MALLA REDDY COLLEGE OF
ENGINEERING & TECHNOLOGY
(An Autonomous Institution – UGC, Govt.of India)
Recognizes under 2(f) and 12(B) of UGC ACT 1956
(Affiliated to JNTUH, Hyderabad, Approved by AICTE –Accredited by NBA & NAAC-“A” Grade-ISO 9001:2015
Certified)
Mathematics-I
B.Tech – I Year – I Semester
DEPARTMENT OF HUMANITIES AND SCIENCES
MATHEMATICS - I
(R18A0021) Mathematics -I
Course Objectives: To learn
1. The concept of rank of a matrix which is used to know the consistency of system of
linear equations and also to find the eigen vectors of a given matrix.
2. Finding maxima and minima of functions of several variables.
3. Applications of first order ordinary differential equations. ( Newton’s law of cooling,
Natural growth and decay)
4. How to solve first order linear, non linear partial differential equations and also
method of separation of variables technique to solve typical second order partial
differential equations.
5. Solving differential equations using Laplace Transforms.
UNIT I: Matrices
Introduction, types of matrices-symmetric, skew-symmetric, Hermitian, skew-Hermitian,
orthogonal, unitary matrices. Rank of a matrix - echelon form, normal form, consistency of
system of linear equations (Homogeneous and Non-Homogeneous). Eigen values and Eigen
vectors and their properties (without proof), Cayley-Hamilton theorem (without proof),
Diagonalisation.
UNIT II:Functions of Several Variables
Limit continuity, partial derivatives and total derivative. Jacobian-Functional dependence and
independence. Maxima and minima and saddle points, method of Lagrange multipliers,
Taylor’s theorem for two variables.
UNIT III: Ordinary Differential Equations
First order ordinary differential equations: Exact, equations reducible to exact form.
Applications of first order differential equations - Newton’s law of cooling, law of natural
growth and decay.
Linear differential equations of second and higher order with constant coefficients:
Non-homogeneous term of the type f(x) = eax, sinax, cosax, xn, eax V and xn V. Method of
variation of parameters.
UNIT IV: Partial Differential Equations
Introduction, formation of partial differential equation by elimination of arbitrary constants
and arbitrary functions, solutions of first order Lagrange’s linear equation and non-linear
equations, Charpit’s method, Method of separation of variables for second order equations
and applications of PDE to one dimensional (Heat equation).
UNIT V: Laplace Transforms
Definition of Laplace transform, domain of the function and Kernel for the Laplace
transforms, Existence of Laplace transform, Laplace transform of standard functions, first
shifting Theorem, Laplace transform of functions when they are multiplied or divided by “t”,
Laplace transforms of derivatives and integrals of functions, Unit step function, Periodic
function.
Inverse Laplace transform by Partial fractions, Inverse Laplace transforms of functions when
they are multiplied or divided by ”s”, Inverse Laplace Transforms of derivatives and integrals
of functions, Convolution theorem, Solving ordinary differential equations by Laplace
transforms.
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TEXT BOOKS:
i) Higher Engineering Mathematics by B V Ramana ., Tata McGraw Hill.
ii) Higher Engineering Mathematics by B.S. Grewal, Khanna Publishers.
iii) Advanced Engineering Mathematics by Kreyszig, John Wiley & Sons.
REFERENCE BOOKS:
i)Advanced Engineering Mathematics by R.K Jain & S R K Iyenger, Narosa Publishers.
ii)Advanced Engineering Mathematics by Michael Green Berg, Pearson Publishers .
iii)Engineering Mathematics by N.P Bali and Manish Goyal.
Course Outcomes: After learning the concepts of this paper the student will be able to
1.Analyze the solution of the system of linear equations and to find the Eigen values and
Eigen vectors of a matrix.
2.Find the extreme values of functions of two variables with / without constraints.
3.Solve first and higher order differential equations.
4.Solve first order linear and non-linear partial differential equations.
5.Solve differential equations with initial conditions using Laplac
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UNIT-I
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MATRICES
Matrix : A system of mn numbers real (or) complex arranged in the form of an ordered set
of ‘m’ rows, each row consisting of an ordered set of ‘n’ numbers between [ ] (or) ( ) (or) || ||
is called a matrix of order m xn.
a11 a12 .........a1n
a
21 a12 .........a 2 n
Eg: .........................
.......................
a m1 a m 2 .......a mn
where 1≤i≤m, 1≤j≤n.
= [aij ]mxn
mxn
Order of the Matrix: The number of rows and columns represents the order of the matrix. It
is denoted by mxn, where m is number of rows and n is number of columns.
Types of Matrices:
Row Matrix: A Matrix having only one row is called a “Row Matrix”.
Eg: 1 2 31x3
Column Matrix: A Matrix having only one column is called a “Column Matrix” .
1
Eg: 1
2 3 x1
Null Matrix: A= aij mxn such that aij=0 i and j. Then A is called a “Zero Matrix”. It is
denoted by Omxn.
0 0 0
Eg: O2x3=
0 0 0
Rectangular Matrix: If A= aij mxn , and m n then the matrix A is called a “Rectangular
Matrix”
1 1 2
Eg :
is a 2x3 matrix
3 4
2
Square Matrix: If A= aij mxn and m = n then A is called a “Square Matrix”.
1 1
Eg :
is a 2x2 matrix
2 2
Lower Triangular Matrix: A square Matrix AnXn aij nxn is said to be lower Triangular
of aij = 0 if i <j i.e. if all the elements above the principle diagonal are zeros.
4 0 0
Eg: 5 2 0
7 3 6
is a Lower triangular matrix
Upper Triangular Matrix: A square Matrix
A aij
nxn
is said to be upper triangular of
aij = 0 if i.>j. i.e. all the elements below the principle diagonal are zeros.
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1 3 8
Eg: 0 4 5
is an Upper triangular matrix
0 0 2
Triangle Matrix: A square matrix which is either lower triangular or upper triangular is
called a triangle matrix.
Principal Diagonal of a Matrix: In a square matrix, the set of all aij, for which i = j are
called principal diagonal elements. The line joining the principal diagonal elements is called
principal diagonal.
Note: Principal diagonal exists only in a square matrix.
Diagonal elements in a matrix: A= [aij]nxn, the elements aij of A for which i = j.
i.e. a11, a22….ann are called the diagonal elements of A
1
2
Eg: A= 4 5
7 8
3
6
9
diagonal elements are 1, 5, 9
Diagonal Matrix: A Square Matrix is said to be diagonal matrix, if aij = 0 for i j
i.e. all
the elements except the principal diagonal elements are zeros.
Note:
1. Diagonal matrix is both lower and upper triangular.
2. If d1, d2…….dn are the diagonal elements in a diagonal matrix it can be
represented as diag d1 , d2 ,., dn
3 0
0
Eg : A = diag (3,1,-2)= 0 1 0
0 0 2
Scalar Matrix: A diagonal matrix whose leading diagonal elements are equal is called a
“Scalar Matrix”.
Eg : A=
2 0 0
0 2 0
0 0 2
Unit/Identity Matrix: If A = aij such that aij=1 for i = j, and aij=0 for i j then A is
nxn
called a “Identity Matrix” or Unit matrix. It is denoted by
Eg:
1 0
I2 =
;
0 1
In
1 0 0
I3= 0 1 0
0 0 1
Trace of Matrix: The sum of all the diagonal elements of a square matrix A is called Trace
of a matrix A, and is denoted by Trace A or tr A.
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a h
Eg : A = h b
g f
g
f then tr A = a+b+c
c
Singular & Non Singular Matrices: A square matrix A is said to be “Singular” if the
determinant of │A│= 0, Otherwise A is said to be “Non-singular”.
Note: 1. Only non-singular matrices possess inverse.
2. The product of non-singular matrices is also non-singular.
Inverse of a Matrix: Let A be a non-singular matrix of order n if there exist a matrix B
such that AB=BA=I then B is called the inverse of A and is denoted by A-1.
If inverse of a matrix exist, it is said to be invertible.
Note: 1. The necessary and sufficient condition for a square matrix to posses inverse is that
| |≠ .
2 .Every Invertible matrix has unique inverse.
3. If A, B are two invertible square matrices then AB is also invertible and
AB
1
4. A1
B 1 A1
AdjA
det A
where detA 0 ,
Theorem: The inverse of a Matrix if exists is Unique.
Note: 1. (A-1)-1 = A
2. I-1 = I
Theorem: If A, B are invertible matrices of the same order, then
(i). (AB)-1 = B-1A-1
(ii). (A1)-1 = (A-1)1
Sub Matrix: - A matrix obtained by deleting some of the rows or columns or both from the
given matrix is called a sub matrix of the given matrix.
1 5
6
7
Eg: Let A = 8 9 10 5 . Then
3 4
5
1
8 9 10
3 4 5
2 x3
is a sub matrix of A obtained by deleting first
row and 4th column of A.
Minor of a Matrix: Let A be an mxn matrix. The determinant of a square sub matrix of A is
called a minor of the matrix.
Note: If the order of the square sub matrix is ‘t’ then its determinant is called a minor of
order ‘t’.
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2
3
Eg: A =
1
5
1 1
1 2
be a 4x3 matrix
2 3
6 7
] is a sub-matrix of order ‘2’
Here B = [
B = 2-3 = -1 is a minor of order ‘2
] is a sub-matrix of order ‘3’
And C = [
det C = 2(7-12)-1(21-10)+(18-5) = -9
Properties of trace of a matrix: Let A and B be two square matrices and be any scalar
1) tr ( A) = (tr A) ; 2) tr(A+B) = trA + trB ; 3) tr (AB) = tr(BA)
Idempotent Matrix: A square matrix A Such that A2=A then A is called “Idempotent
Matrix”.
Eg: A = [
]
Involutary Matrix: A square matrix A such that A2 = I then A is called an Involutary
Matrix.
Eg: A = [
]
Nilpotent Matrix: A square matrix A is said to be Nilpotent if there exists a + ve integer n
such that An = 0 here the least n is called the Index of the Nilpotent Matrix.
Eg: A = [
]
Transpose of a Matrix: The matrix obtained by interchanging rows and columns of the
given matrix A is called as transpose of the given matrix A. It is denoted by AT or A1
Eg: A = [
] Then AT = [
]
Properties of transpose of a matrix: If A and B are two matrices and AT , B T are their
transposes then
1) AT A ; 2) A B AT BT ; 3)
T
T
KA
T
KAT ; 4) AB BT AT
T
Symmetric Matrix: A square matrix A is said to be symmetric if AT A
If A aij then AT a ji nxn where aij a ji
nxn
a
h
Eg: h b
g
f
g
f
c
is a symmetric matrix
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Skew-Symmetric Matrix: A square matrix A is said to be Skew symmetric If AT A .
If A aij then AT a ji where aij a ji
nxn
nxn
.
0
a
0
b c
Eg : a
b
c
0
is a skew – symmetric matrix
Note: All the principle diagonal elements of a skew symmetric matrix are always zero.
Since aij = -aij aij = 0
Theorem: Every square matrix can be expressed uniquely as the sum of symmetric and
skew symmetric matrices.
Proof: Let A be a square matrix, A =
1
1
A A = A AT A AT =
2
2
1
1
1
1
A AT A AT = P + Q, where P = A AT ; Q = A AT
2
2
2
2
Thus every square matrix can be expressed as a sum of two matrices.
T
Consider PT 1 A AT 1 A AT 1 AT AT = 1 A AT =P, since PT P ,
2
2
2
2
T
T
P is symmetric
T
T
T
1
1
1
Consider QT A AT A AT AT AT = - 1 A AT = - Q
2
2
2
2
Since QT Q , Q is Skew-symmetric.
To prove the representation is unique: Let A= R+S 1 be the representation, where R is
symmetric and S is skew symmetric. i.e. RT R, S T S
Consider AT R S RT S T R S 2
T
1 2 A AT
2S S
1
A AT Q
2
Therefore every square matrix can be expressed as a sum of a symmetric and a skew
symmetric matrix
Ex. Express the given matrix A as a sum of a symmetric and skew symmetric matrices
where
2 4 9
A= 14 7 13
9 5 11
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Solution:
2 14 3
A 4 7 5
9 3 11
T
4 10 12
2 5 6
1
T
A A 10 14 18 P A A 5 7 9 ; P is symmetric
2
12 18 22
6 9 11
T
0 18 6
0 9 3
1
A A 18 0 8 Q A A 9 0 4 ; Q is skew symmetric
2
6 8 0
3 4 0
T
2 5
6 0
9 3
0 4
3 4 0
Now A =P+Q = 5 7 9 + 9
6 9 11
Orthogonal Matrix: A square matrix A is said to be an Orthogonal Matrix if AAT=ATA=I,
Similarly we can prove that A=A-1; Hence A is an orthogonal matrix.
Note: 1. If A, B are orthogonal matrices, then AB and BA are orthogonal matrices.
2. Inverse and transpose of an orthogonal matrix is also an orthogonal matrix.
Result: If A, B are orthogonal matrices, each of order n then AB and BA are orthogonal
matrices.
Result: The inverse of an orthogonal matrix is orthogonal and its transpose is also orthogonal
Solved Problems :
cos
1. Show that A =
sin
cos
Sol: Given A =
sin
sin
is orthogonal.
cos
sin
cos
cos
Consider A.AT =
sin
cos
AT =
sin
sin
cos
sin cos
cos sin
sin
cos
then
cos 2 sin 2
=
sin cos cos sin
cos sin cos sin 1 0
=
I
cos 2 sin 2
0 1
A is orthogonal matrix.
1 2
2
2
1
2. Prove that the matrix 1 2 1 2 is orthogonal.
3
2
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1 2 2
1
Sol: Given A = 2 1 2
3
2 2 1
1 2
2 1
2
1 2
1 2
2
2
1
AT = 1 2 1 2
3
Then
2
2
2
9 0
1
0 0 9
0
1 0
0
Consider A .AT = 1 2 1 2 2 1 2 = 1 0 9 0 = 0 1 0 = I
9
9
2
2
0 0 1
⇒ A.AT = I
Similarly AT .A = I
Hence A is orthogonal matrix
0 2b c
3. Determine the values of a, b, c when a b c is orthogonal.
a b c
Sol: - For orthogonal matrix AAT = I
0 2b
a
c 0 a
So, AAT = a b c 2b b b I
a b c c c c
4b 2 c 2
2b 2 c 2
2b 2 c 2
2
2
a2 b2 c2 a2 b2 c2
2b c
2b 2 c 2 a 2 b 2 c 2 a 2 b 2 c 2
=I=
1 0 0
0 1 0
0 0 1
Solving 2b2-c2 =0, a2-b2-c2 =0
We get c = 2b a2 =b2+2b2 =3b2
a = 3b
From the diagonal elements of I
4b2+c2= 1 4b2+2b2=1 (since c2=2b2) b =
a = 3b =
1
;
2
b=
1
6
1
1
; c = 2b =
6
3
3 1
3 1 Orthogonal?
3 1 9
2
4. Is matrix 4
2 3 1
2 4 3
Sol:- Given A= 4 3 1 3 3 1
3 1 9
1 1 9
2 3 1 2 4 3 14 0 0
4 3 1 3 3 1 = 0 26 0 I3
3 1 9 1 1 9 0 0 91
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I3
Matrix is not orthogonal.
Complex matrix: A matrix whose elements are complex numbers is called a complex
matrix.
Conjugate of a complex matrix: A matrix obtained from A on replacing its elements by the
corresponding conjugate complex numbers is called conjugate of a complex matrix. It is
denoted by A
If A aij
mxn
, A aij
mxn
, where aij is the conjugate of aij .
5
5
2 3i
2 3i
Eg: If A=
then A =
6 7i 5 i
6 7i 5 i
Note: If A and B be the conjugate matrices of A and B respectively, then
(i) A =A
(iii) KA = K A
(ii) A B = A + B
Transpose conjugate of a complex matrix: Transpose of conjugate of complex matrix is
called transposed conjugate of complex matrix. It is denoted by A or A* .
Note: If A and B be the transposed conjugates of A and B respectively, then
(i) A = A
(ii) A B A B
(iii) KA K A
(iv) AB A B
Hermitian Matrix: A square matrix A is said to be Hermitian Matrix iff A A .
1 3i
1 3i
1 3i
4
4
4
Eg: A=
then A =
and Aθ=
7
7
7
1 3i
1 3i
1 3i
Note: 1. In Hermitian matrix the principal diagonal elements are real.
2. The Hermitian matrix over the field of Real numbers is nothing but real symmetric matrix.
3. In Hermitian matrix A= aij , aij a ji i, j .
nxn
Skew-Hermitian Matrix: A square matrix A is said to be Skew-Hermitian Matrix iff
A A .
T
3i 2 i
2 i
3i 2 i
3i
A
A
Eg: Let A=
then
=
and
2 i
2 i
i
i
2 i i
A =-A A is skew-Hermitian matrix.
T
Note: 1. In Skew-Hermitian matrix the principal diagonal elements are either Zero or Purely
Imaginary.
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2. The Skew- Hermitian matrix over the field of Real numbers is nothing but real
Skew - Symmetric matrix.
3. In Skew-Hermitian matrix A= aij , aij a ji i, j .
nxn
Unitary Matrix: A Square matrix A is said to be unitary matrix iff
AA A A I or A A1
1 4 2 4i
B
Eg:
4
6 2 4i
Theorem1: Every square matrix can be uniquely expressed as a sum of Hermitian and
skew – Hermitian Matrices.
1
1
1
(2 A) ( A A) ( A A A A )
2
2
2
Proof: - Let A be a square matrix write
1
1
A ( A A ) ( A A )i.eA P Q
2
2
A
Let P
1
1
A A ; Q A A
2
2
1
1
Consider P ( A A ) ( A A ) ( A A ) P
2
2
I.e. P P, P is Hermitian matrix.
1
1
1
Q ( A A ) A A ( A A ) Q
2
2
2
Ie Q Q, Q is skew – Hermitian matrix.
Thus every square matrix can be expressed as a sum of Hermitian & Skew Hermitian
matrices.
To prove such representation is unique:
Let A = R+S------- (1) be another representation of A where R is Hermitian matrix &
S is skew – Hermitian matrix.
R R ; S S
Consider A ( R S ) R S R S . Ie A R A (2)
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1
A A P
2
1
1 2 A A 2S ie S A A Q
2
1 2 A A
2 R ie R
Thus every square matrix can be uniquely expressed as a sum of Hermitian & skew
Hermitian matrices.
Solved Problems :
7 4i 2 5i
3 i then show that A is Hermitian and iA is skew2
2 5i 3 i
4
3
If A= 7 4i
1)
Hermitian.
7 4i 2 5i
3
Given A= 7 4i
3 i then
2
2 5i 3 i
4
Sol:
7 4i 2 5i
7 4i 2 5i
3
3
T
A 7 4i
3 i And A 7 4i
3 i
2
2
2 5i 3 i
2 5i 3 i
4
4
A A
T
Hence A is Hermitian matrix.
Let B= iA
4 7i 5 2i
2i 1 3i
1 3i
4i
3i
i.e B= 4 7i
5 2i
then
4 7i 5 2i
3i
1 3i
B 4 7i
2i
5 2i 1 3i
4i
4 7i 5 2i
3i
4 7i 5 2i
3i
4 7i
2i
1 3i (1) 4 7i 2i 1 3i B
5 2i 1 3i 4i
5 2i 1 3i
4i
B
T
T
B =-B
B= iA is a skew Hermitian matrix.
2). If A and B are Hermitian matrices, prove that AB-BA is a skew-Hermitian matrix.
Sol: Given A and B are Hermitan matrices
A A And B B ------------- (1)
T
T
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Now AB BA AB BA
T
AB BA
T
T
BA B A A B
AB
T
T
T
T
T
T
BA AB (By (1))
AB BA
Hence AB-BA is a skew-Hemitian matrix.
a ic
3). Show that A=
b id
a ic
Sol: Given A=
b id
b id
is unitary if and only if a2+b2+c2+d2=1
a ic
b id
a ic
a ic b id
Then A
b id a ic
T
a ic
Hence A A
b id
b id
a ic
a ic
AA
b id
b id a ic
a ic b id
a 2 b2 c 2 d 2
=
0
b id
a ic
0
a 2 b2 c 2 d 2
AA I if and only if a 2 b 2 c 2 d 2 1
1 2i
0
1
4)Given that A=
, show that I AI A is a unitary matrix.
0
1 2i
1 2i
1 0 0
Sol: we have I A
0
0 1 1 2i
1 2i
1
And
1
1 2i
1 2i
1 0 0
I A
0
0 1 1 2i
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1 2i
1
=
1
1 2i
( I A) 1
1 2i
1
1
2
1
1 4i 1 1 2i
1 2i
1 1
1
6 1 2i
Let B I AI A
1
B
1 2i 1 2i
1 2i 1 1 (1 2i)( 1 2i)
1 2i 1
1 1
1 1 2i
1 6 1 2i 1 2i
(1 2i)(1 2i) 1
6 1 2i
B
1 4 2 4i
4
6 2 4i
Now B
T
B B
T
2 4i
2 4i
1 4
1 4
B
and
4
4
6 2 4i
6 2 4i
1 4 2 4i
4
36 2 4i
B
T
2 4i
4
2 4i 4
1 36 0 1 0
I
36 0 36 0 1
B 1
i.e. B is unitary matrix.
I AI A is a unitary matrix.
1
5) Show that the inverse of a unitary matrix is unitary.
Sol: Let A be a unitary matrix. Then AA I
i.e
AA
1
I 1
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A A1 I
1
A1 A1 I
1
Thus A is unitary.
Rank of a Matrix:
Let A be mxn matrix. If A is a null matrix, we define its rank to be ‘0’. If A is a non-zero
matrix, we say that ‘r’ is the rank of A if
i.
Every (r+1)th order minor of A is ‘0’ (zero) &
ii.
At least one rth order minor of A which is not zero.
It is denoted by (A) and read as rank of A.
Note: 1. Rank of a matrix is unique.
2. Every matrix will have a rank.
3. If A is a matrix of order mxn, then Rank of A ≤ min (m,n)
4. If ρ (A) = r then every minor of A of order r+1, or minor is zero.
5. Rank of the Identity matrix In is n.
6. If A is a matrix of order n and A is non-singular then ρ (A) = n
7. If A is a singular matrix of order n then (A) < n
Important Note:
1. The rank of a matrix is ≤ r if all minors of (r+1)th order are zero.
2. The rank of a matrix is ≥ r, if there is at least one minor of order ‘r’ which is not equal to
zero.
1 2
3
1. Find the rank of the given matrix 3 4 4
7 10 12
1 2
3
Sol: Given matrix A = 3 4 4
7 10 12
det A = 1(48-40)-2(36-28)+3(30-28) = 8-16+6 = -2 ≠ 0
We have minor of order 3
∴ρ (A) =3
2. Find the rank of the matrix
1 2 3 4
5 6 7 8
5
8 7 0
Sol: Given the matrix is of order 3x4
Its Rank ≤ min(3,4) = 3
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Highest order of the minor will be 3.
2
3
8 7
0
1
Let us consider the minor 5 6 7
Determinant of minor is 1(-49)-2(-56) + 3(35-48) = -49+112-39 = 24 ≠ 0.
Hence rank of the given matrix is ‘3’.
Elementary Transformations on a Matrix:
i). Interchange of ith row and jth row is denoted by Ri ↔ Rj
(ii). If ith row is multiplied with k then it is denoted by Ri kRi
(iii). If all the elements of ith row are multiplied with k and added to the corresponding
elements of jth row then it is denoted by Rj Rj +kRi
Note: 1. The corresponding column transformations will be denoted by writing ‘c’. i.e
ci ↔cj,
ci k cj
cj cj + kci
2. The elementary operations on a matrix do not change its rank.
Equivalance of Matrices: If B is obtained from A after a finite number of elementary
transformations on A, then B is said to be equivalent to A.It is denoted as B~A.
Note : 1. If A and B are two equivalent matrices, then rank A = rank B.
2. If A and B have the same size and the same rank, then the two matrices are equivalent.
Elementary Matrix or E-Matrix: A matrix is obtained from a unit matrix by a single
elementary transformation is called elementary matrix or E-matrix.
Notations: We use the following notations to denote the E-Matrices.
1) Eij Matrix obtained by interchange of ith and jth rows (columns).
2) Ei k Matrix obtained by multiplying ith row (column) by a non- zero number k.
3) Eij k Matrix obtained by adding k times of jth row (column) to ith row (column).
Echelon form of a matrix:
A matrix is said to be in Echelon form, if
(i) Zero rows, if any exists, they should be below the non-zero row.
(ii) The first non-zero entry in each non-zero row is equal to ‘1’.
(iii) The number of zeros before the first non-zero element in a row is less than the number of
such zeros in the next row.
Note : 1. The number of non-zero rows in echelon form of A is the rank of ‘A’.
1.
The rank of the transpose of a matrix is the same as that of original matrix.
2.
The condition (ii) is optional.
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MATHEMATICS - I
Eg: 1.
1
0
0
0
0
1
0
0
0 0
0 0
1 1
0 0
is a row echelon form.
1 3 1
2. 0 1 1 is a row echelon form.
0 0
0
Solved Problems :
2
1.
3 7
Find the rank of the matrix A = 3 2 4 by reducing it to Echelon form.
1 3 1
Sol: Given A =
2 3 7
3 2 4
1 3 1
Applying row transformations on A.
R1 ↔ R3
1 3 1
A ~ 3 2 4
2 3 7
R2 → R2 –3R1; R3→ R3 -2R1
1 3 1
~ 0 7 7
0 9 9
R2 → R2/7,R3→ R3/9
1 3 1
~ 0 1 1
0 1 1
R3 → R3 –R2
1 3 1
~ 0 1 1
0 0 0
This is the Echelon form of matrix A.
The rank of a matrix A= Number of non – zero rows =2
2.
4 4 3
1 1 1
For what values of k the matrix
k 2
2
9 9 k
1
0
has rank ‘3’.
2
3
Sol: The given matrix is of the order 4x4
If its rank is 3 det A =0
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MATHEMATICS - I
4 4 3
1 1 1
A=
k 2
2
9 9 k
1
0
2
3
Applying R2 → 4R2-R1, R3 →4R3 – kR1, R4 → 4R4 – 9R1
3
1
4 4
0 0
1
1
We get A ~
0 8 4k 8 3k 8 k
4k 27
3
0 0
Since Rank A = 3 det A =0
0
1
1
4 8 4k 8 3k 8 k 0
0
4k 27 3
1[(8-4k) 3]-1(8-4k) (4k+27)] = 0
(8-4k) (3-4k-27) = 0
(8-4k)(-24-4k) =0
(2-k)(6+k)=0
k =2 or k = -6
3).
Find the rank of the matrix using echelon form
Sol: Given
2
4
A
8
8
2
4
A
8
8
1 3 5
2 1 3
4 7 13
4 3 1
1 3 5
2 1 3
4 7 13
4 3 1
By applying R2 R2 2R1 ; R3 R3 4R1 ; R4 R4 4 R1
R
R
R
R1 1 , R2 2 , R3 3
1
1
3
R3 R3 R2 , R4 R4 R2
2
0
~
0
0
2
0
~
0
0
1
0
0
0
3
5
5
5
5
7
7
7
2
0
~
0
0
1
0
0
0
3
5
0
0
5
7
0
0
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1 3
5
0 5 7
0 5 7
0 15 21
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MATHEMATICS - I
⇒ A is in echelon form
4).
∴ Rank of A = 2
Find the rank of the matrix A =
1
2
3
1
2 0
1 1
3 1
1 1
1
0
1
1
by reducing into echelon form.
Sol: By applying R2 R2 2R1; R3 R3 3R1; R4 R4 R1
A
1 2 0 1
~ 0 3 1 2
0 3 1 2
0 3 1 2
1 2 0 1
3 1 2
0 0 0 0
0 3 1 2
R3 R3 R2
A ~ 0
R3 R4
A ~ 0
R3 R3 R2
A ~ 0
1 2 0 1
3 1 2
0 3 1 2
0 0 0 0
1 2
3
0 0
0 0
0 1
1 2
0 0
0 0
Clear it is in echelon form, rank of A = 2
Normal form/Canonical form of a Matrix:
Every non-zero Matrix can be reduced to any one of the following forms.
Ir
0
0
; Ir
0
I
0 ; r ; I r Known as normal forms or canonical forms by using Elementary
0
row or column or both transformations where I r is the unit matrix of order ‘r’ and ‘O’ is the
null matrix.
Note: 1.In this form “the rank of a matrix is equal to the order of an identity matrix.
2. Normal form another name is “canonical form”
Solved Problems :
1.
1 2 3 4
By reducing the matrix 2 1 4 3 into normal form, find its rank.
3 0 5 10
1 2 3 4
Sol: Given A = 2 1 4 3
3 0 5 10
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MATHEMATICS - I
R2 → R2 – 2R1; R3 → R3 – 3R1
3
4
1 2
5
A ~ 0 3 2
0 6 4 22
R3 → R3/-2
3
1 2
A ~ 0 3 2
0 3
2
4
5
11
R3 → R3+R2
3
1 2
A ~ 0 3 2
0 0
0
4
5
6
c2→ c2 - 2c1, c3→c3-3c1, c4→c4-4c1
0
0 0
A ~ 0 3 2
0 0
0
0
5
6
c3 → 3 c3 -2c2, c4→3c4-5c2
1 0
A ~ 0 3
0 0
0
0
0
0
0
18
c2→ c2 /-3, c4→c4/18
1
A ~ 0
0
0
1
0
0
0
0
0
0
1
c4 ↔ c3
1
A~ 0
0
0
1
0
0
0
1
0
0
0
This is in normal form [I3 0],
∴ Hence Rank of A is ‘3’.
1
2).
1
1
1
2
3 4
2 3 5 5
3 4 5 8
Find the rank of the matrix A = 1
by reducing into canonical form or
normal form.
Sol: Given A
1
1
1 1
= 1 2 3 4
2 3 5 5
3 4 5 8
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MATHEMATICS - I
By applying R2 R2 R1 , R3 R3 2R1 , R4 R4 3R1
R3 R3 R2 , R4 R4 7 R2
R4 R4 6R3
R
R4 4
18
Apply C2 C2 C1 , C3 C3 C1 , C4 C4 C1
C3 C3 2C2 ; C4 C4 5C2
C4 C4 2C3
Clearly it is in the normal form I 4
3).
2 1
4 2
8 4
8 4
1 1 1 1
0 1 2 5
~
0 1 3 7
0 7 8 5
1
0
~
0
0
1
1
0
0
1 1
2 5
1 2
6 30
1
0
~
0
0
1
1
0
0
1 1
2 5
1 2
0 18
1
0
~
0
0
1
1
0
0
1 1
2 5
1 2
0 1
1
0
~
0
0
0
1
0
0
0 0
2 5
1 2
0 1
1
0
~
0
0
0
1
0
0
0 0
0 0
1 2
0 1
1
0
~
0
0
0
1
0
0
0
0
1
0
0
0
0
1
Rank of A = 4
Define the rank of the matrix and find the rank of the following matrix
3
5
1
3
7 13
3 1
2
Sol: Let A= 4
8
8
1 3 5
2 1 3
4 7 13
4 3 1
R2 R2 2 R1
2
0
A~
0
0
R3 R3 4 R1
R4 R4 4 R1
1 3
5
0 5 7
0 5 7
0 15 21
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MATHEMATICS - I
2
0
A~
0
0
R3 R3 R2
R4 R4 3R2
1 3 5
0 5 7
0 0
0
0 0
0
It is in echelon form. So, rank of matrix = no. of non zero rows in echelon form.
Rank
4).
Sol:
( A) 2
Reduce the matrix A to normal form and hence find its rank
2
Given A 0
2
2
1
C1 C1
2
R3 R3 R2
R4 R4 R1
R2 R2 R3
R3 R3 2 R2
R4 R4 4 R2
R4 R4 2R3
C4 4C4 C3
1 3
3 4
3 7
5 11
1
0
A~
1
1
1 3
3 4
3 7
5 11
1
0
A~
0
0
1
3
2
4
3
4
4
8
2
0
A
2
2
1 3
3 4
3 7
5 11
4
1
5
6
4
1
5
6
4
1
5
6
4
1
1
2
1
0
A~
0
0
1
1
2
4
3
0
4
8
4
0
1
2
1
0
A~
0
0
0
1
0
0
0
0
4
8
0
0
1
2
1
0
A~
0
0
0
1
0
0
0
0
4
0
0
0
1
0
1
0
A~
0
0
0
1
0
0
0
0
4
0
0
0
1
0
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MATHEMATICS - I
1
0
A~
0
0
1
C3 C3
3
0
1
0
0
0
0
0
0
0
0
1
0
I
⇒ A~ 3
0
0
0
This is in normal form. Thus Rank of matrix = Order of identify matrix. Rank ( A) 3
0 1 2 2
6 into canonical form and then find its rank.
2 1 3 1
5).
Reduce the matrix A = 4 0 2
Sol:
Apply C1 C2
1 0 2 2
A ~ 0 4 2 6
1 2 3 1
R3 R3 R1
1 0 2 2
A ~ 0 4 2 6
0 2 1 3
1 0 0 0
A ~ 0 4 2 6
0 2 1 3
C3 C3 2C1; C4 C4 2C1
R2
2
1 0 0 0
A ~ 0 2 1 3
0 0 0 0
C2 C3
1 0 0 0
A ~ 0 1 2 3
0 0 0 0
R2
1 0 0 0
A ~ 0 1 0 0
0 0 0 0
C3 C3 2C2 ; C4 C4 3C2
I
Which is in the normal form 2
0
0
,
0
(A) = 2
Note: .If A is an mxn matrix of rank r, there exists non-singular matrices P and Q such that
I r 0
PAQ =
0 0
Suppose we want to find P and Q we have procedure.
Let order of matrix ‘A’ is ‘3
1 0
A = 0 1
0 0
0
1 0
0 A 0 1
0 0
1
i.e. A = I3 A I3
0
0
1
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MATHEMATICS - I
Now we go on applying elementary row operations and column operations on the matrix A
I 0
(L.H.S) until it is reduced to the normal form r
0 0
Every row operations will also be applied to the pre-factor of on R.H.S
Every column operation will also be applied to the post –factor of on R.H.S.
Solved Problems :
1 0
1. Find the non-singular matrices P and Q is of the normal form where A = 2 3
3 3
2
4
6
Sol: Write A = I3 A I3
1 0
~ 2 3
3 3
2
4 =
6
1 0
0 1
0 0
0
1 0
0 A 0 1
0 0
1
0
0
1
R2 → R2-2R1, R3→R3-3R1
1 0
~ 0 3
0 3
2
1 0 0
1 0
0 = 2 1 0 A 0 1
3 0 1
0
0 0
R3 → R3-R2, R2 →1/3 R2
1 0
~ 0 1
0 0
2
0
0
1 0
~ 0 1
0 0
0
0 =
0
c3 → c3-2c1
~[
0
0
1
0
0
1 0
1
2 / 3 1 / 3 0 A 0 1
1
0 0
1
1
0
0
1
1 0
2 / 3 1 / 3 0 A 0 1
1
0 0
1
1
0
0
1
]= PAQ where P = 2 / 3 1 / 3 0
1
1
1
1 0
Q = 0 1
0 0
0
0
1
2
0
1
2
0
1
2. Find the non-singular matrices P and Q such that the normal form of A is P A Q.
1 3
Where A = 1 4
1 5
6
5
4
1
1 . Hence find its rank.
3
Sol: we write A = I3A I4
1 3 6
~ 1 4 5
1 5 4
1 0
1
0 1
=
1
0 0
3
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0
0 A
1
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
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MATHEMATICS - I
R2 →R2- R1, R3→R3-R1
R3 →R3-2R2
1 3
~ 0 1
0 2
6 1
1 0 0
=
1 1 0 A
1 2
1 0 1
2 4
1 3
~ 0 1
0 0
6 1
1 2 =
0 0
1 0 0
1 1 0
1 2
1
0
0
0
A
1
0
1
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
Applying c2→c2-3c1,c3 →c3-6c1, and c4→c4+c1, we get.
1 0
0
2 =
0
0
~ 0 1 1
0 0
0
0
1
1 1
1 2
0
A
0
1
1 3
0
1
0
0
0
0
6 1
0 0
1 0
0 1
Applying c3→c3+c2 and c4→c4-2c2, we get.
~
1 0
0 1
0 0
I 0
2 = P A Q Where P =
0 0
I 0
Here A ~ 2 ,
0 0
0
0
0
0
=
0
0
1 0 0
1 1 0
1 2 1
1 0 0
1 1 0 A
1 2 1
Q=
1 3 9 7
0
1 1 2
0
0 1 0
0 0 1
0
1 3 9 7
0
1 1 2
0
0 1 0
0 0 1
0
∴ Hence ρ(A) =2
System of linear equations: In this chapter we shall apply the theory of matrices to study the
existence and nature of solutions for a system of m linear equations in ‘n’ unknowns.
The system of m linear equations in ‘n’ unknowns x1, x2, x3 xn given by
a11 x1 a12 x2 ...... a1n xn b1
a21 x1 a22 x2 ...... a2 n xn b2
} --------------------------- (1)
..............................................
am1 x1 am 2 x2 ...... amn xn bm
The above set of equations can be written in the Matrix form as A X = B
a11
a
21
an1
a12
a22
an 2
a1n
a2 n
ann
x1
x
2
.
.
xn
b1
b
2
. (2)
.
bm
A-Coefficient Matrix; X-Set of unknowns; B-Constant Matrix
Homogeneous Linear Equations: If b 1 b2 ........... bm 0 then B = 0
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MATHEMATICS - I
Hence eqn (2) Reduces to AX = 0 which are known as homogeneous linear equations
Non-Homogeneous Linear equations:
It at least one of b1, b2----bm is non zero. Then B 0, the system Reduces to AX = B is known
as Non-Homogeneous Linear equations.
Solutions: A set of numbers x1 , x2 xn which satisfy all the equations in the system is known
as solution of the system.
Consistent: If the system possesses a solution then the system of equations is said to be
consistent.
Inconsistent: If the system has no solution then the system of equations is said to be
Inconsistent.
Augmented Matrix: A matrix which is obtained by attaching the elements of B as the last
column in the coefficient matrix A is called Augmented Matrix. It is denoted by [A|B]
a11
A B C a21
am1
a12
a22
am 2
a13
a23
am3
a1n
a2 n
amn
: b1
: b2
: b3
1. If ( A / B) (A) , then the system of equations AX = B is consistent (solution exits).
a). If ( A / B) (A) = r = n (no. of unknowns) system is consistent and have a unique
solution
b). If ( A / B) (A) = r < n (no. of unknowns) then the system of equations AX = B will
have an infinite no. of solutions. In this case (n-r) variables can be assigned arbitrary values.
2. If ( A / B) (A) then the system of equations AX=B is inconsistent (no solution).
In case of homogeneous system AX = 0, the system is always consistent.
(or) x1 = 0, x2 = 0, --------, xn = 0 is always the solution of the system known as the” zero
solution “.
Non-trivial solution:
If ( A / B) (A) = r < n (no. of unknowns) then the system of equations AX = 0 will have
an infinite no. of non zero (non trivial) solutions. In this case (n-r) variables can be assigned
arbitrary values.
Also we use some direct methods for solving the system of equations.
Note: The direct methods are Cramer’s rule, Matrix Inversion, Gaussian Elimination, Gauss
Jordan, Factorization Tridiagonal system. These methods will give a unique solution.
Procedure to solve AX = B (Non Homogeneous equations)
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MATHEMATICS - I
Let us first consider n equations in n unknowns ie. m=n then the system will be of the form
a11x1+a12x2+a13x3+…..+a1nxn =b1
a21x1+a22x2+a23x3+…..+a2nxn =b2
…………………………………..
an1x1+an2x2+an3x3+…..+annxn =bn
The above system can be written as AX = B --------- (1)
Where A is an n n matrix.
Solving AX = B using Echelon form:
Consider the system of m equations in n unknowns given by
a11x1+a12x2+a13x3+…..+a1nxn =b1
a21x1+a22x2+a23x3+…..+a2nxn =b2
…………………………………..
am1x1+am2x2+am3x3+…..+amnxn =bm
We know this system can be we write as AX = B
a11
a
The augmented matrix of the above system is [A / B] = 21
am1
a12
a22
am 2
a1n b1
a2 n b2
amn bm
The system AX = B is consistent if ρ (A) = ρ[A/B]
i). ρ (A) = ρ [A/B] = r < n (no. of unknowns).Then there is infinite no. of solutions.
ii). ρ (A) = ρ [A/B] = number of unknowns then the system will have unique solution.
iii). ρ (A) ≠ ρ [A/B] the system has no solution.
Solved Problems :
1). Show that the equations x+y+z = 4, 2x+5y-2z =3, x+7y-7z =5 are not consistent.
1 1
Sol: Write given equations is of the form AX = B i.e; 2 5
1 7
Consider the Augment matrix is [A /B]
Applying R2 →R2-2R1 and R3 → R3-R1, we get
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1
2
7
x 4
y 3
z 5
1 1
[A/B] = 2 5
1 7
1
2
7
4
3
5
1 1
[A/B] ~ 0 3
0 6
1
4
8
4
5
1
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MATHEMATICS - I
1 1
[A/B] ~ 0 3
0 0
Applying R3→ R3-2R2, we get
1
4
0
4
5
11
∴ρ (A) =2 and ρ (A/B) =3
The given system is inconsistent as ρ (A) ≠ ρ[A/B].
2). Show that the equations given below are consistent and hence solve them
x-3y-8z = -10, 3x+y-4z =0, 2x+5y+6z =3
1 3
Sol: Matrix notation is 3 1
2 5
8
4
6
x 10
y 0
z 3
1 3
1
2 5
8
4
6
10
0
3
Augmented matrix [A/B] is
[A/B] = 3
R2 → R2-3R1 R3 → R3 -2R1
1
~ 0
0
3
10
11
8
20
22
10
3 0
23
R2 →1/10 R2
~ 0
1
0
3
1
11
8
2
22
10
3
23
R3 →R3-11R2
1
~ 0
0
3
1
0
8
2
0
10
3
10
This is the Echelon form of [A/B]
ρ (A) ≠ ρ[A/B].
∴ρ (A) =2, ρ (A/B) =3
The given system is inconsistent.
3). Find whether the following equations are consistent, if so solve them. x+y+2z=4,
2x-y+3z=9, 3x-y-z=2
1
Sol: We write the given equations in the form AX=B i.e; 2
3
1
The Augmented matrix [A/B] = 2
3
1 2 4
1 3 9
1 1 2
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1 2 x 4
1 3 y 9
1 1 z 2
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MATHEMATICS - I
Applying R2→ R2-2R1 and R3→R3-3R1, we get
1
[A/B] ~ 0
0
1
2 4
3 1 1
4 7 10
Applying R3 →3R3-4R2, we get
1
[A/B] ~ 0
0
1
2
3 1
0 17
4
1
34
this matrix is in Echelon form. ρ(A) = 3 and ρ(A/B) = 3
Since ρ (A) =ρ [A/B]. ∴ The system of equations is consistent.
Here the number of unknowns is 3
Since ρ (A) =ρ [A/B] = number of unknowns
The system of equations has a unique solution
1
We have 0
0
1 2 x 4
3 1 y 1
0 17 z 34
-17z = -34 z = 2
-3y-z =1 -3y =z+1 -3y =3 y= -1
and x+y+2z =4 x=4-y-2z =4+1-4=1
x=1, y=-1, z=2 is the solution.
4). Show that the equations x+y+z=6, x+2y+3z=14, x+4y+7z=30 are consistent and solve
them.
Sol: We write the given equations in the form AX=B
1
The Augmented matrix [A/B] = 1
1
1 1
2 3
4 7
6
14
30
1 1
i.e. 1 2
1 4
1
3
7
x 6
y 14
z 30
1
0
0
1 1
1 2
3 6
6
8
24
1
1 1
1 2
0 0
6
8
0
Applying R2→ R2-R1 and R3→R3-R1, we get
[A/B] ~
Applying R3 →R3-3R2, we get
[A/B] ~ 0
0
This matrix is in Echelon form. ρ(A) = 2 and ρ(A/B) = 2
Since ρ (A) =ρ [A/B].
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The system of equations is consistent. Here the no. of unknowns are 3
Since rank of A is less than the no. of unknowns, the system of equations will have infinite
number of solutions in terms of n-r=3-2=1 arbitrary constant.
1 1
The given system of equations reduced form is 0 1
0 0
1
2
0
x 6
y 8
z 0
x+y+z=6……… (1), y+2z=8…….(2)
Let z=k, put z=k in (2) we get y=8-2k
Put z=k y=8-2k in (1), we get
x=6-y-z=6-8+2k=-2+k
∴ x=-2+k, y=8-2k, z=k is the solution, where k is an arbitrary constant.
5). Show that x 2 y z 3 ; 3 x y 2 z 1 ; 2 x 2 y 3z 2; x y z 1 are consistent
and solve them
1 2 1
3
3 1 2 x
y 1
Sol: The above system in matrix notation is 2 2 3
2
z 31
1 1 1 43
1 41
A
X B
1 2 1 3
3 1 2 1
The Augmented matrix is AB
2 2 3 2
1 1 1 1
R2 R2 3R1
R3 R3 2 R1
R4 R4 R1
R2 R2 R3
R3 R3 3R1
R4 R4 3R2
R
R3 3
5
1 2 1 3
0 7 5 8
~
0 6 5 0
0 3 2 4
1 2 1 3
0 1 0 4
~
0 6 5 4
0 3 2 4
1 2 1 3
0 1 0 4
~
0 0 5 20
8
0 0 2
1 2 1 3
0 1 0 4
~
0 0 1 4
0 0 2 8
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1 2 1 3
0 1 0 4
~
0 0 1 4
0 0 0 0
R4 R4 2R3
∴ρ (A) = 3 = ρ (A/B
∴ρ (A) =ρ (A/B) = No. of unknowns = 3
The given system has unique solution.
The systems of equations equivalent to given system are
x 2y z 3
y 4; z 4
x84 3
y 4; z 4
x 3 4 1
x 1, y 4, z 4.
6). Solve x y z 3; 3x 5 y 2 z 8; 5 x 3 y 4 z 14
1 1 1 x 3
Sol: - 3 5 2 y 8
5 3 4 z 14
1 1 1 3
AB 3 5 2 8
5 3 4 14
Augmented Matrix is
1 1 1 3
~ 0 8 1 1
0 8 1 1
R2 R2 3R1
R3 R3 5R1
1 1 1 3
~ 0 8 1 1
0 0 0 0
R3 R3 R2
A AB 2 Number of unknowns (3)
The system has infinite number of solutions.
x y z 3, 8 y z 1 8 y z 1
Let z k y
1 k
8
and
x 238 87 k
X y 18 8k
z 0 k
x 3
238
X 81
1
1 k k 24 1 k 8k 23 7k
8
8
8
where k is any real number.
7). Find whether the following system of equations is consistent. If so solve them.
x 2 y 2 z 2, 3x 2 y z 5,
2 x 5 y 3z 4, x 4 y 6 z 0.
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1 2 2
2
x
3 2 1
y 5
Sol: In Matrix form it is 2 5 3 4
z
1 4 6
0
AX B
R4 R4 R1
1 2 2 2
0 8 5 1
~
0 9 1 8
0 2 4 2
R2 R2 R3
1 2 2 2
0 1 4 7
~
0 9 1 8
0 2 4 2
R2 R2 3R1
R3 R3 2 R1
1
0
~
0
0
2 2
2
1 4 7
0 37 55
0 12 16
1
R4
4
1
~ 0
0
0
2 2 2
1 4 7
0 37 55
0 3 4
R4 37 R4 3R3
1
~ 0
0
0
2 2 2
1 4 7
0 37 55
0 0 17
R3 R3 9 R2
R4 R4 2 R2
R4
is in echelon form
A 3 and AB 4 A AB .∴The given system is in consistent.
8). Discuss for what values of ,
x+2y+ z =
the simultaneous equations x+y+z = 6, x+2y+3z =10,
have
(i). No solution
(ii). A unique solution
(iii). An infinite number of solutions.
1 1
Sol: The matrix form of given system of Equations is A X = 1 2
1 2
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1
3
x 6
y 10 = B
z
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1 1
The augmented matrix is [A/B] = 1 2
1 2
1 6
3 10
1
[A/B] ~ 0
0
R2 → R2 – R1, R3 → R3-R1
1
~ 0
0
R3 →R3 – R2
1
1
0
1
1
1
1
2
1
6
4
6
6
4
3 10
1
2
Case (i): let ≠ 3 the rank of A = 3 and rank [A/B] = 3
Here the no. of unknowns is ‘3’
∴ρ (A) =ρ (A/B)= No. of unknowns
The system has unique solution if ≠3 and for any value of ‘ ’.
Case (ii): Suppose =3 and ≠10.
We have ρ (A) = 2, ρ (A/B) = 3
The system has no solution.
Case (iii): Let =3 and =10.
We have ρ (A) = 2 ρ (A/B) = 2
Here ρ (A) = ρ (A/B) ≠ No. of unknowns =3
The system has infinitely many solutions.
9). Find the values of a and b for which the equations x+y+z=3; x+2y+2z=6;x+ay+3z=b
have
(i) No solution
(ii) A unique solution
(iii) Infinite no of solutions.
1 1 1 x 3
Sol: The above system in matrix notation is 1 2 2 y 6
1 a 3 z b
Augmented matrix
R2 R2 R1
R3 R3 R1
1 1 1 3
AB 1 2 2 6
1 a 3 b
1
1
3
1
~ 0
1
1
3
0 a 1 2 b 2
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R3 R3 R2
For a 3 & b 9
1
1
3
1
~ 0
1
1
3
0 a 3 0 b 9
1 1 1 3
~ 0 1 1 3
0 0 0 0
A AB 2 3 ⇒ It has infinite no of solutions.
For
a 3 & b any value
1
1 3
1
~ 0
1
1 3
0 a 3 0 b 9
A AB 3 ⇒ It has a unique solution.
For
a 3&b 9
1 1 1 3
~ 0 1 1 3
0 0 0 b 9
A 2 AB 3 ⇒ Inconsistent no solution
10). Solve the following system completely. x y z 1; x 2 y 4 x ; x 4 y 10 z 2
1 1 1 x 1
1 2 4 y
Sol: The above system in matrix notation is
1 4 10 z 2
A
Augmented Matrix is
R2 R2 R1
R3 R3 R1
R3 R3 3R2
X
B
1 1 1 1
AB 1 2 4
1 4 10 2
1
1 1 1
~ 0 1 3 1
0 3 9 2 1
1 1 1
1
1
~ 0 1 3
0 0 0 2 3 2
Here A 2 and AB 3 ⇒ The given system of equations is consistent if
2 3 2 0 2 2 2 0 2 1 0 2, 1
Case (i): When 1
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A AB 2 Number of unknowns.
The system has infinite number of solutions.
1 1 1 1
The equivalent matrix is 0 1 3 0
0 0 0 0
The equivalent systems of equations are x y z 1; y 3z 0
Let z k
y 3k
and
x (3k ) K 1 x 2k 1 x 1 2k
x 1 2k
1
2
X y 0 3k X 0 k 3 where k is any arbitrary constant.
z 0 k
0
1
Case (ii): When 2
A AB 2 no. of unknowns.
The system has infinite number of solutions.
1 1 1 1
The equivalent matrix is ~ 0 1 3 1
0 0 0 0
The system of equations equivalent to the given system is x + y + z = 1; y + 3z = 1
Let z k
y 1 3k and x (1 3K ) k 1 x 2k
x 0 2k
0
2
X y 1 3k X 1 k 3
z 0 k
0
1
where k is any arbitrary constant.
11). Show that the equations 3x 4 y 5 z a; 4 x 5 y 6 z b;5 x 6 y 7 z c don’t have
a solution unless a c 2b. solve equations when a=b=c= -1.
3 4 5 x a
4 5 6 y b
Sol: The Matrix notation is
5 6 7 z c
A X B
Augment Matrix is
R2 3R2 4 R1
R3 3R3 5R1
3 4 5 a
AB 4 5 6 b
5 6 7 c
a
3 4 5
~ 0 1 2 3b 4a
0 2 4 3c 5a
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R3 R3 2R2
a
3 4 5
~ 0 1 2
3b 4a
0 0 0 3a 6b 3c
Here A 2 and AB 3
The given system of equations is consistent if 3a 6b 3c 0 3a 3c 6b a c 2b
Thus the equations don’t have a solution unless a c 2b, when a b c 1
3 4 5 1
The equivalent matrix is 0 1 2 1
0 0 0 0
A AB 2 No. of unknowns.
The system has infinite number of solutions. The system of equations equivalent to the
given system 3x 4 y 5 z 1; y 2 z 1 y 2 z 1
Let z k y 1 2k and 3x 4 8k 5k 1 x 1 k
x 1 k 1
1
X y 1 2k 1 k 2
z 0 k 0
1
Linearly dependent set of vectors: A set { x1,x2,--------------xr} of r vectors is said to be a
linearly dependent set, if there exist r scalars k1,k2---kr not all zero, such that k1x1+k2x2+-------+krxr = 0
Linearly independent set of vectors: A set { x1, x2,-------------xr} of r vectors is said to be a
linearly independent set, if k1x1+k2x2+--------------+krxr =0 then k1 =0, k2 = 0------kr =0.
Linear combination of vectors:
A vector x which can be expressed in the form x = k1x1+k2x2+--------------+knxn is said
to be a linear combination of x1,x2------xn here k1,k2-----------kn are any scalars.
Linear dependence and independence of Vectors:
Solved Problems :
1). Show that the vectors (1, 2, 3), (3,-2, 1), (1,-6,-5) from a linearly dependent set.
1
3
1
Sol: The Given Vector X 1 2 X 2 2 X 3 6
3
1
5
The Vectors X1, X2, X3 from a square matrix.
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1 3 1
1 3 1
Let A 2 2 6 Then A 2 2 6
3 1 5
3 1 5
= 1(10+6)-2(15-1) + 3(-18+2)
= 16+32-48 = 0
The given vectors are linearly dependent |A| = 0
2). Show that the Vector X1=(2,2,1), X2=(1,4,-1) and X3=(4,6,-3) are linearly dependent.
Sol: Given Vectors X1=(2,-2,1) X2=(1,4,-1) and X3=(4,6,-3) The Vectors X1, X2, X3 form a
square matrix.
2 1 4
2 1 4
Let A 2 4 6 Then A 2 4 6
1 1 3
1 1 3
= 2(-12+6) + 2(-3+4)+1(6-16)
= -20≠ 0
The given vectors are linearly dependent |A|≠0
Consistency of system of Homogeneous linear equations:
A system of m homogeneous linear equations in n unknowns, namely
a11 x1 a12 x2 a13 x3 .. a1n xn 0
a21 x1 a22 x2 a23 x3 .. a2 n xn 0
(1)
am1 x1 am 2 x2 am3 x3 .. amn xn 0
i.e.
a11 a12 a13 ....a1n
a 21 a 22 a 23 ....a 2 n
a
m1 a m 2 a m3 ....a mn
0
x1
x 0
2
...
..
x n .0
Here A is called Co –efficient matrix.
Note: 1. Here x1= x2 = --------- xn = 0 is called trivial solution or zero solution of AX = 0
2. A zero solution always linearly dependent.
Theorem: The number of linearly independent solutions of the linear system AX = 0 is
(n-r), r being the rank of the matrix A and n being the number of variables.
Note: 1.if A is a non-singular matrix then the linear system AX = 0 has only the zero
solution.
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2. The system AX =0 possesses a non-zero solution if and only if A is a singular
matrix.
Working rule for finding the solutions of the equation AX = 0
(i). Rank of A = No. of unknowns
i.e. r = n
∴ The given system has zero solution.
(ii). Rank of A < No of unknowns (r<n) and No. of equations < No. of unknowns (m<n) then
the system has infinite no. of solutions.
Note: If AX =0 has more unknowns than equations the system always has infinite solutions.
Solved Problems :
1). Solve the system of equations x+3y-2z = 0, 2x-y+4z = 0, x-11y+14z = 0
Sol: We write the given system is AX = 0
1 3 2
i.e. 2 1 4
1 11 4
R2 →R2 -2R1; R3→R3-R1
1 3 2
A ~ 0 7 8
0 14 16
x 0
y 0
z 0
1 3 2
A ~ 0 7 8
0 0 0
R3 →R3 -2R2
The Rank of the A = 2 i.e. ρ (A) = 2 < No. of unknowns = 3
We have infinite No. of solution
Above matrix can we write as x+3y-2z = 0 -7y+8z = 0, 0 = 0
Let z = k then y=8/7k & x= -10/7 k
Giving different values to k, we get infinite no. of values of x,y,z.
2). Find all the non-trivial solution 2 x y 3z 0;3x 2 y z 0; x 4 y 5 z 0.
2 1
Sol: In Matrix form it is 3 2
1 4
AX
3 x 0
1 y 0
5 z 0
0
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2 1 3 0
The Augumented matrix A O 3 2 1 0
1 4 5 0
R1 R3
1 4 5 0
~ 3 2 1 0
2 1 3 0
1 4 5 0
~ 0 14 14 0
0 7 7 0
1 4 5 0
R3 2 R3 R2
~ 0 14 14 0 it is echelon form.
0 0
0 0
R2 R2 3R1
R3 R3 2 R1
The Rank of the A = 2 i.e. ρ(A) = 2 < No. of unknowns = 3
Hence the system has non trivial solutions. From echelon form, reduced equations are
x 4 y 5 z 0 and 14 y 14 z 0
Let z k then y k and x 4k 5k 0 x k .
x k
1
Thus, the solution set is X y k k 1 K .
z k
1
3). Show that the only real number for which the system x+2y+3z = x, 3x+y+2z= y,
2x+3y+z = z, has non-zero solution is 6 and solve them.
1
Sol: Above system can we expressed as AX = 0
i.e. 3
2
2 3
1 2
3 1
x 0
y 0
z 0
Given system of equations possess a non –zero solution i.e. ρ (A) < no. of unknowns.
For this we must have det A = 0
1
2
3
3 1
2 0
2
3 1
R1 R1 R2 R3
6 6 6
3
1
2 0
2
3
1
1
1
1
2 0
(6 ) 3 1
2
3 1
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→
−
,
→
−
−
|
(6- )[(-2- )(-1- )+1] =0
− −
−
− −
(6- ) ( 2+3 +3) = 0
|=
= 6 only real values.
When = 6, the given system becomes
5 2 3 x 0
3 5 2 y 0
2 3 5 z 0
R2 → 5R2+3R1, R3→5R3+2R1
~
5 2
0 19
0 19
5
R3 →R3+R2
2
~ 0 19
0
0
3
19
19
x 0
y 0
z 0
3
19
0
x 0
y 0
z 0
-5x+2y+3z = 0 and -19y+19z = 0 y = z
Let z = k y = k and x = k.
∴ The solution is x = y = z = k.
Eigen Values and Eigen vectors:
Let A= aij be a square Matrix. Suppose the linear transformation Y = AX transforms X
nxn
into a scalar multiple of itself i.e. AX = Y = X, Then the unknown scalar is known as an
“Eigen value” of the Matrix A and the corresponding non-zero vector X is known as “Eigen
Vector” of A. Corresponding to Eigen value . Thus the Eigen values (or) characteristic
values (or) proper values (or) latent roots are scalars which satisfy the equation.
AX = X for X 0, AX IX 0 ( A I ) X 0
Which represents a system of ‘n’ homogeneous equations in ‘n’ variables x1, x2, ----, xn this
system of equations has non-trivial solutions If the coefficient matrix (A- I) is singular i.e.
a11
A I 0
a21
an1
a12
a22
an 2
a1n
a2 n
0
ann
Expansion of the determinant is (1) n n K1 n 1 K 2 n 2 K n is the nth degree of
a polynomial
Pn ( )
which is known as “Characteristic Polynomial”. Of A
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(1) n n K1 n 1 K 2 n 2 K n =0 is known as “Characteristic Equation”. Thus
the Eigen values of a square Matrix A are the roots of the characteristic equation.
Eg: Let A= [
AX = [
][
−
]
X=[
]
−
] = [ ] = .[ ] = .
-
Here Characteristic vector of A is [ ] and Characteristic root of A is “1”.
−
Eigen Value: The roots of the characteristic equation are called Eigen values or characteristic
roots or latent roots or proper values.
Eigen Vector: Let A= aij be a Matrix of order n. A non-zero vector X is said to be a
nxn
characteristic vector (or) Eigen vector of A if there exists a scalar such that AX = X.
Method of finding the Eigen vectors of a matrix.
Let A=[aij] be a nxn matrix. Let X be an eigen vector of A corresponding to the eigen value .
Then by definition AX = X.
⇒
AX = IX
⇒
(A- I)X = 0 ------- (1)
AX – IX = 0
⇒
This is a homogeneous system of n equations in n unknowns.
Will have a non-zero solution X if and only |A- I| = 0
A- I is called characteristic matrix of A
|A- I| is a polynomial in of degree n and is called the characteristic polynomial of A
|A- I|=0 is called the characteristic equation
Solving characteristic equation of A, we get the roots , 𝜆 , 𝜆 , 𝜆 , … … . 𝜆 , These are
called the characteristic roots or eigen values of the matrix.
Corresponding to each one of these n eigen values, we can find the characteristic
vectors.
Procedure to find Eigen values and Eigen vectors
……
…..
Let A = [… … .
𝑖
𝑖
…….
… … .]
…….
ℎ
𝑖 𝑖
𝑖
𝑖
− λI
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a11
a
i.e., A I 21
an1
a12
a22
an 2
a1n
a2 n
ann
Then the characteristic polynomial is A I
a11
a21
say A I
...
an1
a12
a22
...
an 2
...
a1n
...
a2 n
...
...
... ann
we solve the ∅ λ = |A − λI| = , we get n roots,
The characteristic equation is |A-λI| =
these are called eigen values or latent values or proper values.
Let each one of these eigen values say their eigen vector X corresponding the given value
is obtained by solving Homogeneous system
a11
a21
an1
a12
a22
an 2
a1n x1 0
a2 n x2 0
and determining the non-trivial solution.
ann xn 0
Solved Problems
1.
Sol:
Find the eigen values and the corresponding eigen vectors of [
= [
−
]
−λ
Characteristic matrix = [ − 𝜆 ] = [
Characteristic equation is | − I|=0
⟹ |
⟹
−λ
−λ
−λ +
=
λ+
=
+λ −
⟹ λ −
⟹ λ−
−
|=
−λ
λ−
λ+
−
]
−
]
−λ
=
=
⟹ λ = , are eigen values of A
Consider the system [
Eigen vector corresponding to 𝛌 =
−λ
−
]
−λ
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MATHEMATICS - I
Put λ =
−
−
in the above system, we get
⟹
−
Let x1 =
=
−−−
ℎ
=
=
−
=
−−−
x
1
Eigen vector is 1
1
x2
[ ]𝑖
𝑖
𝑖
,
𝑖
Eigen Vector corresponding to 𝛌 =
λ=
⟹
−
−
in the above system, we get
−
=
−−−
𝑖
=
−
from (1) and (2) we have x1 = 2x2
𝜆=
=
−−−
Let x2 x1 2
2
2
Eigen vector
1
[ ]𝑖
𝑖
𝑖
𝑖
𝑖
𝜆=
Find the eigen values and the corresponding eigen vectors of matrix [
2.
Sol: Let A = [
]
]
The characteristic equation is |A- I|=0
i.e. |A- I| = |
−λ
−λ
−λ
−λ
− + [−
⟹ λ − [– λ −
− ]=
⟹
⟹
−λ
−λ — λ−
⟹ λ−
⟹ λ−
=1, 2, 3
|=
=
−λ ]=
[−λ + λ − ] =
λ−
λ−
=
The eigen values of A is 1, 2, 3.
⟹ [
𝑖
𝑖
−λ
𝑖
−λ
ℎ
−λ
] [ ]=[ ]
𝑖
− λI X =
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[
] [ ]=[ ]
+
=
+
=
=
=− ,
=𝛼
⇒
= −𝛼
[ ]= [
[
−
]𝑖
−𝛼
𝛼
=
= ,
] = 𝛼[
−
]
𝛌=
] [ ]=[ ]
=
=
[ ] = [𝛼 ] = 𝛼 [ ]
𝑖
−
−
𝑖
𝑖 [ ]
−
ℎ
=𝛼
𝑖
[
[
𝛌=
+
−
=
−
=
=∝ ,
=
−
=
1
x1
x 0 0
2
1
x3
,
=∝
=
=𝛼
𝛌=
] [ ]=[ ]
𝑖
𝑖.
,
=
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𝑖
𝑖 [ ]
3 6 2
3. Find the Eigen values and Eigen vectors of the matrix is 6 7 4
2 4 3
8 6 2
Sol: Let A 6 7 4
2 4 3
Consider characteristic equation is
A I 0
8 6
2
i.e. 6 7 4 0
2
4 3
8 7 3 16 6 6 3 8 2 24 2 7 0
8 21 7 3 2 16 6 18 6 8 2 24 14 2 0
8 2 10 5 6 6 10 2 10 2 0
8 2 80 40 3 10 3 5 36 60 20 4 0
3 18 2 45 0
2 18 45 0
0 (OR) 2 18 45 0
0, 3, 15
Eigen Values
= , ,
Case (i): If
[−
[−
=
−
−
−
−
− ]
=
− ][ ] = [ ]
8 x1 6 x2 2 x3 0 (1)
6 x1 7 x2 4 x3 0 (2)
2 x1 4 x2 3 x3 0 (3)
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Consider (2) & (3)
x1
x2
x3
6
2
7 4
4 3
x3
x1
x2
k
21 16 18 8 24 14
x
x
x2
1
3 k
10 10
5
x1 k , x2 2k , x3 2k
x1 k
1
Eigen Vector is x2 2k k 2
x
2
k
3
2
Case (ii): If
=
2
6
8
0
6
7
4 x 0
2
0
3
4
2 x1 0
6
5
6
4
4 x2 0
2
4 0 x3 0
5 x1 6 x2 2 x3 0 (1)
6 x1 4 x2 4 x3 0 (2)
2 x1 4 x2 0 0 (3)
Consider (2) & (3)
x1
x2
6
4
2
4
x3
4
0
x3
x2
x1
k
0 16
08
24 8
x
x2
x1
3 k
16
8
16
x3
x2
x1
k
2
1
2
x1
k , x2 k , x3 2k
2
x1 2k , x2 k , x3 2k
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x1 2k
2
Eigen Vector is x2 k k 1
x3
2k
2
Case (iii): If
7
6
2
7
6
2
=
6
8
4
2
4 x 0
12
6
8
4
2 x1 0
4 x2 0
12 x3 0
7 x1 6 x2 2 x3 0 (1)
6 x1 8 x2 4 x3 0 (2)
2 x1 4 x2 12 x3 0 (3)
Consider (2) & (3)
x1
x2
x3
6 8 4
2 4 12
x3
x1
x2
k
96 16 72 8 24 16
x
x
x
1 2 3 k
80 80 40
x
x1
x
k, 2 k, 3 k
2
2
1
x1 2k , x2 2k , x3 k
x1 2k 2
Eigen Vector is x2 2k 2 k
x3 k 1
4. Find the Eigen values and the corresponding Eigen vectors of the matrix.
2
2
1
2
1
2
3
6
0
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Sol: Let
2
A
2
1
2
1
2
3
6
0
2
The characteristic equation of A is A I 0 i.e.
2
1
2
1
2
3
6 0
2
1 12
2 2 6 3
2 2 1
0
3 2 21 45 0
3 3 5 0
3, 3, 5
The Eigen values are -3,-3, and 5
Case (i): If
=−
3 x1 0
2 3 2
We get
1 3 6 x2 0
2
1
2 0 3 x3 0
2 3 0
1
The augment matrix of the system is 2 4 6 0
1 2 3 0
1 2 3 0
Performing R2 2 R1 , R3 R1 , we get 0 0 0 0
0 0 0 0
Hence we have x1 2 x2 3xy 0 x1 2 x2 3x3
Thus taking x2 k1 and x3 k2 , we get x1 2k1 3k2 ; x2 k1; x3 k2
x1
2
3
Hence x2 k1 1 k2 0
x3
0
1
2
3
So 1 and 0 are the Eigen vectors corresponding to 3
0
1
Case (ii): If
=
7 2 3 x1 0
We get 2 4 6
x2 0
1 2 5 x 0
3
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7 x1 2 x2 3x3 0 (1)
2 x1 4 x2 6 x3 0 (2)
x1 2 x2 5 x3 0 (3)
Consider (2) & (3)
x1
x2
2
1
4
2
x3
6
5
x3
x1
x2
k3
20 12 10 6 4 4
x
x2 x3
1
k3
8 16 8
x
x2 x3
1
k3
1
2
1
x1 1
Eigen vector is x2 2 k3
x3 1
5. Find the Eigen values and Eigen vectors of the matrix A and it’s inverse where
1 3 4
A 0 2 5
0 0 3
1 3 4
Sol: Given A 0 2 5
0 0 3
The characteristic equation of “A” is given by A I 0
1
0
0
3
4
2
5 0
0
3
1 2 3 0
1, 2,3 i.e. EigenValues are 1, 2,3
Note:
In upper le (or) Lower lar of a square matrix the Eigen values of a diagonal matrix
are just the diagonal elements of the matrix.
Case (i): If
=
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∴ A I x 0
3
4 x1 0
1
2
5 x2 0
0
0
0
3 x3 0
0 3 4 x1 0
0 1 5 x2 0
0 0 2 x3 0
3x2 4 x3 0; x2 5x3 0; 2 x3 0 x1 k1; x2 0; x3 0
x1 k1 1
X x2 0 0 k1
x3 0 0
Case (ii): If
=
3
4 x1 0
1
0
2
5 x 2 0
0
0
3 x3 0
1 3 4 x1 0
0 0 5 x2 0
0 0 1 x3 0
1 3 4 x1 0
0 0 5 x2 0
0 0 1 x3 0
x1 3x2 4 x3 0; 5 x2 0; x3 0
x1 3k 4(0) 0 x1 3k 0 x1 3k
x1 3k
3
X x2 k k 1
x3 0
0
Case (iii): If
=
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3
4 x1 0
1
2
5 x2 0
0
0
0
3 x3 0
2 3 4 x1 0
0 1 5 x2 0
0 0 0 x3 0
2 x1 3 x2 4 x3 0; x2 5 x3 0; x3 0
x3 k
Let
x2 5 x3 0 x2 5k
and 2 x1 3 x2 4k 0 2 x1 15k 4k 0
2 x1 19 x 0 x1
19
k
2
19
x1 19
2 k
2
X x2 5k k 5
1
x3 k
Note: Eigen Values of A-1 are
1
,
1
1 2
,
1
1 1
i.e1, , and the Eigen vectors of A-1 are same as
2 3
3
Eigen vectors of the matrix A
6. Determine the Eigen values and Eigen vectors of
B 2 A2
8 4
1
A 3I where A
2
2 2
8 4
1
Sol : Given that B 2 A A 3 A
2
2 2
8 4 8 4 56 40
we have A2 A. A
2 2 20 4
2
2
B 2 A2 1 A 3I
2
56 40 1 8 4
1 0
2
3
20 4 2 2 2
0 1
112 80 4 2 3 0
40 8 1 1 0 3
111 78
39 6
Characteristic equation of B is
B I 0
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111
39
78
0
6
i.e. 2 105 2376 0
33 72 0
33 or 72
Eigen Values of B are 33 and 72.
Case (i): If
= 33
111
39
78
0
6
78 78 x1 0
39 39 x2 0
39 x1 78 x2 0 x1 2 x2
x1 x2
k ( say )
2 1
x 2
1 k
x2 1
Case (ii): If
= 72
111
39
78
0
6
78
111 72
X 0
6 72
39
39 78
X 0
39 78
39
39
78 x1 0
78 x2 0
39 x1 78 x2 0 x1 2 x2
x1 x2
k ( say )
2 1
x 2
1 k
x2 1
Properties of Eigen Values:
Theorem 1: The sum of the eigen values of a square matrix is equal to its trace and product
of the eigen values is equal to its determinant.
Proof: Characteristic equation of A is |A- I|=0
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a11
a
i.e. 21
an1
a1n
a2 n
ann
a12
a22
an 2
𝑖
ℎ𝑖
a11 a22 ann a12 (a polynomial of degree n – 2)+ a13 (a polynomial of degree n
-2) + … = 0
(1) n n (a11 a22 .... ann )n1 a polynomial of deg ree (n 2) 0
(1) n n (1) n1 (Trace A)n1 a polynomial of deg ree (n 2) in 0
ℎ
λ ,λ …..λ
− 𝑛+
=
− 𝑛
ℎ
−
𝜆=
ℎ
=
ℎ𝑖
s
|A − λI| = −
ℎ
λ +a
−
ℎ
λ + ⋯ . +a
| |=
λ
−
+a
= | | = det
𝑖
=
−
λ
−
−
−
+ ….+ a =
=
ℎ
is an eigen value of A corresponding to the eigen vector X, then
Theorem 2: If
is eigen
value An corresponding to the eigen vector X.
Proof: Since
is an eigen value of A corresponding to the eigen value X, we have
AX=
----------(1)
Pre multiply (1) by A, A(AX) = A( X)
(AA)X = (AX)
A2X= ( X)
A2X=
2
2
X
is eigen value of A2 with X itself as the corresponding eigen vector. Thus the
theorm is true for n=2
Let we assume it is true for n = k
i.e. AKX =
K
X------------(2)
Premultiplying (2) by A, we get
A(AkX) = A( KX)
(AAK)X=
K
(AX)=
K
( X)
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K+1
AK+1X=
K+1
X
K+1
is eigen value of
with X itself as the corresponding eigen vector.
Thus, by Mathematical induction.
n
is an eigen value of An.
Theorem 3: A Square matrix A and its transpose AT have the same eigen values.
Proof: We have (A- I)T = AT- IT
= AT - I
|(A- I)T|=|AT- I| (or)
T
|A- I|=|AT- I| A A
|A- I|=0 if and only if |AT- I|=0
Hence the theorem.
Theorem 4: If A and B are n-rowed square matrices and If A is invertible show that A-1B and
B A-1 have same eigen values.
Proof: Given A is invertble i.e, A-1 exist
We know that if A and P are the square matrices of order n such that P is non-singular
then A and P-1 AP have the same eigen values.
Taking A=B A-1 and P=A, we have
B A-1 and A-1 (B A-1 ) A have the same eigen values
ie.,B A-1 and (A-1 B)( A-1 A) have the same eigen values
ie.,B A-1 and (A-1 B)I have the same eigen values
ie.,B A-1 and A-1 B have the same eigen values
Theorem 5: If 1 , 2 ,...........n are the eigen values of a matrix A then k
1,
k
2,
….. k
n
are the eigen value of the matrix KA, where K is a non-zero scalar.
Proof: Let A be a square matrix of order n. Then |KA- KI| = |K (A- I)| = Kn |A- I|
Since K≠0, therefore |KA- KI| = 0 if and only if A I 0
i.e., K is an eigen value of KA if is an eigen value of A
Thus k
1,
1,
2…
Theorem 6: If
2…
k
n
k
n
are the eigen values of the matrix KA if
are the eigen values of the matrix A
is an eigen values of the matrix A then +k is an eigen value of the matrix
A+KI
Proof: Let
be an eigen value of A and X the corresponding eigen vector. Then by definition
AX= X
Now (A+KI) X
AX IKX X KX
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= ( +K) X
+k is an eigen value of the matrix A+KI.
Theorem 7: If
1,
2…
n
1–
are the eigen values of A, then
K,
2
– K, …
n–
K,
are the eigen values of the matrix ( A KI ), where K is a non zero scalar
Proof: Since
The characteristic polynomial of A is
|A –
I| = (
1–
)(
2
– ) …(
n–
)-----------------------1
Thus the characteristic polynomial of A-KI is
|(A – KI) –
=[
1-
I| = |A – (k+ )I|
( +K)][
+K)]................................[
2-(
= [ ( 1-K)- )][ ( 2-K)- ]................................[(
n-(
+K) ].
n-K)-
].
Which shows that the eigen values of A-KI are
are the eigen values of A, find the eigen values of the matrix
Theorem 8: If
Proof: First we will find the eigen values of the matrix ASince
are the eigen values of A
The characteristics polynomial is
| A-
(
(
The characteristic polynomial of the matrix (A|A-
-KI| = |A-( +K)I|
= [
( +K)]
= [
K)]
Which shows that eigen values of (A- I) are
We know that if the eigen values of A are
then the eigen values of A2 are
Thus eigen values of ( A I ) 2 are (1 ) 2 , (2 ) 2 , .....(n ) 2
Theorem 9: If
vector X, then
is an eigen value of a non-singular matrix A corresponding to the eigen
–1
is an eigen value of A–1 and corresponding eigen vector X itself.
Proof: Since A is non-singular and product of the eigen values is equal to |A|, it follows that
none of the eigen values of A is 0.
If
s an eigen value of the non-singular matrix A and X is the corresponding eigen
vector
≠0 and AX=
.
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premultiplying this with A–1, we get A–1(AX) = A–1( X)
( A1 A) X A1 X IX A1 X
X = A1 X A1 X 1 X ( 0)
Hence 1 is an eigen value of A1
Theorem 10: If is an eigen value of a non-singular matrix A, then
| |
is an Eigen value of
the matrix AdjA.
is an eigen value of a non-singular matrix, therefore ≠ 0
Proof: Since
Also
is an eigen value of A implies that there exists a non-zero vector X such that
AX =
-----(1)
(adjA) A A I
A
A
X (adj A) X or (adj A) X X
Since X is a non – zero vector, therefore the relation (1)
it is clear that
A
is an eigen value of the matrix Adj A
Theorem 11: If is an eigen value of an orthogonal matrix A, then is also an Eigen value A
Proof: We know that if
is an eigen value of a matrix A, then is an eigen value of A–1
Since A is an orthogonal matrix, therefore A–1 = A1
∴
is an eigen value of
But the matrices A and A1 have the same eigen values, since the determinants |A- I|
and
| A1 - I| are same.
Hence
Theorem 12: If
a0
2
is also an eigen value of A.
is eigen value of A then prove that the eigen value of B = a0A2+a1A+a2I is
+a1 +a2
Proof: If X be the eigen vector corresponding to the eigen value , then AX = X --- (1)
Premultiplying by A on both sides
2
This shows that is an eigen value of A2
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We have B = a0A2+a1A+a2I
BX = a0A2+a1A+a2I) X
= a0A2 X+a1AX+a2 X
= a0
(a0
2
2
X+a1 X+a2X = (a0
2
+a1 +a2 ) X
+a1 +a2 ) is an eigen value of B and the corresponding eigen vector of B is X.
Theorem 13: Suppose that A and P be square matrices of order n such that P is non singular.
Then A and P-1AP have the same eigen values.
Proof: Consider the characteristic equation of P-1AP
1
It is | ( P-1AP)- I| = | P-1AP- P-1IP| ( I P P)
= | P-1 (A- I) P| = | P-1 | |A- I| |P|
= |A- I| since |P-1 | |P| = 1
Thus the characteristic polynomials of P-1AP and A are same. Hence the eigen values of
P-1AP and A are same.
Corollary 1: If A and B are square matrices such that A is non-singular, then A-1B and BA-1
have the same eigen values.
Corollary 2: If A and B are non-singular matrices of the same order, then AB and BA have
the same eigen values.
Theorem 14: The eigen values of a triangular matrix are just the diagonal elements of the
matrix.
a11 a12 ... ...
0 a22 ......
Proof: Let A =
...... ......
0
0 ......
a1n
a2 n
be a triangular matrix of order n
......
ann
The characteristic equation of A is |A- I|=0
i.e.,
i.e, (a11- ) (a22- ) ….. (ann- )=0
a11 , a22 ,…. ann
Hence the eigen values of A are a11 , a22 ,…. ann which are just the diagonal elements of A.
Note: Similarly we can show that the eigen values of a diagonal matrix are just the diagonal
elements of the matrix.
Theorem 15: The eigen values of a real symmetric matrix are always real.
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Proof: Let
be an eigen value of a real symmetric matrix A and Let X be the corresponding
eigen vector then AX=
Take the conjugate
Taking the transpose
Since
Post multiplying by X, we get
------- (2)
T
T
, we get X AX X X ------ (3)
Premultiplying (1) with
– (3) gives X X 0 but
(1)
T
0
is real. Hence the result follows
Theorem 16: For a real symmetric matrix, the eigen vectors corresponding to two distinct
eigen values are orthogonal.
Proof: Let
1,
2
be eigen values of a symmetric matrix A and let X1, X2 be the
corresponding eigen vectors.
Let
1≠
2. We
want to show that X1 is orthogonal to X2 (i.e.,
Sice X1, X2 are eigen values of A corresponding to the eigen values
AX1=
1X1
----- (1)
AX2 =
2 X2
1,
2 we
have
------- (2)
Premultiply (1) by
Taking transpose to above, we have
X 2T AT X 2T
T
1 X 1T X 2T
T
-------- (3) s
Premultiplying (2) by
Hence from (3) and (4) we get
( 1 2 )
Note: If is an eigen value of A and f(A) is any polynomial in A, then the eigen value of
f(A) is f( ) .
Theorem17: The Eigen values of a Hermitian matrix are real.
Proof: Let A be Hermitian matrix. If X be the Eigen vector corresponding to the eigen value
of A, then AX = X -------------------- (1)
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Pre multiplying both sides of (1) by X ,we get
X AX X X ----------------------- (2)
Taking conjugate transpose of both sides of (2)
We get X AX X X
i.e X A X
X X ABC C B A and KA K A
X
(or) X A X X X
X , A A -------------------- (3)
From (2) and (3), we have
X X X X
i.e
X
X 0 0
X X 0
Hence is real.
Note: The Eigen values of a real symmetric are all real
Corollary: The Eigen values of a skew-Hermitian matrix are either purely imaginary (or)
Zero
Proof: Let A be the skew-Hermitian matrix
If X be the Eigen vector corresponding to the Eigen value of A, then
AX X (or ) iA X i X
From this it follows that i is an Eigen value of iA
Which is Hermitian (since A is skew-hermitian)
A A
iA
Now
i A iA i A iA
Hence i is real. Therefore must be either
Zero or purely imaginary.
Hence the Eigen values of skew-Hermitian matrix are purely imaginary or zero
Theorem 18: The Eigen values of an unitary matrix have absolute value l.
Proof: Let A be a square unitary matrix whose Eigen value is with corresponding eigen
vector X
AX X 1
T
AX X X A X 2
T
T
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Since A is unitary, we have A
T
T
and (2) given X A
i.e
X X X X From (3)
T
A I 3
AX X
(1)
T
T
T
X
T
X X 1 0
T
Since X X 0 ,we must have 1 0
1
Since =
We must have =1
Note 1: From the above theorem, we have “The characteristic root of an orthogonal matrix is
of unit modulus”.
2. The only real eigen values of unitary matrix and orthogonal matrix can be 1
Theorem 19: Prove that transpose of a unitary matrix is unitary.
Proof: Let A be a unitary matrix, then A. A A . A I
where A is the transposed conjugate of A.
AA A A I
T
T
T
AA A A I
T
T
T
A
AT AT A I
A
AT AT AT I
T
T
T
Hence AT is a unitary matrix.
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Solved Problems:
1 2 3
3
2
1. For the matrix A 0 3 2 find the Eigen values of 3 A 5 A 6 A 2 I
0 0 2
1
2
3
Sol: The Characteristic equation of A is A I 0 i.e. 0
3
2 0
0
0
2
⇒ 1 3 2 0
Eigen values are 1, 3,-2.
If is the Eigen value of A. and F (A) is the polynomial in A then the Eigen value of
f (A) is f ( )
3
2
Let f A 3 A 5 A 6 A 2I
∴ Eigen Value of f (A) are f (1), f (-2), f (3)
f (1) = 3+5-6+2 = 4
f (-2) = 3(-8)+5(4)-6(-2)+2 = -24+20+12+2 = 10
f (3) = 3(27)+5(9)+6(3)+2 = 81+45-18+2 = 110
The Eigen values of f (a) are f ( ) = 4, 10,110
2. Find the eigen values and eigen vectors of the matrix A and its inverse, where
A=
Sol: Given A =
The characteristic equation of A is given by |A- I| = 0
Charecetstic roots are 1, 2, 3.
Case (i): If
=
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and x1
1
X 0 0 is the solution where is arbritary constant
0
0
1
X 0 is the eigen vector corresponding to 1
0
Case (i): If
=
5 x3 0 x3 0
x1 3x2 0 x1 3x2
Let x2 k
x1 3k
3k
3
X k k 1
0
0
is the solution where k is arbitrary constant
3
X 1 is the eigen vector corresponding to 2
0
Case (iii): If
=
2 3 4 x1 0
For 3, becomes 0 1 5 x2 0
0
0 0 x3 0
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X=
is the solution, where k/2 is arbitrary constant.
19
X 10 is the eigen vector corresponding to 3
2
Eigen values of A –1are 1, ,
We know Eigen vectors of
.
−
are same as eigen vectors of A.
3i
2 i
2 i i
3. Find the eigen values of A=
3i
2 i
Sol: we have A=
2 i i
3i
So A =
2 i
2 i
and AT 3i 2 i
i
i
2 i
T
A = A
Thus A is a skew-Hermitian matrix.
The characteristic equation of A is A I 0
AT
3i 2 i
0
2i i
2 2i 8 0
4i, 2i are the Eigen values of A
1
i
4.Find the eigen values of A 2
3
2
1
3
i
2 and
Now A 2
3 1 i
2
2
3
2
1
i
2
1
3
i
2
A 2
3 1 i
2
2
T
T
1 0
We can see that A . A
I
0 1
Thus A is a unitary matrix
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The characterstic equation is A I 0
1
i
2
3
2
Which gives
3
2
1
i
2
0
3
1
3 1
i and
i and
2
2
2
2
Hence above values are Eigen values of A.
Cayley-Hamilton Theorem: Every Square Matrix satisfies its own characteristic equation
To find Inverse of matrix: If A is non-singular Matrix, then A-1 exists, Pre multiplying (1)
above by A-1 we have a0 An 1 a1 An 2 an 1I an A1 0 ,
A1
1
a0 An 1 a1 An 2 an 1 I
an
To find the powers of A: - Let K be a +ve integer such that K n
Pre multiplying (1) by Ak-n we get
AK
a0 AK a1 AK 1 an Ak n 0
,
1
a1 Ak 1 a2 AK 1 an AK n
a0
Solved Problems :
satisfies its characteristic equation and hence find A-1
1. S.T the matrix A =
Sol: Characteristic equation of A is det (A- I) = 0
C2 → C2+C3
⇒
−
+ −
=
By Cayley – Hamilton theorem, we have A3-A2+A-I = 0
1 2 2
1 0 0
1 2 2
2
3
A 1 2 3 A 1 1 2 A 2 2 1
0 1 2
1 0 1
1 1 0
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1 2 2 1 0 0 1 2 2 1 0 0
A A A I 2 2 1 1 1 2 1 2 3 0 1 0 =
1 1 0 1 0 1 0 1 2 0 0 1
3
2
Multiplying with A–1 we get A2 – A + I =A–1
1 0 0 1 2 2 1 0 0 1 2 2
A 1 1 2 1 2 3 0 1 0 2 2 1
1 0 1 0 1 2 0 0 1 1 1 0
1
2. Using Cayley - Hamilton Theorem find the inverse and A4 of the matrix
A =[−
−
−
Sol: Let A = [-
−
]
-
]
-
7
2
2
2 0
The characteristic equation is given by |A- I|=0 i.e., 6 1
6
2
1
⇒
−
+
−
=
By Cayley – Hamilton theorem we have A3-5A2+7A-3I=0…..(1)
Multiply with A-1 we get
−
= [
−
+
]
25 8 8
79 26 26
3
A 24 7 8 A 78 25 26
24 8 7
78 26 25
2
3 2 2
1
A 6 5 2
3
6 2 5
1
multiplying (1) with A,we get,
A4-5A3+7A2-3A= 0
A4 = 5A3-7A2+3A
395 130 130
390 125 130
390 130 125
175 56 56 21 6 6 241 80 80
168 49 56 18 3 6 240 79 80
168 56 69 18 6 3 240 80 79
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2 1 2
3. If A 5 3 3 Verify Cayley-Hamilton theorem hence find A-1
1 0 2
2 1
Sol: - Given that A 5 3
1 0
2
3
2
The characteristic equation of A is |A- I|=0
i.e. |
−
−
−
− −
|=0
2 6 3 2 2 1 10 5 3 2 0 3
2 2 6 1 5 7 2 3 x 0
2 2 2 12 3 2 6 5 7 6 2 0
3 3 2 7 1 0
3 3 2 7 1 0 (1)
According to Cayley Hamilton theorem. Square matrix ‘A’ satisfies equation (1)
Substitute A in place of
2
Now A 5
1
1
3
0
2
3
2
2 1 2 2 1 2 7 5 3
A 5 3 3 5 3 3 22 14 13
1 0 2 1 0 2 0 1 2
2
7 5 3 2 1 2 36 22 23
A A . A 22 14 13 5 3 3 101 64 60
0 1 2 1 0 2 7 3 7
3
2
3
2
Now A 3 A 7 A I 0
36 22 23 21 15 9 14 7 14 1 0 0 0 0 0
101 64 60 66 42 39 35 21 21 0 1 0 0 0 0 0
7 3 7 0
6 7
3
0
14 0 0 1 0 0 0
Cayley –Hamilton theorem is verified.
To find A-1
⇒
A3 3 A2 7 A I 0
Multiply A-1, we get
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A1 A3 3 A2 7 A I 0
A2 3 A 7 I A1 0
A1 A2 3 A 7 I
0
7 5 3 6 3 6 7 0
A 22 14 13 15 9 9 0 7 0
0 1 2 3 0
6 0
0 7
1
6 2 3
A 7 2 4
3 1 1
1
Check A.A-1= I
2 1 2 6 2 3 1
A.A = 5 3 3 7 2 4 0
1 0 2 3 1 1 0
-1
0 0
1 0 I
0 1
1 2
2 1
4. Using Cayley – Hamilton theorem, find A8, if A
1 2
Sol: Given A
2 1
Characteristic equation of A is A I 0
1 1 4 0
2 5 0 (1)
Substitute A in place of
A2 5 I 0 A 2 5 I
find A8
A8 5 A6 5( A2 )( A2 )( A2 )
5 5 I 5 I 5 I
625I
A8 625I
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Diagonalization of a Matrix by similarity transformation:
Similar Matrix: A matrix A is said to be similar to the Matrix B if there Exists a nonsingular matrix P such that B= P 1 AP . This transformation of A to B is known as “Similarity
Transformation”
Diagonalization of a Matrix:
Let A be a square Matrix. If there exists a non-singular Matrix P and a diagonal Matrix D
such that P-1AP=D, then the Matrix A is said to be diagonalizable and D is said to be
“Diagonal” form (or) canonical diagonal form of the Matrix A
Modal Matrix:The modal matrix which diagonalizes A is called the modal Matrix of A and
is obtained by grouping the Eigen vectors of A into a Square Matrix.
Spectral Matrix: The resulting diagonal Matrix D is known as Spectral Matrix.
In this spectral Matrix D whose principal diagonal elements are the Eigen values of the
Matrix.
Calculation of powers of a matrix:
We can obtain the power of a matrx by using diagonalization
Let A be the square matrix then a non-singular matrix P can be found such that D = P-1AP
D2= (P–1AP) (P–1AP)
= P–1A (PP–1) AP
= P–1A2P
(since PP–1=I)
Simlarly D3 = P–1A3P
In general Dn = P–1AnP……..(1)
To obtain An, Premultiply (1) by P and post multiply by P–1
n
n 1
Then PDnP–1 = P(P–1AnP)P–1 = (PP–1)An (PP–1) = An A PD P
n
Hence A = P
1n
0
0
0
n
2
0
0
0
0
0
0 1
P
nn
Diagonalization of a matrix:
Theorem: If a square matrix A of order n has n linearly independent eigen vectors
(X1,X2…Xn) corresponding to the n eigen values
1, 2…. n
respectively then a matrix P can
be found such that P-1AP is a diagonal matrix.
Note: 1. If X1,X2…Xn are not linearly independent this result is not true.
2. Suppose A is a real symmetric matrix with n pair wise distinct eigen values 1 , 2
n then
the corresponding eigen vectors X1,X2…Xn are pairwise orthogonal.
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Hence if P = (e1,e2…en)
Where e1 = (X1 / ||X1||), e2 = (X2 / ||X2||)….en = (Xn)/ ||Xn|| then P will be an orthogonal
matrix.
i.e, PTP=PPT=I
Hence P–1 = PT
−
∴
=
Solved Problems :
. Verify that P-1AP is a diagonal
1. Determine the modal matrix P of A=
matrix.
Sol: The characteristic equation of A is |A- I| = 0 i.e.
which gives ( -5) ( +3)2= 0
Thus the eigen values are =5, =-3 and =-3
When =5
By solving above we get X1 =
Similarly, for the given eigen value = -3 we can have two linearly independent eigen vectors
X2 =
P=[
P=[
]
−
−
] = modal matrix of A
Now det P=1(-1)-2(2)+3(0-1) = -8
=
−
= [
−
]
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=[
∴
−
−
−
] = diag [5,-3,-3].
= diag [5,-3,-3].
−
2. Find a matrix P which transform the matrix A = [
calculate A4.
] to diagonal form. Hence
Sol: Characteristic equation of A is given by |A- I| = 0 i.e.
=> (1=> 9 −
[
−
−
−
=> = , = , =
−
− ]−
=
−[ −
−
]=
Thus the eigen values of A are 1, 2, 3.
If x1, x2, x3 be the components of an eigen vector corresponding to the eigen value , we have
[A- I] X =
Case (i): If
=
i.e, 0.x1+0.x2+0.x3=0 and x1+x2+x3=0
x3=0 and x1+x2+x3=0
x3=0, x1=-x2
x1=1, x2=-1, x3=0
Eigen vector is [1,-1,0]T
Also every non-zero multiple of this vector is an eigen vector corresponding to =1
For =2, =3 we can obtain eigen vector [-2,1,2]T and [-1,1,2]T
P=
The Matrix P is called modal matrix of A
0 2 1
1
P = 2 2 0
2
2 2 1
-1
0
Now P 1 AP 1
1
1
1
1
1
1
2
0 1
1
2
2
0
2
2
1 1
1 1
3 0
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2
1
1
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A4 = PD4P-1
1 2
1 1
0
2
1 1 0
1 0 16
2 0 0
0 0
0 1
81 2
1
2
0
1
1
1
2
3. Determine the modal matrix P for [
] and hence diagonalize A
]
Sol: Given that[
1
1
3
5
1 0
The characteristic equation of A is A I 0 i.e. 1
3
1
1
1 5 1 1 1 1 3 3 1 3 5 0
1 5 5 2 1 2 3 1 15 3 0
1 4 6 2 2 3 14 3 0
4 6 2 4 6 2 3 2 42 9 0
3 7 2 9 9 36 0
3 7 2 36 0
2,3, 6
The Eigen Values are -2, 3, and 6
Case (i): If
=−
[A- I]X =
1
3 x1 0
1
⇒ 1 5
1 x2 0
3
1
1 x3 0
3 1 3 x1
0
3 1 3 x3
0
⇒ 1 7 1 x 0
2
3x1 x2 3x3 0 (1)
x1 7 x2 x3 0 (2)
3x1 x2 3x3 0 (3)
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From (2) & (3)
x1
x2
x3
1
3
7
1
1
3
x
x1
x2
3 k
1 21 3 3 21 1
x
x
x
1 2 3 k
0
20
20
x1 20k , x2 0, x3 20k
x1 20k
1
X x2 0k 20k 0
x3 20k
1
=
Case (ii): If
[A- I]X =
2 1 3 x1 0
⇒ 1 2 1 x2 0
3 1 2 x3 0
2 x1 x2 3 x3 0 (1)
x1 2 x2 x3 0 (2)
3 x1 x2 2 x3 0 (3)
Consider (1) & (2)
x1
x2
x3
2 1 3
1 2 1
x
x2
x1
3 k
1 6 2 3 4 1
x
x
x
1 2 3 k
5 5 5
x1 5k
1
X x2 5k 5k 1
x3 5k
1
Case (iii): If
=6
[A- I]X =
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1
3 x1 0
1
1
5
1 x2 0
3
1
1 x3 0
3 x1 0
5 1
1 11 1 x2 0
3
1 5 x3 0
5 x1 x2 3x3 0 (1)
x1 x2 x3
0 (2)
3x1 x2 5 x3 0 (3)
Consider (2) & (3)
x1
x2
x3
1 1 1
3 1 5
x
x2
x1
3 k
5 1 5 3 1 3
x
x x
1 2 3 k
4
8
4
x
x x
1 2 3 k
1 2 1
x1 1
X x2 2 k
x3 1
1 1 1
0 1 2
1 1 1
1(1 2) 1(0 2) 1(0 1)
(1)(3) 1(2) 1 3 2 1 6
3 2
0 2
3 2
3
Adj ( ) 0
3
1
1
2
1
2 1
2 2
2 1
Adj
3
1
Cofactor of 2
6
1
0
2
2
3
12
1
2
3
1
1
6
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0
1
3
1
3
1
6
1
2
1
3
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D 1 A
12
D 13
16
12
13
16
12
13
16
.𝑰 𝑨 = [
1 1 3 1 1
1 5 1 0 1
1
1 1 1 1
6 3
1
2 1 0 3 1 1 3
1
3 1 0 1 1 5 1
1
3 1 1
6 3 0 1
1
2
0
13
1
3
1
3
0
13
1
3
1
2
1
1 2 3
1 10 1
3 2 1
3 6 2 0 0
2
0 3 12 0 3 0
1
2
3
6
0
0
6
6
1
2
0
13
1
3
1
3
] Find (a) A8 (b) A4
−
]
Sol: Given that A=[
−
The Characteristic equation of A is A I 0
i.e.
1
1
1
0
2
1 0
4
4
3
1 2 3 4 1 0 4 1 0 4 2 0
1 6 2 3 2 4 4 8 4 x 0
1 2 5 2 4 4 0
2 5 2 3 5 2 2 4 4 0
3 6 2 11 6 0
1, 2,3
The Eigen values are 1, 2, and 3
Case (i): If
−
=>[
=>[
−
−
=
[ − I]X =
−
−
]
=
][ ] = [ ]
x1+x2=0, x1+x2=0, -4x1+4x2+2x3=0
Let x3 = k, x2 + k=0, x2 = -k
4 x1 4( k ) 2 K 0 4 x1 2k 0 4 x1 2k x1
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2
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MATHEMATICS - I
x1 2k 12
1
x k 1 k 2 k
2
2
x3 k 1
2
1
X 1 2
2
=
Case (ii): If
[ − I]X =
=>
=> [
=> [
−
−
−
−
−
]
=
][ ] = [ ]
−
x1 x2 x3 0 (1)
x3 0 (2)
4 x1 4 x2 x3 0 (3)
Consider (1) & (3)
x1
x2
x3
1 1 1
4 4 1
x3
x2
x1
k
1 4 1 4 4 4
x
x
x
1 2 3 k
3
3
0
x1 k ; x2 k
x1 k 1
x2 k 1 k
x3 0 0
Case (iii): If
=>
=> [
=> [
=
[ − I]X =
−
−
−
−
1
X 2 1
0
−
−
−
]
=
][ ] = [ ]
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2 x1 x2 x3 0
x2 x3 0
4 x1 4 x2 0
Let x1 k
x3 k
and
2 x1 x2 x3 0 2k x2 k 0 x2 k
x1 k 1
X 3 x2 k 1 k
x3 k 1
1
p X 1 X 2 X 3 2
2
1
1
4
2
D
1
18
D8 0
0
( a ).
1
3
2
1
AP 0
0
0
28
0
1
1
0
1
1
1
0
1
1
0
2
0
0 1
0 0
3 0
0
2
0
0
0
3
0 1 0
0
0 0 256
0
38 0 0 6561
A8 PD8 P 1
1 1
2 1
2 0
12099
12100
13120
14
(b). D 4 0
0
1 1
0
0 1
1 0 256
0 4
1 0
0
6561 2
12355 6305
12356 6305
13120 6561
0
24
0
1
3
2
0
1
1
0 1 0 0
0 0 16 0
34 0 0 81
1 1 1 1 0 0 1 1 0
A =PD P 2 1 1 0 16 0 4 3 1
2 0 1 0 0 81 2 2 1
4
4 -1
1 64 162 1 48 162 0 16 81
1 16 81 1 1 0
A 2 16 81 4 3 1 2 64 162 2 48 162 0 16 81
2 0 162 2 0 162 0 0 81
2 0 81 2 2 1
4
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99
A
100
160
4
115
116
160
65
65
81
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UNIT-II
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FUNCTIONS OF SEVERAL VARIABLES
is a function where ‘ ’ is dependent variable and ‘ ’ is
=
Introduction: We know that
independent variable. We are going to expand the idea of functions to include functions for more than
one independent variable. In day to day life we deal with things which depend on two or more
quantity. For example, the area of the room which is a rectangle consists of two variables: length(say
=
a) and breadth (say b) is given by
. Similarly the volume of the rectangular parallelepiped
consists of three variables a, b, c i.e., length, breadth, height is given by V = a b c.
In this chapter we say that z is a function of two variables x , y and write
dependent variable and ‘ &
are independent variables.
=
,
where ‘ ’ is
Limit of a function of two variables:
A function f(x, y) is said to tend to the limit l as (x, y) tends to (a,b) i.e., x→
corresponding to any given positive number ∈ ∃ a positive number 𝛿
all points (x,y) whenever | − |
𝛿, | − |
𝛿.
ℎ ℎ
|
,
→
if
− | <∈ for
In other words the variable value (x ,y) approaches a finite fixed value l when the variable
value (x, y) approaches a fixed value (a, b) i.e.,
lim
→
,
→
=
,
lim
→
,
,
Continuity of a function of two variables at a point:
=
A function f(x,y) is continuous at a point (a, b) if, corresponding to any given positive
number ∈ ∃ a positive number 𝛿
|
ℎ ℎ
(x-a)2+(y-b)2<𝛿
,
−
,
| <∈ for all points (x,y) whenever 0<
Note: Every differentiable function is always continuous, but converse need not be true.
Solved Problems:
1. Evaluate lim
Sol. lim
→
→
+
→
→
+
+
+
= lim {lim [
→
→
= lim
→
=
+
+
]}
+
=
(or)
lim
→
→
+
+
= lim {lim [
→
→
= lim
→
+
+
+
]}
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MATHEMATICS - I
=
=
−
+
2. If f(x ,y) =
Sol. lim {lim
→
→
,
show that lim {lim
→
→
−
} = lim {lim
→
= lim
→
=
lim {lim
→
,
→
}
+
→
−
} = lim {lim
→
→
+
Hence the result follows.
}
= lim
= -1
→
,
} ≠ lim {lim
→
,
→
}
−
3. Discuss the continuity of the function
f(x,y) = {
,
+
,
,
,
≠
=
,
,
Sol. Let us consider the limit of the function for testing the continuity along the line y = mx.
Now lim
, = lim
→
→
+
→
= lim
→
→
+
= +
Which is different for the different m selected.
∴ lim
, does not exist.
→
→
Consider
lim
,
→
lim
,
→
= lim
→
= lim
→
+
+
=lim
→
=lim
→
=
=
=
,
=
,
∴ f(x,y) is continuous for given values of x and y but it is not continuous at (0,0)
Partial Differentiation:
Let z = f x, y be a function of two variables x and y. Then lim
→
+∆ ,
∆
−
,
, if it exists ,
is said to be partial derivative or partial differential coefficient of z or f(x,y) , w.r.t.x . It is
denoted by the symbol
or
i.e, for the partial derivative of
or
=
Similarly, the partial derivative of
as lim
→
, +∆ −
∆
,
.
=
and is denoted by
,
w.r.t. ‘x’ , ‘y’ is kept constant.
,
or
w.r.t. ‘y’ , ‘x’ is kept constant and is defined
or
.
Higher order Partial Derivatives:
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In general the first order partial derivatives
and
are also functions of x and y and they
can be differentiated repeatedly to get higher order partial derivatives,
So
∂
∂
∂
∂
∂
∂
∂
=∂
∂
∂
=
∂
∂
,
∂
,
∂
∂
∂
∂
∂
=∂
∂
∂
∂
=
∂
∂
,
,
∂
∂
∂
∂
∂
∂
=∂
∂
∂
=
∂
=
∂
∂
,
∂
∂
∂
=∂
∂
∂
,
, and so on.
∂ ∂
The chain rule of Partial Differentiation:
Let
∂
,∂
∂
where u = ∅ x, y and v = g(x,y) . Then
∂
=
∂
∂
+
∂
∂
∂
∂
and
∂
Total differential coefficient:
∂
∂
=
∂
∂
∂
∂
∂
+
∂
∂
∂
Let z = f x, y where x = ∅ t and y = g(t)
Substituting x and y in z = f(x,y) , z becomes a function of a single variable t.
Then the derivative of z w.r.t. ‘t’ i.e,
is called the total differential coefficient or total
derivative of z.
∴
∂
∂
=
∂
∂
∂
∂
∂
+
∂
∂
∂
=
Note: In the differential form , this result can be written as
Here, du is called the total differential of u.
∂
. dx +
∂
∂
∂
. dy
Solved Problems:
+
1. If U = log(
+
−
, prove that
Sol: Given that U = log( x + y + z − xyz
∴
=
=
=
∴
Now
+
⟹
∂
∂
−
+
+
−
+
+
−
−
−
+
+
+
∂U
∂
∂
+
∂
+∂ +∂
𝛛
𝛛
𝐔=
−
+ +
( x and z are constant)
( y and x are constant)
=
∂
𝛛
+𝛛 +𝛛
( y and z are constant)
−
∂U
𝛛
+
−
+
+
∂U
=
=
∂
+
−
−
+
+
+
−
+
−
+
−
x + y + z − xy − yz − zx
x + y + z x + y + z − xy − yz − zx
U =
=
+ +
∂
∂
……..(1)
∂
∂
+∂ +∂
∂
∂
∂
∂U
∂
+∂ +∂
∂
DEPARTMENT OF HUMANITIES & SCIENCES
∂U
∂U
+∂ +∂
+ +
[from
]
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MATHEMATICS - I
=
∂
∂
+ +
=-
+ +
=2. If
-
+
+ +
∂
∂
+ +
−
+ +
𝛛
= e show that at x = y = z , 𝛛
Sol: Given that x y z = e
+
𝛛
∂
∂
+ +
+ +
=− 𝐥
−
Taking logarithm on both sides, we get
x log x + y log y + z log z = log e
⟹ z log z = 1 - x log x – y log y
∂
z. + . log z
+l
⇒∂ =-
+l
Differentiating partially w. r. t , ′x ′ , we get
∂
= - x. + . log x
∂
……..(1)
+l
∂
Similarly ∂ = -
……..(2)
+l
When x = y = z , we have
= -1 and
= -1
Now differentiating (2) partially w.r.t, ‘x’, we get
∂
∂
∂
=∂
∂ ∂
∂
∂
+l
= ∂ [−
]
+l
+ log z
= - (1+ log y)[−
∂
−
When x = y = z from (3) , we have
∂
∂ ∂
+l
=
=-
+l
= -
l
since
(-1)
+l
=-
l
+l
= − x log ex
∂
∂
∂
+l
]=
∂
+l
…….(3)
∂
= −
(since log e =1)
−
3. If u = f( y-z ,z-x, x-y) prove that
+
+
=0
Sol: Let r = y - z , s = z - x , t = x – y .Then u = f (r,s,t)
∂
Now
∂
=
=− ,
and
∂
∂
∂
∂
= ,
=
,∂ = ,∂ = −
∂
∂
=1, ∂ = -1 , ∂ = 0
∴ By chain rule of partial differentiation , we have
DEPARTMENT OF HUMANITIES & SCIENCES
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MATHEMATICS - I
=
+
=
+
=
=
and
+
+
−
+
+
−
+
+
−
∂u ∂u ∂r ∂u ∂s ∂u ∂t
=
+
+
∂z
∂r ∂z ∂s ∂z ∂t ∂z
=
∂
(1) + (2) + (3) gives
∂
∂
+
∂
+
∂
∂
∂
= −
∂
∂
+
∂
∂
+
∂
∂
−
∂
−
∂
∂
+
∂
+
∂
+ −
∂
∂
+
∂
∂
=−
+
…(1)
=
−
…(2)
∂
∂
∂
…(3)
=−∂ +∂
∂
=0
Jacobian :
Let u = u (x , y) , v = v(x , y) are two functions of the independent variables x , y.
The jacobian of ( u , v ) w .r .t (x , y ) or the jacobian transformation is given by the
∂
∂
∂
∂
∂
determinant |∂
∂
∂
|
(or)
The determinant value is denoted by J (
) or
Similarly if u = u(x, y, z ) , v = v (x, y , z) , w = w(x, y , z) ,then the Jacobian of u , v , w
w.r.to x , y , z is given by
J(
) =
=
Properties of Jacobians
1. If J
(u, v)
( x, y )
and J 1
then JJ 1 1
( x, y )
(u, v)
2. If u,v are functions of r, s and r, s are functions of x ,y , then
Solved Problems:
1. If x + y2 = u , y + z2 = v , z + x2 = w find
( x, y , z )
(u , v, w)
,
,
=
,
,
.
,
,
Sol : Given x + y2 = u , y + z2 = v , z + x2 = w
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MATHEMATICS - I
We have
=
=
= 1(1-0) – 2y(0 – 4xz) + 0
= 1 – 2y(-4xz)
= 1 + 8xyz
=
=
2. Show that the functions u = x + y + z , v = x2 + y2 + z2 -2xy – 2yz -2xz and w = x3
+ y3 + z3 -3xyz are functionally related.
Sol: Given u = x + y + z
v = x2 + y2 + z2 -2xy – 2yz -2xz
w = x3 + y3 + z3 -3xyz
we have
=
=
=6
c1 c1 c2
c2 c2 c3
0
6
2x 2 y
2
x yz y 2 xz
0
1
2 y 2z
z yx
2
2
y xz z xy z 2 xy
=6[2(x - y) (y2 + xy – xz -z2 )-2(y - z)(x2 + xz – yz - y2)]
=6[2(x - y)( y – z)(x + y + z) – 2(y – z)(x – y)(x + y + z)]
=0
Hence there is a relation between u,v,w.
3. If x + y + z = u , y + z = uv , z = uvw then evaluate
Sol: x + y + z = u
y + z = uv
z = uvw
y = uv – uvw = uv (1 – w)
x = u – uv = u (1 – v)
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MATHEMATICS - I
=
=
R2 R2 R3
=
= uv [ u –uv +uv]
= u2v
4. If u = x2 – y2 , v =2xy where x = r cos
u = x2 – y2
Sol: Given
, y = r sin
,
v = 2xy
– r2sin2
=r2cos2
= 2rcos
– sin2
= r2 (cos2
= 4r3
S.T
r sin
= r2 sin2
= r2 cos2
=
=
= (2r)(2r)
= 4r2 [rcos22 + r sin22 ]
=4r2(r)[ cos22 + sin22 ]
=4r3
5.
If u =
,v=
Sol: Given u =
,w=
,v=
find
,w=
We have
=
ux = yz(-1/x2) =
=
=
,
, uz =
xz(-1/y2) =
,
,
uy =
=
,
,
= xy (-1/z2)
DEPARTMENT OF HUMANITIES & SCIENCES
=
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86
MATHEMATICS - I
=
=
.
.
=
= 1[-1(1-1) -1(-1-1) + (1+1) ]
= 0 -1(-2) + (2)
=2 + 2
=4
6.
If x = er sec , y = er tan P.T
.
=1
Sol: Given x = er sec , y = er tan
=
,
=
= er sec
=x ,
= ersec tan
= er tan
=y ,
= er sec2
x2 – y2 = e2r (sec2 - tan2 )
2r = log (x2 – y2 )
r = ½ log (x2 – y2 )
rx
x
1
1
(2 x) 2
2
2
(x y2 )
2x y
ry
y
1
1
(2 y ) 2
2
2
(x y2 )
2x y
=
=
=
=
,
x
= sin-1( )
y
1
y 2
y x x x2 y2
1 2
x
1
2
=
=
(1/x)
=
= e2r sec2 - y er sec tan
= e2r sec [sec2 - tan2 ] = e2r sec
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MATHEMATICS - I
y
(x y2 )
1
x
(x y2 )
r ,
y
( x, y )
x x2 y2
2
2
x2 y2
-
=[
]
=
=
.
=
=1
Functional Dependence:
Two functions u and v are functionally dependent if their Jacobian i.e,
J(
) =
=
=0
If the Jacobian of u, v is not equal to zero then those functions u, v are functionally
independent.
Solved Problems :
=
1. If
+
−
= 𝐚
,
−
+ 𝐚
−
,
. Find
. Hence prove that u and v are
,
functionally dependent. Find the relation between them.
Sol : Given u =
∴
∴
∂
∂
∂
=
,
∂ ,
+
−
=
+
−
∂
∂
∂
|∂
∂
,∂ =
∂
∂
∂
∂
and v = tan− x + tan− y
+
∂
,∂ =
−
+
−
| =|
+
+
+
−
+
∴ u and v are functionally dependent .
Now v = tan− x + tan− y = tan−
∂
and ∂ =
|=
+
−
−
+
-
−
=0
= tan− u
∴ v = tan− u is the functional relation between u and v.
2. Determine whether the following functions are functionally dependent or not. If
they are functionally dependent , find a relation between them.
i)
u=
,v=
ii) u =
∂
Sol: i) Jacobian = ∂
,
,
u
= |v
u
v |
DEPARTMENT OF HUMANITIES & SCIENCES
,v=
+
−
e sin y
=|
e cos y
e cos y
|
− e sin y
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MATHEMATICS - I
= e ( -sin y − cos y = −
≠
∴ u,v are functionally independent .
ii)
u= ,v=
+
,
,
∴ J( ) =
,
∂
−
=|
∂ ,
−
−
−
−
∴ u and v are functionally dependent ,
Now v =
+
∴v=
−
+
−
+
=
−
=
| =
-
−
=0
−
+
−
is the functional relation between u and v .
3. Show that the functions u= xy + yz + zx , v =
+
functionally related .find the relation between them.
+
and w= x + y + z are
Sol: We have
u = xy + yz + zx , v = x + y + z , w= x + y + z
∴
∂
∂
∂
∂
∂
∂
∂
∂
∂
, ,
∂ , ,
=2|
∂
+
|∂
= ∂
|
∂
∂
∂
+
∂
∂
∂
= 2(x + y + z) |
= 2(x + y + z) (0)
=0
∂
+
y+z
|
=| x
|
|
z+x x+y
y
z |
( Applying
→
+
)
|
( since R andR are identical )
Hence u,v and w are functionally dependent . That is , the functional relationship
exists between them.
Now w = x + y + z
=x +y +z +
xy + yz + zx = v + u
∴ w = v + u is the functional relation between u ,v and w.
4. Verify if u = 2x – y + 3z , v = 2x – y – z , w = 2x – y + z are functionally dependent
and if so , find the relation between them.
Sol: Given u = 2x – y + 3z , v = 2x – y – z , w = 2x – y + z
The functions u,v,w are functionally dependent if and only if J(
DEPARTMENT OF HUMANITIES & SCIENCES
, ,
,
=0
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MATHEMATICS - I
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
|∂
= ∂
|
, ,
Now J(
∂
,
∂
∂
∂
∂
−
−
−
|
=|
|
∂
∴ u, v, w are functionally dependent
− | = 2(-1) |
− | = (-2)(0) =0
Now u + v -2w = (2x-y +3z) + (2x – y –z) – 2 (2x – y + z )
= (4x – 2y + 2z) – (4x – 2y +2z) = 0
Hence u + v -2w = 0 is the functional relationship between u, v and w.
5. Show that the functions u = x+y+z, v = x2+y2+z2-2xy-2yz-2zx and w = x3+y3+z3-3xyz are
functionally related.
Sol. Given u = x+y+z, v = x2+y2+z2-2xy-2yz-2zx and w = x3+y3+z3-3xyz
, ,
Now
→
=6 |
−
|
|
=
, ,
|
=|
|
−
+
−
, ,
∴
+
= 12 |
, ,
−
−
−
=6|
−
−
−
+
−
+
−
−
−
−
−
−
−
+
+
= 12 (x-y)(y-z) |
−
−
−
−
−
−
−
| C1 →
−
+
−
+
|
+ +
+ +
= 12 (x-y)(y-z) (0) [C1 and C2 are identical]
=0
Hence the functional relationship exists between u, v, w.
6. Prove that u =
them.
−
+
,v=
Sol. We are given u =
∴
=
=
(
(
+
=2 y [
(
=2 x [
(
).
+
). −
+
−(
+
−(
+
). − .
+
+ ). − .
+
−
−
−
+
).
]=
).
]=
,v=
=
+
+
−
−
|
−
−
|
and C2
|
are functionally dependent and find the relation between
+
=
−
−
−
+
+
−
+
+
and
+
+
−
=
+
=
+
−
+
−
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MATHEMATICS - I
Thus
,
,
|=|
=|
=
+
+
=
−
+
(
∴ u,v are functionally dependent.
u2 + v2 =
2
−
+
+
+
2
=
−
−
−
+
+
+
+
)−
+
+
|
−
+
−
−
=0
=1
Hence u + v = 1 is the functional relation between u and v.
Maxima & Minima for functions of two Variables:
Definition : Let f(x,y) be a function of two variables x and y.
At x = a ; y = b , f(x ,y) is said to have maximum or minimum value , if f (a ,b) > f( a +h ,
b +k) or f (a ,b) < f( a +h , b +k) respectively where h and k are small values.
Extremum : A function which have a maximum or minimum or both is called
‘extremum’
Extreme value :- The maximum value or minimum value or both of a function is
Extreme value.
Stationary points: - To get stationary points we solve the equations
= 0 and
= 0 i.e the pairs (a1, b1), (a2, b2) ………….. are called
Stationary.
Working procedure:
1.
Find
and
Equate each to zero. Solve these equations for x & y we get the pair
of values (a1,b1) (a2,b2) (a3 ,b3) ………………
2.
Find l =
2 f
2 f
2 f
,
n
=
,m
x 2
x y
y 2
3 i) If l n –m2 > 0 and l < 0 at (a1,b1) then f(x ,y) is maximum at (a1,b1) and maximum
value is f(a1,b1)
ii) If l n –m2 > 0 and l > 0 at (a1,b1) then f(x ,y) is minimum at (a1,b1) and minimum
value is f(a1,b1) .
iii) If l n –m2 < 0 and at (a1, b1) then f(x, y) is neither maximum nor minimum at (a1, b1).
In this case (a1, b1) is saddle point.
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MATHEMATICS - I
iii)
If l n –m2 = 0 and at
(a1, b1) , no conclusion can be drawn about maximum or
minimum and needs further investigation. Similarly we do this for other stationary
points.
Solved Problems:
1.
Locate the stationary points & examine their nature of the following functions.
u =x4 + y4 -2x2 +4xy -2y2, (x > 0, y > 0)
Sol: Given u(x ,y) = x4 + y4 -2x2 +4xy -2y2
For maxima & minima
u
= 0, u = 0
x
y
= 4x3 -4x + 4y = 0 x3 – x + y = 0
-------------------> (1)
= 4y3 +4x - 4y = 0 y3 + x – y = 0
-------------------> (2)
Adding (1) & (2),
x3 + y3 = 0
x = – y -------------------> (3)
x3 – 2x x = 0, 2, 2
(1)
y = 0, - 2, 2
Hence (3)
l=
2u
2u
2
=
12x
–
4,
m
=
=
x 2
xy
(
) =4&n=
2u
= 12y2 – 4
y 2
ln – m2 = (12x2 – 4 )( 12y2 – 4 ) -16
At (
), ln – m2 = (24 – 4)(24 -4) -16 = (20) (20) – 16
,
The function has minimum value at (
,
> 0 and l=20>0
)
At (0,0) , ln – m2 = (0– 4)(0 -4) -16 = 0
(0,0) is not a extreme value.
2. Investigate the maxima & minima, if any, of the function f(x) = x3y2 (1-x-y).
f(x) = x3y2 (1-x-y)
Sol: Given
= 3x2y2 – 4x3y2 -3x2y3
For maxima & minima
= x3y2- x4y2 – x3y3
= 2x3y – 2x4y -3x3y2
= 0 and
=0
3x2y2 – 4x3y2 -3x2y3 = 0
=> x2y2(3 – 4x -3y) = 0 ---------------> (1)
2x3y – 2x4y -3x3y2
=> x3y(2 – 2x -3y) = 0 ----------------> (2)
From (1) & (2)
= 0
4x + 3y – 3 = 0
2x + 3y - 2 = 0
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MATHEMATICS - I
2x = 1 => x = ½
4 ( ½) + 3y – 3 = 0 => 3y = 3 -2 , y = (1/3)
2 f
x 2
l=
2 f
2
x
= 6xy2-12x2y2 -6xy3
(1/2,1/3) = 6(1/2)(1/3)2 -12 (1/2)2(1/3)2 -6(1/2)(1/3)3 = 1/3 – 1/3 -1/9 = -1/9
m=
2 f
xy
=
f
= 6x2y -8 x3y – 9x2y2
x y
2 f
(1/2 ,1/3) = 6(1/2)2(1/3) -8 (1/2)3(1/3) -9(1/2)2(1/3)3 =
x
y
n=
=
2 f
= 2x3 -2x4 -6x3y
y 2
2 f
2
y
(1/2,1/3)
= 2(1/2)3 -2(1/2)4 -6(1/2)3(1/3) = - -
ln- m2 =(-1/9)(-1/8) –(-1/12)2 =
-
=
=
= -
> 0 and l =
1
0
9
The function has a maximum value at (1/2 , 1/3)
1 1 1 1
2 3 8 9
Maximum value is f , 1
3.
1 1 1 1 1
1
2 3 72 2 3 432
Find three positive numbers whose sum is 100 and whose product is maximum.
Sol: Let x ,y ,z be three +ve numbers.
Then x + y + z = 100
z = 100 – x – y
Let f (x,y) = xyz =xy(100 – x – y) =100xy –x2y-xy2
For maxima or minima
= 0 and
=0
=100y –2xy-y2 = 0 => y(100- 2x –y) = 0 ----------------> (1)
= 100x –x2 -2xy = 0 => x(100 –x -2y) = 0 ------------------> (2)
100 -2x –y = 0
200 -2x -4y =0
-----------------------------100 + 3y = 0 => 3y =100 => y =100/3
100 – x –(200/3) = 0
=> x = 100/3
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MATHEMATICS - I
=
2 f
x 2
2 f
2
x
=- 2y
(100/3 , 100/3 ) = - 200/3
2 f
f
= 100 -2x -2y
=
x y
xy
2 f
(100/3 , 100/3 ) = 100 –(200/3) –(200/3) = -(100/3)
xy
=
=
2 f
y 2
= -2x
2 f
2 (100/3 , 100/3 ) = - 200/3
y
ln -m2 = (-200/3) (-200/3) - (-100/3)2 = (100)2 /3
The function has a maximum value at (100/3 , 100/3)
z = 100
i.e. at x = 100/3, y = 100/3
100 100 100
3
3
3
The required numbers are x = 100/3, y = 100/3, z = 100/3
4. Find the maxima & minima of the function f(x) = 2(x2 –y2) –x4 +y4
Sol: Given f(x) = 2(x2 –y2) –x4 +y4 = 2x2 –2y2 –x4 +y4
For maxima & minima
= 0 and
=0
= 4x - 4x3 = 0 => 4x(1-x2) = 0 => x = 0 , x = ± 1
= -4y + 4y3 = 0
2 f
l = 2
x
= 4-12x2
2 f
=
m =
x
y
2 f
n = 2
y
=> -4y (1-y2) = 0 =>y = 0, y = ± 1
f
x y
=0
= -4 +12y2
we have ln – m2 = (4-12x2)( -4 +12y2 ) – 0
= -16 +48x2 +48y2 -144x2y2
= 48x2 +48y2 -144x2y2 -16
i)
At ( 0 , ± 1 )
ln – m2 = 0 + 48 - 0 -16 =32 > 0
l = 4-0 = 4 > 0
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MATHEMATICS - I
f has minimum value at ( 0 , ± 1 )
f (x ,y ) = 2(x2 –y2) –x4 +y4
f ( 0 , ± 1 ) = 0 – 2 – 0 + 1 = -1
The minimum value is ‘-1 ‘.
ii)
At ( ± 1 ,0 )
ln – m2 = 48 + 0 - 0 -16 =32 > 0
l = 4-12 = - 8 < 0
f has maximum value at ( ± 1 ,0 )
f (x ,y ) = 2(x2 –y2) –x4 +y4
f ( ± 1 , 0 ) =2 -0 -1 + 0 = 1
The maximum value is ‘1 ‘.
iii)
At (0,0) , (± 1 , ± 1)
ln – m2 < 0
l = 4 -12x2
(0 , 0) & (± 1 , ± 1) are saddle points.
f has no max & min values at (0 , 0) , (± 1 , ± 1).
Lagrange’s method of undetermined multipliers
Suppose it is required to find the extremum for the function f(x , y , z) subject to condition
( x , y , z) = 0 ------------- (1)
Step 1 : Form lagrangean function F(x , y , z) = f(x , y , z) +
( x , y , z) where
is called
Lagrange’s constant, which is determined by the following conditions.
Step 2: Obtain the equations
F
= 0 =>
x
+
= 0 --------------- (2)
F
= 0 =>
y
+
= 0 --------------- (3)
F
= 0 =>
z
+
= 0 --------------- (4)
Step 3: Solving the equations (1) (2) (3) & (4) we get the stationary point (x, y, z).
Step 4 : Substitute the value of x , y , z so obtained in equation (1) we get the extremum.
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Solved Problems:
1. Find the minimum value of x2 +y2 +z2, given x + y + z =3a
Sol: u = x2 +y2 +z2
= x + y + z - 3a = 0
Using Lagrange’s function
F(x , y , z) = u(x , y , z) +
( x , y , z)
For maxima or minima
F
=
x
+
= 2x + = 0 ------------ (1)
F
=
y
+
= 2y +
F
=
z
+
= 2z + = 0 ------------ (3)
= 0 ------------ (2)
(1) , (2) & (3)
= -2x = -2y = -2z
= x + x + x - 3a = 0
=a
= y =z = a
Minimum value of u = a2 + a2 + a2 =3 a2
2. Find the minimum value of
Sol: Let u = x + y + z
+
, given that xyz = 𝐚
…… (1)
=0
And ∅ = xyz -
+
…… (2)
Consider the lagrangean function F(x,y,z) = u( x,y,z ) + λ ϕ x, y, z
i.e , F(x,y,z) = x + y + z + λ xyz − a )
∂F
∂
……(3)
∂∅
……(4)
= 0 ⇒ ∂ + λ ∂ = 2y + λxz = 0
Now ∂ = 0 ⇒ ∂ + λ ∂ = 2x + λyz = 0
∂F
∂
∂F
∂
∂
∂∅
……(5)
∂
∂∅
……(6)
= 0 ⇒ ∂ + λ ∂ = 2z + λyx = 0
From (4) , (5) and (6) ,we have
=
From the first two members , we have
From the last members , we have
=
=
=
= −
λ
……(7)
⇒x =y
⇒y =z
From (8) and (9) , we have x = y = z ⇒ x = y = z
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…(8)
……(9)
……(10)
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on solving (2) and (10) , we get , x = y = z = a
∴ Minimum value of u = a + a = a
3. Find the maximum value of u =
if 2x + 3y + 4z = a
.….. (1)
Sol: Given u = x y z
Let ϕ x, y, z = 2x + 3y + 4z - a = 0
……(2)
i.e, , F(x,y,z) = x y z + λ (2x + 3y + 4z – a)
……(3)
Consider the lagrangean function F(x,y,z) = u( x,y,z ) + λ ϕ x, y, z
for maxima or minima
Now
∂F
∂
∂F
and
∂
∂F
∂
= 0 ⇒ 2x
∂F
∂
=0 ,
∂F
=0,
∂
+ 2λ = 0 ⇒ x
∂F
∂
=0
……(4)
=−λ
…….(5)
= 0 ⇒ 3x y z +3λ = 0 ⇒ x y z = − λ
…….(6)
= 0 ⇒ 4x y z + λ =0 ⇒ x y z = − λ
From (4) and (5) , we have x = y
…….(7)
From (5) and (6) , we have y = z
…….(8)
Hence from (7) and (8) , we get x = y = z
On solving (2) and (9) ,we get x = y = z =
…….(9)
a
a
a
∴ Maximum value of u =
a
=
a
4. Show that the rectangular solid of maximum volume that can be inscribed in a
sphere is a cube.
Sol: Let 2x ,2y, 2z are the length , breadth and height of rectangular solid
Then its volume V = 8 xyz
……..(1)
Let the sphere have a radius of ‘r’ so that x + y + z = r
……(2)
Consider the lagrangean function F(x,y,z) = u( x,y,z ) + λ ϕ x, y, z
i.e, F(x,y,z) = V + 𝜆 x + y + z − r
= 8xyz + 𝜆 x + y + z − r
∂F
∂F
For maxima or minima ∂ = 0 , ∂ = 0 ,
∂F
∂
∂F
∂
∂F
∂
= 0 ⇒ 8yz +2 λx = 0
∂F
∂
…….(3)
=0
……(4)
……(5)
= 0 ⇒ 8zx +2 λy = 0
= 0 ⇒ 8xz +2 λz = 0
……(6)
From (4) ,(5) and (6) we have 2
𝜆 = - 8xyz = - 2
⇒x=y=z
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Thus for a maximum value x = y = z which shows that the rectangular solid is a cube.
Taylor’s series for a function of two variables:
Consider a function f(x,y) defined in a region enclosing (a,b) and having successive
partial derivatives ,then taylor’s series gives an expansion of f(x,y) in powers of (x-a) and
(y-b) and partial derivatives of f at (a,b) and is expressed in ascending powers of (x-a) and
(y-b) .
f(x,y) = f(a,b) + (x-a) f a, b + y − b f a, b +
y−b f
b f
a, b + (y − b f
a, b +
!
[ x−a f
a. b ] + [ x − a f
!
x−a y−b f
a, b +
a, b +
x−a
a, b ] + …….
a, b + (y − b f
x−a
y−
Note: The above expansion is called the expansion of f(x,y) at (a,b) or in the
neighbouhood of (a,b) or in powers of (x-a) and (y-b) .
Solved Problems:
1. Expand
near (1,
𝝅
Sol: Let f(x,y) = e cos y ⇒ f(1,
f x, y = e cos y ⇒ f(1,
π
f x, y = −e sin y ⇒ f(1,
f
f
f
π
x, y = −e cos y ⇒ f(1,
x, y = −e sin y ⇒ f(1,
x, y = −e cos y ⇒ f(1,
e cos y =
2. Expand
√
[
+
𝜋
=
=
√
=
π
√
=
π
. Then
−
=
π
√
=
−
√
−
√
−
√
By substituting above values in taylor ′ s series , we get
+ x−
− y−
−
Sol: Let f(x,y) = x y + y −
π
+
−
−
!
𝐚
− x−
+
y−
π
−
−
π
!
+⋯]
using taylor’s thorem.
, a = 1 , b = -2
Now f(a,b) = f(1,-2) = -10
f a, b = 2xy ⇒
f
f
f a, b = x +
a, b = 2y ⇒ f
a, b = 2x ⇒ f
f
⇒
,−
= −
f
,−
,−
=4
,−
=4
=-4
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f
f
a, b = 0 ⇒ f
a, b = 0 ⇒ f
f
a, b = 2 ⇒ f
f
a, b = 0 ⇒ f
f
a, b = 0 ⇒ f
,−
=0
,−
=0
,−
=0
,−
=2
,−
=0
All other partial derivatives of higher order will vanish
By substituting above values in taylor ′ s series , we get
x y+ y−
y+
+ y+
]
=
+[ x−
+ [ x−
−
= 10 - 4(x - 1) + 4( y + 2) - 2(x −
+ y+
+3(x −
]+
[ x−
y+
+
+ 2 (x - 1) (y + 2) + (x −
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−
+
x−
x−
y+
y+
+
(y + 2)
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UNIT – III
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ORDINARY DIFFERENTIAL EQUATIONS
I.Differential equations of first order and first degree
Definition: An equation involving variables and its differentials is called a Differential
equation.
They are of two types
1. Ordinary differential equations
2. Partial differential equations
Ordinary differential equation: An equation is said to be ordinary if one or more
variables are differentiated w.r.to only one independent variable.
Ex . (1)
dy
7xy x 2
dx
(2)
d2 y
dy
3 2y e x
2
dx
dx
A Differential equation is said to be partial if the
Partial Differential equation:
derivatives in the equation have reference to two or more independent variables.
2
z z
E. g: 1.
4z
x y
z
z
2.
x y 2z
x
y
Order of a Differential equation: It is the order of the highest derivative occurring in
2
the Differential equation. Differential equation is said to be of order ‘n’ if the
derivative is the highest derivative in that equation.
E. g : (1). (x2+1) .
dy
+ 2xy = 4x2
dx
Order of this Differential equation is 1.
d2 y
dy
x 1 y e x
(2) x 2 2x 1
dx
dx
Order of this Differential equation is 2.
2
(3)
d2 y
dy
5 2y 0 .
2
dx
dx
Order=2
(4)
2u 2u
0.
x 2 y2
Order is 2.
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Degree of a differential Equation is the highest
Degree of a Differential equation:
degree of the highest derivative in the equation, after the equation is made free from
radicals and fractions in its derivations.
2
dy
dy
1 on solving we get
E.g : 1 ) y = x .
dx
dx
2
dy
dy
2
1 x 2 dx
2xy. 1 y 0 . Degree = 2
dx
32
2
d 2 y dy
1 on solving . we get
2) a.
dx 2 dx
3
2
d 2 y dy
a . 2 1 . Degree = 2
dx dx
2
2
The general form of first order, first degree differential equation is
, ,
′
=
+
[i.e
=
Where
are functions of
dy
f x, y or
dx
and ]. There
is no general method to solve any first order differential equation The equation which
belong to one of the following types can be easily solved.
In general the first order first degree differential equation can be classified as:
(1) Exact equations
(2) Non exact equations (reducible to exact equations).
Exact Differential Equations
Def: Let M(x, y) dx +N(x, y) dy =0 be a first order and first degree Differential Equation
where M & N are real valued functions of x, y . Then the equation M dx + N dy =0 is said to
be an exact Differential equation if a function f .
d f x, y
f
f
dx dy
x
y
Eg : d ( x 2 y) 2 xydx x 2 dy
Condition for Exactness: If
,
&
,
are two real functions which have
continuous partial derivatives then the necessary and sufficient condition for the Differential
equation
M N
y x
+
=
is to be exact if
Hence solution of the exact equation
,
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+
,
=
is
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Mdx
Ndy c .
(y is taken as constant)
(terms free from x are taken).
Solved Problems :
1. Solve
dy y cos x sin y y
0.
dx sin x x cos y x
Sol : Given equation can be written as
( y cos x sin y y )dx (sin x x cos y x)dy 0 …(1)
It is of the form
Here
+
= .
M y cos x sin y y
N sin x x cos y x
M
cos x cos y 1
y
N
cos x cos y 1
x
Clearly
M N
y x
⟹Equation is exact.
The general solution is given by Mdx Ndy c
(y constant) (terms independent of x)
( y cos x sin y y )dx (0)dy c
y sin x (sin y y ) x c .
2.
x
x
x
Solve 1 e y dx e y 1 dy 0
y
x
Sol : Here M 1 e & N e 1
y
x
y
x
y
x N
1 x 1
M
ey 2 &
e y 1 e y
y
y x
y y y
x
x
x N
x
M
ey 2 &
ey 2
y
y x
y
x
x
x
M N
equation is exact
y x
General solution is
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Mdx
(y constant)
Ndy
c.
(terms free from x)
x
1 e y dx 0 dy c.
x
ey
x
c
1
y
x
y
x ye c
3. Solve the D.E (x+y-1) dy-(x-y+2) dx=0
Sol : Here M x y 2 ;
N x y 1
M
N
1;
1
y
x
Clearly
M N
y x
Thus the equation is exact.
General solution is Mdx
(y constant)
Ndy
c.
(terms free from x)
⟹ ( x y 2)dx ( y 1)dy c
x2
y2
xy 2 x
yc
2
2
x 2 y 2 2 xy 4 x 2 y c1
4. Solve e y 1 .cos xdx e y sin xdx 0 .
Sol.
M e y 1 cos, N e y sin x
M
N
e y cos x;
e y cos x
y
y
M N
e y c os x
y x
Equation is exact.
Gen Sol. is
Mdx
Ndy
c.
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(y constant)
e
y
(terms free from x)
1 cos x dx 0 dy c
e y sin x c
1
5. Solve y 1 cos y dx x log x x sin y dy 0 .
x
1
Sol : Here M y 1 cos y, N x log x x sin y
x
1
1
M
N
1 sin y
1 sin y
x
x
y
x
M N
y x
so the equation is exact
General sol Mdx
(y constant)
Ndy
c.
(terms free from x)
y
y cos y dx 0.dy c.
x
y x log x x cos y c.
6. Solve (cosx-xcosy)dy – (siny+ysinx)dx =0
Sol : N cos x x cos y & M sin y y sin x
N
sin x cos y
x
M
cos y sin x
y
M N
the equation is exact.
y x
General sol
Mdx
(y constant)
Ndy
c.
(terms free from x)
sin y y sin x .dx 0.dy c
x sin y y cos x c
ycosx - xsiny = c
7. Solve r sin cos dr r sin cos d 0
Sol : Given equation is r sin cos dr r sin cos d 0............ 1
This is of the form Md Ndr 0
Where M r sin cos ; N r sin cos
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We have
Clearly
M
N
sin cos ;
cos sin
r
M N
r
The given equation is exact.
The general solution is given by Md Ndr c
(r constant)
r (sin cos )d rdr c
r (sin cos )
(terms independent of 𝜃)
r2
c
2
The general solution is r 2 2r sin cos c1 .
REDUCTION OF NON-EXACT DIFFERENTIAL EQUATIONS TO EXACT
FORM USING INTEGRATING FACTORS
Definition: If the Differential Equation M(x,y) dx + N (x,y ) dy = 0 be not an exact
differential equation it can be made exact by multiplying with a suitable function
u (x,y)
0 called an Integrating factor(I.F).
Note: There may exits several integrating factors.
Some methods to find an I.F to a non-exact Differential Equation Mdx+N dy =0
Case -1: Integrating factor by inspection/ (Grouping of terms).
Some useful exact differentials
1.
d (xy)
= xdy +y dx
2.
x
d
y
ydx xdy
y2
3.
x
d
y
ydx xdy
x2
4.
x2 y 2
d
2
= x dx + y dy
5.
x
d log
y
=
xdy ydx
xy
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6.
x
d log
y
=
ydx xdy
xy
7.
x
d tan 1
y
=
ydx xdy
x2 y 2
8.
x
d tan 1
y
=
9.
d log xy
10. d log x 2 y 2
ex
11. d
y
xdy ydx
x2 y 2
=
xdy ydx
xy
=
2 xdx ydy
x2 y 2
=
ye x dx e x dy
y2
Solved Problems
1) Solve
y( xy e x )dx e x dy
0.
y2
Sol : It can be written as
( xy 2 ye x )dx e x dy
0.
y2
xy 2
e x ydx e x dy
2 dx
0
y
y2
xdx
ye x dx e x dy
0
y2
ex
xdx d
y
0
x2 ex
c
2
y
This is the required solution.
On integrating, we get
2) Solve the differential equation y x 2 dx x 2 cot y x dy 0
Sol : Given equation can be written as ydx xdy x 2 dx x 2 cot y dy
Dividing with x 2 , we get
ydx xdy
xdy ydx
dx cot y dy or
dx cot y dy
2
x
x2
y
i.e , d dx cot y dy
x
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y
x cos ec 2 y cos y x x cos ec 2 y cx
x
which is the required solution.
Integrating, we get
3) Solve xdx ydy
xdy ydx
0
x2 y 2
Sol : Given equation is xdx ydy
xdy ydx
0
x2 y 2
x2 y 2
1 y
d
d tan 0 on Integrating we get
x
2
x2 y 2
y
tan 1 c .
2
x
4) Solve y x3 .e xy y dx x y x 3 .e xy dy 0 .
Sol : Given equation is on regrouping
We get yx3e xy dx y 2 dx xydy x 4e xy dy 0
x3e xy ydx xdy y xdx ydx 0
Dividing by x3
y xdy ydx
e xy ydx xdy .
0
x2
x
y y
d e xy .d 0
x x
on Integrating
2
y
e xy 1 c is required G.S.
2 x
5) Solve (1+xy) x dy + (1- yx ) y dx = 0
Sol: Given equation is (1+xy) xdy + (1-yx) y dx =0.
( xdy + ydx ) + xy ( xdy – ydx ) = 0.
xdy ydx xdy ydx
Divided by x2y2
0
2 2
xy
x y
1 1
1
d dy dx 0 .
x
xy y
On integrating we get
1
log y log x log c
xy
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1
log x log y log c .
xy
6) Solve : ydx xdy a x 2 y 2 dx
Sol : Given equation is ydx ydx xdy a x 2 y 2 dx
ydx xdy
a dx
x 2 y2
x
d tan 1 a dx
y
On Integrating tan 1
x
ax c where c is an arbitrary constant.
y
Method -2: If M x, y dx N x, y dy 0 is a homogeneous differential equation and
1
is an integrating factor of M dx+ N dy =0.
Mx Ny
Mx Ny 0 then
Solved Problems :
1 . Solve x 2 ydx x 3 y3 dy 0
Sol : Given equation is x 2 ydx x 3 y3 dy 0 -----------------(1)
Where M x 2 y
Consider
&
N x 3 y3
M
x2 &
y
N
3x 2
x
M N
y x
equation is not exact .
But given equation (1) is homogeneous differential equation then
Mx Ny x x 2 y y x 3 y3 y 4 0 .
I.F =
1
1
4
Mx Ny
y
Multiplying equation (1) by
1
y4
x2 y
x 3 y3
dx
dy 0 ---------------------- (2)
y4
y4
x2
x 3 y3
dx
dy 0
y4
y3
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This is of the form M1dx N1dy 0
x 2
x 3 y3
&
N
1
y3
y4
For M1
M1 3x 2
N1 3x 2
4 &
4
y
y
y
x
M1 N1
equation (2) is an exact D.E.
y
x
General sol M1dx N1dy c
(y constant)
(terms free from x in N1)
x 2
1
3 dx dy c.
y
y
x3
log y c
3y3
2. Solve y y 2 2x 2 dx x 2y 2 x 2 dy 0
Sol : Given equation is y y 2 2x 2 dx x 2y 2 x 2 dy 0 -----------(1)
It is the form Mdx + Ndy = 0
Where M y y 2 2x 2 , N x 2y 2 x 2
Consider
M
3y2 2x 2 &
y
M N
y x
equation is not exact .
N
2y 2 3x 2
x
Since equation(1) is Homogeneous differential equation then
Mx Ny x y y 2 2x 2 y x 2y 2 x 2
Consider
3xy y 2 x 2 0 .
I.F.
1
3xy y 2 x 2
Multiplying equation (1) by
y y2 2x 2
3xy y 2 x 2
dx
1
we get
3xy y 2 x 2
x 2y 2 x 2
3xy y 2 x 2
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y(y2 - 2x 2 )
x(2y2 - x 2 )
M1 =
; N1 =
3xy(y2 - x 2 )
3xy(y2 - x 2 )
M1
N1
2xy
=
=
2
2 2
y
3(y - x )
x
Now it is exact
y
2
x2 x2
3x y 2 x 2
dx
y2 x y2 x 2
3y y 2 x 2
dy 0
dx
xdx
ydy
dy
2
2
0.
2
2
x y x
y x
y
dx dy
2ydy
2xdx
0
2
2
y 2 y x 2 y2 x 2
x
On integrating we get
1
1
log x log y log y 2 x 2 log y 2 x 2 log c xy c
2
2
Method- 3: If the equation Mdx Ndy 0 is of the form
y.f (xy)dx x.g(xy)dy 0 & Mx Ny 0 then
1
is an integrating factor of
Mx Ny
Mdx+ Ndy =0.
Problems :
1 . Solve xysin xy cos xy ydx xysin xy cos xy xdy 0 .
Sol : Given equation xysin xy cos xy ydx xysin xy cos xy xdy 0 -------(1).
Equation (1) is of the form y. f xy dx x.g xy dy 0 .
Where M xysin xy cos xy y
N xysin xy cos xy x
M N
y x
⟹ equation (1) is not exact
Now consider Mx-Ny
Mx-Ny =2xycosxy ≠ 0
Integrating factor =
1
2xy cos xy
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So equation (1) x I.F gives
xysin xy y dx xysin xy cos xy x dy 0
2xy cos xy
2xy cos xy
1
1
y tan xy dx y tan xy dy 0
x
y
M1dx N1dy 0
M1
N1
= tanxy + xysec2 xy =
y
x
Now the equation is exact.
General solution is
M dx N dy c. .
1
(y constant)
1
(terms free from x in N1)
1
1
y tan xy dx dy c.
x
y
y.log sec xy
log x log y log c
y
log sec xy log
x
log c.
y
x
.sec xy c.
y
2. Solve 1 xy ydx 1 xy xdy 0
Sol : Here M 1 xy y : N 1 xy x
M
N
= 1+ 2xy;
= 1- 2xy
y
x
Hence, the equation is not exact
Also Mx - Ny= 2 x 2 y 2 0
I.F
1
1
2 2 0
Mx Ny 2x y
Multiply the given equation by I.F, we get
1
1
1
2
dy 0
dx
2y
2x y 2x
M1
N
-1
= 2 2 = 1
y
2x y
x
⟹Equation is exact.
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On integrating, we get
1
1
1
2x y 2x dx 2y dy c
2
1
1 1
log x log y c .
2xy 2
2
x
1
log c1
xy
y
where
= 2c.
Method 4: If there exists a continuous single variable function
such that
M N
f x dx
y x
f x , then I.F. of Mdx Ndy 0 is e
N
Solved Problems :
1 . Solve 3xy 2ay 2 dx x 2 2axy dy 0
Sol : Given equation is 3xy 2ay 2 dx x 2 2axy dy 0
This is of the form Mdx+ Ndy = 0
M 3xy 2ay2 & N x 2 2axy
M
3x 4ay
y
M N
y x
&
N
2x 2ay
x
equation is not exact .
M N
y x 3x 4ay 2x 2ay
Now consider
N
x x 2ay
M N
y x 1
f x .
N
x
1
e x x is an Integrating factor of (1)
dx
Multiplying equation (1) with I.F we get
3xy 2ay xdx x
2
2
1
3x y 2ay x dx x
2
2
3
2axy
1
xdy 0
2ax 2 y dy 0
It is the form M1dx + N1dy =0
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M1 3x 2 y 2ay 2 x,
N1 x3 2ax 2 y
M1
N1
3x 2 4axy,
3x 2 4axy
y
x
M1 N1
y
x
Equation is exact
General sol M1dx N1dy c.
(y constant)
(terms free from x in N1)
3x 2 y 2ay 2 x dx 0dy c
x 3 y ax 2 y2 c ..
2. Solve ydx xdy 1 x 2 dx x 2 sin ydy 0
Sol: Given equation is y 1 x 2 dx x 2 sin y x dy 0.
M y 1 x 2 & N x 2 sin y x
M
N
1,
2x sin y 1
y
x
M N
the equation is not exact.
y x
M N
y x 1 2x sin y 1 2x sin y 2 2 x sin y 1 2
2
So consider
N
x 2 sin y x
x sin y x
x x sin y 1
x
1
2 dx
f x dx
1
e x e2log x 2
I.F e
x
Equation (1) x I.F gives
y 1 x2
x 2 sin y x
dx
dy 0
x2
x2
It is the form of M1dx+ N1dy =0.
M1 1 N1
= 2=
y
x
x
Equation is exact
Gen. sol. is thus
1
y
2 2 1dx sin ydy 0
x
x
y 1
x cos y c .
x x
x 2 y 1 x cos y cx.
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3. Solve 2xy dy – (x2+y2+1) dx = 0
Sol. Given equation is 2xy dy – (x2+y2+1) dx = 0
This is of the form M dx + N dy = 0, where N = 2xy, M = -x2 - y2 – 1
We have
= -2y and
= 2y, so that
≠
∴ The given equation is not exact.
−
We have
−
=
(-2y-2y) = = f(x)
∴ I.F. =
∫
=
Multiplying (1) with
− ∫
𝑥
=
, we get
−
=
−
−
+
=
+
This is of the form M1 dx + N1 dy = 0, where M1 = − −
We have
=
Since
−
=
and
=
−
………(1)
….(2)
−
and N1 =
, ∴ (2) is exact.
General solution is given by
M1dx N1dy c
(y constant)
⇒ ∫ − −
(y constant)
(terms free from x in N1)
−
+ ∫
=
⇒ -x + + = c
This is the general solution o (2) and hence of (1)
M N
y x
Method -5: For the equation Mdx + Ndy = 0 if
g ( y ) (is a function of y alone)
M
then e g ( y ) dy is an integrating factor of M dx + N dy =0.
Solved Problems :
1 . Solve (3x2y4+2xy)dx +(2x3y3-x2) dy = 0
Sol :
Given equation (3x2y4+2xy)dx +(2x3y3-x2) dy =0 -----------------(1).
Equation is of the form Mdx + Ndy =0.
where M =3x2y4+2xy & N = 2x3y3-x2
M
N
12x 2 y3 2x ;
6x 2 y3 2x
y
x
M N
equation(1) is not exact.
y x
M N
2
y x
g ( y) g y
So consider
M
y
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I .F e
g ( y )dy
e
2
1
dy
y
e
2 log y
1
.
2
y
3x 2 y4 2xy
2x 3 y3 x 2
Equation (1) x I.F
dx
dy 0
y2
y2
2x
x2
3x 2 y2 dx 2x 3 y 2 dy 0
y
y
It is the form M1dx N1dy 0
M1
x N1
= 6x 2 y - 2 2 =
y
y
x
Equation is exact
General sol. is M1dx N1dy c
(y constant)
(terms free from x in N1)
2x
3x 2 y 2
dx 0dy c .
y
3x 3 y2 2x 2
c.
3
2y
x 3 y2
x2
c.
y
2 . Solve (xy3+y) dx + 2(x2y2+x+y4) dy =0
Sol : Here M xy3 y ; N 2 x 2 y 2 x y 4
M
3xy 2 1;
y
N
4xy 2 2
x
We see equation is not exact.
M N
y x
Also
M N
y x
g y
M
xy 2 1
1
g(y)
2
y
y xy 1
1
g ( y ) dy
dy
e y y.
Thus I .F e
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M1
N
= 4xy3 + 2y = 1
y
x
Gen Sol:
xy
4
where M1 = xy 4 + y 2 ; N1 = 2x 2 y3 + 2xy + 2y 5
y 2 dx 2y5 dy c
x 2 y4
2y6
y2 x
c.
2
6
3. Solve y 4 2y dx xy3 2y 4 4x dy 0
Sol: The given equation is not exact.
M N
y 3 4 4 y 3 2 3
y x
g ( y)
g y .
Also
y
M
y4 2 y
I .F e
g y dy
Here M1 = y +
e
3
1
dy
y
1
y3
2
4x
; N1 = x + 2y - 3
2
y
y
M1
2 N1
= 1- 3 =
y
y
x
∴Equation is exact.
2
Gen sol is y 2 dx 2 ydy c.
y
2
2
y 2 x y c.
y
APPLICATIONS OF DIFFERENTIAL EQUATIONS OF FIRST ORDER FIRST
DEGREE
ORTHOGONAL TRAJECTORIES (O.T)
Def: A curve which cuts every member of a given family of curves at a right angle is an
orthogonal trajectory of the given family.
Orthogonal trajectories in Cartesian co-ordinates:
Working rule: To find the family of O.T in Cartesian form : Let f(x,y,c) =0 ……(1)
be the given equation of family of curves in Cartesian form.
1) Differentiate (1) with respect to ‘x’ and obtain F(x, y, y’) = 0 ----------(2)
of the given family of curves.
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2) Replace
dx
dy
by
in (2)
dx
dy
Then the Differential Equation of family of O.T is
F(x, y,
dx
) =0 ---------------- (3).
dy
3) Solve equation (3) to get the equation of family of O.T’s of equation(1).
Solved Problems
1 . Find the O.T’s of family of semi-cubical parabolas ay2=x3 where a is a parameter.
Sol : The given family of semi-cubical parabola is ay2=x3 ---------------(1)
Differentiating with respect to ‘x’ ⟹ a 2y
dy
= 3 x 2 -------------(2)
dx
x3
dy
2
Eliminating ‘a’ from (1) and (2) ⟹ y 2 2y dx 3x
2 x 3 dy
⟹
3x 2
y dx
Replace
2x
dx
dy
by ⟹
y
dx
dy
dx
3x 2
dy
3
-2 dx
x
=y
3 dy
-2
xdx - ydy = c
3
-x 2 y 2
=c
3 2
x 2 y2
+ =1
3c 2c
2. Find the O.T of the family of circles x2+y2+2gx+c =0, Where g is the parameter
Sol : x2+y2+2gx+c =0. --------- (1)
represents a system of co- axial circles with g as parameter.
Differentiating with respect to ‘x‘ ⟹ 2x+ 2y
dy
+ 2g =0--------------(2)
dx
Substituting equation from (2) in (1)
⟹x2+y2 -(2x+ 2y
dy
)x +c =0.
dx
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⟹y2-x2-2xy
Replace
dy
+ c=0
dx
dx
dy
by
dx
dy
dx
⟹y2-x2-2xy + c=0
dy
dx
+ c=0
dy
⟹y2-x2+2xy
This can be written as
2x
dx 1 2 -(c + y 2 )
x =
dy y
y
This is a Bernoulli’s equation in x
So put x2=u 2x
dx du
=
dy dy
-(c + y2 )
du 1
- u=
dy y
y
which is a linear equation in ‘u’
1
I.F = e
y dy
e log y
1
y
General solution is u(I.F) = Q(y).I.Fdy + k
x2
-(c + y2 ) 1
1
=
dy k
y
y
y
-2
= -c - y + k
y
x2 c
= -y+k
y y
2
3
2
3
3. Find orthogonal trajectories of the family of curves x + y = a
2
2
2
3
2
Sol : The equation of the given family of curves is x 3 + y 3 = a 3 …(1)
-1
3
-1
3
-1
3
2
2
dy
dy y
x + y
=0
= -1
3
3
dx
dx
x3
This is the differential equation of the given family of curves.
Differentiating (1) w.r.t x, we get
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-1
3
Changing
-1
3
dy
dx
dx
y
dx y
in (2), we get = - -1
= -1
to dy
dy
dx
dy
3
x
x3
1
3
…(3)
1
3
Separating the variables in (3), we get x dx = y dy
1
Integrating, we get
1
x 3dx = y 3dy + c
4
4
3 4 3 4
x 3 = y 3 + c or x 3 - y 3 = c1
4
4
This is the orthogonal equation.
4.Find the O.T’s of the family of parabolas through origin and foci on y –axis.
Sol : The equation of the family of parabolas through the origin and foci on y-axis is
x2=4ay where ‘a’ is parameter.
Differentiating with respective ‘x’ 2x =4a
x
dy
=
dx 2a
O.T ⟹ -
dy
dx
dx x
=
dy 2a
dx dy
=
x 2a
On integrating, we get -logx =
y
+c
2a
y + 2alogx = c1 is the equation of required O.T.
5.Find the O.T of the one parameter family of curves e x + e-y =c.
Sol : Given equation is e x + e-y =c.
Differentiating with respective ‘x’ e x + e-y (Its O.T e x + e-y (
dy
)=0
dx
dx
) = 0 ⟹ e-x dx + e ydy = 0
dy
On integrating, we get, -e-x + e y = c which is the required O.T
ORTHOGONAL TRAJECTORIES IN POLAR FORM
Working Rule: To find the O.T of a given family of curves in polar-co ordinates.
Let f(r, θ, c)=0-----(1) be the given family of curves in polar form.
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1.) Differentiating (1) with respect to
obtain F [r, θ,
dr
] =0 by eliminating the
dθ
parameter c.
2.) Replace
dr
dθ
by r2
then the Differential Equation of family of O.T is
dθ
dr
F [ r, θ,- r2
dθ
] =0
dr
3.) Solve the above equation to get the equation of O.T of (1)
Self-orthogonal
A family of curves is said to be self orthogonal when the differential equation of the
family of O.T is same as that of the original family.
Solved Problems
1. Find the orthogonal trajectories of the family of curves r n cosθ = a n .
Sol : Given family of curves is r n cosθ = a n
Taking logarithms on both sides,
nlogr + logcosθ = nloga
Differentiating w.r.t θ, we get
n dr
1
+
(-sinθ) = 0
r dθ cosθ
n dr
- tanθ = 0
…(2)
r dθ
is the differential equation of family of curves.
To get the differential equation of orthogonal trajectories, replace
Now, we have
dr
dθ
with -r 2
dr
dθ
n 2 dθ
-r
- tanθ = 0
r dr
dθ
+ tanθ = 0
…(3)
dr
This is differential equation of orthogonal trajectories.
Separating variables, we get
dr
ncotθdθ + = 0
r
Integrating, we get
nlogsinθ + logr = log logsin n θ + logr = logc
nr
or log(rsin n θ) = logc
c = rsin n θ which is the required O.T.
2. Find the Orthogonal trajectories of the family of cardioids r = a(1- cosθ) where
‘a’ is the parameter.
Sol : Given r = a(1- cosθ)
…(1)
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Differentiating the given equation w.r.t ‘θ’, we get
dr
1 dr
= asinθ a =
dθ
sinθ dθ
Eliminating
‘a’
from
(1)
and
(2)
θ
θ
2rsin cos
dr
rsinθ
2
2
=
=
dθ 1- cosθ
θ
2sin 2
2
dr
θ
= rcot
dθ
2
(2)
we
get
r=
1- cosθ dr
sinθ dθ
…(3)
The differential equation of required O.T is obtained by replacing
dr
dθ
with -r 2
in (3)
dr
dθ
dθ
θ dr
θ
= rcot Þ = -tan dθ
dr
2 r
2
On integrating, we get
logr = 2log | cos(θ / 2) | +log(2c) = log(2ccos 2 (θ / 2))
-r 2
or r = 2ccos 2 (θ / 2) r = c(1+ cosθ)
This is the equation of family of orthogonal trajectories.
3. Prove that the system of parabolas y2 = 4a(x + a) is self orthogonal.
Sol : Given parabola is
y2 = 4a(x + a)
…(1)
y' = 4ax + 4a 2
Differentiating (1) with respect to x,we get 2yy1 = 4a a =
yy1
…(2)
2
yy1
y 2 y12
x+4
Substituting (2) in (1), y = 4
2
4
2
2
2
or y = 2xyy1 + y y1
….. (3)
2
Equation (3) is the differential equation of the given system of parabolas. Replacing
with
−
, we get equation of the orthogonal trajectories as
2
-1
-1
2xy y 2
y = 2xy + y 2 y 2 =
+ 2
y
y
y
y1
1
1
1
2
y 2 y12 = -2xyy1 + y 2 or y 2 = 2xyy1 + y 2 y12 …(4)
which is differential equation of the orthogonal trajectories of the given family.
Equations (3) and (4) are same, hence the given system is self orthogonal.
x2
y2
+
= 1 where is a
4. Show that the system of confocal conics 2
a + λ b2 + λ
parameter,
is self orthogonal.
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x2
y2
Sol : Given equation of family of confocal conics is 2
+
= 1 …(1)
a + λ b2 + λ
Differentiating (1) w.r.t x, we get
2x
2y dy
+ 2
= 0. For convenience, we write
a + λ b + λ dx
2
x
y
+ 2
p = 0 x(b 2 + λ) + py(a 2 + λ) = 0
a +λ b +λ
-(b 2 x + a 2 yp)
λ=
x + yp
=
2
(a 2 - b 2 )x
a + λ =
x + yp
2
and b2 + λ =
-(a 2 - b2 )yp
x + yp
…(2)
Eliminating from (1) and (2), we get
x(x + yp) y(x + yp)
x + yp y
- 2 2 = 1 2 2 x - = 1
2
2
a -b
(a - b )p
a - b p
…(3)
y
(x + yp) x - = a 2 - b 2
p
This is the differential equation of the family of curves (1). We get the differential
equation of the family of orthogonal trajectories by replacing
p=
dy
dx
1
1
with ==- .
dx
dy
p
dy
dx
Hence the differential equation of orthogonal trajectories is
y
2
2
x - (x + py) = a - b …(4)
p
which is same as (3). Thus we see that the differential equation of the family of
orthogonal trajectories is same as that of the original family. Hence the given family of
curves is orthogonal to itself. Hence it is a self orthogonal family of curves.
NEWTON’S LAW OF COOLING
STATEMENT: The rate of change of the temperature of a body is proportional to the
difference of the temperature of the body and that of the surrounding medium.
Let
be the temperature of the body at time ‘t’ and
θ0 be the temperature of its
surrounding medium(usually air). By the Newton’s law of cooling , we have
dθ
dθ
α(θ - θ 0 )
= -k(θ - θ 0 ) k is +ve constant
dt
dt
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dθ
= -k dt
(θ - θ 0 )
log (θ - θ0 ) = -kt +c.
If initially θ= θ1 is the temperature of the body at time t=0 then
c = log(θ1 - θ0 )
log (θ - θ0 ) = -kt+ log(θ1 - θ0 )
θ - θ0
log
= -kt.
θ1 - θ 0
θ - θ0
-kt
=e
θ1 - θ 0
θ = θ 0 + (θ1 - θ 0 ) = e -kt
which gives the temperature of the body at time‘t’.
Solved Problems
1. A pot of boiling water 1000 C is removed from the fire and allowed to cool at
30 0 C room temperature. 2 mins later, the temperature of the water in the pot is
90 0 C .What will be the temperature of water after 5 mins?
Sol : We have θ - 30 = ce -kt
…(1)
When t = 0,θ = 100
from (1), we get c = 70
θ - 300 = 70e -kt
…(2)
When t = 2,θ = 90
From (2), 90 - 30 = 70e -2k
60 = 70e -2k
6
-2k = log
7
= 0.1542
k = 0.0771
When t = 5,θ - 30 = 70e-5k
θ = 77.460
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0
2. A body is originally at 800C and cools down to 60 C in 20 min . If the
temperature of the air is 400 C find the temperature of body after 40 min.
Sol : By Newton’s law of cooling we have
dθ
= -k (θ - θ 0 ) where θ0 is the temperature of the air.
dt
dθ
= -k dt log(θ - θ 0 ) = -kt + logc
(θ - θ 0 )
Here θ 0 = 400 C
log(θ - 40) = -kt + log c
θ - 40
= -kt
c
log
θ - 40
-kt
=e
c
θ = 40 + ce-kt --------------(1)
When t=0, θ = 800 C
80 = 40 +c c = 40 ------------(2).
When t=20, θ = 600 C 60 = 40+ ce-20k ------------(3).
Solving (2) & (3) ce-20k =20
40e 2 k =20
k = -20log2
When t = 40 C then equation (1) is θ = 40 + 40e
0
1
- log2 40
20
= 40 +40 e -2log2
1
= 40 + 40X
4
θ = 500 C
3. An object whose temperature is 750C cools in an atmosphere of constant
temperature C, at the rate of k𝛉, being the excess temperature of the body over
that of the temperature. If after 10min, the temperature of the object falls to 65 0
C , find its temperature after 20 min. Also find the time required to cool down to
550C.
Sol : We will take one minute as unit of time.
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It is given that
dθ
= -kt
dt
θ = ce -kt ------------(1).
Initially when t=0 θ = 750 - 250 = 500
c= 500
Hence c = 50 θ = 50e-kt -----------------(2)
When t= 10 min θ = 650 - 250 = 400
40= 50 e-10k
e -10k =
4
---------------------(3).
5
The value of θ when t=20 θ = ce -kt
θ = 50e -k
θ = 50(e-10k )2
4
50
5
2
When t=20 θ= 320 C .
Hence the temperature after 20min = 320 C + 250 C = 57 0 C
When the temperature of the object = 550C
θ = 550 C - 250 C = 300 C
Let t, be the corresponding time from equation (2)
30 = 50e -kt -----------(4)
1
10
From equation (3) e(-k)
4
4 10
= i.e e-k =
5
5
t1
t
4
3
4 10
From equation (4), we get 30 = 50 1 log = log
10
5
5
5
3
log 5
= 22.9min
t1 = 10
log 4
5
4. A body kept in air with temperature 250C cools from 1400C to 800C in 20 min.
Find when the body cools down in 350C.
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Sol : By Newton’s law of cooling
dθ
dθ
= -k(θ - θ 0 )
= -kdt
dt
θ - θ0
log(θ - θ0 ) = kt + c . Here, θ 0 = 250 C
log(θ - 25) = kt + c -------------(1).
When t=0, θ = 1400 C
log (115) =c
c =log (115).
kt = - log (θ - 25) + log 115--------(2)
When t=20, θ = 800 C
log (80-25)= -20k + log 115
20 k =log (115) - log(55) ---------(3)
Divide equation (2) by (3), we get
⟹
kt
log115 - log(θ - 25)
=
20k
log115 - log55
t
log115 - log(θ - 25)
=
20
log115 - log55
When θ = 350 C
t
log115 - log(10)
=
20
log115 - log55
t
log(11.5)
=3.31
=
20
28
log
11
temperature = 20 3.31 = 66.2
The temp will be 350 C after 66.2 min.
5.
The temperature of the body drops from 1000 C to 75 0 C in 10 min. When the
surrounding air is at 20 0 C temperature. What will be its temp after half an
hour. When will the temperature be 25 0 C .
Sol :
dθ
= -k(θ - θ 0 )
dt
log(θ - 20) = -kt + logc
when t=0, θ = 1000 ⟹c=80
when t=10, θ = 750 ⟹ e-10k =
11
.
16
1331
0
when t =30min⟹ θ = 20 + 80
= 46 C
4096
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when θ = 250 C t = 10
log5 - log80
= 74.86min
(log11- log18)
6. If the air is maintained at 150 C and the temperature of the body drops from
700 C to 400 C in 10 minutes. What will be its temperature after 30 minutes?
Sol : If θ be the temperature of the body at time t, then
dθ
= -k(θ -15) , where k is
dt
constant
Integrating,
dθ
θ -15 = -k dt + logc
…(1)
i.e log(θ -15) = -kt + logc i.e , θ -15 = ce-kt
When t 0, 70 C and when t 10, 40 C
70 15 ce 0 c 55
40 15 ce 10 k
and
25
5
= e-10k or e -10k =
…(2)
55
11
Then (1) becomes θ -15 = 55e-kt
When t=30 min, θ = 15 + 55e-30k
0
0
3
5
θ = 15 + 55(e ) = 15 + 55 using (2)
11
625 2441
= 15 +
=
= 20.160 C .
121 121
7. In a pot of boiling water 1000 C is removed from the fire and allowed to cool at
30 0 C room temperature. Two minutes later, the temperature of the water in the pot
-10k 3
is 90 0 C . What will be the temperature of the water after 5 minutes?
Sol : We have θ - 300 C = ce-kt
when t = 0,θ = 100 c = 70
θ - 300 C = 70e -kt
when, t = 2,θ = 900
from (2), 60 = 70e-2k
6
-2k = log = -0.1542
7
k = 0.0771
when t = 5,θ - 300 = 70e-5k
θ = 77.460
…(1)
…(2)
8. The temperature of a cup of coffee is 92 0 C when freshly poured, the room
temperature being 24 0 C . In one min it was cooled to 80 0 C . How long a period must
elapse, before the temperature of the cup becomes
65 0 C .
Sol : By Newton’s Law of cooling,
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dθ
= -k(θ - θ 0 ) ;k>0
dt
θ 0 = 240 C log(θ - 24) = -kt + log c--------------(1).
When t=0; θ = 92 c =68
When t =1; θ = 800 C e-k =
k = log
68
56
56
.
68
When θ = 650 C , t =
65x41
= 0.576min
682
LAW OF NATURAL GROWTH OR DECAY
Statement : Let x(t) or x be the amount of a substance at time ‘ t’ and let the substance be
getting converted chemically . A law of chemical conversion states that the rate of change
of amount x(t) of a chemically changed substance is proportional to the amount of the
substance available at that time
dx
αx
dt
Note: a) In case of Natural growth we take
dx
= kx
dt
(k > 0)
b) In case of Natural decay, we take
dx
= -kx (k>0)
dt
where k is a constant of proportionality
RATE OF DECAY OF RADIO ACTIVE MATERIALS
Statement : The disintegration at any instant is proportional to the amount of material
present in it.
If u is the amount of the material at any time ‘t’ , then
du
= - ku , where k is any
dt
constant (k >0). i.e Law of Natural Decay is applied.
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MATHEMATICS - I
Solved Problems
1. The number N of bacteria in a culture grew at a rate proportional to N . The
value of N was initially 100 and increased to 332 in one hour. What was the value
1
of N after 1 hrs .
2
Sol : The differential equation to be solved is
dN
= kN
dt
dN
= kdt
N
dN
= kdt
N
logN = kt + log c
N = c e -kt ------------(1).
When t= 0sec, N =100 100 = c c =100
When t =3600sec, N =332 332 =100 e3600k
e3600k =
Now when t =
332
100
3
hours = 5400 sec then N=100 e5400k
2
N =100 e
3600k
3
2
3
332 2
N=100
= 605.
100
N = 605.
2. A bacterial culture, growing exponentially, increases from 100 to 400 gms in 10
hrs. How much was present after 3 hrs, from the initial instant?
Sol : Let N be the weight of bacteria culture at any > .
N = ce -kt
Then
…(1)
By data, when t = 0, N = 100g
100 = c
Substituting in (1), we get N = 100e -kt …(2)
When t = 100, N = 400g
from (2), 400 = 100e-10k
4 = e-10k
-10k = log4
1
1
k = - log22 = - log(2)
10
5
When t = 3, N = 100e
…(3)
-3k
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= 100e
1
-3 - log2
5
3
= 100e(log2)
3
5
1
= 100X(2) 5 = 100X8 5 = 100X1.414
= 141.4gms
3.If a radioactive Carbon-14 has a half life of 5750 years, what will remain of one
gram after 3000years?
Sol : Let mass of radioactive Carbon-14 at any time be denoted by
Then it is known that
.
dx
= -kt where k is a constant
dt
x = Ae -kt where A is also a constant.
It is known that at t=0, we have 1gm of Carbon-14
1 = Ae0 A = 1
x = e-kt
However when t=5750 years, we have 1/2gm of Carbon-14.
1
1
= e-k(5750) k =
log2
2
5750
Suppose t=3000years, we have to find x.
-
3000
x = e-kt = e-3000k = e 5750
-
3000
x = (2) 5750
log2
gms
4. If 30% of a radioactive substance disappears in 10 days, how long will it take for
90% of it to disappear?
Sol : The differential equation of the diffusing radioactive material is,
dm
= -km
…(1)
dt
Separating the variables and integrating, we get
m = ce -kt
…(2)
When t = 0 , let m = m1
m1 = c
…(3)
70m
By data, when t = 10, m =
100
70m1
= ce-10k = m1e-10k
100
7
1
7
e-10k = k = - log
10
10
10
1
10
k = log
10
7
…(4)
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Required time at is
10m1
1
= ce-kt = m1e-kt = e-kt
100
10
1
t = log(10)
k
10log(10)
64.5 days.
log10 log 7
II .LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER
Definition: An equation of the form
dn y
d n-1y
d n-2 y
+
P
(x)
+
P
(x)
+…+ Pn (x)y = Q(x)
1
2
dx n
dx n-1
dx n-2
where P1 (x), P2 (x), P3 (x)…Pn (x) and Q(x) (functions of x) are continuous is called a linear
differential equation of order n.
LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
Def:
An
P1 , P2 , P3
equation
of
the
form
dn y
d n-1y
d n-2 y
+
P
+
P
+…+ Pn y = Q(x) where
1
2
dx n
dx n-1
dx n-2
Pn , are real constants and Q(x) is a continuous function of x is called an linear
differential equation of order ‘ n’ with constant coefficients.
Note:
d2
dn
d
2
n
1. Operator D =
; D = 2 ; …………………… D = n
dx
dx
dx
Dy=
2. Operator
d2 y
dn y
dy
; D2y= 2 ;…………………… Dn y= n
dx
dx
dx
1
Q = Qdx i e D-1Q is called the integral of Q.
D
To find the general solution of f(D).y = 0 :
Here f(D) = D n + P1D n-1 + P2 D n-2 +…+ Pn is a polynomial in D.
Now consider the auxiliary equation: f(m) = 0
i.e f(m) = m n + P1m n-1 + P2 m n-2 +…+ Pn 0
where P1 , P2 , P3 …Pn are real constants.
Let the roots of f(m) =0 be m1 , m2 , m3 …mn .
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Depending on the nature of the roots we write the complementary function
as follows:
Consider the following table
S.No
Roots
of
A.E
Complementary function(C.F)
f(m) =0
1.
m1, m2, ..mn are
y c = c1e m1x + c 2e m2 x +…c n e mn x
real and distinct.
2.
m1, m2, ..mn and
two
roots
are
yc = (c1 + c 2 )e m1x + c3e m3x +…c n e mn x
equal i.e., m1,
m2 are equal and
real (i.e repeated
twice) &the rest
are
real
and
different.
3.
4.
m1, m2, ..mn are
real and three
roots are equal
i.e., m1, m2 , m3
are equal and
real (i.e repeated
thrice) &the rest
are real and
different.
Two roots of
A.E
are
complex say +i
-i and rest
are real and
distinct.
5.
If
±i
are
repeated twice
& rest are real
and distinct
6.
If
±i
are
repeated thrice
& rest are real
and distinct
If roots of A.E.
7.
yc = (c1 + c2 x + c3 x 2 )e m1x + c 4e m4 x +…c n e mn x
yc = eαx (c1cosβx + c 2sinβx) + c3e m3x +…c n e mn x
yc = eαx (c1 + c2 x)cosβx + (c3 + c4 x)sinβx + c5em5x +…cn emn x
yc = eαx (c1 + c 2 x + c3 x 2 )cosβx + (c 4 + c5 x + c6 x 2 )sinβx
+c7 e m7 x +…c n e mn x
yc = eαx c1cosh βx + c2sinh βx + c3em3x +.......+ cn e mn x
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MATHEMATICS - I
irrational
say
and rest
are real
distinct.
and
Solved Problems
1. Solve
d3y
dy
-3
+ 2y = 0
3
dx
dx
Sol : Given equation is of the form f(D).y = 0
Where f(D) = (D3 - 3D + 2)y = 0
Now consider the auxiliary equation f (m) = 0
f(m) = (m3 - 3m + 2)y = 0 (m-1) (m-1) (m+2) = 0
m =1, 1, -2
Since m1 and m 2 are equal and m 3 is -2
We have yc = (c1 + c 2 )e x + c3e-2x
2. Solve (D4-2D3-3D2+4D+4)y = 0
Sol : Given f(D) = (D4 -2 D3 - 3 D2 + 4D +4) y = 0 …(1)
Auxiliary equation is f(m)=0
m 4 - 2m3 - 3m 2 + 4m + 4 = 0
…(2)
By inspection m+1 is its factor.
(m +1)(m3 - 3m2 + 4) = 0
…(3)
By inspection m+1 is factor of (m3 - 3m2 + 4) .
(3) is m +1 m +1 m 2 - 4m + 4 = 0
(m +1) 2 (m - 2) 2 = 0
m = -1, -1, 2, 2
Hence general solution of (1) is
y = (c1 + c 2 x)e-x + (c3 + c 4 x)e 2x
3.Solve (D4 +8D2 + 16) y = 0
Sol : Given f(D) = (D4 +8D2 + 16) y = 0
Auxiliary equation f(m) = (m4 +8 m2 + 16) = 0
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(m2 + 4)2 = 0
(m+2i)2 (m+2i)2 = 0
m= 2i ,2i , -2i , -2i
Here roots are complex and repeated
Hence general solution is
yc = [(c1+c2x)cos x + (c3+c4x) sin x)]
4.Solve y11+6y1+9y = 0 ; y(0) = -4 , y1(0) = 14
Sol :
Given equation is y11+6y1+9y = 0
Auxiliary equation f(D) y = 0 (D2 +6D +9) y = 0
A.equation f(m) = 0 (m2 +6m +9) = 0
m = -3 ,-3
yc = (c1+c2x)e-3x -------------------> (1)
Differentiate of (1) w.r.to x y1 =(c1+c2x)(-3e-3x ) + c2(e-3x )
Given y1 (0) =14 c1 = -4 & c2 =2
Hence we get y =(-4 + 2x) (e-3x )
5.Solve 4y111 + 4y11 +y1 = 0
Sol : Given equation is 4y111 + 4y11 +y1 = 0
That is (4D3+4D2+D)y=0
Auxiliary equation f(m) = 0
4m3 +4m2 + m = 0
m(4m2 +4m + 1) = 0
m (2m +1) 2 = 0
m = 0 , -1/2 ,-1/2
y =c1 + (c2 + c3x) e-x/2
6.Solve (D2 - 3D +4) y = 0
Sol : Given equation (D2 - 3D +4) y = 0
A.E. f(m) = 0
m2-3m + 4 = 0
3 ± 9 -16 3 ± i 7
m=
=
2
2
3
7
α ± iβ i
2
2
3
7
7
y = e 2 (c1cos
x + c 2sin
x)
2
2
To Find General solution of f(D) y = Q(x)
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It is given by y = yc + yp
i.e. y = C.F+P.I
Where the P.I consists of no arbitrary constants and P.I of f (D) y = Q(x)
1
Is evaluated as
P.I =
Qx
f(D)
Depending on the type of function of Q(x) , P.I is evaluated .
1.
1 2
(x )
D
Find
Sol :
1 2
x3
(x ) = x 2dx =
D
3
2. Find Particular value of
Sol :
1
(x)
D +1
1
(x) = e-x xe x dx
D +1
(By definition)
= e-x (xe x - e x )
= x-1
General methods of finding Particular integral :
P.I of f(D)y=Q(x), when
1
is expressed as partial fractions.
f(D)
Q. Solve (D2 + a 2 )y = secax
Sol : Given equation is …(1)
Let f(D) = D2 + a 2
The AE is f(m) = 0 i.e m 2 + a 2 = 0
…(2)
The roots are m= -ai , -ai
yc = c1cosax + c2sinax
yp =
1
1 1
1
secax =
secax …(3)
2
D +a
2ai D - ai D + ai
2
1
cosax - isinax
secax = eiax secaxdx = eiax
dx
D - ai
cosax
i
= eiax (1- itanax)dx = eiax x + logcosax …(4)
a
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Similarly we get
1
i
secax = e-iax x - logcosax
D + ai
a
…(5)
From (3),(4) and (5), we get
1 iax
i
i
yp =
e x + logcosax - e-iax x - logcosax
2ai a
a
x(eiax - e-iax ) 1
(eiax + e-iax )
=
+ 2 (logcosax)
2ai
a
2
x
1
= sinax + 2 cosaxlog(cosax)
a
a
The general solution of (1) is
x
1
y = y c + y p = c1cosax + c 2sinax + sinax + 2 cosaxlog(cosax)
a
a
RULES FOR FINDING P.I IN SOME SPECIAL CASES;
Type 1. P.I of f (D)y=Q(x) where Q(x) =eax , where ‘a’ is constant.
1
1 ax
1 ax
.Q(x) =
e =
e
f(D)
f(D)
f(a)
provided f(a) ≠ 0
Case1.P.I =
i.e In f(D), put D=a and Particular integral will be calculated.
Case 2: If f(a)=0 then the above method fails. Then if f D = D - a D (i.e ‘a’ is
k
repeated root k times).
Then P.I =
1 ax 1 k
e . x provided (a) 0
k!
(a)
Type 2. P.I of f(D)y = Q(x) where Q(x) = sinax or Q(x) = cosax where ‘a’ is constant
then P.I=
1
Q(x) .
f(D)
Case 1: In f(D) put D2 = -a 2 f(-a 2 ) 0 then P.I =
sinax
f(-a) 2
Case 2: If f(-a2)=0 then D2 + a2 is a factor of (D 2 ) and hence it is a factor of f(D). Then
let f(D)=(D2 + a2) f(D) = (D2 + a 2 ) (D2 ) .
Then
sinax
sinax
1
sinax
1 -xcosax
= 2 2
=
=
2
2
2
2
f(D) (D + a ) (D ) (-a ) D + a
-a 2 2a
cosax
cosax
1
cosax
1 xsinax
= 2 2
=
=
2
2
2
2
f(D) (D + a ) (D ) (-a ) D + a
-a 2 2a
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Type 3.P.I for f(D)y=Q(x) where Q(x)=xk where k is a positive integer .f(D) can be
expressed as f(D) = [1 (D)]
Express
1
1
[1 (D)]1
f(D) [1 (D)]
Hence P.I =
1
Q(x)
[1 (D)]
= [1 (D)]1 x k
Type 4.P.I of f(D)y=Q(x) when Q(x)=eax V where ‘a’ is a constant and V is function of
x. where V =sinax or cosax or xk
Then P.I =
=
1
Q(x)
f(D)
1 ax
e V
f(D)
1
1
V &
= eax
V is evaluated depending on V.
f(D + a) f(D + a)
Type 5. P.I of f(D)y=Q(x) when Q(x)=xV where V is a function of x.
Then P.I =
=
1
Q(x)
f(D)
1
V
f(D)
1
1
f'(D)
V
= x f(D)
f(D)
Type 6. P.I. of f(D)y = Q(x) where Q(x)= xm v where v is a function of x.
When P.I. =
1
1 m
×Q(x) =
x v, where v = cosax or sinax
f(D)
f(D)
i.
P.I. =
1 m
1 m iax
x sinax = I.P.of
x e
f(D)
f(D)
ii.
P.I. =
1 m
1 m iax
x cosax = R.P.of
x e
f(D)
f(D)
Formulae
1.
= (1 – D)-1 = 1 + D + D2 + D3 + ------------------
2.
= (1 + D)-1 = 1 - D + D2 - D3 + ------------------
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3.
= (1 – D)-2 = 1 + 2D + 3D2 + 4D3 + ------------------
4.
= (1 + D)-2 = 1 - 2D + 3D2 - 4D3 + ------------------
5.
= (1 – D)-3 = 1 + 3D + 6D2 + 10D3 + ------------------
6.
= (1 + D)-3 = 1 - 3D + 6D2 - 10D3 + ------------------
Solved Problems
1. Solve (4D2 - 4D +1)y = 100
1 -1
Sol : A.E is 4m 2 - 4m +1 = 0 (2m -1) 2 = 0 m = ,
2 2
x
C.F = (c1 + c2 x)e 2
Now P.I =
100
100e0x
100
=
=
100 { since 100e0x 100 }
2
2
2
4D - 4D +1 (2D -1)
(0 -1)
x
Hence the general solution is y = C.F + P.F = (c1 + c2 x)e 2 100
2.
Solve the differential equation (D2 + 4)y = sinh2x + 7 .
Sol : Auxillary equation is m 2 + 4 = 0
m 2 = -4 m = ±2i
C.F is yc = c1cos2x + c2sin2x …(1)
To find P.I :
1
yp = 2
(sinh2x + 7)
D +4
1 e 2x + e-2x
= 2
+ 7e0
D +4
2
1 e 2x
1 e-2x
e0
= . 2
+
+
7
2 D + 4 2 D2 + 4
(D 2 + 4)
=
e 2x
e-2x
7
+
+
2(4 + 4) 2(4 + 4) (0 + 4)
e2x + e-2x 7 1
7
+ = sinh2x +
16
4 8
4
y = yc + yp
…(2)
=
1
7
= c1cos2x + c 2sin2x + sinh2x +
8
4
3.
Solve (D + 2)(D -1)2 y = e-2x + 2sinhx
Sol : The given equation is
(D + 2)(D -1)2 y = e-2x + 2sinhx
…(1)
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This is of the form f(D)y = e-2x + 2sinhx
A.E is f(m) = 0 (m + 2)(m -1)2 = 0 m = -2,1,1
The roots are real and one root is repeated twice.
C.F is y c = c1e-2x + (c 2 + c3 x)e x .
e-2x + 2sinhx
e-2x + e x - e-x
=
= yp1 + yp2 + y p3
(D + 2)(D -1) 2 (D + 2)(D -1) 2
e-2x
Now yp1 =
(D + 2)(D -1) 2
Hence f(-2) = 0. Let f(D) = (D -1)2 . Then (2) 0 and m 1
P.I =
e-2x x xe-2x
y p1 =
=
9
9
ex
and yp2 =
. Here f(1)=0
(D + 2)(D -1) 2
e x x 2 x 2e x
=
=
(3)2!
6
-x
e
and yp3 =
(D + 2)(D -1) 2
e-x
e-x
Putting D=-1, we get yp3 =
=
(1)(-2)2
4
The general solution is y = yc + y p1 + y p2 + y p3
i.e y = c1e-2x + (cc + c3 x)e x +
xe-2x x 2 ex e-x
+
9
6
4
4. Solve the differential equation (D2 + D +1)y = sin2x .
Sol : A.E is m 2 + m +1 = 0
-1± 1- 4 -1± 3i
m=
=
2
2
-x
x 3
x 3
+ c 2sin
yc = e 2 c1cos
2
2
…(1)
To find P.I :
sin2x
sin2x
yp = 2
=
D + D +1 -4 + D +1
sin2x (D + 3)sin2x (D + 3)sin2x
=
=
=
D-3
D2 - 9
-4 - 9
Dsin2x + 3sin2x 2cos2x + 3sin2x
=
=
-13
-13
-x
x 3
x 3 1
y = y c + y p = e 2 c1cos
+ c 2sin
- (2cos2x + 3sin2x)
2
2 13
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5.
Solve (D2 - 4)y = 2cos2 x
Sol : Given equation is (D2 - 4)y = 2cos2 x
Let f(D) = D2 - 4 A.E is f(m) = 0 i.e m 2 - 4 = 0
The roots are m=2,-2. The roots are real and different.
C.F = y c = c1e 2x + c 2e -2x
1
1
P.I = y p = 2 (2cos 2 x) = 2 (1+ cos2x)
D -4
D -4
0x
e
cos2x
= 2 + 2
= P.I1 + P.I 2
D -4 D -4
e0x
e0x
1
P.I1 = y p1 = 2
=[Put D=0] =
D -4
-4
4
cos2x cos2x
P.I 2 = y p2 = 2
=
[Put D2 = -22 = -4 ]
D -4
-8
The general solution of (1) is y = y c + y p1 + y p2
…(1)
1 cos2x
y = c1e 2x + c 2e-2x - 4
8
2
6. Solve (D + 1)y = sinxsin2x
Sol : Given D.E is (D2 +1)y = sinxsin2x
A.E is m 2 +1 = 0 m = ±i
The roots are complex conjugate numbers.
C.F is yc = c1cosx + c2sinx
w.k.t 2sinAsinB=cos(A-B)–cos(A+B)
sinxsin2x 1 cosx - cos3x
P.I =
=
= P.I1 + P.I 2
(D2 +1)
2 (D2 +1)
1 cosx
Now P.I1 =
2 D 2 +1
Put D2 = -1 we get D 2 +1 = 0
1 xsinx xsinx
cosax
x
P.I1 =
=
Case of failure : 2
=
sinax
2 2
4
D
+
a
2a
1 cos3x
and P.I 2 = 2 D 2 +1
Put D2 = -9, we get
1 cos3x cos3x
P.I 2 = =
2 -9 +1
16
General solution is
xsinx cos3x
y = y c + y p1 + y p2 = c1cosx + c 2sinx +
+
4
16
3
2
7. Solve the differential equation (D - 3D -10D + 24)y = x + 3 .
Sol : The given D.E is (D3 - 3D2 -10D + 24)y = x + 3
A.E is m3 - 3m 2 -10m + 24 = 0
⟹m=2 is a root.
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The other two roots are given by m 2 - m - 2 = 0
(m - 2)(m +1) = 0
⟹m=2 (or) m = -1
One root is real and repeated, other root is real.
C.F is yc = e 2x (c1 + c 2 x) + c3e-x
x +3
1
yp = 3
=
2
(D - 3D -10D + 24) 24
x3 + 3
D3 - 3D 2 -10D
1+
24
-1
1 1+ D3 - 3D 2 -10D
=
(x + 3)
24
24
=
1 D3 - 3D 2 -10D
1-
(x + 3)
24
24
1
10 24x + 82
x +3+ =
24
24
576
General solution is y = yc + y p
24x + 82
y = e 2x (c1 + c 2 x) + c3e -x +
576
8. Solve the differential equation (D2 - 4D + 4)y = e2x + x2 + sin3x .
=
Sol : The A.E is (m2 - 4m + 4) = 0 (m - 2)2 = 0 m = 2, 2
yc = (c1 + c 2 x)e 2x
…(1)
1
(e 2x + x 2 + sin3x)
To find y p : y p = 2
D - 4D + 4
=
=
e 2x
x2
sin3x
+
+ 2
2
2
(D - 2) (D - 2) D - 4D + 4
x 2 2x
x2
sin3x
e +
+
2
2!
D -9 - 4D + 4
4 1-
2
-2
x2
1 D
(4D - 5)sin3x
= e 2x + 1- x 2 2
4 2
(5 + 4D)
=
x 2 2x 1 2D 3D 2 2 (4D - 5)sin3x
e + 1+
+
x 2
4
2
4
16D 2 - 25
=
x 2 2x x 2 x 3 (12cos3x - 5sin3x)
e + + + 2
4 2 8
-144 - 25
x 2 2x x 2 x 3 (12cos3x - 5sin3x)
e + + + +
…(2)
2
4 2 8
169
x2
x 2 x 3 (12cos3x - 5sin3x)
y = yc + y p = (c1 + c2 x)e 2x + = e 2x + + + +
2
4 2 8
169
2
9. Solve the differential equation (D + 4)y = xsinx .
=
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Sol : Auxiliary equation is m2 + 4 - 0 m2 = (2i) 2
m = ±2i. The roots are complex and conjugate.
Hence Complementary Function, yc = c1cos2x + c2sin2x
1
xsinx
Particular integral, y p = 2
D +4
1
xeix
D +4
1
1
= I.P ofeix
x = I.P of eix 2
x
2
(D + i) + 4
D + 2Di + 3
= I.P of
2
-1
D 2 + 2Di
1+
x
3
ix
2
e D + 2Di
= I.P of
+ ... x
13
3
eix
= I.P of
3
2
2
1- Di x D (x) = 0, etc
3
1
2
= I.P of (cosx + isinx) x - i
3
3
1 2
= - cosx + xsinx
3 3
= I.P of
eix
3
Hence the general solution is
1
2
y = y c + y p = c1cos2x + c 2sin2x + xsinx - cosx
3
3
where c1 and c2 are constants.
1
xsinx
Other Method (using type 5): y p = 2
D +4
2D sinx
= x - 2
2
D + 4 D + 4
xsinx 2(Dsinx)
=
3
3(D 2 + 4)
xsinx 2cosx
=
3
9
Hence the general solution is
1
2
y = y c + y p = c1cos2x + c 2sin2x + xsinx - cosx
3
3
10. Solve the Differential equation(D2+5D+6)y=ex
Sol : Given equation is (D2+5D+6)y=ex
Here Q(x) =e x
Auxiliary equation is f(m)=m2+5m+6=0
m2+3m+2m+6=0
m(m+3)+2(m+3)=0
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m=-2 or m=-3
The roots are real and distinct
C.F=yc= c1e-2x + c 2e-3x
Particular Integral=yp=
=
1
Q(x)
f(D)
1
1
ex =
ex
D + 5D + 6
(D + 2)(D + 3)
2
Put D = 1 in f(D)
P.I=
1
ex
(3)(4)
Particular Integral = yp =
1 x
e
12
General solution is y = yc + y p
y = c1e-2x + c 2e-3x +
ex
12
11. Solve y''- 4y' + 3y = 4e3x , y(0) = -1, y'(0) = 3
Sol : Given equation is y'' - 4y' + 3y = 4e3x
i.e
d 2 y dy
- 4 + 3y = 4e3x it can be expressed as
dx 2 dx
D2 y - 4Dy + 3y = 4e3x
(D2 - 4D + 3)y = 4e3x
Here Q(x) = 4e3x; f(D)= D2-4D+3
Auxiliary equation is f(m)=m2-4m+3 = 0
m2- 3m - m+3=0
m(m-3) -1(m-3)=0⟹m=3 or 1
The roots are real and distinct.
C.F= y c = c1e3x + c 2 e x
P.I= y p =
1
Q(x)
f(D)
=
1
4e3x
D - 4D + 3
=
1
4e3x
(D -1)(D - 3)
2
Put D=3
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4e3x
4 e3x
x'
yp =
=
= 2 e3x = 2xe3x
3 -1 D - 3 2 D - 3 1!
General solution is y = yc+yp
…(3)
y = c1e3x + c 2e x + 2xe3x
Equation (3) differentiating with respect to ‘x’
y' = 3c1e3x + c 2e x + 2e3x + 6xe3x
…(4)
By data, y(0) = -1 , y1(0)=3
From (3),
-1=c1+c2
From (4),
3=3c1+c2+2
3c1+c2=1
…(5)
…(6)
Solving (5) and (6) we get c1=1 and c2 = -2
y=-2ex + (1+2x) e3x
12. Solve y'' + 4y' + 4y = 4cosx + 3sinx, y(0) = 0, y'(0) = 0
Sol : Given differential equation in operator for
(D2 + 4D + 4)y = 4cosx + 3sinx
A.E is m2+4m+4 = 0
(m+2)2=0 then m=-2, -2
C.F is yc = (c1 + c 2 x)e -2x
P.I is= y p =
4cosx + 3sinx
put D2 = -1
(D 2 + 4D + 4)
4cosx + 3sinx (4D - 3)(4cosx + 3sinx)
=
(4D + 3)
(4D - 3)(4D + 3)
(4D - 3)(4cosx + 3sinx)
=
16D 2 - 9
yp =
yp =
=
(4D - 3)(4cosx + 3sinx)
-16 - 9
-16sinx +12cosx -12cosx - 9sinx -25sinx
=
= sinx
-25
-25
General equation is y=yc+yp
y = (c1 + c 2 x)e-2x + sinx
…(1)
By given data y(0) = 0, c1 = 0 and
Differentiating (1) w.r.t ‘x’, y' = (c1 + c 2 x)(-2)e-2x + e-2x (c 2 ) + cosx
…(2)
given y'(0) = 0
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Substitute in (2) -2c1 + c2+1=0
c2 = -1
Required solution is y = -xe-2x + sinx
13. Solve (D2+9)y = cos3x
Sol : Given equation is (D2+9)y = cos3x
A.E is m2+9 = 0
m = ±3i
yc = C.F = c1cos3x + c2sin3x
y p = P.I =
=
cos3x
cos3x
= 2 2
2
D +9 D +3
x
x
sin3x = sin3x
2(3)
6
General equation is y=yc+yp
x
y = c1cos3x + c 2sin3x + sin3x
6
14. Solve y''' + 2y''- y'- 2y = 1- 4x3
Sol : Given equation can be written as
(D3 + 2D2 - D - 2)y = 1- 4x 3
A.E is m3 + 2m 2 - m - 2 = 0
(m2 -1)(m + 2) = 0
m2 = 1 or m = -2
m = 1, -1, -2
C.F= c1e x + c2 e-x + c3e-2x
P.I =
1
(1- 4x 3 )
2
(D + 2D - D - 2)
3
=
-1
(D + 2D 2 - D)
2 1(1- 4x 3 )
2
3
-1
-1 (D3 + 2D 2 - D)
(1- 4x 3 )
= 1
2
2
-1 (D3 + 2D2 - D) (D3 + 2D2 - D) 2 (D3 + 2D2 - D)3
+
+
+… (1- 4x 3 )
= 1+
2
2
4
8
=
-1 1 3
1
1
1+ D + 2D 2 - D + D 2 - 4D3 + -D3 1- 4x 3
2 2
4
8
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=
-1 5 3 5 2 1
1- D + D - D (1- 4x 3 )
2 8
4
2
=
-1
5
5
1
(1- 4x 3 ) - (-24) + (-24x) - (-12x 2 )
2
8
4
2
=
-1
-4x 3 + 6x 2 - 30x +16
2
= [2x 3 - 3x 2 +15x -8]
The general solution is
y= C.F + P.I
y = c1e x + c 2e-x + c3e-2x +[2x 3 - 3x 2 +15x - 8]
15. Solve (D3 - 7D2 +14D - 8)y = excos2x
Sol : Given equation is
(D3 - 7D2 +14D -8)y = e x cos2x
A.E is (m3 - 7m2 +14m -8) = 0
(m-1)(m-2)(m-4)=0
Then m=1,2,4
C.F= c1e x + c 2 e 2x + c3e 4x
P.I=
ex cos2x
(D3 - 7D2 +14D -8)
= ex
1
cos2x
(D +1) - 7(D +1) 2 +14(D +1) -8
3
1 ax
1
e v = eax
v
P.I =
f(D)
f D + a
1
cos2x
(D - 4D2 + 3D)
1
= ex 3
cos2x
(D - 4D2 + 3D)
= ex
3
= ex
1
cos2x (Replacing D2 with -22)
(-4D + 3D +16)
= ex
1
cos2x
(16 - D)
= ex
16 + D
cos2x
(16 - D)(16 + D)
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16 + D
cos2x
256 - D 2
16 + D
= ex
cos2x
256 - (-4) 2
= ex
ex
=
(16cos2x - 2sin2x)
260
2ex
=
8cos2x - sin2x
260
ex
=
8cos2x - sin2x
130
General solution is y = yc + yp
ex
y = c1e x + c 2e2x + c3e 4x +
8cos2x - sin2x
130
16. Solve (D2 - 4D + 4)y = x 2sinx + e 2x + 3
Sol : Given (D2 - 4D + 4)y = x 2sinx + e 2x + 3
A.E is (m2 - 4m + 4) = 0
(m - 2)2 =0 then m=2,2
C.F= (c1 + c 2 x)e 2x
x 2sinx + e2x + 3
1
1
1
P.I=
=
(x 2sinx) +
e2x +
(3)
2
2
2
(D - 2)
(D - 2)
(D - 2)
(D - 2) 2
1
1
Now
(I.P of e ix )
(x 2sinx) =
(x 2 )
2
2
(D - 2)
(D - 2)
1
= I.P of
(x 2 )eix
(D - 2) 2
1
= I.P of (eix )
(x 2 )
2
(D + i - 2)
1
I.P of (eix )
(x 2 )
2
(D + i - 2)
On simplification, we get
1
1
(x 2sinx) =
(220x + 244)cosx + (40x + 33)sinx
2
(D + i - 2)
625
1
x 2 2x
2x
and
e
=
e ,
(D - 2)2
2
1
3
(3) =
2
(D - 2)
4
1
x2
3
P.I=
(220x + 244)cosx + (40x + 33)sinx + e2x +
625
2
4
y=yc+yp
1
x2
3
y = (c1 + c2 x)e 2x +
(220x + 244)cosx + (40x + 33)sinx + e2x +
625
2
4
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Linear equations of second order with variable coefficients
d2 y
dy
+ P x + Q(x)y = R (x), where P(x), Q(x), R(x) are real
An equation of the form
2
dx
dx
valued functions of ‘x’is called linear equation of second order with variable coefficients.
Variation of Parameters :
This method is applied when P,Q in above equation are either functions of ‘x’ or real
constants but R is a function of ‘x’.
Working Rule :
1. Find C.F. Let C.F= y c = c1u(x) + c2 u(x)
2.
Take P.I= y p =Au+Bv where A= -
vRdx
uRdx
and B =
uv' - vu'
uv' - vu'
3. Write the G.S. of the given equation y = yc + y p
1. Apply the method of variation of parameters to solve
d2y
+ y = cosecx
dx 2
Sol : Given equation in the operator form is (D2 +1)y = cosecx
…(1)
A.E is (m2 +1) = 0
m = ±i
The roots are complex conjugate numbers.
C.F is yc = c1cosx + c2sinx
Let yp= A cosx+ B sinx be P.I. of (1)
dv du
u
-v
= cos 2 x + sin 2 x = 1
dx dx
A and B are given by
vRdx
sinxcosecx
= -
dx = - dx = -x
A= -
uv' - vu'
1
uRdx
= cosx.cosecxdx = cotxdx = log(sinx)
B= 1
uv - vu1
y p = -xcosx + sinx.log(sinx)
General solution is y= yc + yp.
y = c1cosx + c2sinx - xcosx + sinx.log(sinx)
2. Solve (D2 - 2D + 2)y = ex tanx by method of variation of parameters.
Sol : A.E is m 2 - 2m + 2 = 0
2 ± 4 - 8 2 ± i2
m =
=
= 1± i
2
2
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We have yc = e x (c1cosx + c 2sinx) = c1e x cosx + c 2e x sinx
= c1 (u) + c2 (u)
where u = e x cosx, v = e xsinx
du
dv
= e x (-sinx) + e x cosx,
= e x cosx + e x sinx
dx
dx
dv du
u
-v
= e x cosx(e x cosx + e x sinx) - e x sinx(e x cosx - e x sinx)
dx dx
= e2x (cos2 x + cosxsinx - sinxcosx + sin 2 x) = e2x
Using variation of parameters,
e x tanx
vR
A = -
= - 2x (e x sinx)dx
dv du
e
u
-v
dx dx
sin 2 x
(1- cos 2 x)
= - tanxsinxdx =
dx =
dx
cosx
cosx
= (secx - cosx)dx = log(secx + tanx) - sinx
uR
dx
dv du
u
-v
dx dx
e x cosx.e x tanx
=
dx = sinxdx = -cosx
e2x
General solution is given by = +
+
x
x
i.e y = c1e cosx + c 2e sinx +[log(secx + tanx) - sinx]e x cosx - e x cosxsinx
B =
or y = c1e x cosx + c 2e x sinx +[log(secx + tanx) - 2sinx]e x cosx
3. Solve the differential equation (D2 + 4) y = sec2x by the method of variation of
parameters.
Sol. Given equation is (D2 + 4) y = sec2x
…..(1)
∴ A.E is m2 +4 = 0 ⇒ m = ± 𝑖
The roots are complex conjugate numbers.
∴ yc = C.F = c1 cos2x + c2 sin2x
Let yp = P.I = A cos2x + B sin2x
Here u = cos2x, v = sin2x and R = sec2x.
∴ = -2 sin2x and = 2 cos2x
= (cos 2x) (2 cos2x) – (sin2x) (-2sin2x)
= 2 cos22x + 2 sin22x = 2(cos22x + sin22x) = 2
A and B are given by :
𝑖
A=-∫
dx = - ∫
= − ∫
=
∴u
⇒A=
B=∫
−
𝑥
|
𝑥
−
−
𝑥
|
𝑥
dx = ∫
|
|
=
∫
|
|
=
∴ yp = P. I =
(cos2x) + (sin2x)
∴ The general solution is given by :
y = yc + yp = C.F. + P.I
|
|
i.e., y = c1 cos2x + c2 sin2x +
(cos2x) + (sin2x)
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UNIT-IV
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PARTIAL DIFFERENTIAL EQUATIONS
Definition:
A Differential equation involves a dependent variable and its derivatives with respect to two
or more independent variables is called partial differential equation.
Ex:
+
=
+
Linear & non linear p.d.e:
If the partial derivatives of
the dependent variable occur in first degree only and
separately, Such a P.D.E is said to the linear P.D.E, otherwise it is said as non –linear P.D.E
Homogeneous & non homogeneous p.d.e:
A P.D.E is said to the Homogeneous if each term of the equation contains either the
dependent variable or one of its derivatives. Otherwise it is said to be Non - Homogeneous
Formation & partial differential equations:
Partial Differential equations can be formed by two methods
1.By the elimination of arbitrary constants
2.By the elimination of arbitrary functions
1.By elimination of arbitrary constants
Let the given function be
constants.
, , , ,
=
……….
To eliminate a and b, differentiating (1) partially w.r.t. ‘ ’
f f z
f f
. 0
. p 0 ……………….(2)
x z x
x z
f f z
f f
. 0 .q 0
y z y
y z
where a and b are arbitrary
‘ ’
and
……………….(3)
Now eliminate the constants a and b from (1), (2) and (3). We get a partial differential
equation of the first order of the form. x, y, z, p, q 0
Note : 1. If the number of arbitrary constants is equal to the number of variables, a partial
differential equation of first order can be obtained.
2.If the number of arbitrary constants is greater than the number of variables, a partial
differential equation of order higher than one can be obtained.
Solved Problems
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1. Form the partial differential equation by eliminating the arbitrary constants
from (i)
Sol: we have
=
=
+
+
+
+
.......
Differentiating (1) partially w.r.t. ‘ ’
‘ ’, we get
z
a p a ……… (2) and z b q b ………..(3)
x
y
Putting the values of a and b from equation (2) and (3) in (1), we get
=
+
+
This is the required partial differential equation
2. Form the partial differential equation by eliminating the arbitrary constants a and b
from (a) z ax by a 2 b 2 (b) z ax by
a
b
b
Sol. (a) we have z ax by a 2 b 2 ....... (1)
Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get
z
z
a p a ……… (2) and
b q b ………..(3)
x
y
Putting the values of a and b from equation (2) and (3) in (1), we get
z px qy p 2 q 2
This is the required partial differential equations
(a) We have z ax by
a
b ……………(1)
b
Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get
z
z
a p a ..... (2) and
b q b ………… (3)
x
y
Putting the values of a and b from equation (2) and (3) in (1), we get
z px qy
p
q
q
This is the required partial differential equation.
3. Form the partial differential equation by eliminating the arbitrary constants from
x a
2
y b z2 r 2
2
(OR )
Find the differential equation of all spheres of fixed radius having their centre on the
xy-plane.
Sol. The equation of sphere of radius r having their centers on xy-plane is
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x a
2
y b z 2 r 2 ……….(1)
2
Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get.
2 x a 2 z.
z
0 x a zp 0 or x a zp (2)
x
And 2 y b 2 z.
z
0 or y b zq 0 or y b zq (3)
2y
Putting the values of (x-a) and (y-b) from (2) and (3) in (1), we get
zp
2
zq z 2 r 2
2
is the required partial differential equation.
4. Form the partial differential equation by eliminating the arbitrary constants a and b
=
from
+
+
=
Sol:-The given equation
Differentiating (1) w.r.t., x
P=
+
= 1.
+
+
−−−−−−−−−−
−−−−−−−−−−
Differentiating (1) w.r.t., x
q=
from (2
= 1.
=
+
+
−−−−−−−−−
=
Substituting in (1) we get
+
= .
This is the required partial differential equations
5. Form the partial differential by eliminating the arbitrary constants from
log az 1 x ay b
Sol. We have log az 1 x ay b ………….(1)
Differentiating (1) partially
. . .‘ ’
‘ ’, we get
z
1
1
ap 1or ap az 1 …………….(2)
.a. 1or
az 1 x
az 1
and
z
1
a. a aq az 1 a ………….(3)
az 1 y
(3) (2), gives
q
a ap q …………..(4)
p
Putting (4) in (2), we get
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q
q
z 1or pq qz p or p q 1 q 2
p
is the required partial differential equation.
1
1
6.Form the differential equation by eliminating a and b from 2z x a 2 y a 2 b
1
1
Sol: We have 2z x a 2 y a 2 b …………..(1)
Differentiating (1) partially w.r.t. ‘ ’
2
z
1
2p
x
2 xa
or
xa
or x a
And 2
‘ ’, we have,
1
4p
xa
1
4p
1
(2)
16 p 2
1
1
1
z
or 2q
or y a
4q
y 2 y a
2 ya
yq
1
………… (3)
16q 2
Adding (2) and (3), we get
x y
1 1
1
2 2
16 p
q
or 16 x y p2q2 p2 q2
is the required partial differential equation.
7.Form the partial differential equation by eliminating the arbitrary constants a and b
from z ax3 by 3
Sol. We have z ax3 by3 (1)
Differentiating (1) partially w.r.t. ‘ ’
‘ ’, we get
p
z
3ax 2 or p 3ax 2 a 2 (2)
3x
x
And
z
q
3by 2 or q 3by 2 b 2 (3)
y
3y
Putting the values of ‘a’ and ‘b’ from (2) and (3) in (1), we get
z
p
q
x y
3
3
Or
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3z px qy
8.Form the partial differential equation by eliminating the arbitrary constants a and b
=
from
+
Sol:-The given equation
+
=
+
Differentiating (1) w.r.t.,
P=
+
=
+
= 2y
−−−−−−−−−−
−−−−−−−−−−
Differentiating (1) w.r.t. , ,we get
q=
+
Substituting in (1) we get
∴
+
=
−−−−−−−−−
∴
=
+
implies that
=
−
=
is the required partial differential equation.
9.Form the partial differential equation by eliminating the arbitrary constants from
−
Sol:Given
−
+
−
=
=
𝜶
−
=
𝜶
−
=
𝜶
𝜶……..(1)
Differentiating (1) w.r.t.,
−
+
Differentiating (1) w.r.t.,
Substituting (2),(3) in (1),we get
𝜶
+
𝜶
=
𝜶
∴The required Partial differential equation is
+
= 𝐚
𝜶
Formation of the partial differential equation by the elimination of arbitrary functions:
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Derive a p.diff.eqn by the elimination of the arbitrary function from u, v 0 where u,
v are functions of x, y and z.
u, v 0 …. (1)
Differentaite partially equation (1) w.r.to. ,
u u z v v z
.
. 0
u x z x v x z x
i.e.,
u
u v
v
p
p 0 ……(2)
u x
z v x
z
and
4
4 v
v
q
q 0 ……(3)
4 y
z v y
z
Eliminating
and
from (2) and (3)
u
v
u u v v u
u v v
p q q
z x z
y z y z y
p
u v u v
u v u v
u v u v
i.e.
p
q
x y y x
z x x z
y z z y
is the P.D.E after the elimination of from u, v 0. Written in a simpler form
u, v
u, v
u, v
p
q
y, z
z, x
x, y
Above equation is generally written as pP+qQ=R where
P
u v u v
u v u v
u v u v
,Q
and R
x y y x
y z z y
z x x z
Solved Problems
1.Form the partial differential equation by eliminating a,b,c from
Sol. Given
x2 y 2 z 2
1
a 2 b2 c2
x2 y 2 z 2
1 ….. (1)
a 2 b2 c2
Differentiating (1) partially w.r.t. ‘x’ and ‘y’.
2x 2z
x z
2 . p 0 or 2 2 . p 0 (2)
2
a c
a c
And
2 y 2z
y z
2 .q 0 or 2 2 .q 0 (3)
2
b c
b c
Since it is not possible to eliminate a, b,c from eqn (1), (2) and (3). We require one more
relation.
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Differentiating (2), partially w.r.t. ‘x’, we get
1 1 p
1 1 2 z 1
z
z
.
p
.
0
or
.z.
.p
a 2 c 2 x
a 2 c 2 x 2 c 2
x
1 1
p2
.
zr
0
a2 c2
c2
(4)
Multiplying (4) by ‘x’ and then subtracting (2) from it, we get
xz
xp 2 z
1
.r 2 2 . p 0 or 2 xzr xp 2 zp 0
2
c
c
c
c
pz xp 2 xzr
is the required partial differential equation.
2.
Form a partial differential equation by eliminating the arbitrary
+
function𝝋
Sol:-Given 𝜑
, −
+
=
, −
This can be written as −
=
= f(
+
--------------(1)
Now we have to eliminate f from (1)
Differentiating (1) w.r.t., x
−
=
−
′
=
′
+
+
--------(2)
Differentiating (2) w.r.t., y
−
=
′
+
--------(3)
Dividing (2) by (3)
py–qx=
−
is the required partial differential equation.
3. Form a partial differential equation by eliminating the arbitrary function
from z f x 2 y 2
Sol. We have z f x 2 y 2 (1)
Put u x 2 y 2 , we have z f u (2)
Differentiating (2) partially w.r.t. ‘x’ and ‘y’,
z
u
f 1 u . f 1 u .2 x
x
x
p f 1 u 2 x (3)
Similarly we get
=−
(u)
(4)
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3 4 , gives
=
∴
−
+
=
is the required partial differential equation.
4. Form the partial differential equation by eliminating the arbitrary functions from
xyz f x 2 y 2 z 2
Sol. We have xyz f x 2 y 2 z 2 (1)
Differentiating (1) partially w.r.t. x and y
z
yz xy. p f 1 x 2 y 2 z 2 . 2 x 2 z.
x
(or) yz xyp f 1 x 2 y 2 z 2 . 2 x 2 zp (2)
And xz xy.q f 1 x 2 y 2 z 2 . 2 y 2 z.q (3)
2 3 , gives
yz xyp 2 x 2 zp
xz xyq 2 y 2 zq
yz xyp y zq xz xyq x zp
y 2 z z 2 yq xy 2 p xyzpq x 2 z x 2 zp x 2 yq xyzpq
x y2 z 2 p y z 2 x2 q x2 y 2 z
is the required partial differential equation.
5. Form the partial differential equation by eliminating the arbitrary functions
=
From
+
Sol: Given equations
+
=
+
Differentiating (1) partially w.r.t. ‘x’
y(xp+z) =
′
+
---------(1)
+
+
+
----------(2)
+
+
+
------------(3)
Differentiating (1) partially w.r.t. ‘x’
x(y q + z ) =
′
Dividing (2) by (3)
Y xp + z
+
+
+
=
+
+ q = x yq + z ( +
−
+
−
−
=
+
−
−
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=
−
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This is the required partial differential equation.
6. Form the partial differential equation by eliminating the arbitrary function
z
from xy yz zx f
x y
z
(1)
x y
Sol. We have xy yz zx f
Differentiating (1) partially w.r.t. ‘x’ and ‘y’, we get
z x y . p z
y y. p z x. p f 1
(2)
2
x y
x y
z x y q z
And x z yq xq f 1
(3)
2
x y x y
Dividing (2) by (3), we get
x y p y z x y p z
x y q x z x y q z
is the required partial differential equation.
7.
Form the partial differential equation by eliminating the arbitrary function
from
z f x e y .g x
Sol. We have z
f x e y .g x
(1)
Differentiating (1) partially w.r.t. ‘x’ and y, we get
z 1
f x e y .g 1 x or p f 1 x e y .g 1 x (2)
x
And
q e y .g x or
z y
e .g x (3)
y
Differentiating (3), partially w.r.t. ‘y’, we get
2 z y
z
e .g x [using (3)]
2
y
y
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2 z z
2 0
y y
t q 0
Which is the required P.D.E.
8.Form a partial differential equation by eliminating the arbitrary function
=
Sol.
We
=
+
have
+
…..
Put u x 2 y 2 , we have z f u (2)
Differentiating (2) partially w.r.t. ‘x’ and ‘y’,
z
u
f 1 u . f 1 u .2 x
x
x
p f 1 u 2 x (3)
And
z
u
f 1 u . f 1 u .2 y
y
y
q f 1 u 2 y (4)
1
p f u .2 x x
1
3 4 , gives
q
f u 2 y y
py qx 0
This is the required partial differential equation.
9.Form a partial differential equation by eliminating the arbitrary
function 𝝋
+
+
,
Sol:Given function can be written as
+
+
+
=
+
+
=
+
Differentiating (1) partially w.r.t. ‘x’ and ‘y’,we get
implies
……………………
+
=
+
+
+
….
+
=
+
+
+
….
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+
+
=
+
+
Which is the required complete solution of given Partial differential equation.
SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS :
Complete integral:
A solution in which the number of arbitrary constants is equal to the number of independent
variables is called complete integral or complete solution of the given equation.
Particular integral :
A solution obtained by giving particular values to the arbitrary constants in the complete
integral is called a particular integral or particular solution.
Singular integral:
Let f x, y, z, p, q 0 (1) be the partial differential equation.
Let x, y, z, a, b 0 (2)
Be the complete integral of (1). Where a and b are arbitrary constants.
Now find
0 (3)
0 (4)
a
b
Eliminate a and b between the equations(2), (3) & (4) When it exists is called the singular
integral of (1).
General integral : In the complete integral (2). Assume that one of the constant is a function
of the other i.e. b=f(a) Then (2), becomes x, y, z, a, f a 0 (5)
Differentiating (5) partially w.r.t. ‘a’, we get
1
. f a 0 (6)
a f
Eliminate ‘a’ between (5) and (6), when it exists is called the general integral or general
solution of (1).
LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER:
A differential equation involving partial derivatives p and q only and no higher order
derivatives is called a first order equation. If p and q occur in the first degree, it is called a
linear partial differential equation of first order; otherwise it is called a non-linear partial
differential equation of the first order.
For example: px qy 2 z is a linear p.d.e of first order and p 2 q 2 1 is non-linear
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Lagrange’s linear equation:
A linear partial differential equation of order one involving a dependent variable z and two
independent variables x and y of the form Pp Qq R
Where P, Q, R are functions of x, y, z is called Lagrange’s linear equation.
Lagrange’s auxiliary equations or Lagrange’s subsidiary equations
dx dy dz
are called Lagrange’s auxiliary equations.
P Q R
The equations
WORKING RULE TO SOLVE LAGRANGE’S LINEAR EQUATION
Step 1: Write down the auxiliary equations
+
dx dy dz
P Q R
=
Step 2 : Solve the auxiliary equations by the method of grouping or the method of multipliers
or both to get two independent solutions u=a and v=b where a,b are arbitrary constants
Step 3: Then Q(u,v) =0 or u=f(v) is the general solution of the equation
To solve
dx
dy
dz
…… (1)
P( x, y, z ) Q( x, y, z ) R( x, y, z )
+
=
(i) Method of grouping : In some problems, it is possible that two of the equations
dy dz
Q R
or dx =
P
dz
R
dx = dy
P
Q
or
are directly solvable to get solutions u(x,y) = constant or v(y,z)=constant
or w(x,z)=constant. These give the complete solutions of (1)
Sometimes one of them, say
dx dy
P Q
may give rise to solution u(x,y)=c1
From this we may express y, as a function of x. Using this in dy dz and integrating we get
Q
,
=
. ℎ
𝑖
=
,
=
𝑖
R
ℎ
2. Method of multipliers: This is base on the following elementary result.
𝑖
If a1 a2 a3 ...... an then each ratio is equal to l1a1 l2 a2 .... ln an
b1
b2
b3
l1b1 l2b2 .... lnbn
bn
Consider dx dy dz
P
Q
R
If possible identity multipliers l, m, n, not necessarily constant, so that each ratio
ldx mdy ndz
lP mQ nR
Where
+
+
Integrating this we get
=
, ,
Similarly we get another solution
Then
=
.
, ,
+
=
We have the complete solution of (1) constituted
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+
=
𝑖ndependent of the earlier one.
=
=
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Linear Partial P.D.E:
Solved Problems
+
=
1.
Solve
Sol.
The given equations can be written as tan xp tan yq tan z (1)
= , we have P tan x, Q tan y, R tan z
+
Comparing with
dx
dy
dz
tan x tan y tan z
The auxiliary equations are
Taking the first two members, we have
dx
dy
tan x tan y
Integrating log sin x log sin y l og c1
or log
sin x
sin x
log c1 or
c1 (2)
sin y
sin y
Taking the last two members, we have
𝑖
Integrating,
or log
=
𝑖
dy
dt
tan y tan z
+
sin y
sin y
log c2 or
c2 (3)
sin z
sin z
From (2) and (3). The general solution of (1) is
c1 , c2 0
sin x sin y
i.e.
,
0
sin y sin z
is the required Complete Solution.
2.
Find the general solution of y 2 zp x 2 zq y 2 x
Sol.
We have y 2 zp x 2 zq y 2 x (1)
Comparing 𝑖 ℎ
+
P y 2 z, Q x 2 z, R y 2 x
= ,
The auxiliary equations are
ℎ
dx
dy
dz
2 2
2
y z x z y x
Taking the first two members, we have
dx
dy
dx dy
2 2 2 or x 2 dx y 2 dy
2
y z x z
y
x
Integrating,
x3
x3 y 3
y 3 c1 or
c1 (2)
3
3
3 3
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Taking the first and last two members, we have
dx
dz
2 or xdx zdz
2
y z y x
Integrating
x2 z 2
x2 z 2
c2 or
c 2 (3)
2
2
2 2
From (2) and (3) The general solution of (1) is
c1 , c2 0 i.e.
x3 y 3 x 2 z 2
, 0
3 3 2 2
is the required Complete Solution.
3.
Solve p x q y z
Sol.
The given equation can be written as
x p yq z (1)
Comparing with Pp+Qq=R, we have
P x,Q y, R z
The auxiliary equations are
dx
dy
dz
x
y
z
From the first two members, we have
dx dy
x
y
Integrating, 2 x 2 y c1 or 2 x 2 y c1 or
From the last two members, we have
x y a (2)
dy
dz
y
z
Integrating, 2 y 2 z c2 or 2 y 2 z c2
or y z b (3)
From (2) and (3). The general solution of (1) is
a, b 0 i.e.,
x y, y z 0
is the required Complete Solution.
4.
Solve x y z p y z x q z x y
Sol. We have x y z p y z x q z x y (1)
Comparing with
+
= , we have
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P x y z , Q y z x , R z x y
The auxiliary equations are
dx
dy
dz
x y z y z x z x y
Using l=1,m=1, n=1 as multipliers, we get
dx
dy
dz
dx dy dz
x y z y z x z x y
0
x y z y z x z x y 0
dx dy dz 0
Integrating, x y z a (2)
1
1
1
Again using l , m , n as multipliers, we get
x
y
z
1
1
1
dx dy dz
x
y
z
Each fraction
=k(say)
0
1
1
1
dx dy dz 0
x
y
z
Integrating, logx+logy+logz=logb. or xyz=b……. (3)
From (2) and (3). The general solution of (1) is
a, b 0 i.e.,
x y z, xyz 0
is the required Complete Solution.
5.
Solve x2 y z p y 2 z x q z 2 x y
Sol.
Given x2 y z p y 2 z x q z 2 x y (1)
Comparing with
+
= , we have
P x2 y z , Q y 2 z x , R z 2 x y
The auxiliary equations are
Using l
dx
dy
dz
2
2
x y z y z x z x y
2
1
1
1
, m 2 , n 2 as multipliers, we get
2
x
y
z
1
1
1
dx 2 dy 2 dz
2
x
y
z
Each fraction
=k(say)
0
1
1
1
dx 2 dy 2 dz 0
2
x
y
z
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1 1 1
Integrating , a
x y z
or
1 1 1
c1 (2)
x y z
1
1
1
Again using l , m , n as multipliers, we get
x
y
z
1
1
1
dx dy dz
x
y
z
Each fraction
=k(say)
0
1
1
1
dx dy dz 0
x
y
z
Integrating log x log y log z log c2
or xyz c2 (3)
From (2) and (3), The general solution of (1) is .
1
x
1
y
1
z
, xyz 0
is the required Complete Solution.
6.
Solve mz ny p nx lz q ly mx
Sol.
Given eqn is mz ny p nx lz q ly mx (1)
Comparing with
=
−
,
=
+
−
= , we have
,
The auxiliary equations are
=
−
dx
dy
dz
mz ny nx lz ly mx
Using l=x, m=y, n=z as multipliers, we get
Each fraction
xdx ydy zdz
0
xdx ydy zdz 0
Integrating,
x2 y 2 z 2
a or x 2 y 2 z 2 c1 (2)
2
2 2
Again using ,
Each fraction
, as multipliers, we get
ldx mdy nz
=k(say)
0
ldx+mdy+ndz=0
Integrating,
lx my nz c2
(3)
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From (2) and (3), the general solution of (1) is
x 2 y 2 z 2 , lx my nz 0
is the required Complete Solution.
7.
Solve xp yq y 2 x 2
Sol.
Here P x, Q y, R y 2 x 2
The auxiliary eqn’s are
dx dy
dz
2
x y y x2
From the first two members,
dx dy
x y
Integrating, log x log y log c1 or xy c1 (1)
Using l=x, m=y, n=1 as multipliers, we get
Each fraction
xdx ydy dz
0
xdx ydy dz 0
Integrating,
1 2 1 2
x y z c or x 2 y 2 2 z c2 (2)
2
2
From (1) and (2), The general solution is
xy, x 2 y 2 2 z 0
is the required Complete Solution.
8. Find the integral surface of x y 2 z p y x 2 z q x 2 y 2 z
Which contains the straight line x+y=0, z=1
Sol. Given that x y 2 z p y x 2 z q x 2 y 2 z ……..(1)
Comparing with
+
= , we have
P x y2 z , Q y x2 z , R x2 y 2 z
The auxiliary equations are
dx
dy
dz
2
2
2
x y z y x z x y2 z
1
1
1
Using l , m , n as multipliers, we get
x
y
z
1
1
1
dx dy dz
y
z
Each fraction = x
0
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1
1
1
dx dy dz 0
x
y
z
Integrating, log x log y log z log a
or xyz a (2)
Again using l=x, m=y, n=-1 as multipliers, we get
Each fraction
xdx ydy dz
=k(say)
0
xdx ydy dz 0
Integrating,
x2 y 2
z c or x 2 y 2 2 z b (3)
2
2
Given that z=1, using this (2) and (3), we get
xy=a and x 2 y 2 2 b
Now b+2a= x 2 y 2 2 2 xy x y 2 =0-2 x y 0 =-2
2
2a b 2 0
Hence the required surface is
x 2 y 2 2 z 2 xyz 2 0
is the required Complete Solution.
9.Solve
Sol:Given
+
=
+
=
is a Lagrange’s linear equation
The Auxillary equations are
=
By Consider first group, we get
= …..(1)
log
=
∫
=∫
∫
=∫
= log + log
By Consider second group, we get
log
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= …..(2)
∴
,
=
is the required solution.
−
10. Solve
−
+
−
Sol: The auxiliary equations are
−
−
=
=
−
−
−
=
−
−
Taking 1,-1-1 multipliers,we get
(
−
−
−
−
+
−
+
−
=
) (
+
−
−
)
−
Integrating,we get
−
−
=
Taking , − ,
………(1)
=
∴
11. Solve
=
+
+
as multipliers,we get
−
−
−
…………………
−
−
Sol: The auxiliary equations are
−
−
−
−
−
𝐥
)
−
=
−
+
=
−
DEPARTMENT OF HUMANITIES & SCIENCES
−
=𝐥
+
=
=
−
𝝋
+
−
=
−
−
− ,
+
=
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Taking , , , multipliers, we get
(
−
−
+
−
+
+
+
=
)
+
Taking − ,
+
-
(
−
+
)
+
+
=
=
………………….
=
…….(2)
, multipliers, we get
+
=
Integrating, we get
From (1),(2),
∴
−
12.Solve
𝝋
=
,
+
+
=
−
Sol:Comparing with Pp+Qq=R, we have
The auxiliary equations are
=−
=
−
From the first two members, we have Type equation here.
=
Integrating,we get
+
−
=
…..(2)
From the last two members, we have
−
−
=
………..(3)
−
=
=
−
−
−
=
From (2) and (3). The general solution of (1) is
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i.e., ∅
12. Solve
−
+
,
+
=0
+
+
=
+
Sol:Comparing with Pp+Qq=R, we have
The auxiliary equations are
=
+
=
+
+
Taking 1,1,1 and 1,-1,0 and 0,1,-1 as multipliers , we have
+
+
+ +
=
−
−
=
−
−
From the last two members, we have
−
−
Integrating,we get
log
−
−
−
= log
…..(1)
=
−
−
−
=
From the first two members, we have
𝑰
log
+
+
−
+
+
+ +
+
+
=
,
= log
−
−
−
+ log
………..(2)
=
From (2) and (1). The general solution of given pde is
i.e., ∅
13. Solve
−
−
,
−
+
Sol:Comparing with
+
=
+
−
−
=0
= , we have
The auxiliary equations are
=−
=
−
From the first two members, we have
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=
Integrating,we get
−
…..(1)
+ =
Taking 1,1,0 as multipliers,we get
+
=
−
+
+
−
+
+
Integrating,we get
+
=
−
−
=
=
….
From (2) and (1). The general solution is
+
i.e., ∅
, +
−
14. Solve
=0
+
Sol: The auxiliary equations are
(
)
−
=(
)
−
=
−
=
−
−
Taking 1,-1,0 and 0,-1,-1 as multipliers,we get
(
−
−
∴(
−
−
−
−
−
−
and also
)
−
=
+ +
=
)
−
−
(−
−
+
+
−
+
−
+
−
+
−
−
)
)
+ +
solving it,we get
−
(−
= …..(1)
𝑖
+
+
+ +
+
−
, ,
+
+
=
+
+
+
−
, ,
+
−
+
−
−
+
+
multipliers,we get
+
−
=
+
+
−
+
+
+
−
+
+
+
−
+
+
−
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=
=
+
+
+
+
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Integrating,we get
+
+
+
∴
+
…..(2)
=
+
+
∴
+
=
=
𝑖
+
+
+
+
𝑖
+
𝑖 𝜑
+
+
,
−
−
=
NON-LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER
A partial differential equation which involves first order partial derivatives
𝑖 called a non-linear partial differential
degree higher than one and the products
equations.
CHARPIT’S METHOD
, , , ,
In this method give D.E of the form
𝜑
, , , ,
=
which is compatible with ℎ
=
these values in the relation
+
=
.
to find another relation of the form
, , , ,
−
Solved Problems
1.Solve
Sol: Given
+
=
=
+
=
=
−
−
−
−
−
=
=
, , , ,
Which on integration gives the required solution of
Charpit’s equation :
with
[
+
]
[
+
]
=
the we solve for , and substitute
=
[
+
]
[
+
]
−
The auxillary equations are
Here
= ;
−
= ;
=
=
−
;
=
=
Substituting them in above ,we get
−
=
−
By considering first and last groups,
;
=
=
−
=
=
+
=
−
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=
=
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Integrating,we get log
Integrating,we get log
= log + log
∴
=
= log + log
Substitute (1),(2) in given pde ,we get
∴
=
But we have
=
=
=
…..
…..
+
;
+
=
=
+
+
+
+
Integrating it we get required solution as
2.Solve
Sol: Given
=
=−
Here
=
=
−
−
;
=
−
=−
;
=
−
=−
Substituting them in above ,we get
=
Considering
=−
+
Taking
,
−
=−
and
𝑖 𝑖
+
+
Solving ,
+
=
−
+
+
=
;
=
[
=−
−
+
+
+
;
]
=
=
=
[
−
+
]
+
+
𝑖
=
−
+
,
, ℎ
𝑖
ℎ
+
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=
Integrating,we get
log
= log
=
Substituting in given pde,we get
∴
=
√
;
But we have
+log
=
= √
=
=
∫
Integrating,we get,
+
+
√
= ∫ √
√
+∫
∴ Required solution is
=
√
√
√
Now let us start solving some standard forms of first order partial differential equations by
using Charpit’s method
STANDARD FORM I:
Equation of the form f(p,q)=0
i.e., equations containing p and q only.
Given partial differential equation is
The auxillary equations are
Here
=
;
=
−
;
=
−
=
=
−
,
=
−
……
=
[
+
]
=
[
+
]
Substituting above and considering last group, we get
−
∴
=
−
= ,𝑖
=
−
−
𝑖
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=
=
=
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=
Put
,
But we have
=
+
=
,
+
.
……(2)
+ 𝝋
=
𝝋
Integrating (2),we get required complete solution of (1) is
z ax a y c
Which contains two arbitrary constants a and c.
PROCEDURE:
Given partial differential equation is
=
STEP1:Put
value.
:
.
,
=
𝑖
, ℎ
𝑖
+
,
=
……
𝑖
..then we can obtain ‘p’
+
=
STEP3:Integrating it ,we get required complete solution of (1) .
Solved Problems
= , where k is a constant.
1.Solve
=
….
Put p=a in (1),we get
=
Sol. Given that
Since (1) is of the form
But we have
=
=
,
=
+
+
Integrating,we get ,
z ax
k
yc
a
which contains two arbitrary constants a and c.
2.Solve
+
Sol : Given that
=
+
=
… … … … . . 1)
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,
Since (1) is of the form
Put p=a in (1),then we get
But we have
+
=
=
+ [ ±√
Integrating,we get ,
z ax
=
=
= [ ±√
− ]
− ]
+ [ ±√
∫
a
n n2 4 y c
2
− ]∫
This is the complete integral of (1), which contains. two arbitrary constants a and c.
+
3.Find the complete integral of
+
Sol. Given that
=
,
Since (1) is of the form
=
Put
𝑖
,
But we have
……….
=
=√
…….(2)
+
=
=
−
Put the values of p,q in (2), we get
z ax
m2 a 2 y c
Which is the complete integral of (1)
STANDARD FORM II :
, ,
Equation of the form
= (i.e., not containing x and y)
PROCEDURE
, ,
Given partial differential equation is
STEP1:Put
.
:
=
,
+
𝑖
=
=
=
𝑖
, ℎ
+
……
𝑖
, .then
STEP3:Integrating it ,we get required complete solution of (1) .
Solved Problems :
Solve the following partial differential equations
1.
=
+
2.
+
=
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3.
=
+
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Sol.
=
1. We have
Since (1) is of the form
=
Put
𝑖
, ℎ
+
, ,
=
+
√ +
Integrating ,we get
∴ √ =
√
=
=√
∴
𝑖
Putting the values
√
…….
= √
+
=
,
∫
√
+
This is the required solution of (1)
=
+
Since (1) is of the form
=
Put
𝑖
, ℎ
, ,
=
, we get
∫
√ +
=
+
Integrating,we get
=
+
+
=
∴
Putting the values of p and q in
+
+
=
+
+
∫
+
=
,we get
∴ a tan−
which is the required complete solution of (1)
Given that
3.
=
Since (1) is of the form
Put
+
Given that p 2 z 2 q 2 p 2 q (1)
2.
∫
+
=
𝑖
, ℎ
+
+
, ,
=
Putting the values
=
+ …..(1)
,
𝑖
+
=
=
+
+
=
∴
=
+
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+
, we get
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Integrating ,we get
∴
+
=
∫
+
+
=
+
∫
+
This is the required solution of (1)
STANDARD FORM III :
Equation of the form f1 x, p f 2 y, q i.e. Equations not involving z and the terms
containing x and p can be separated from those containing y and q.
We assume that these two functions should be equal to a constant say k.
f1 x, p f 2 y, q k
Solve for p and q from the resulting equations
f1 x, p k and f 2 y, q k
Solve for p and q, we obtain
p F1 x, k and q F2 y, k
Since z is a function of x and y
dz
z
z
dx dy [By total differentiation]
x
y
=
+
dz F1 x, k dx F2 y, k dy
Integrating on both sides
z F1 x, k dx F2 y, k dy c
Which is the complete solution of given equation
Solved Problems:
1. Solve p 2 q 2 x y
Sol :. Given that p 2 q 2 x y ………..(1)
Separating
p 2 x q 2 y
, the given equation can be written as
Let p 2 x q 2 y k (constant)
p 2 x k and q 2 y k
p 2 k x and q 2 y k
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p k x and q y k
Since dz
z
z
dx dy pdx qdy
x
y
dz k x dx y k dy
Integrating on both sides
1
1
z k x 2 dx y k 2 dy c
z
3
3
2
2
k x2 y k 2 c
3
3
Which is the complet solution of (1)
2. Solve xp yq y 2 x 2
Sol: Given that xp yq y 2 x 2 (1)
Separating
and
from
and . The given equation can be written as.
xp x 2 yq y 2
Let xp x 2 yq y 2 k (arbitrary constant)
xp x 2 k and yq y 2 k
p
k x2
k y2
and q
x
y
We have dz
z
z
dx dy pdx qdy
x
y
k
k
dz x dx y dy
x
y
Integrating on both sides
k
k
z x dx y dy c
x
y
k log x
x2
y2
k log y c
2
2
z k log xy
1 2
x y2 c
2
Which is the complete integral of (1)
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2
2
p
q
3. Solve x y 1
2
2
, the given equation can be written as.
Sol: Separating
2
p
q
x 1 y
2
2
2
2
2
p
q
Let x 1 y k 2 (arbitrary constant)
2
2
2
2
p
q
x k 2 and 1 y k 2
2
2
2
p
q
q
x k and y 1 k 2 or y 1 k 2
2
2
2
p 2 k x and q 2 1 k 2 y
dz
We have
z
z
dx dy pdx qdy
x
y
dz 2(k x)dx 2 1 k 2 y dy
Integrating on both sides
z 2 (k x)dx 2 1 k 2 y dy c
z 2(kx
x2
y2
) 2 1 k 2 y c
2
2
z 2kx x 2 2
1 k 2 y y2 c
This is the complete solution of (1)
−
4.Solve
Sol:Let
Then
−
∴
=
−
=
=
=
+
+
+
and
=
+
=
=
+
But we have
dz
z
z
dx dy pdx qdy
x
y
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Integrating,we get
=
+
+
+
+c
is the required complete solution.
−
5. Solve
=
Sol: Let
=
Then
−
−
+
=
- =
=√ +
But
dz
z
z
dx dy pdx qdy
x
y
Integrating,we get
=
+
+
+
+C
is the required complete solution.
6.Solve
=
=
Sol:Let
+
Then we get
+
=
+
Solving,we get
But
=
dz
− ±√
+
−
=
=
=
and
z
z
dx dy pdx qdy
x
y
Integrating,we get
=−
+ [ √
+
is the required complete solution.
STANDARD FORM IV:
=
−
+
+
+
√
)+k +C
,
An equation analogous to the clairaut’s equation it is complete solution is
which is obtained by writing
=
+
+
,
The differential equation which satisfies some
specified conditions known as the boundary conditions.The differential equation together with these
boundary conditions, constitute a boundary value problem
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Solved Problems:
1.
=
Solve
+
+
Sol : The given PDE is form IV
Therefore complete solution is given by
2.
=
+
+
Find the solution of p q z px qy 1
Sol. The given equation can be written as
z px qy
1
pq
z px qy
1
(1)
pq
Hence the complete solution of (1) is given by
z ax by
3.
1
ab
Solve pqz p 2 qx p 2 q 2 py q 2
Sol. The given equation can be written as
p2 2
q2
pqz p 2 q x
q
p
y
q
p
p2
q2
z p x
q
y
q
p
p3 q3
z px qy
(1)
p
q
Since it is in the form z px qy f p, q
Hence the complete solution of (1) is given by
z ax by
a 3 b3
b a
4. Solve =
+
+
+
Sol. We have =
+
+
+ …………….
Since (1) is of the form =
+
+
, .
Hence the complete solution of (1) is given by
=
+
+
+ …….
For singular solution, differentiating (2) partially w.r.t. a and b, we get
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= ,
Implies that
= ,
= + ….
= + +
… … … … 4)
Eliminating a, b between (2), (3) and (4), we get
=
− −
∴ =
−
−
is the singular solution
EQUATIONS REDUCIBLE TO STANDARD FORMS:
+
EQUATIONS OF THE FORM
,
=
The above form of the equation of the type can be transformed to an equation of the form f(p,q)=0
By substitutions given below.
Case (i):- when ≠
≠
Put
−
=
=
=
−
=
𝑖
−
ℎ
=
=
Now the given equation reduces to [
Case(ii):- when = , =
Put =
=
ℎ
p=
=
𝑖
=
=
−
=
−
−
]=
=
,
= = ℎ
=
𝑖 𝑖
−
−
=
now the given equation reduces to the form
,
=
EQUATIONS OF THE FORM
,
, =
This can be reduced to an equation of the form
, ,
,
, = as above.
−
ℎ
ℎ𝑖 ℎ 𝑖
ℎ
=
ℎ
=
=
ℎ
→
=
,
=
:
by the substitutions given for the equation
Solved Problems:
1.
Solve the partial differential equation
Sol.
Given equation can be written as
x2 y 2
z
p q
x 2 p 1 y 2 q 1 z or x 2 p y 2 q z (1)
1
1
This is of the form f x m p, y n q, z 0 with
=− ,
Put X x1m x1 2 x3 and Y y1n y1 2 y 3
=− .
z z X
z
.
P.3x 2 where P
x X x
X
2
x p 3P
Then p
and q
z
z z Y
Q.3 y 2 where Q
.
Y
y Y y
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=
−
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y 2 q 3Q
Now equation (1), becomes.
3P
1
3Q z (2)
1
Since (2) is of the form f P, Q, z 0
Put
=
𝑖
, ℎ
Putting the values of Pand Q in
=
+
+
+
=
∴
=
=
+
+
,we get
Integrating,we get
∫
∫
=
+
+∫
a 1 3
3
3z 2 2
x ay c1
a
=
,taking
=
+
+
+
+
Which is the required solution of (1)
2.
Solve the partial differential equation
p q
z
x2 y 2
Sol. The given equation can be written as
px 2 qy 2 z (1)
Since (1) is of the form f x m p, y n q, z 0 With
Put X x1m x3 , and Y y1n y 3
z z X
z
.
P.3x 2 where P
x X x
X
2
x p 3P
=− ,
=−
Now p
and q
z
z z Y
Q3 y 2 where Q
.
Y
y Y y
y 2 q 3Q
Equation (1) becomes, 3P 3Q z (2)
Since (2) is of the form f P, Q, z 0
Put
=
𝑖
, ℎ
Putting the values of Pand Q in
=
=
∴
+
=
+
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+
,we get
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MATHEMATICS - I
=
+
Integrating,we get
∫
=
∫
+
log z
+
+∫
log =
+
1
x3 ay 3 c
3 1 a
+
+
This is the complete solution of (1)
3.
Solve q2 y 2 z z px
Sol.
Given equation can be written as
q 2 y 2 z 2 zpx or xp z qy z 2 (1)
2
Since (1) is of the form f x m p, y n q, z 0 with
Put X log x and Y log y
Now p
z z X
1
z
.
P. where P
x X x
x
X
=
=
xp P
z
1
z z Y
.
Q. where Q
Y
y
y Y y
qy Q
Equation (1), becomes, Pz Q 2 z 2 (2)
and q
Since (2) is of the form f P, Q, z 0
=
Put
𝑖
, ℎ
Putting the values of Pand Q in
= [−a ± √a + ]
Integrating,we get
∫
=
log =
[−a ± √a + ]
∴ log =
[−a ± √a + ]
∫
[−a ± √a + ]
+
+
∴
= [−a ± √a + ]
=
[−a ± √a + ]
=
+
,we get
+∫
+
+
+
is the complete integeral of (1)
4.
Solve the partial differential equation p 2 x 4 y 2 zq 2 z 2
Sol.
Given that p 2 x 4 y 2 zq 2 z 2
Then given equation can be written as
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px qy z 2z
2 2
2
2
(1)
Since (1) is of the form f x m p, y n q, z 0 with m=2 and n=2
1
1
and Y y 1
x
x
z
z z X
1
P. 2 , where P
.
Now P
X
x X x
x
Put X x1 m x1 2 x 1
x2 p P
and q
1
z
z z Y
.
Q. 2 , where Q
Y
y Y y
y
y 2 q Q
Now equation (1) becomes, P 2 Qz 2 z 2 or P 2 Qz 2 z 2 (2)
Since (2) is of the form
=
Put
𝑖
, ℎ
, ,
=
Putting the values of Pand Q in
[ ±√ a + ]
=
Integrating,we get
∫
=
log =
∴ log =
[ ±√ a + ]
[ ±√ a + ]
∫
[ ±√ a + ]
+
∴
+
=
=
=
[ ±√ a + ]
[ ±√ a + ]
+
,we get
+∫
+
+
Which is the complete integral of (1).
5. Solve x 2 p 2 xpq z 2
+
Sol. The given equation can be written as
xp xp q z 2 (1)
2
Since (1) is of the form f x m p. y n q, z 0 with m=1 and n=0
Put X log x
Now P
1
z z X
.
P , where
x
x X x
z
X
xp P
P
Equation (1) becomes, P 2 Pq z 2 (2)
Since (2) is of the form
Put =
𝑖
,
, ,
=
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=
,
+
√
P=a
√
But we have
=
+
Substuting P,q ,we get
=
+
+
√
+
Integrating on both sides
∫
+
√
log =
+
/ =
+
∫
+
√
+
be the complete integral of (1)
6.Solve z p 2 x q 2 y
Given that z p 2 x q 2 y
Sol.
The given equation can be written as
p x q y
2
2
2
2
1 1
z or px 2 qy 2 z (1)
This is of the form
1
Put X x1m x
Now p
1
2
1
1
x 2 and Y y
1
2
1
y2
1 1
z z X
z
.
P x 2 , where P
x X x
X
2
1 21
z z Y
z
and q
Q y , whereQ
.
y Y x
Y
2
1
px 2
1
P
Q
and qy 2
2
2
2
2
P Q
Then equation (1) becomes, z i.e.P 2 Q 2 4 z (2)
2 2
This if of the form f P, Q, z 0
Put
=
=√
𝑖
+
,
, P=a√
+
+
=
,
But we have
=
+
Substuting P,Q ,we get
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= √
+
+
√
Integrating on both sides
∫
+
√
+
√ =
=
/√ =
+
√
+
√
+
+ √ =( √ + √ ) +
√
Which is the complete integral of (1)
7.Solve
+
=
Sol: Given
+
= ……..
+
Since (1) is of the form f x p, y q, z 0 with
m
n
Put X log x and Y log y
Now p
1
z z X
z
.
P. where P
x
x X x
X
+
∫
=
+ ∫
=
=
xp P
z
1
z z Y
.
Q. where Q
Y
y
y Y y
qy Q
Equation (1), becomes
and q
+
=
……(2)
=
But we have
=
+
Substuting P,Q ,we get
=
√
𝑖
+
+
,
=
Integrating on both sides
∫
√
√
+
+
log =
log =
+
/ =
+
log + log
is the Complete solution of (1)
=
√
+
√
+
√
+
;
=
√
+
+
∫
+ ∫
+
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+
8. Solve
Sol: Given
+
=
=
……..
Since (1) is of the form
,
Put X log x and Y log y
Now p
=
+
with
z z X
1
z
.
P. where P
x X x
x
X
=
=
=
xp P
and q
qy Q
+
z
z z Y
1
Q. where Q
.
Y
y Y y
y
Equation (1), becomes
= …… 2
=
But we have
=
+
Substuting P,Q ,we get
+√
=
𝑖
,
=√ −
−
Integrating on both sides
∫
z =(
+√ −
z=( log + √ −
=
∫
+√ −
∫
)+
log ) +
is the Complete solution of (1)
EQUATIONS OF THE FORM
,
=
:
Use the following substitution to reduce the above form to an equation of the form f(P,Q)=0
+
𝑖
≠−
={
log ,
𝑖
=−
EQUATIONS OF THE FORM
,
=
,
:
An equation of the above form can be reduced to an equation of the form f(P,Q)=0
by the substitutions given for the equation
,
= as above
Solved Problems :
1. Solve z 2 p 2 q 2 x 2 y 2
Sol. Given that z 2 p 2 q 2 x 2 y 2
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The given equation can be written as
z 2 p 2 z 2 q 2 x 2 y 2 or z 2 p 2 x 2 y 2 z 2 q 2
Or zp x 2 y 2 zq (1)
2
2
Since (1) is the of the form f x, pz n g y, qz n . with n=1
put Z z n1 z11 z 2
Z
z
z
2 z. P 2 zp where P
x
x
x
P
pz
2
Q
Z
z
z
2 z. Q 2 zq where Q qz
and
2
y
y
y
Then
P2
Q2
x2 y 2
4
4
2
2
2
2
i.e., P 4 x 4 y Q (2)
Equation (1) becomes,
This is of the form f1 x, P f 2 y, Q
Let P 2 4 x 2 4 y 2 Q 2 4k 2 (say)
P 2 4 x 2 4k 2 and 4 y 2 Q2 4k 2
P 2 4 x 2 4k 2 and Q2 4 y 2 4k 2
P 2 x 2 k 2 and Q 2 y 2 k 2
Z
Z
.dx
dy
x
y
Pdx Qdy [By total differentiation]
We have dZ
dZ 2 x 2 k 2 dx 2 y 2 k 2 dy
Integrating on both sides
Z 2 x 2 k 2 dx 2 y 2 k 2 dy
x 2
x
k2
k2
x
y
x k 2 sinh 1 2
y 2 k 2 cosh 1 c
2
2
2
k
k
2
2
x
y
x x 2 k 2 k 2 sinh 1 x y 2 k 2 k 2 cosh 1 c
k
k
x
y
or z 2 x x 2 k 2 y y 2 k 2 k 2 sinh 1 cosh 1 c
k
k
x x2 k 2
or z x x k y y k k log
y y2 k 2
2
2
2
2
2
2
c
This is the complete solution of (1)
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2. Solve the partial differential equation. p 2 z 2 sin 2 x q 2 z 2 cos2 y 1
Sol. Given that p 2 z 2 sin 2 x q 2 z 2 cos2 y 1
The given equation can be written as
pz
2
sin 2 x qz cos2 y 1or pz sin 2 x 1 qz cos 2 y (1)
2
2
Since (1) is of the form f x, pz
n
2
g y, qz with n=1.
n
Put Z z n1 z 2
P
Z
z
2 z. P 2 zp or pz where
Now
2
x
x
Z
z
Q
2 z. Q 2 zq or qz
and
y
y
2
=
2
;
=
2
P
Q
Then equation (1) becomes, sin 2 x 1 cos 2 y
2
2
2
2
P
Q
sin 2 x 1
cos 2 y (2)
4
4
This is of the form f1 x, p f 2 y, q
i.e.
2
2
P
Q
sin 2 x 1
cos 2 y k 2 (constant)
Let
4
4
2
2
P
Q
cos 2 y k 2
sin 2 x k 2 and 1
4
4
P 2 sin 2 x 4k 2 and Q 2 cos 2 y 4 1 k 2
P
2k
2 1 k 2
and Q
sin x
cos y
We have dZ
Z
Z
dx
dy [By total differential]
x
y
dZ Pdx Qdy
dZ
2k
2 1 k 2
dx
dy
sin x
cos y
Integrating on both sides
=
∫ csc
+ √ −
2k log cos ecx cot x 2 1 k 2 log sec y tan y c
∫ sec
z 2 2k log cos ecx cot x 2 1 k 2 log sec y tan y c
This is the required complete solution of (1)
3.Solve +
+ +
=
Sol:Given +
+ +
= …….(1)
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𝑖
Put
𝑖
+
=
ℎ
=
,
Differentiating partially w.r.t ‘x’,we get
But
=
=
=
implies
Substitute in (1),we get
+
+
Separating
+
=
=
+
+
=
+
=1
−
+
=
, ,
=
=
; 𝑖 𝑖
𝑖
=
𝑖
ℎ
=
=
, the given equation can be written as.
=
−
=
+
Implies that
=
=
√
−
)−
−
dZ
We have
Z
Z
dx
dy
x
y
dz 2(k x)dx 2 1 k 2 y dy
Integrating on both sides
z 2 (k x)dx 2 1 k 2 y dy c
x2
y2
2
z 2(kx ) 2 1 k y c
2
2
z 2kx x 2 2
1 k 2 y y2 c
This is the complete solution of (1).
4. Solve ( −
Sol: Given
−
Put
=
+
)=
=
=
−
− ….. 2
𝑖
(
𝑖
Differentiating partially w.r.t ‘x’,we get
ℎ
−
)=
− …………
,
, ,
=
=
=
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But
=
=
=Pimplies
𝑖
; 𝑖 𝑖
𝑖
ℎ
=
−
=
=
Substitute in (2),we get
−
, the given equation can be written as.
Separating
−
Solving, we get
= √ +
We have dZ
and
= √ +
=− +
=
Z
Z
dx
dy [By total differential]
x
y
dZ Pdx Qdy
= [√ +
+√ +
Integrating on both sides
]
=
[∫ √ +
=
+∫√ +
+
+
This is the required complete solution of (1)
+
+
Methods Of Separation Of Variables:
This method is used to reduce one partial differential equation to two or more ordinary
differential equations,each one involving one of the independent variables.This will be done
by separating these variables from the beginning. This method is explained through following
examples.
=
1. Solve by the method of separation of variables
Solution:- Given equation is
=
+ ----------(1)
+
where U(x,0)=6
−
Let U(x,t)= X(x) T(t) =XT ---------------(2)
be a solution of (1)
Differentiating (2) partially w.r.t x ant t
=
′
,
Put these values in equation (1), we have
′
′
=
+
Dividing by XT
′
=
′
=
′
+ --------(3)
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Since L.H.S is a function of ‘x’ and the R.H.S is a function of ‘t’ where x and t are
independent variables, the two sides of (3) can be equal to each other for all values of ‘x’ and
‘t’ if and only if both sides are equal to a constant.
Therefore
′
′
=
+ =k--------(4) where k is a constant
′
Now from (4)
′
Now consider (5)
′
= k----(5) and
X kX 0
= k
′
Now consider (6)
+ =k
+ =k-------(6)
X C1ekx
k 1
t
2
k 1
T
T
T
C
e
0
2
2
(8)
Substituting the values of X and T in (2) we get
U ( x, t ) X C1e C2e
kx
k 1
t
2
k 1
t
kx 2
U ( x, t ) X Ae e
( where A=C1 C2 )
kx
Put t=0 in the above equation ,we have U(x,0) = A e -----(9)
but given that U(x,0)=6 − --------(10)
from (9) and (10) we have A e kx =6 −
A=6 and k=-3 the solution of the given equation becomes
U ( x, t ) X 6e 3 x e 2 t 6e (3 x 2t )
2. Solve the equation by the method of separation of variables
Sol: Given equation is
=
+
----------(1)
=
+
Let U(x,y)= X(x) Y(y) =X Y ---------------(2)
be a solution of (1)
Differentiating (2) partially w.r.t x ant y
=
′
,
Put these values in equation (1), we have
′′
Dividing by XY on both sides we have
′′
′′
−
′
=
=
=
=
′′
′
+2
′′
+
=
′′
-----------------(3)
Since L.H.S is a function of ‘x’ and the R.H.S is a function of ‘y’ where x and y are
independent variables, the two sides of (3) can be equal to each other for all values of ‘x’ and
‘y’ if and only if both sides are equal to a constant.
′′
−
Now from (4)
=
′′
= --------(4)
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′′
′′
And
−
=
=
−−−−−−
′′
From (5) ′′ −
=
−
Which is second order differential equation
Auxiliary equation is
−
+
Solution of the given equation (5) is
Now consider equation (6)
′
=
=
=
→
−−−−−−
+
′
=
→ m=±√
√ +
+
=
Integrating on both sides we get logy = ky + log
Y
log ky Y C3eky (8)
C3
Substituting the values of X and Y in (2) we have
U C1e (2 k ) x C2 e (2 k ) x C3e ky
U Ae (2 k ) x Be (2 k ) x e ky
Where
A C1C3
and
B C1C2
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UNIT V
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LAPLACE TRANSFORMS
INTRODUCTION
Laplace Transformations were introduced by Pierre Simmon Marquis De Laplace
(1749-1827), a French Mathematician known as a Newton of French. Laplace
Transformations is a powerful technique, it replaces operations of calculus by operations of
algebra. An Ordinary (or) Partial Differential Equation together with Initial conditions is
reduced to a problem of solving an Algebraic Equation by this method.
USES
Particular Solution is obtained without first determining the general solution.
Non-Homogeneous Equations are solved without obtaining the complementary
integral.
L.T is applicable not only to continuous functions but also to piecewise continuous
functions, complicated periodic functions, step functions and impulse functions.
APPLICATIONS:
L.T is very useful in obtaining solution of linear differential equations, both ordinary
and partial, solution of system of simultaneous differential equations, solution of
integral equations, solution of linear difference equations and in the evaluation of
definite integrals.
DEFINITION:
Let f (t) be a function of‘t’ defined for all positive values of t. Then Laplace
transforms of f (t) is denoted by L {f (t)} is defined by
L f t e st f t dt f s (1)
0
provided that the integral exists. Here the parameter‘s’ is a real (or) complex number.
The relation (1) can also be written as f t L1 f s
In such a case the function f(t) is called the inverse Laplace transform of f s .The
symbol ‘L’ which transform f(t) into f s is called the Laplace transform operator. The
symbol ‘L-1’ which transforms
operator.
̅
(s)
to f (t) can be called the inverse Laplace transform
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Conditions for Laplace Transforms
Exponential order: A function f (t) is said to be of exponential order ‘a’ If lt e st f t a
t
finite quantity.
Ex: (i). The function t2 is of exponential order
3
(ii). The function et is not of exponential order (which is not finite quantity)
Piece – wise Continuous function: A function f (t) is said to be piece-wise continuous over
the closed interval [a,b] if it is defined on that interval and is such that the interval can be
divided into a finite number of sub intervals, in each of which f (t) is continuous and has both
right and left hand limits at every end point of the subinterval.
Sufficient conditions for the existence of the Laplace transform of a function:
The function f (t) must satisfy the following conditions for the existence of the L.T.
(i).The function f(t) must be piece-wise continuous (or sectionally continuous) in any limited
interval 0 a t b .
(ii).The function f (t) is of exponential order.
Laplace Transforms of standard functions:
1.
Prove that L 1
1
s
Proof: By definition
e st
e e0
1
L 1 e .1dt
s s 0 s if s 0
s
0
0
L 1 1
2.
st
s
e
0
Prove that L t 1
s2
Proof: By definition
e st
L t e .tdt t.
0
s
st
e st
1.
dt
s 0
e st
e st
t .
1 2
2
s
s
s 0
3.
Prove that L t n
Proof: By definition L t
n
n!
where n is a +ve integer
s n 1
0
e st n 1 e st
e .t dt t n .
n.t . s dt
s 0 0
st
n
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00
n st n 1
e t dt
s 0
n
L t n 1
s
Similarly L t n 1
n 1
L t n 2
s
L t n 2
n2
L t n 3
s
By repeatedly applying this, we get
n n 1 n 2
2 1
L t n .
.
..... . L t n n
s s
s
s s
n!
n! 1 n!
L 1 n . n 1
n
s
s s s
Note: L t n can also be expressed in terms of Gamma function.
n ! n 1
n 1 n!
s n 1
s n 1
i.e., L t n
Def: If n>0 then Gamma function is defined by n e x x n1dx
0
We have L t n e st .t n dt
0
Putting x=st on R.H.S, we get
L t t
n
0
L t n
xn 1
e . n . dx
s s
x
1
s n 1 0
1
s n 1
e x .x n dx
x st
1 dx dt
s
When t 0, x 0
When t , x
. n 1
If ' n 'is a +ve integer then n 1 n !
L t n n !
s n 1
Note: The following are some important properties of the Gamma function.
1. n 1 n. n if n 0
2. n 1 n! if n is a +ve integer
3. 1 1, 1
2
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Note: Value of n in terms of factorial
2 1 1 1!
3 2 2 2!
4 3 3 3!
and so on.
In general n 1 n! provided ‘n’ is a +ve integer.
Taking n=0, it defined 0! = 1 1
4.
Prove that L e at
1
sa
Proof: By definition,
L eat e st .eat dt e s a t dt
0
0
e s a t
s a 0
e
e0
1
if s a
sa sa sa
Similarly L e at
5.
1
if s a
sa
a
s a2
Prove that L sinh at
2
1
eat e at
at
at
Proof: L sinh at L
L e L e
6.
2
2
a
1 1
1 1 s a s a
2a
2 s a s a 2 s 2 a 2 2 s 2 a 2 s 2 a 2
Prove that L cosh at
s
s a2
2
eat e at
Proof: L cosh at L
2
7.
1
1
1
1
L e at L e at
2
2 s a s a
1 s a s a
2s
s
2 s 2 a 2 2 s 2 a 2 s 2 a 2
Prove that L sin at
a
s a2
2
Proof: By definition, L sin at e st sin atdt
0
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e st
2
s sin at a cos at
2
s a
0
ax
e sin bxdx
8.
eax
a sin bx b cos bx
2
2
a b
a
s a2
2
Prove that L cos at
s
s a2
Proof: We know that L e at
2
1
sa
Replace ‘a’ by ‘ia’ we get
L eiat
1
s ia
s ia s ia s ia
i.e., L cosat i sin at
s ia
s2 a2
Equating the real and imaginary parts on both sides, we have
L cos at
s
a
and L sin at 2
2
s a
s a2
2
Solved Problems :
1.
Find the Laplace transforms of (t 2 1)2
Sol:
Here
f(t) (t 2 1)2 t 4 2t 2 1
L{(t 2 + 1)2 } = L{t 4 + 2 t 2 + 1} = L{t 4 } + 2 L{t 2 } + L{1}
2.
Sol:
4!
2! 1 4!
2! 1
2. 3 5 2. 3
4 1
s
s s s
s
s
24 4 1 1
3 5 (24 4 s 2 s 4 )
5
s
s s s
e at 1
Find the Laplace transform of L
a
e a t 1 1
1
at
at
L
= L e 1 L e L 1
a
a a
1 1
1
1
a s a s
s(s a)
3.
Find the Laplace transform of Sin2tcost
Sol:
W.K.T sin 2t cos t [2 sin 2t cos t ] [sin 3t sin t ]
1
2
1
2
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1
1
L{sin 2t cos t} L [sin 3t sin t ] L sin 3t L sin t
2
2
1 3
1
2( s 2 3)
2 s 2 9 s 2 1 ( s 2 1)( s 2 9)
4.
Find the Laplace transform of Cosh22t
Sol:
w.k.t cosh 2 2t
Lcosh 2 2t
1
1 cosh 4t
2
1
L(1) L{cosh 4t}
2
s2 8
1 1
s
2 s s 2 16 s( s 2 16)
5.
Find the Laplace transform of Cos33t
Sol:
Since cos9t=cos3(3t)
cos9t=4cos3 3t-3cos3t (or) cos3 3t=
L{cos3 3t} =
∴
1
cos 9t 3cos 3t
4
1
3
L{cos 9t} + L{cos 3t}
4
4
1
3 s
s
= . 2
. 2
4 s 81 4 s 9
s s 2 63
s 1
3
= 2
4 s 81 s 2 9 s 2 9 s 2 81
6.
Find the Laplace transforms of sin t cos t
Sol:
Since sin t cos t sin 2 t cos 2 t 2sin t cos t 1 sin 2t
2
2
L{(sin t + cos t )2 } = L{1 + sin 2t}
= L{1} + L{sin 2t}
7.
Sol:
s 2 2s 4
1
2
2
s s 4
s s2 4
Find the Laplace transforms of cost cos2t cos3t
1
cos t cos 2t cos 3t .cos t 2.cos 2t.cos 3t
2
=
1
1
cos t[cos 5t + cos t ] = [cos t cos 5t + cos 2 t ]
2
2
1
1
2 cos t cos 5t 2 cos 2 t cos 6t cos 4t 1 cos 2t
4
4
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1
[1 cos 2t cos 4t cos 6t ]
4
=
1
L{1 cos 2t cos 4t cos 6t}
4
L{cos t cos 2t cos 3t}
1
[ L{1} L{cos 2t} L{cos 4t} L{cos 6t}]
4
=
= 1 1 2 s 2 s 2 s
4 s s 4 s 16 s 36
8.
Find L.T. of Sin2t
1 cos 2t
L{sin 2 t} L
2
Sol:
1
1 1
s
[L{1} L{cos 2t}] 2
2
2 s s 4
9.
Find L(√ )
L
Sol:
t L t
1
1
1
2
2
1 where n is not an integer
1
s2
1 1
2 2
s
10.
Sol:
Find 𝑳 {
{ 𝑖
3
2
2s
n 1 n. n
3
2
+ 𝜶 }, where a constant is
+𝛼 }= { 𝑖
=
=
𝛼+
𝛼 { 𝑖
𝛼
}+ 𝑖 𝛼 {
+ 𝑖 𝛼
+
+
𝑖 𝛼}
}
Properties of Laplace transform:
Linearity Property:
Theorem1: The Laplace transform operator is a Linear operator.
i.e. (i). L cf t c.L f t
(ii). L f t g t L f t L g t Where
‘c’
is
constant
Proof: (i) By definition
L cf t e cf t dt c e st f t dt cL f t
0
st
0
(ii) By definition
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MATHEMATICS - I
L f t g t e st f t g t dt
0
e
st
0
f t dt e st g t dt L f t L g t
0
Similarly the inverse transforms of the sum of two or more functions of ‘s’ is the sum of the
inverse transforms of the separate functions.
Thus, L1 f s g s L1 f s L1 g s f t g t
Corollary: L c1 f t c2 g t c1L f t c2 L g t , where c1, c2 are constants
Theorem2: If a, b, c be any constants and f, g, h any functions of t, then
L{af (t ) + bg (t ) - ch(t )} = a.L{ f (t )} + b.L{g (t )} - cL{h(t )}
Proof: By the definition
L{af (t ) bg (t ) ch(t )} e st {af (t ) bg (t ) ch(t )}dt
0
0
0
0
a. e st f t dt b e st g t dt c e st h t dt
a.L{ f (t )} bL{g (t )} cL{h(t )}
Change of Scale Property:
1 s
If L{ f (t )} = f ( s) then L{ f (at )} . f
a a
Proof: By the definition we have
L{ f (at )} e st f (at )dt
0
Put at u dt
du
a
when t then u and t = 0 then u = 0
L{ f (at )} e
su
a
0
s
1
du 1 a .u
f u du . f s
. e
f (u )
a
a 0
a
a
Solved Problems :
1.
Sol:
Find 𝑳{
}
{sinh } =
∴ {sinh
}=
−
=
⁄ (Change of scale property)
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1
1
3
2
2
3 s
1 s 9
3
2.
Sol:
}
Find 𝑳{
{cos } =
{cos
}=
{cos
}=
First shifting property:
+
=
( ⁄ ) (Change of scale property)
⁄
( ⁄ ) +
=
+
at
If L{ f (t )} = f ( s) then L{e f (t )} = f ( s - a)
Proof: By the definition
L{e f (t )} e st eat f (t )dt
at
0
e
s a t
f t dt
0
e ut f t dt where u s a
0
= f (u ) f s a
- at
Note: Using the above property, we have L{e f (t )} = f (s+ a)
Applications of this property, we obtain the following results
1. L{e at t n }
n!
n!
L(t n ) n 1
n 1
( s a)
s
2. L{e at sin bt}
b
b
L(sinbt) 2
2
2
( s a) b
s b 2
3. L{e at cos bt}
sa
s
L(cosbt) 2
2
2
( s a) b
s b 2
4. L{e at sinh bt}
b
b
L(sinhbt) 2
2
2
( s a) b
s b 2
5. L{e at cosh bt}
sa
s
L(coshbt) 2
2
2
( s a) b
s b 2
Solved Problems :
1.
Find the Laplace Transforms of t 3e 3t
Sol:
Since L{t 3} = 3!4
s
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Now applying first shifting theorem, we get
3!
( s + 3) 4
L{t 3 e- 3t } =
2.
Find the L.T. of e t cos 2t
Sol:
Since L{cos
}=
+
Now applying first shifting theorem, we get
3. Find L.T of
Sol: -
−
{
L[
+
]=L[
= { [
=
−
=
−
=
}=
cos
−
]
]+ [
+ { [
+
−
+
−
−
−
+
+
]}
+
=
+
+
+
+
]}
→ −
+
Second translation (or) second Shifting theorem:
If {
}=
Proof: By the definition
{
}=
=∫
∫
∞
∞
−
−
={
−
.
+∫
, >
<
{
ℎ
−
=∫
∞ −
−
}=
+∫
−
=∫
∞ −
∞
−
−
Let t-a = u so that dt = du And also u = 0 when t = a and u → when t →
∴ {
}=∫
∞
−
−
=
+
{
}=
=
−
−
∫
∞ −
=
−
∫
∞
−
Another Form of second shifting theorem:
If
{
}=
where H (t) = {
,
,
>
>
<
{
ℎ
−
−
}=
−
and H(t) is called Heaviside unit step function.
Proof: By the definition
{
−
−
}=∫
∞
−
−
−
→
Put t-a=u so that dt= du and also when t=0, u=-a when t → , u→
Then {
−
−
}=∫
∞ −
+
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]
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0
e
s (u a )
a
e
F u H u du e s (u a ) F u H u du
0
−
= ∫−
=∫
∞
sa
+
−
e
.
+∫
+
st
∞
−
=
−
+
.
[Since By the definition of H (t)]
∫
∞
−
F t dt by property of Definite Integrals
0
−
=
{
−
}=
Note: H t a is also denoted by u t a
Solved Problems
cos t
3
Find the L.T. of g (t) when g t
0
1.
if t
if t
3
3
Let f t cos t
Sol.
∴ {
}= {
={
}=
=
+
( − 𝜋⁄ ) =
( − 𝜋⁄ ), 𝑖
> 𝜋⁄
< 𝜋⁄
, 𝑖
Now applying second shifting theorem, then we get
{
}=
−𝜋
=
+
−
.
−𝜋
+
− ) (ii)
−
−
2.
Find the L.T. of (ii)
Sol:
(i). Comparing the given function with f(t-a) u(t-a), we have a=2 and f(t)=t3
∴ {
}= { }=
!
}=
−
=
=
Now applying second shifting theorem, then we get
{
−
(ii).
{
−
−
=
−
−
}= {
then
=
−
−
−
=
−
.
−
+
}=
−
{
−
−
−
}
Now applying second shifting theorem then, we get
{
−
−
}=
−
.
−
+
=
−
+
+
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Multiplication by‘t’:
}=
Theorem: If 𝑳{
𝑳{
Proof: By the definition f s e st f t dt
}=
−
0
d
d
f s e st f t dt
ds
ds 0
By Leibnitz’s rule for differentiating under the integral sign,
d
f s e st f t dt
ds
s
0
te st f t dt
0
= −∫
Thus {
}=
∴ {
𝛼
{∫
−
{
𝑛
}= −
Note: Leibnitz’s Rule
If f x, and
−
∞
𝑛
}
=− {
}
=
f x, be continuous functions of x and then
,𝛼
}=∫
,𝛼
𝛼
Where a, b are constants independent of
Solved Problems:
1.
Find L.T of tcosat
Sol:
Since {
{
+
−
+
}=−
=
2.
Find t2sin at
Sol:
Since { 𝑖
{ . 𝑖
}=
}=
}= −
+
[
− .
+
]
=
−
+
+
+
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2
2
d 2as 2a 3s a
=
2
2 3
ds s 2 a 2 2
s
a
3.
Sol:
4.
Sol:
Find L.T of tet sin 3t
Since { 𝑖
{
∴ { 𝑖
−
Find 𝑳{
𝑖
}=
}=
}=
Since { 𝑖
L e 2t sin 3t
{
−
[
+
+
}
}=
+
+
3
s 2
2
=
s
9
[
5.
Find the L.T. of 1 tet
Sol:
Since
−
+
=
+
+
2
−
4s 13
+
Now using the shifting property, we get
3
s 4s 13
3 2s 4
2
+
=
+
}= −
𝑖
]=
+
]= −
+
2
s
6 s 2
2
−
[
4s 13
−
−
+
]
2
2
−
+
+
−
L 1 tet L 1 2L tet L t 2e2t
2
2
1
d 1
2 d 1
2 1
1
s
ds s 1
ds 2 s 2
1
2
d 1
2
s s 1 ds s 2 2
1
2
2
2
s s 1 s 2 3
6.
Sol:
Find the L.T of t3e-3t (already we have solved by another method)
{
−
}= −
=−
=
7.
Sol.
Find 𝑳{
{cosh
sin
+
!
{
−
=
+
}
}= {
}
− ! −
+
+ −
. sin
}
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MATHEMATICS - I
= [ {
−
}+ {
sin
sin
1
a
a
2 s a 2 a 2 s a 2 a 2
f t t 1 ,
t 1
0
0 t 1
2
8.
Find the L.T of the function
Sol:
By the definition
{
}=∫
∞ −
−
=∫
+∫
e st 0dt e st t 1 dt
1
0
=∫
}]
2
∞
−
1
∞
0
−
−
2 st
e t 1 dt
s 1
=[
−
−
−
∞
] −∫
∞
−
−
−
st
e
e st
2
dt
t 1
1
s
s
s
1
st
2 st
2 1
2 e
0 e st dt 2
3 e 1
1
s
s
s
s s 1
9.
2
2
0 e s 3 e s
3
s
s
Find the L.T of f (t) defined as f (t ) 3 ,
t >2
0,
Sol:
{
}=∫
∞ −
0<t<2
e st f t dt e st f t dt
2
0
=∫
2
−
.
+∫
0 e st 3dt
2
∞
−
3 st 3
e 2 s 0 e2s
s
3
e 2 s
s
10.
Sol:
Find 𝑳{
{
+
+
}
} = {cos
= cos . {cos
= cos .
sin }
cos − sin
+
} − sin
− sin .
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}
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{ .
[cos .
s
1
2
a
2 2
]
Find L.T of L [t
We know that L[sint] =
Sol: -
− sin .
+
]
+
2
s 2 a 2 .0 a.2s
s a 2 .1 s.2s
sin b
= cos b.
2
2 2
2
2 2
s a
s a
11.
−
}=
+
s 2 a 2 2 cos b 2as sin b
+
L[tsint] = (-1)
=
L[sint] = -
(
+
+
=-
−
+
By First Shifting Theorem
𝑖
L [t
Division by‘t’:
]=[
+
]
→ −
}=
Theorem: If 𝑳{
=
−
𝑳{
−
=
+
}=∫
−
−
+
∞
We have f s e st f t dt
Proof:
0
Now integrating both sides w.r.t s from s to , we have
0
st
f ( s)ds e f (t )dt ds
s 0
0
f t e
st
s
dsdt (Change the order of integration)
f t e st ds dt (
s
0
0
=∫
∞
e st
f t
dt
t s
−
{
Solved Problems:
1.
Sol:
Find 𝑳 {
}
t is independent of‘s’)
Since { 𝑖 } =
+
}
=
Division by‘t’, we have
{
i
}=∫
∞
=∫
∞
+
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−
=[
]∞ =
−
Tan 1s cot 1 s
2
2.
Find the L.T of
Sol:
Since {sin
sin at
t
}=
+
Division by t, we have
{
i a
}=∫
−
∞−
∞
=
=∫
∞
+
1
a. Tan 1 s Tan 1 Tan 1 s
as
a
a
Tan1 s cot 1 s
2
a
a
3.
Sol:
Evaluate 𝑳 {
Since
{
−c
−
𝐚
}
{ − cos a } = { } − {
a
}=∫
∞
−
+
}= −
+
1
log s log s 2 a 2
2
s
1
1 s2
2 log s log s 2 a 2 log 2
s
2
2 s a2 s
1
1
1
s2
l og
log1 log 2
1 a2
s a 2
2
2
2
s s
1
s2
s2 2
s2 a2
1
l og 2
log
log
2
2
2
s2
2
s a
s a
Note: 𝑳 {
4.
Sol:
−
Find 𝑳 {
s2 1
} = log
(Putting a=1 in the above problem)
s
−
{
− −
−
}
− −
}=∫
∞
+
−
+
s a
log s a log s b s log
s b s
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a
1 s
sa
l t log
log
s
sb
1 b
s
sb
log1 log( s a) log( s b) log
sa
1 cos t
2
t
Find L
5.
1 cos t
1 1 cos t
L .
..... 1
2
t
t
t
Sol: L
s
1
1 cos t 1
2
Now L
s 2 ds log s log s 1
2
t
s s 1
s
1
s2
s2 1
s2 1
1
log 2 log 2 log 2
2
s 1 s
2
s 1 2
s
1
s 1
1 cos t
log 2 ds
L
2
s
2
s
t
2
2
s
1 s 2 1
2
log 2 .s 2 3 .sds
s s 1 s
2 s
s
ds
s2 1
1
1
lt s.log 1 2 s log 2 2 2
s
s
2
s 1
s
s
1
1
1
s 2 1
1
lt s 2 4 6 .... s log 2 2Tan 1s
2 s s 2s 3s
s
s
1
1
0 s log 1 2 2 Tan1s
2
s 2
cot 1 s
6.Find L.T of
Sol:
W.K.T
−
∴
L[
−
1
1
s log 1 2
2
s
− −
L[
L[
x 2 x3 x 4
.....
x
x
log
1
2 3 4
−
]=∫
−
]=
∞
−
̅
+
]=∫
−
,L[
∞
= [log
= log
+
+
+
+
]=
−
∞
+
+
− log
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]∞
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+
= log
+
∞
=log (1)-log (
=0- log (
+
+
+
)
) = log (
+
Laplace transforms of Derivatives:
If f 1 t be continuous and {
}=
+
)
+
{
ℎ
}=
−
Proof: By the definition
{
}=∫
∞
−
e st f t s e st f t dt
0
0
(Integrating by parts)
e st f t s e st f t dt
0
0
lt e
= t
st
f (t ) f (0) + . {
}
Since f (t) is exponential order
lt
t
e
∴ {
st
f (t ) =0
}=
=
−
−
{
+
}
The Laplace Transform of the second derivative f11(t) is similarly obtained.
∴ {
}= . {
}−
s. s f s f 0 f 1 0
s 2 f s sf 0 f 1 0
∴ {
}= . {
= [
=
{
}=
{
{
}−
}−
}−
Proceeding similarly, we have
{
Note 1: {
Note 2: Now |
}=
We have |
|
−
}−
.
𝑖
|=
−
=M.
]−
−
−
−
−
−
=
|
− −
−
……
= ,
|
→
.
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→ ∞ if s>a
−
=
…
−
=
.
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∴
→∞
Solved Problems:
−
=
>
Using the theorem on transforms of derivatives, find the Laplace Transform of the
following functions.
(i). eat (ii). cosat
(iii). t sin at
(i). Let f t eat Then f 1 t a.eat and f 0 1
{
}= . {
𝑖. . , {
}= . {
𝑖. . , {
}− . {
𝑖. . ,
(ii).
}−
−
∴ {
}=
=
∴ {
{
−
}=−
}=−
ℎ
}=
}−
=−
{
𝑖
}− .
=−
−
Now f 0 cos0 1and f 1 0 a sin 0 0
{−
ℎ
⟹−
}=
cos
{cos
⟹−
}−
{cos
{cos
{cos
+
}− . −
}=−
} = − ⇒ {cos
}=
(iii). Let f t t sin at then f 1 t sin at at cos at
+
f 11 t a cos at a cos at at sin at 2a cos at a 2t sin at
Also f
{
𝑖. . , {
=
}=
cos
−
{cos
𝑖. . ,
𝑖. . , −
and
{
}−
=
sin
{t sin
+
}−
}=
{ sin
}=
−
}−
+
Laplace Transform of Integrals:
If
{
}=
ℎ
Proof: Let g t f x dx
t
{∫
−
{tsin
{tsin
}−
⇒ {t sin
−
}=
}=
+
}=
0
Then
=
[∫
]=
Taking Laplace Transform on both sides
{
}= {
}
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But {
}=
∴ {
}= {
{
⟹
{
}
}= {
{
=
} ⇒ {
=∫
∴ {∫
}−
}−
}=
{
[ 𝑖
= ]
}
}=
Solved Problems:
1.
Find the L.T of
Sol:
L{sin
}=
t
sin atdt
0
=
+
Using the theorem of Laplace transform of the integral, we have
{∫
2.
Sol:
}=
∴ {∫ sin
}=
{sin } =
+
∴ {
i
}=∫
𝑖. . , {
i
}=
t
i
→
∞
=
{sin }
=∫
∞
𝑖
+
Tan1s Tan1 Tan1s Tan1s cot 1 s (or ) Tan1 1
s
2
s
i
3.
∴
{∫
Find L.T of
Sol:
L[
−
∫
sin t
dt
0
t
Find the L.T of
+
−
−
}=
−
𝑖
−
( ⁄)
∫
( ⁄)
−
]
We know that
L {sint} =
L{
𝑖
}=∫
+
∞
=(
=
∴ L{
𝑖
}=
̅
−
−
= ̅
∞
∞−
−
=∫
∞
−
+
𝜋
= −
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Hence
L {∫
𝑖
−
}=
By First Shifting Theorem
L[
−
∴ L[
−
∫
∫
𝑖
𝑖
]= ̅
+
]=
−
=
+
−
→ +
+
Laplace transform of Periodic functions:
If f (t) is a periodic function with period ‘a’. i.e, f t a f t then
L f t
1
1 e sa
a
0
e st f t dt
Eg: sin x is a periodic function with period 2
i.e., sin x sin 2 x sin 4 x .............
Solved Problems:
1.
A function f (t) is periodic in (0,2b) and is defined as f t 1 if 0 t b
1 if b t 2b
Find its Laplace Transform.
Sol:
L f t
1
1 e 2bs
0
e st f t dt
2b
1 b st
e f t dt e st f t dt
2 bs 0
b
1 e
2b
1 b st
e dt e st dt
2 bs 0
b
1 e
1
1 e2bs
L f t
2.
2b
e st b e st 2b
s 0 s b
1
e sb 1 e 2bs e sb
2 bs
s 1 e
1
1 2e sb e 2bs
2 bs
s 1 e
Find the L.T of the function f t sin t if 0 t
0if
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2
t
where f t has period
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Sol:
2
Since f (t) is a periodic function with period
L f t
L f t
1
1 e sa
0
1
1 e
s 2
1
1 e
2 s
a
e st f t dt
2
e st f t dt
0
2
st
st
e
t
dt
sin
e .0dt
0
b
a
1
1 e
2 s
eat sin bt
eat
a sin bt b cos bt
a 2 b2
1
1 e
e st s sin t cos t
s2 2
0
2 s
s
1
s 2 2 e .
Laplace Transform of Some special functions:
1.
The Unit step function or Heaviside’s Unit functions:
−
It is defined as
={
<
>
Laplace Transform of unit step function:
−
To prove that {
−
}=
{
−
}=∫
−
Proof: Unit step function is defined as
Then
∞ −
={
−
<
>
e st u t a dt e st u t a dt
a
0
a
a
0
a
e st .0dt e st .1dt
a
∴ {
}=
−
e st
1 as
e as
e dt
s . e e s
s a
st
−
Laplace Transforms of Dirac Delta Function:
The Dirac delta function or Unit impulse function
2.
Prove that 𝑳{
∈
}=
− − ∈
∈
∈
= { ⁄∈
hence show that 𝑳{𝜹
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Proof: By the definition
And
∈
Hence {
∈
⁄
={ ∈
}=∫
=∫
=∫
Now
∴ {𝛿
∴ {
{𝛿
}=
} =∈→
{
𝑖
∞
∈
∈
−
−
−
−
∈
∈
∈
∈
∈
>∈
∞
+ ∫∈
∞
−
+ ∫∈
= ∈ [ − ] = − ⁄∈ [
}=
∈
− − ∈
} = ∈→
∈
∈
.
−
− ∈
∈
−
]=
− − ∈
∈
− − ∈
L-Hospital rule.
∈
Properties of Dirac Delta Function:
∞
1. ∫ 𝛿
=
∞
2. ∫ 𝛿
∞
3. ∫ 𝛿
=
−
where G(t) is some continuous function.
=
where G(t) is some continuous function.
4. G (t ) 1 t a G1 (a)
0
Solved Problems:
1.
Sol:
Prove that 𝑳{𝜹
By Translation theorem
L{𝛿
2.
Sol:
−
}=
−
−
}=
Evaluate ∫
=
∞
−
−
{𝛿
}
{𝛿
[sin
𝜹( − 𝝅⁄ )
}= ]
By using property (3) then we get
∞
∫ 𝛿
Here
−
= 𝜋⁄ ,
=
= cos
3 cos 2 3 12
G a G
∞
3.
∴ ∫ cos
𝛿( − 𝜋⁄ )
Evaluate e 4t 1 t 2 dt
= cos 𝜋⁄ = −𝜋⁄
0
Sol:
By the 4th Property then we get
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t a G t dt G (a)
1
1
0
G t e4t and a 2
G1 t 4.e4t
G1 a G1 2 4.e8
e 4t 1 t 2 dt G1 a 4.e 8
0
Inverse Laplace Transforms:
If f s is the Laplace transforms of a function of f (t) i.e. {
is called the inverse Laplace transform of f s and is written as
−
∴
is called the inverse L.T operator.
}=
−
=
then f (t)
{
}
Table of Laplace Transforms and Inverse Laplace Transforms
{
S.No.
1.
2.
4.
5.
{ }=
6.
{
!
−
+
−
{cos
8.
{sinh
9.
11.
{
{sin
7.
10.
{ }= ⁄
{
3.
{
{
{cosh
sin
cos
−
}=
}=
}=
𝑖
}=
}=
}=
}=
−
+
}=
!
−
{ ⁄ }=
−
+
−
+
−
−
−
−
−
−
−
+
+
DEPARTMENT OF HUMANITIES & SCIENCES
−
−
{
{
{
{
{
{
−
−
{ ⁄ }=
{ ⁄ + }=
−
𝑖
−
−
{ ⁄ − }=
−
+
}=
}=
−
}=
{
{
+
−
−
+
+
−
}=
!
!
,
= , , …
. sin
} = cos
}=
sinh
} = cosh
−
−
}=
+
+
}= .
}=
sin
cos
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MATHEMATICS - I
12.
{
13.
{
14.
{
15.
{
16.
}=
sinh
{
{
+
}=
cos
}=
−
+
}=
−
−
−
−
}=
sin
−
17.
}=
cosh
−
−
−
−
−
+
+
−
+
−
−
+
{
{
{
{
−
{
−
−
+
{
+
+
−
+
−
−
−
+
}=
+
}=
.
}=
.
}=
−
}=
}=
sinh
cosh
−
−
−
−
{
sin
cos
−
}
{
}
Solved Problems :
1.
Sol:
−
{
−
+
−
}=
{ ⁄ − . ⁄
Sol:
+ ⁄ }
s 3L 1 s L 4 s
L1 1
1
1 3t 4.
2.
1
2
Find the Inverse Laplace Transform of
−
{
−
+
+
}=
=
−
−
{
{
−
−
Sol:
+
−
+
+
}=
−
}+ .
{
−
{
−
−
−
}=
−
=
{
{
−
−
−
}−
−
}
−
{
1
2.cosh 2t 5. sinh 2t
2
4.
Find 𝑳− {
+
+
−
+
− +
−
{
−
−
+
+
}
+
}
2s 5
s2 4
Find the Inverse Laplace Transform of
−
3
t2
1 3t 2t 2
2!
4
e 2t cos 3t e 2t sin 3t
3
3.
s 2 3s 4
s3
Find the Inverse Laplace Transform of
}
}
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Sol:
−
+ +
{
+
=
5.
Find 𝑳− {
Sol:
𝑳− {
−
+
−
}=
−
−
+
}=
=
{
+
{
−
{
−
}+
}
−
+ }
+
{
{ }=
−
}−
+
−
}−
+( ⁄ )
−
{
3
5 8 2
5
.cos t . sin t
4
2 4 5
2
+
{
}
+
+( ⁄ )
}
3
5 4
5
cos t sin t
4
2 5
2
6.
s
Find the Inverse Laplace Transform of
Sol:
−
{
}=
+
−
{
+ −
+
}=
=
−
−
−
=
−
−
[
−
{
s a
−
}
{ − }
{ }− .
eat 1 at
7.
Find 𝑳− {
Sol:
Let
−
{
8.
Sol:
+
3s 7
A
B
s 2s 3 s 1 s 3
+
+
= ,
= − ,−
2
+
}=
−
=
=
=
⇒
⇒
+
=
−
=−
−
+
}=−
+
−
{
e t 4.e3t
Find 𝑳− {
+
{ }]
}
−
3s 7
1
4
s 2s 3 s 1 s 3
−
−
2
−
−
2
+
+
=
(
+
+ )
+
}
+
+
−
{
+
}+
−
{
−
}
+
+
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+
+
+
+
+
+
+
Equating Co-efficient of s3, A+C=0……..(1)
=
Equating Co-efficient of s2, A+B+2C+D=0…….(2)
Equating Co-efficient of s, A+C+2D=1…….(3)
put s 1, 2 B 1 B
1
2
1
2
Substituting (1) in (3) 2 D 1 D
Substituting the values of B and D in (2)
− +
i.e.
+ =
⇒
+
}−
−
1
1
s
2 2
2
2
1
1
s
s
s
12 s 2 1
−
{
+
}= [
+
{
= [sin −
Find 𝑳− {
+
−
−
1
sin t te t
2
9.
−
}
+
{ }]
Sol: Since s 4 4a 4 s 2 2a 2 2as
Let
=
2
+
−
+
Solving we get A 0, C 0, B
+
}=
−
{
+
−
+
+
1
1
,D
4a
4a
}+
−
{
=
⇒
= ,
=
2
+
+
+
}]
+
−
+
+
+
s
As B
Cs D
2
2
4
2
s 4a
s 2as 2a
s 2as 2a 2
4
+
{
{
= ,
−
+
=
}
1
1 1
1
1
.. L1
a.L
2
2
2
2
4
( s a ) a 4a
( s a) a
1 1 at
1 1
. .e sin at . e at sin at
4a a
4a a
1
1
1
sin at e at e at 2 .sin at.2sinh at 2 sin at sinh at
2
4a
4a
2a
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Sol:
i.
ii.
−
Find i. 𝑳− {
10.
−
−
{
−
{
+
−
}=
(
− )
{ }−
=
−
=
−
−
}=
{
+
(
(
ii. 𝑳− {
}
{ −
−
=
+
+ }=
−
− )
!
{ −
−
{ }+
{ −
−
=
−
+
}+ {
}=
−
− )
+
{
{ }
−
+
−
}
+ }
}
{ }−
−
{ }+
−
{ }}
]=
−
[
3
t 2 4t 4 3
t4 1 4
2
1
2
1 4
t
t 6t 2 6
2
2! 4! 2
6 4
Find 𝑳− [
11.
Sol:
−
[
−
−
]=
]
−
[
−
]=
−
[
−
+
−
+
]
1 at at
e e cosh at
2
4
L1
( s 1)(s 2)
12.
Find
Sol:
4
1
1
1
4 L1
4 L1
4[et e2t ]
L1
s 1 s 2
( s 1)(s 2)
( s 1)(s 2)
13.
1
Find L1
2
2
( s 1) ( s 4)
Sol:
+
1
A
B
Cs D
2
2
2
( s 1) ( s 4) s 1 ( s 1)
s 4
2
A
2
1
2
3
, B ,C
,D
25
5
25
25
2 1 1 1 1 1 2 1 s 3 1 1
1
L
L1
L 2
L
L 2
2
2
2
s 4 25
s 4
( s 1) ( s 4) 25 s 1 5 ( s 1) 25
2 t 1 1 1 t 1 1 2
3 1
e L e L 2 cos 2t . sin 2t
25
25 2
s 5
s 25
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14.
Sol:
Find
2 t 1 t
2
3
e e .t cos 2t sin 2t
25
5
25
50
s2 s 2
L1
s( s 3)(s 2)
s2 s 2
A
B
C
s s 3 s 2 s s 3 s 2
Comparing with s2, s, constants, we get
A 1 , B 4 ,C 2
3
15
5
s2 s 2
4
2
1 1
L
L
3s 15( s 3) 5( s 2)
s( s 3)(s 2)
1
1 2
4
1
L1 L1
L 5( s 2)
3s
15( s 3)
1 4
2
e 3t e 2t
3 15
5
15.
s 2 2s 4
Find L1 2
( s 9)(s 5)
Sol:
A
Bs C
s 2 2s 4
2
=
2
( s 9)(s 5) s 5 s 9
Comparing with s2, s, constants, we get
A 31
34
,B 3
34
, C 83
34
2
s 2 2s 4
1 s 2 s 4
L1 2
L
2
( s 9)(s 5)
( s 9)(s 5)
31 1
1
3
83
L1
L
L
2
2
34(s 5)
34( s 9)
34( s 9)
31 5t 1
83
e
3cos 3 t sin 3t
34
34
3
First Shifting Theorem:
If L1 f ( s) f (t ), thenL1 f (s a) e at f (t )
Proof: We have seen that L eat f (t ) f ( s a ) L1 f ( s a) eat f (t ) eat L1 f (s )
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Solved Problems :
1.
1
1
Find L1
L f ( s 2)
2
( s 2) 16
Sol:
1
1
2 t 1
L1
e L 2
2
s 16
( s 2) 16
1
e2t sin 4t
e2t . sin 4t
4
4
3s 2
Find L1 2
s 4 s 20
2.
3s 2 1 3s 2 1 3( s 2) 4
Sol: L1 2
L
L
2
2
2
s 4s 20
( s 2) 16
( s 2) 4
1
s2
3L1
4 L1
2
2
2
2
( s 2) 4
( s 2) 4
s
1
3e 2t L1 2
4e 2t L1 2
2
2
s 4
s 4
3e 2t cos 4t 4e 2t
1
sin 4t
4
s3
Find L1 2
s 10s 29
3.
1 s 5 8
s3
s3
1
L
Sol: L1 2
L
2
2
2
2
s 10s 29
( s 5) 2
( s 5) 2
1
s 8 5t
e5t L1 2
e cos 2t 8. sin 2t
2
2
s 2
Second shifting theorem:
f t a if t a
If L1 f ( s) f (t ), then L1{e as f (s)} G (t) , where G(t )
if t a
0
f t a if t a
Proof: We have seen that G(t )
if t a
0
then L G t e as . f (s)
L1 e as f ( s) G t
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Solved Problems :
1.
3 s
1 e s
1 e
Evaluate (i) L1 2
(ii)
L
2
( s 4)
s 1
s
1 e s
1
1 e
1
Sol: (i) L1 2
=
+
L
L
2
2
s 1
s 1
s 1
1
sin t f (t ) , say
2
s 1
Since L1
e s sin(t ) , if t
∴ By second Shifting theorem, we have L 2
, if t
s 1 0
1
e s
or L1 2 =sin(t-π)H(t-π)= -sint. H(t-π)
s 1
1 e s
Hence L1 2
=sint-sint. H (t-π) =sint [1- H (t-π)]
s 1
Where H (t-π) is the Heaviside unit step function
1 4t 1 1
e L 2
(ii) Since L1
2
s
( s 4)
= e4t .t f (t ) , say
e3s e4(t 3) .(t 3) , if t 3
2
(
4)
s
0
, if t 3
∴ By second Shifting theorem, we have L1
e3s 4(t 3)
or L
=e
.(t 3) H(t-3)
2
( s 4)
1
Where H (t-3) is the Heaviside unit step function
Change of scale property:
If L f t f s , Then L1 f as
1 t
f ,a 0
a a
Proof: We have seen that L f t f s
f as
Then
L1 f as
1
L f
a
t
, a 0
a
1 t
f ,a 0
a a
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Solved Problems :
1.
s
1
8s
If L1 2
= t sin t , find L1
2
2
2
( s 1) 2
(4s 1)
1
s
= t sin t ,
2
( s 1) 2
Sol: We have L1
2
Writing as for s,
1 1 t
as
t
t
t
= . . sin = 2 .sin , by change of scale property.
L1 2 2
2
a 2a
a
(a s 1) 2 a a
Putting a=2, we get
2s
t
8s
1
t
t
L1 2
= sin or L1
= sin
2
2
2
2
2
(4s 1) 8
(4s 1) 2
Inverse Laplace Transform of derivatives:
n
n
Theorem: L1 f ( s) f (t ) , then L1 f ( s ) (1) n t n f (t ) where f ( s)
Proof: We have seen that L t n f (t ) (1) n
n
dn
f (s)
ds n
n
d
f (s)
ds n
L1 f ( s ) (1) n t n f (t )
Solved Problems :
1.
Sol:
s 1
Find L1 log
s 1
s 1
Let L1 log
f (t )
s 1
L f (t ) log
s 1
s 1
L tf (t )
d
ds
s 1
log
s 1
L tf (t )
1
1
s 1 s 1
1
1
tf (t ) L1
s 1 s 1
1
1 1
tf (t ) 1. L1
L
s 1
s 1
=
−
+
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t f t 2sinh t f t
2sinh t
t
s 1 2sinh t
L1 log
s 1
t
t
1 s 1 e
Note: L1 log
=
s
t
2.
Sol:
Find L1 cot 1 ( s )
Let L1 cot 1 ( s) f (t )
L f (t ) cot 1 (s)
L tf (t )
d
1
1
[cot 1 ( s )
2
2
ds
1 s 1 s
1
tf (t ) L1 2 sin t
s 1
f t
sin t
t
1
L1 cot 1 ( s ) sin t
t
Inverse Laplace Transform of integrals:
1
Theorem: L
f (t )
f ( s) f (t ) , then L f ( s )ds
t
s
1
f (t )
f ( s)ds
t s
Proof: we have seen that L
f (t )
L f ( s )ds
t
s
1
Solved Problems :
s 1
2
( s 2s 2)
1.
Find L1
Sol:
Let f ( s )
1
Then L
2
s 1
( s 2s 2) 2
2
s 1
f ( s) = L 2
ds
2
s ( s 2 s 2)
1
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MATHEMATICS - I
s 1
2
2
[( s 1) 1]
= L1
s
= et L1 2
, by First Shifting Theorem
2
( s 1)
et
t
s
sin at
L1 2
2 2
( s a ) 2a
t
t
sin t e t sin t
2
2
Multiplication by power of’s’:
Theorem: L1 f ( s) f (t ) , and f (0),then L1 s f ( s) f 1 (t )
Proof: we have seen that L f 1 (t ) s f ( s ) f (0)
L f 1 (t ) s f ( s ) [
s f (s) f
f (0) 0] or
L1 s f ( s ) f 1 (t )
Note: L1
n
n
(t ),if f n (0) 0forn 1, 2,3.........n 1
Solved Problems :
1.
s
s
(ii) L1
2
2
( s 2)
( s 3)
Find (i) L1
Sol: Let f ( s)
1
Then
( s 2) 2
1 2t 1 1 2t
L1 f ( s ) L1
= e L 2 = e .t f (t ) ,
2
s
( s 2)
Clearly f (0) =0
s 1
s 1
Thus L1
= L s.
= L s. f ( s) = f 1 (t )
2
2
s
(
2)
(
2)
s
=
d
(te 2t ) = t(2e2t ) e2t .1= e2t (1 2 t)
dt
s 3t
Note: in the above problem put 2=3, then L1
= e (1 3 t)
2
( s 3)
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Division by S:
t
f s
Theorem: If L1 f ( s) f (t ) , Then L1
f u du
s 0
t
f s
Proof: We have seen that L f u du
s
0
f s t
L1
f u du
s 0
1
Note: If L
t t
f s
f ( s) f (t ) , then L 2 f u du.du
s
00
1
Solved Problems :
1. Find the inverse Laplace Transform of
1
s (s a 2 )
2
2
1
1
Sol: Since L1 2
= sinat , we have
2
(s a ) a
t1
1
L1 2
= sin atdt
2
s( s a ) 0 a
t
=
1 cos at
1
1
= a 2 (cosat 1) = a 2 (1 cosat)
a a 0
t 1
1
Then L 2 2
= 2 (1 cos at )dtdt
2
s (s a ) 0 a
1
t
=
1 sin at 1 sin at
t
= t a
a2
a 0 a 2
1 sin at
1
L1 2 2
= 2 t
2
a
s (s a ) a
Convolution Definition:
If f (t) and g (t) are two functions defined for t 0 then the convolution of f (t) and g (t) is
defined as f t * g t f u g t u du
t
0
f t * g t can alsobe written as f * g t
Properties:
The convolution operation * has the following properties
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MATHEMATICS - I
f * g t g * f t
1. Commutative i.e.
2. Associative f * g * h t f * g * h t
3. Distributive f * g h t f * g t f * h t for t 0
Convolution Theorem: If f (t ) and g(t ) are functions defined for t 0 then
L f t * g t L f t .L g t f s .g s
i.e., The L.T of convolution of f(t) and g(t) is equal to the product of the L.T of f(t) and g(t)
Proof: WKT L t e st
0
0
t
e
st
0
f u g t u dudt
t
0
f u g t u du dt
The double integral is considered within the region enclosed by the line
u=0 and u=t
On changing the order of integration, we get
L t
0
u
e st f u g t u dt du
e
f u e
e su f u
0
e su
0
sv
0
s t u
u
g t u dt du
g v dv du
put t u v
e su f u g s du g s e su f u du g s . f s
0
0
L f t * g t L f t .L g t f s .g s
Solved Problems :
1.
s
Using the convolution theorem find L1 2
2 2
(s a )
Sol:
1 s
s
1
L1 2
=
. 2
2 2 L 2
2
2
s a s a
(s a )
Let f s
s
1
and g s 2
2
s a
s a2
2
s
So that L1 f ( s ) L1 2
cos at f (t ) say
2
s a
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MATHEMATICS - I
s 1
L1 g ( s ) L1 2
sin at g (t ) say
2
s a a
By convolution theorem, we have
t
s
1
cos au. .sin a(t u)du
L1 2
2 2
a
(s a ) 0
t
1
sin(au at au) sin(au at au) du
2a 0
t
1
sin at sin(2au at ) du
2a 0
t
1
1
sin at.u .cos(2au at )
2a
2a
0
1
1
1
t
sin
at
cos
2
at
at
cos at
2a
2a
2a
1
1
1
t sin at
cos at cos at
2a
2a
2a
t
sin at
2a
2.
s2
Use convolution theorem to evaluate L 2
2
2
2
( s a )(s b )
Sol:
s2
s
s
L1 2 2 2 2 L1 2 2 . 2 2
s a s b
( s a )( s b )
1
Let f s
s
s
and g s 2
2
s a
s b2
2
s
So that L1 f ( s ) L1 2
cos at f (t ) say
2
s a
s
L1 g ( s) L1 2 2 cos bt g (t ) say
(s b )
By convolution theorem, we have
t
s
s
cos au.cosb(t u )du
L 2
. 2
2
2
s a s b 0
1
t
1
cos(au bu bt ) cos(au bu bt ) du
2 0
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MATHEMATICS - I
t
1 sin(au bu bt ) sin(au bu bt )
2
a b
ab
0
1 sin at sin bt sin at sin bt a sin at b sin bt
2
a b
ab
a 2 b2
3.
1
Use convolution theorem to evaluate L1 2
2
s( s 4)
Sol:
1 1
1
s
L 2. 2
L1 2
2
2
s( s 4)
s ( s 4)
Let f s
1
s
and g s
2
2
2
s
s 4
1
So that L1 g ( s ) L1 2 t g (t ) say
s
t.sin 2t
ts in 2t
s
s
f (t ) say L1 2
L1 f ( s) L1 2
2
2 2
s
a
4
(
)
2a
( s 4)
1
tu
s
L 2 . 2
sin 2u(t u )du
2
s ( s 4) 0 4
1
t
t
4
u sin 2udu
0
t
1 2
u sin 2udu
4 0
t
1
u
t cos 2u sin 2u
4 2
4
0
t
1 u 2
1
u
cos 2u sin 2u cos 2u
4 2
2
4
0
4.
Sol:
1
1 t sin 2t cos 2t
16
1
Find L1
2
( s 2)( s 1)
1
1
1
L1
L1
. 2
2
s 2 s 1
( s 2)( s 1)
Let f s
1
1
and g s 2
s2
s 1
1
2t
So that L1 f ( s ) L1
e f (t ) say
s 2
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MATHEMATICS - I
1
L1 g ( s ) L1 2 sin t g (t ) say
s 1
1
1
. 2
L1
s 2 s 1
t
f (u).g(t u) du
(By Convolution theorem)
0
t
t
e sin(t u) du (or) sinu.e2(t u ) du
2u
0
0
t
e
2t
sin ue
2u
du
0
t
e 2u
e 2 2sin u cos u
2 1
0
2t
1
e 2t e 2t 2sin t cos t 1 1
5
5
1 2t
e 2sin t cos t
5
5.
1
Find L1
( s 1)( s 2)
Sol:
1 1
1
1
L1
.
L
s 1 s 2
( s 1)( s 2)
Let f s
1
1
and g s
s 1
s2
1 t
So that L1 f ( s ) L1
e f (t ) say
s 1
1
2t
L1 g ( s ) L1
e g (t ) say
s 2
By using convolution theorem, we have
t u 2(t u )
1
L
du
e e
( s 1)(s 2) 0
1
t
e e
0
2 t 3u
t
du e
2t
e
0
6.
1
Find L1 2 2
2
s (s a )
Sol:
1
1
1
L1 2 2
L1 2 . 2
2
2
s s a
s (s a )
t
3u
e 3u
1 2t
t
du e
e e
3 0 3
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2t
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MATHEMATICS - I
Let f s
1
1
and g s 2
2
s
s a2
1
So that L1 f ( s ) L1 2 t f (t ) say
s
1 1
L1 g ( s ) L1 2
sinh at g (t ) say
2
s a a
By using convolution theorem, we have
t 1
1
L1 2 2 2 u. sinh a(t u )du
s (s a ) 0 a
t
1
u sinh(at au)du
a0
sin at au
1 u
cosh at au
a a
a2
0
t
1 t
1
cosh(at at ) 0 2 [0 sinh at ]
a
=a a
3.
Sol:
1 t 1
sinh at
a a a 2
1
at sinh at
a3
Using Convolution theorem, evaluate 𝑳− {
−
̅
̅
{
+
=
.
=
+
+
+
}=
−
= {
{
.
+
}⇒
= {
+
}⇒
}=
=
−
=
By Convolution theorem we have
Where
∴
−
{
∗
+
.
+
−
{ ̅
}=∫
−
=∫
=
=
=
=
−
−
−
[
{ ̅
−
{
−
+
{
. ̅
−
∫
.
[
−
}=
+
−
[
−
DEPARTMENT OF HUMANITIES & SCIENCES
+
}
}
----------------- (1)
}=
----------------- (2)
}=
−
+
. ̅
+
∗
−
−
−
𝑖
𝑖
𝑖
]
]−
]
−
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MATHEMATICS - I
Application of L.T to ordinary differential equations:
(Solutions of ordinary DE with constant coefficient):
1. Step1: Take the Laplace Transform on both the sides of the DE and then by using the
formula
L{f n (t )} s n L{f(t)} s n1 f (0) s n1 f 1 (0) s n2 f 2 (0) ............. f n1 (0)
and
apply
given initial conditions. This gives an algebraic equation.
2. Step2: replace f (0), f 1 (0) , f 2 (0) ,……… f n 1 (0) with the given initial conditions.
Where f 1 0 s f 0 – f 0
f 2 (0) s 2 f s – s f 0 f 1 0 , and so on
3. Step3: solve the algebraic equation to get derivatives in terms of s.
4. Step4: take the inverse Laplace transform on both sides this gives f as a function of t
which gives the solution of the given DE
Solved Problems :
1.
Solve y111 2 y11 y1 2 y 0 using Laplace Transformation given that
y (0) y1 (0) 0 and y11 (0) 6
Sol:
Given that y111 2 y11 y1 2 y 0
Taking the Laplace transform on both sides, we get
L y111 (t ) 2L y11 (t ) L y1 2L y 0
s 3L y (t ) s 2 y (0) sy1 (0) y11 (0) 2 s 2 L y (t ) sy (0) y1 (0)
sL y(t ) y(0) 2L y(t ) 0
s 3 2s 2 s 2 L y (t ) s 2 y (0) sy1 (0) y11 (0) 2sy (0) 2 y1 (0) y (0)
0 0 6 2.0 2.0 0
s 3 2s 2 s 2 L y (t ) 6
L y (t )
6
6
2
s 2s s 2 ( s 1)( s 1)( s 2)
3
A
B
C
s 1 s 1 s 2
A( s 1)( s 2) B( s 1)( s 2) C ( s 1)( s 1) 6
A(s 2 3s 2) B(s 2 s 2) C (s 2 1) 6
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MATHEMATICS - I
Comparing both sides s2,s,constants,we have
A B C 0,3 A B 0, 2 A 2 B C 6
A BC 0
2A 2B C 6
______________
3A B 6
3A B 0
_____________
6A 6 A 1
3 A B 0 B 3 A B 3
A B C 0 C A B 1 3 2
L y (t )
1
3
2
s 1 s 1 s 2
1
1 1
1 1
t
t
2 t
y (t ) L1
3.L
2.L
= e 3e 2.e
s 1
s 1
s 2
Which is the required solution
2.
Solve y11 3 y1 2 y 4t e3t using Laplace Transformation given that
y 0 1and y1 0 1
Sol:
Given that y11 3 y1 2 y 4t e3t
Taking the Laplace transform on both sides, we get
L{ y11 (t ) 3L y1 (t ) 2L y (t ) 4L t L e3t
s 2 L{ y (t ) sy (0) y1 (0) 3 sL{ y (t ) y (0) 2L{ y (t )
( s 2 3s 2)L{ y (t )
4
1
s4
2
s
s 3
( s 2 3s 2)L{ y(t )
4s 12 s 4 s 2 3s3 4s3 12s 2
s 2 (s 3)
L{ y(t )
s 4 7s3 13s 2 4s 12
s 2 (s 3)(s 2 3s 2)
L{ y(t )
s 4 7s3 13s 2 4s 12
s 2 (s 3)(s 1)( s 2)
4
1
2
s
s 3
s 4 7s3 13s 2 4s 12 As B
C
D
E
2
2
s ( s 3)(s 1)(s 2)
s
s 3 s 1 s 2
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MATHEMATICS - I
( As B)(s 1)(s 2)(s 3) C (s 2 )(s 1)(s 2) D(s 2 )(s 2)(s 3) E (s 2 )(s 1)(s 3)
s 2 (s 3)(s 1)(s 2)
s 4 7 s3 13s 2 4s 12 ( As B)( s3 6s 2 11s 6)
C ( s 2 )( s 2 3s 2) D( s 2 )( s 2 5s 6) E.s 2 ( s 2 4s 3)
Comparing both sides s4,s3,we have
A C D E 1......................(1)
6 A B 3C 5D 4 E 7......................(2)
put s 1, 2 D 1 D
1
2
put s 2, 4 E 8 E 2
put s 3,18C 9 C
from eq.(1) A 1
1
2
1 1
2 A3
2 2
3 5
from eq.(2) B= -7+18+ 8 3 1 2
2 2
3 2
1
1
2
y (t ) L1 2
s s 2( s 3) 2( s 1) s 2
1
1
y t 3 2t e3t et 2.e 2t
2
2
3. Using Laplace Transform Solve
Sol:
Given equation is
d2y
dy
dy
0 when t=0
2 3 y sin t , given that y
2
dt
dt
dt
d2y
dy
2 3 y sin t.
2
dt
dt
L y11 t 2L y1 t 3L y t L sin t
s 2 L y t sy 0 y1 0 2 sL y t y 0 3.L y t
s 2 2s 3 L y t
1
s 1
2
1
s 1
2
1
L y t 2
s 1 s 2 2s 3
1
y t L1
s 1 s 3 s 2 1
Now consider
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MATHEMATICS - I
1
A
B
Cs D
2
2
s 1 s 3 s 1 s 1 s 3 s 1
A s 3 s 2 1 B s 1 s 2 1 Cs D s 1 s 3 1
Comparing both sides s3,we have
put s 1,8 A 1 A
1
8
put s 3, 40 B 1 B
1
40
1 1
A B C 0 C 0
8 40
C
5 1 4 1
40
40 10
3 1 1
3 A B 2C D 0 D
8 40 5
D
15 1 8 8 1
40
40 5
1 1 1
1
s
1
8
40
10
y t L
2 5
s 1
s 1 s 3
1
1 1 1 1 1 1 s 1 1 1
L1
L
L 2 L 2
8
s 1 40
s 3 10
s 1 5
s 1
1
1
1
1
y t et e 3t cos t sin t
8
40
10
5
dx
x sin t, x 0 2
dt
4.
Solve
Sol:
Given equation is
dx
x sin t
dt
L x1 t L x t L sin t
s.L x t x 0 L x t
s.L x t 2 L x t
s 1 L x t
s 2
2
s 2
2
2
s 2
2
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MATHEMATICS - I
2
x t L1
2
2
s 1 s s 1
1 1
2L1
L
2
2
s 1
s 1 s
(By using partial fractions)
s
2
2
2
2et L1 1 12 2 12 2
s
s 1 s
2e t
1
2
et
1
cos t
. sin t
2
1
1 2
5.
Solve D 2 n 2 x a sin nt given that x=Dx=0, when t=0
Sol:
Given equation is D 2 n 2 x a sin nt
x11 t n2 x t a sin nt
L x11 t n 2 L x t L a sin nt cos a cos nt sin
s 2L x t sx 0 x1 0 n2L x t a cos L sin nt a sin L cos nt
s 2 n 2 L x t a cos
L x t a cos
s
n
2
n
n
s
a sin . 2
2
s n
s n2
2 2
2
a sin
s
s
2
n2
2
(By using convolution theorem I –part, partial fraction in II-part)
1
1
1
a sin 1
d
L
.sin nx. sin n t x dx
2
2
0 n
2
n
ds
s n
na cos
t
a cos
2n
a cos t
a
cos n t 2 x cos nt dx
sin t sin nt
2n 0
2n
a cos
2n
cos nt 2nx cos nt dx
t
0
a sin 1
t sin nt
n
2
1
at sin
2n .sin n t 2 x x cos nt 2n sin nt
0
t
a cos
2n
sin nt
at sin
2n t cos nt 2n sin nt
a cos sin nt at
cos cos nt sin sin nt
2n 2
2n
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MATHEMATICS - I
a cos sin nt at
cos nt
2n 2
2n
−
6.
Solve
Sol:
Given equation is
+
−
=
−
using L.T given that y (0) = y1 (0) = 1.
+
Applying L.T on both sides we get
−
=
−
+
⇒ {s2L[y] –s y (0) – y1 (0)} – 4{s L[y] – y (0)} + 3L{y} =
⇒ (s2 + 4s +3) L{y} –s-1-4 =
⇒ (s2 + 4s +3) L{y} =
−
y=
+
Let us consider
−
[
+
[
+
+
+
−
=
+
−
=
−
=−
−
−
−
[
[
[
+
+
+
+
+
+
+
]=
=
−
−
] =
+
[
−
+
+
[
+
+
−
−
−
]+
]+
From (1) & (2)
∴
=−
−
+
−
+
−
]=−
−
−
−
[
+
+
[
−
+
−
+
−
[
[
−
− − −→
+
−
DEPARTMENT OF HUMANITIES & SCIENCES
+
+
+
+
+
]
−
+
+
+
+
+
+
−
+
−
]+
−
+
+
+
+
+
[
]
+
+
]
+
+
+
+
+
[
+
−
=
=
[
=
+
+
−
]+
+
]=
+
+
+
+
+
+
+
+
+
+s+5
+
L{y} =
−
+s +5
+
⇒ (s2 + 4s +3) L{y} =
+
=
[
−
+
−
+
]
+
+
]
]+
− −−→
−
[
+
]
]
−
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MATHEMATICS - I
7.
Solve
Sol:
Given
+
+
=
[ ]−
⇒
⇒(
[ ]=
+
−
−
=
−
[
[ ]=
+
[
−
+
+
−
=
+
+
[
𝜋
+
+
+
⇒
∴
.
x=
[
{
⇒{
[ ]−
⇒
⇒
=
−
]−
[
+
+
−
[
+
+
[
]+
𝜋
+
−
+
]+
L[y]-
+
]+
𝑖
+
𝑖
𝜋
−
−
+
+
−
+
[
+
]
𝑖
------------------→
𝜋
+
𝑖
+
𝜋
𝑖
From (1)
=
Using L.T given y (0) =1,
−
[
=
]− [ ]= [
}− [ ]= [
+
+
+
+
+
+
−
−
+
−
+
[ ]−
−
+
[ }=
+
=
]+
−
=−
+
]+
-1= - − +
Given
Sol:
𝜋
+
−
]−
Given x ( ) = -1.
∴ − =
]
−
[ ]− −
+
⇒(
=
using L.T. given x (0) =1, x ( ) = -1.
]+ [ ]= [
L[
X=
𝝅
=
]
[ ]−
=
=
−
]
}− {
−
−
−
+ +
DEPARTMENT OF HUMANITIES & SCIENCES
+
[ ]−
−
= ,
−
= −
=−
}+
[ ]
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MATHEMATICS - I
⇒
−
L[y] =
[ ]=
=
−
−
=
W.K.T
−
[
−
[
[
−
−
−
]
]=
−
=
∴
=
!
+
−
=
−
[
!
[ ]=
]=
+
−
=
−
[
−
=
Consider
−
+
−
!
−
[
−
−
]+
]+
]+
!
−
[
−
−
+
−
−
[
−
−
−
[
]−
−
]−
−
−
]
[
−
−
−
[
]
−
[ ]
]
=
+
!
−
−
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