e3z
dx
=∫
3e 5 z
1
= ∫ 1 + e − 2 z dz
3
1
e −2 z 
= z +

3
−2 

= −∫

[

]

1
1
z − e −2 z
3
6
f1 Z
dz
f1 f 2' − f 1' f 2
=

Q=

∫

e 4 z (e 4 z + e 3 z )
∫ − 3e 5 z dz
1 e8z + e 6 z
dz
= − ∫
3
e5z
1
= − ∫ e 3 z + e z dz
3

1  e3z
= − 
+ ez 
3 3


=

(

)

1
1
= − e3z − e z
9
3
1
1 
1

 1
∴ P
...
F + P
...
+ a n y = X
...

dny
d n −1 y
dy
a n x n n + a n −1 n −1 +
...
It is also known
as Euler’s equation
...

x = e z , (or ) z = log x
Then
dy dy dz 1 dy
=
Dy =
=
dx dz dx x dz
d
d

 D = dx , θ = dz 


d
d
x
= xD =
=θ
dx
dz
Similarly
d 2 y d  dy  d  1 dy 
=  = 

dx 2 dx  dx  dx  x dz 
1 dy 1 d  dy 
1 dy 1 d  dy  dz
=− 2
+
+
 =− 2
 
x dz x dz  dz  dx
x dz x dx  dz 
an

d2y
1 dy 1 d 2 y
=− 2
+
dx 2
x dz x 2 dz 2
d 2 y d 2 y dy
∴ x2 2 = 2 −
dz
dx
dz

21

(

)

∴ x 2 D 2 = θ 2 − θ = θ (θ − 1)
Similarly,
x 3 D 3 = θ (θ − 1)(θ − 2)

x 4 D 4 = θ (θ − 1)(θ − 2)(θ − 3)

and soon
...


d2y
dy
+ x + y = 4 sin(log x)
2
dx
dx
Solution: Consider the transformation

1
...
H
...
E : m 2 + 1 = 0, m = ±i
C
...
F = A cos (log x) +B sin (log x)
1
4(sin z )
P
...
I = − 2 log x cos(log x)
∴ Complete solution is:
y = A cos (log x) +B sin (log x) − 2 log x cos(log x)
y = ( A − 2 log x ) cos(log x) − 2 log x cos(log x)
d2y
dy
+ 4 x + 2 y = x log x
2
dx
dx
Solution: Given (x 2 D 2 + 4 xD + 2)y = x log x ……………(1)
Consider: x = e z , (or ) z = log x
∴ xD = θ
x 2 D 2 = θ (θ − 1)
(1) : (θ (θ − 1) + 4θ + 2) y = e z z

2
...
E : m 2 + 3m + 2 = 0
M = - 2, -1
−2
−1
C
...
F = 2 +
x
x
1
P
...
F +P
...
Solve ( ( x 2 D 2 − 3xD + 4) y = x 2 , giventhat y(1) = 1 and y’(1) = 0
Solution: Given x 2 D 2 − 3xD + 4 y = x 2 …………………(1)
Take x = e z , (or ) z = log x
∴ xD = θ
x 2 D 2 = θ (θ − 1)
(1) : θ 2 − 4θ + 4 y = e 2 z

(

)

(

)

A
...
F = ( A + Bz )e 2 z = ( A + B log x )x 2
1
1
P
...
I = e 2 z

z2
2

23

=

x 2 (log x )
2

2

…………………(2)

Complete solution is : y = ( A + B log x )x 2 +

x 2 (log x )
2

2

Apply conditions: y(1) =1,y’(1) = 0 in (2)
A = 1, B = -2

x 2 (log x )
Complete solution is y = (1 − 2 log x )x +
2

2

2

EQUATION REDUCIBLE TO THE HOMOGENEOUS LINEAR
FORM (LEGENDRE LINEAR EQUATION)
It is of the form:

(a + bx )n d

n

y

dx n

+ A1 (a + bx )

n −1

d n −1 y
+
...
(
dx n −1

1)

A1 , A2,
...

Solve (2 x + 3) y ' '−(2 x + 3) y '−12 y = 6 x
Solution: This is Legendre’s linear equation:
(2 x + 3)2 y' '−(2 x + 3) y '−12 y = 6 x ………………
...
H
...
E : 4m 2 − 6m − 12 = 0
3 − 57
3 + 57
, m2 =
m1 =
4
4
C
...
F = A(4 x + 3) m1 + B(2 x + 3) m2
2

(

(

)

)

24
3
3e z
= − (2 x + 3)
2
14
4θ − 6θ − 12
θz
9e
P
...
I 1 =

Solution is

(3+
y = A(2 x + 3)

57 / 4

)


+ B(2 x + 3)

 3− 57


4


−

3
(2 x + 3) − 3
4
14

SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH
CONSTANT COEFFICIENTS

dx
dy
+ 2 x + 3 y = 2e 2t , + 3x + 2 y = 0
dt
dt
dx
dy
Solution: Given
+ 2 x + 3 y = 2e 2t , + 3x + 2 y = 0
dt
dt
d
Using the operator D =
dt
2t
(D + 2)x + 3 y = 2e ………………(1)
3 x + (D + 2 ) y = 0 ………………
...
(3)
A
...
F = Ae t + Be −5t
− 6e 2t
6
= − e 2t
P
...
Solve the simultaneous equations,

8
x = − Ae t + Be −5t + e 2t
7
6
y = Ae t + Be −5t − e 2t
7
dx
dy
+ y = sin t , + x = cos t , giventhat t=0, x = 1, y =0
dt
dt
Solution: Dx + y = sin t ……………
...
Solve

25

(D

)

− 1 y = −2 sin t
...
F = Ae t + Be − t
sin t
sin t
P
...
(1)
- 2x +Dy = cos t ……………
...
F = A cos 2t + B sin 2t
3 sin t
− 3 sin t
P
...
Solve

x = A cos2t +Bsin2t – cost
y = A cos 2t + B sin 2t - sin t

26
dx dy
dx dy
−
+ 2 y = cos 2t , +
− 2 x = sin 2t
dt dt
dt dt
Solution: Dx + (-D +2)y = cos 2t …………
...
H
...
F = e ( A cos t + B sin t )
(− 2 sin 2t ) = sin 2t
P
...
I 2 = 2
D − 2D + 2
(cos 2t + 2 sin 2t )
= −
10

4

---

Title: Mathematics notes for Engineering
Description: this notes consists of allmost all the chapters belonging to mathematics 1.for the first year students of engineering.the notes contain the following chapters. 1.Matrices. 2.Differential Calculus Methods. 3.Improper,Multiple Integration And Applications. 4.Differential Equations Applications. 5.Laplace Transforms. i hope this information will be helpful for you to download the notes..

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