MATHEMATICS-II
(CONTENT MANAGEMENT)
UNIT-I
LINEAR SYSTEMS
MATRIX: A Set of mn elements arranged in the form of a rectangular array having m
rows and n columns is called mxn matrix.
ELEMENTARY TRANSFORMATIONS :
1) INTERCHANGE OF TWO ROWS:
Ri↔Rj
2) MULTIPLICATION OF EACH ELEMENT OF A ROW
WITH A NON-ZERO SCALAR :
Ri→ kRi
3) MULTIPYING EVERY ELEMENT OF A ROW WITH A
NON-ZERO SCALAR AND ADDING TO THE CORRESPONDING
ELEMENTS OF ANOTHER ROW :
Rj→ Rj+kRi
RANK :
The no of non zero rows in a echolen form of a matrix is called a rank.
NORMAL FORM:
Every non-zero mxn matrix can be reduced to the form
 Ir 0 0
 0 0 0 called the normal form.


 0 0 0
Where ‘r’ is called renk of the matrix.
EX:
2
0
Reduce the matrix A= 
2

2
4
1 
to the normal form.
5

5 11 6
1
3
3
3
4
7
Solution:
By applying elementary transformations to the above matrix
R1↔R3
1  1
4 2

1  2

1  2
0 3
0 2
0 6

1 2
R2→R2-4R1
R3→R3-2R1
R4→R4-R1
3 
1  1 0
0 6 0  10


0 0 0
0 


0  1 1  1 
C2C2+C3
3 
1  1 0
0 6 0  10


0 0 0
0 


0 0 1  1 
We proceede like this until weget
1
0

0

0
0 0 0
1 0 0
0 1 0

0 0 0
This is in Normal Form where rank is ‘3’.
ECHELON FORM :
A Matrix A is Said to be in echelon form if
1) Every row of A which has all its entries 0 occures below every row
Which has a non-zero entry.
2) The first non-zero entry in each non-zero row is equal to 1
3) The no.of zeros before the first non-zero element in a row is less than
the no.of such zeros in the next row.
EX:
1
0

0

0
1 2 2
1 0 1 
is in Echolen form
0 1 0

0 0 0
Here the rank of the Matrix is ’3’
EX:
1 0 1
0 0 0 is in echolen form


0 0 0
Here rank of the Matrix is ‘1’
SOLUTIONS OF SYSTEM OF LINEAR HOMO. EUATIONS :
CONSISTENT:
A System of equations posses at least one solution .
INCOSISTENT:
A System has no solution.
AUGUMENTED MATRIX :
Let ‘A’ be a given matrix . ‘B’ be a column matrix .
Attaching the elements of ‘A’ to elements of ‘B’ has last column , then [A/B] is called
Augumented matrix.
WORKING RULE TO FIND THE SOLUTION OF THE EQ’NS AX=0 :
Find the rank ‘r’ of of the coef. Matrix ‘A’ by reducing it to Echolen form by
Applying elementary row operations.
I)
i) If r=n,the system of eqn’s have only trivial solution.
ii) If r<n , the system of eqn’s have an infinite no.of non-trivial solution
here we have (n-r) linearly independent solutions.
II)
If the no.of eqn’s is less then no.of va’s the solution is always other than
Trivial solution.
III)
If the no.of eqn’s = no.of va’s the necessary and sufficient condition
For solutions other than a trivial solution is that the determinant of the
Coeff. Matris is zero.
EX:
Solve
x+y-2z+3w=0, x-2y+z-w=0, 4x+y-5z+8z=0, 5x-7y+2z-w=0
Solution:
The given system in matrix form is
1 1  2 3 
1  2 1  1

AX= 
4 1  5 8 


5  7 2  1
By applying elementary operations
We get
1 1  2
0  3 3
AX= 
0 0
0

0
0 0
3
 4
0 

0 
x
 y
  =O
z
 
 w
x
 y
  =O
z
 
 w
It is in echelon form rank is ‘2’
No.of independent solutions =(n-r)=4-2=2
This gives the equartions
x+y-2z+3w=0
-3y+3z-4w=0
(1)
(2)
Let z=k1 , w=k2
in (1) and (2)
we get x=k1-(5/3)k2 , y=k1-(4/3)k2, w=k2
EX:
S.T the only real number ‘λ’ for which the system
X+2y+3z=λx
, 3x+y+2z=λy
, 2x+3y+z=λz
solve them When λ=6
has non-zero solution is 6 and
Solution: The given system in matrix form is
AX=0
2
3 
1  

2 
Where A=  3 1  
 2
3 1   
Here n=3
 x
X=  y 
 z 
0 
0 
O=  
0 
 
0 
Det(A)=0
If
we get λ= 6 and other values complex
λ= 6
0 
 x  
 y  = 0 
  0 
 z   
0 
By applying elementary operations
We get
0 
3   x  
 5 2
 0  19 19  y  = 0 

   0 
 0
0
0   z   
0 
This gives the following equations
3
 5 2
 3 5 2 


 2
3  5
-5x+2y+3z=0
-19y+19z=0
(1)
(2)
Let z=k and solve (1) and (2)
We get
x=k
y=k
z=k
SOLUTIONS OF SYSTEM OF N0N-HOMO. EQUATIONS :
PROCEDURE:
The system AX=B is consistent if r(A)=r(A/B)
1) If r(A)=r(A/B)<n the system has infinitely many solutions and the system has
(n-r) independent solutions.
2) If r(A)=r(A/B)=n the system has unique solution
3) If r(A)  r(A/B) the system is inconsistent and it has no solution
EX:
Solve x+y+z=9, 2x+5y+7z=52, 2x+y-z=0
Solution:
The above equation is non-homo. Equation
The given system in matrix form is
AX=B
1 1 1   x   0 
2 5 7   y  =  0 

   
2 1 1  z  0 
By applying elementary operations
We get
1 1 1   x   9 
0 3 5   y  =  34 

  

0 0  4  z   20
The above system in echolen form r=3, n=3
r(A)=r(A/B)=n
the system is consistent and it has unique solution
This gives the euations
X+y+z=9
(1)
3y+5z=34
(2)
-4z=-20=> z=5
Solving (1) and (2) we get
X=1
Y=3
Z=5
SOME EXAMPLES:
0 1 2  2 

 and hence
1. Reduce the matrix A to its normal form. Where A  4 0 2 6


2 1 3 1 
find the rank
2. Define the rank of the matrix and find the rank of the following matrix
2
4

8

8
5
3 
13 

4  3  1
 2  3 1
4
3 9 orthogonal
3. Is the matrix

 3 1 9
1
2
4
3
1
7
4. Express the matrix A as a sum of symmetric and skew symmetric matrix where
3  2 6 
A  2 7  1
5 4
0 
 1 2 3 

3  1 and I is a unit matrix
5. Find A  3 A  9 I where A  2

 3 1
2 
2
6. Find the values of x such that the matrix A is singular where
2
2 
3  x

A 2
4 x
1 
  2
 4  (1  x)
2  2 0
4 2 0
7. Find the rank of the matrix A  
1  1 0

1  2 1
1

8. Find the value of k such that the rank of 2

3
6
2
by reducing it to the normal form
3

2
2 3
k 7  is 2
6 10
UNIT-II
EIGEN VALUES AND EIGEN VECTORS
DEFINITION:
Let A=[aij] be an nxn matrix. A non-zero vector X is said to be a characterstic vector
Of A if there exists a scalar λ such that AX=λX
If AX=λX, (X  0) we say that X is eigen vector or characterstic vector of A corr. To
the Eigen value or characterstic value λ of A.
PROCEDURE TO FIND CHARECTERSTIC VECTORS OF THE MATRIX :
Let A=[aij] be nxn matrix . Let X be an eigen vector of A corre. To the eigen value λ
Then by the definition AX=λX
i.e
AX=λIX
i.e
AX-λIX=0
i.e
(A-λI)X=0
Note that this is a Homo. System of n equations and n unknowns
This will have a non-zero solution X, if and only if | A-λI|=0
| A-λI| is called characterstic polynomial of A
( A-λI) is called characterstic matrix of A
| A-λI|=0 is called characterstic equation of A
EX:
Find the eigen values and eigen vectors of the following matrix
1 2  1
A= 0 2 2 
0 0  2
SOLUTION:
If λ is an eigen value of A and X is the corr. Eigen vector ,then by def’n we have
(A-λI)X=0
2
 1   x  0 
1  
2
2   y  = 0 
i.e.  0
 0
0
 2     z  0 
the character equation of A
1 
2
1
0
0
2
0
2
2
|A-λI|=0
= 0
We get λ=1,2,-2
If λ=1:
We get
0 2  1  x  0 
0 1 2   y  =  0 

   
0 0  3  z  0 
From this we get the equations
2y-z=0
Y+2z=0
-z=0
Solving these equations we get
x=k, y=0,z=0
Eigen vector corr. To eigen value 1 is
X=
k 
0 
 
 0 
If λ=2:
 1 2  1 
0 0 2 


 0 0  4
 x  0 
 y  = 0 
   
 z  0 
From this we get the equations
-x+2y-z=0
2z=0
-4z=0
Solving these equations we get
x=2k, y=k,z=0
Eigen vector corr. To eigen value 2 is
 x  2
 y  =k 1 
   
 z  0 
If  =-2 :
Similarly we get
 x   4 / 3
 y  =k  1 
  

 z    2 
Properties of eigen values and eigen vectors:
Property1:
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 deteriminent.
Proof:
Characteristic equation of A is |A-λI|=0
a11  
a12
a1n
a 21
a 22  
a 2n
(i.e.)
=0
an1
an2
ann  
Expanding this we get
(a11-λ)(a22-λ)………(ann-λ)-a12(a polynomial of degrww n-2)+a13(polynomial of
degrww n-2)+……=0
 (-1) n λ n +(-1) n 1 (Trace A) λ n 1 + a polynomial of degree (n-2) in λ =0
If λ1,λ2,………λn are the roots of this equation
Sum of the roots= (-1) n 1 (Trace A)/ (-1) n = Trace(A)
Further |A-λI|= (-1) n λ n +………….ao=0
Put λ=0 then |A|=ao
 Product of the roots= (-1) n ao/ (-1) n =ao=|A|
Hence the result
Property2:
If λ1,λ2,………λn are the eigen roots of A then A 3 has eigen roots λ1-k,
λ2-k,………λn-k are the eigenvalues of the matrix (A-KI)
proof:
Since λ1,λ2,………λn are the eigen roots of A
The characterstic polynomial of A is
|A-  I|=( λ1-  )(λ2-  ),………(λn-  )
(1)
Thus the characterstic polynomial of A-KI is
(A-KI-  I)X=| (A-(K+  )I)|
=[ λ1-(  +k)][λ2-(  +k)],………[λn-(  +k)]
This shows that the eigen values of A-kI are
λ1-k,λ2-k,………λn-k
Property3:
The eigen values of a triangular matrix are just the diagonal elements of the matrix.
Proof:
a11 a12
 0 a 22

Let A= 


 0
0
a1n 
a 2n
 be a triangular matrix of order n


ann 
The cha. Equation of A is |A-  I | =0
i.e. (a11-  )(a22-  )………..(a33-  )=0
  =a11,a22,………..ann.
Which are just the diagonal elements
Propert4:
If  is an eigen value of a non-singular matrix A, then |A|/  is an eigen value of adj(A)
Proof:
Since  is an eigen value of a non-singular matrix A   0
Also  is an eigen value of a non-singular matrix A umplies there exists a non-zero
vector X such that
AX=  X
(1)
(AdjA)(AX)=(adjA)(  X)
[ (AdjA)(A)]X=(adjA)  X
|A|IX=(adjA)  X
|A|/(  ) X=(adjA)X
Since X is a non-zero vector ,therefore |A|/(  ) is an eigen value of adj(A).
THE CAYLEY-HAMILTON THEOREM
STATEMENT:
Every square matrix satisfies its charecterstic equation
By using cayley-hamilton theorem we can get inverse of the matrix and we can find
powers of the matrix .
EX;
2  2
7

Find charecterstic polynomial of A=  6  1 2 


 6
2  1
Solution:
find inverse of A and A 4
2  2
7

Let A=  6  1 2 
 6
2  1
The char. Equation is given by |A-  I|=0
7-
i.e.
-6
6
2
-2
-1- 
2 =0
2
-1- 
=>  3 -5  2 +7  -3=0
By Cayley-hamilton theorem, we must have
A 3 -5A 2 +7A-3I=0
8  8
 25

A =  24  7 8 
 24
8  7
2
Multiflying (1) with A 1
A 1 [ A 3 -5A 2 +7A-3I=0
(1)
26  26
 79

A =  78  25 26 
 78
26  25
3
3 A 1 =A 2 -5A+7I
A 1 =(A 2 -5A+7I)/3
8  8  35 10  10 7 0 0
 25

A 2 -5A+7I=  24  7 8  -  30  5 10  + 0 7 0

 
 

 24
8  7  30 10  5  0 0 7
 3  2 2 
= 6
5  2

 6  2 5 
 3  2 2 
A =(1/3)  6
5  2

 6  2 5 
1
Multiflying (1) with A we get
A 4 -5A 3 +7A 2 -3A=0
=> A 4 =5A 3 -7A 2 +3A
80  80
 241

=  240  79 80 
 240
80  79
DIAGONALIZATION OF A MATRIX
If a square matrix A of order n has n linearly independent eigen vectors (X1, X2,……Xn)
Corr. to the n eigen values λ1,λ2,………λn respectively then a matrix P can be found
such that P 1 AP is a Diagonal Matrix.
MODEL AND SPECTRAL MATRICES
The matrix P in the above result which diadonalise the square matrix A is called the
‘MODEL MATRIX’ of A and the resulting diagonal matrix D is known as
‘SPECTRAL MATRIX’ of A.
P=(e1, e2, ……..en)
Where e1=
X1
|| X 1 ||
e2=
X2
|| X 2 ||
………
en=
Xn
|| Xn ||
Then P will ba an orthogonal matrix
P 1 AP=D
 P T AP =D
CALCULATION OF POWERS OF A MATRIX
We can obtain the powers of a matrix by using diagonalisation
Let A be the square matrix . then a non-singular matrix P can be found such that

P 1 AP=D
P 1 A n P=D n
A n =PD n P 1
EX:
1 0  1
Find a matrix P which transform the matrix A= 1 2 1  to Diagonal form .


2 2 3 
Hence calculate A 4
Solution:
i.e.
1 
0
1
2
2
2
1
1
= 0
3
=>  =1,2,3
If  =1 :
0 0  1  x  0 
1 1 1   y  =  0 

   
2 2 2   z  0 
From this we get the following equations
-z=0
x+y+z=0
solving these equations
x=1 , y= -1, z=0
If  =2 :
0 0  1  x  0 
1 1 1   y  =  0 

   
2 2 2   z  0 
Here we get x=-2,
Similarly for  =3
y=1,
z=2
We get x=-1,
y=1,
z=2
 1  2  1
The MODAL MATREIX P=  1 1
1 

 0
2
2 
2  1
0
P 1 = (-1/2)  2
2
0 

 2  2  1
1 0 0
P 1 AP= 0 2 0


0 0 3
There fore
 49  50  40
A =P D P =  65
66
40 
 130 130 81 
4
SOME EXAMPLES
4
1
1.
1 1 1


Find the Eigen values and the corresponding Eigen vectors of 1 1 1


1 1 1
2. Prove that the product of the Eigen values is equal to the determinant of the matrix
 1 1 1

3. Diagonalize the matrix A  0
2 1 and hence find A4

 4 4 3
 2 2  3

4. Determine the modal matrix P of  2
1  6 vertices that P-1AP is a diagonal

  1  2 0 
matrix
 8 6 2 

7  4
5. Find the Eigen values and Eigen vectors of  6

 2  4 3 
8  8 2 


6. Verify Cayley Hamilton theorem for the matrix A  4  3  2


3  4 1 
7. State and prove Cayley Hamilton theorem
1 2  1 

1  2 verify Cayley Hamilton theorem. Find A4 and A-1 using Cayley
8. If A  2

2  2 1 
Hamilton theorem
UNIT-III
COMPLEX MATRICES
DEFINATIONS:
HERMITIAN MATRIX:
A Square matrix A such that A  =A is called ‘Hermitian Matrix’
SKEW-HERMITIAN MATRIX:
A Square matrix A such that A  =-A is called ‘Skew-Hermitian Matrix’
Here A  is transeposed conjugate of A
UNITARY- MATRIX:
A Square matrix A such that A  A=I is called ‘Unitary Matrix’
PROBLEMS:
Find the gigen values and eigen vectors of the matrix
3  4i 
 2
A= 
2 
3  4i
Solution:
The char. Equation is given by |A-  I|=0
=>  = -3,7
Eigen vectors:
If  = -3
 3  4i 
The gigen vector is X= 

 5 
If  =7
3  4i 
The gigen vector is X= 

 5 
PROPERTIES:
PROPERTY1:
Prove that transpose of a unitary matrix is unitary
Proof:
Let A be a unitary matrix
We know that A  A=A A  =I
 (A  A) T =(A A  ) T =(I) T
 (A T )  A T = A T (A T )  =I
Hence (A T ) is a unitary matrix.
PROPERTY2:
The eigen values of an unitary matrix have absolutely1
Proof:
We know that AX=  X…………..(1)

X  A  =  X  ……….(2)
Multyplying 1 and 2
We get

X  X(1-   )=0

=>   =1
Hence the proof
PROPERTY3:
The eigen values of a skew-hermitian matrix are purly imaginary or zero.
PROPERTY4:
The eigen values of a Hermitian matrix real..
UNIT-IV
QUADRATIC FORMS
DEF:
An expression of the form Q=X T AX=   a ij x i x j where i=1,2,….n , j=1,2,…..n and
a ij ’ s are constants is called a quadratic form in n var’s x1,x2, …… xn.
PROBLEMS:
1) Find the symmetric matrix corr.to the quadratic form x 12 +6x1x2+5 x 22
Solution:
1 3
A= 

3 5
2) Find the quadratic form corr.to the symmetric matrix
1 2 3
2 1 3


3 3 1
Solution:
x 2 +y 2 +z 2 +4xy+5xz+6yz
CANNONICAL FORM(NORMAL FORM) :
The Q.F. X T AX transformed to the form Y T BY= 1 y 12 + 2 y 22  ........n y n2 under the
transformation X=PY is called canonical form.where B is diagonal matrix whose
diagonal elements are eigen values of A.
RANK OF Q.F. :
The total no. of terms in a Q.F. is called rank of the matrix.
It is denoted by ‘r’ .
INDEX OF Q.F. :
The no.of positive terms in a normal form is called a index of a matrix.
It is denoted by ‘s’ .
SIGNATURE OF Q.F. :
Difference between no. of positive and –ve terms is called signature.
NATURE OF A Q.F. :
1) POSSITIVE DEFINITE: if r=n and s=n
2) NEGETIVE DEFINITE: if r=n and s=0
3) POSSITIVE SEMI DEFINITE: if r<n and s=r
4) NEGETIVE SEMI DEFINITE: if r<n and s=0
5) INDEFINITE : In all other cases
SYLVESTERS LAW OF INERTIA:
The signature of Q.F. is invariant for all normal reductions.
PROBLEMS:
Reduce the quadratic form 3x 2 +5y 2 +3z 2 -2yz+2zx-2xy to canonical form .
And also find rank, index, signature of a Q.F.
SOLUTION:
 3 1 1 
The matrix of the above Q.F. is given by  1 5  1
 1  1 3 
The char. Equation is given by |A-  I|=0
The eigen values are  =2,3,6
The eigen vectors are
If
 =2
 1
X1=  0 
 
 1 
If  =3
1
X2= 1
1
If  =6
1 
X3=   2
 1 
 1 1 1 
MODAL MATRIX P : =[X1 X2 X3]=  0 1  2
 1 1 1 
 1
 2


NORMALISED MODAL MATRIX B=  0
 1
 2


This is an orthogonal matrix
Diagonalized matrix D= B 1 AB
 2 0 0
 D= 0 3 0


0 0 6
 Q= Y T AY=2x 2 +3y 2 +6z 2
Which is a required canonical fom.
Here rank(r)=3
Index(s)=3
Sign=3-0=3
r=n=s ,so it is possitive definite.
1
1
3
6

1  2
3
6
1
1
3
6


SOME EXAMPLES :
 1 1 1 1 


1  1 1 1 1 
1. Show that A 
is orthogonal
2  1 1 1 1 


 1 1 1  1
2. Reduce the quadratic form x2+y2+2z2-2xy+4xz+4yz to the canonical form
3. Prove that the product of two orthogonal matrices is orthogonal
 i 0 0


4. Show that A  0 0 i is a skew Hermitian and also unitary . Find Eigen values


0 i 0
and the corresponding Eigen vectors of A
5. Show that the matrix
3  4i 
 2
A
Hermitian. Find its Eigen values and the
2 
3  4i
corresponding Eigen vectors
6. Find the orthogonal transformation which transforms the quadratic form 6x 12+3x22+3x322x2x3 to the canonical form
7. Define i) Spectral matrix ii) Quadratic form iii) Canonical form
8. Reduce the quadratic form 3x2+5y2+3z2-2yz+2zx-2xy to the canonical form
9. Identify the nature of the quadratic form 3 x12 +3x22+3x32+2x1x2+2x1x3-2x2x3
UNIT-V
FOURIER SERIES
PROBLEMS:
1)Obtain the fourier series for f(x)=x-x 2 in [-  ,  ]
Solution:

The forier expansion is given by f(x)=ao/2 +  (a cos nx  bn sin nx)
n 1
Where a0=
1




f (x) =
 2 2
3

 4(1)
an=  (x - x )cosnx dx =
 
n2
1
n
2
n
(n not 0)

 2(1) n
bn=  (x - x )sinnx dx =
 
n2
1
There fore
2
 2
f(x)=
+
3

 4(1)
 2(1) n
{
cosnx+
sinnx)
n2
n2
n
2) Obtain the fourier series for f(x)=x in [0 , 2  ]
Solution:

The forier expansion is given by f(x)=ao/2 +  (a cos nx  bn sin nx)
n 1
n
Where a0=
an=
1

bn=
1

1

2
 f (x) =2 
0
2
 (x )cosnx dx =
0
0
2
 (x )sinnx dx
=
0
There fore f(x)=  +
2
n

2
sinnx
n
3) find half range fourie cosine series for f(x)=x in in [0 ,  ]
solution:

forier cosine expansion is given by f(x)=ao/2 +  (a cos nx)
n 1
Where a0=
an=
2

2

n

 f (x) = 
0

 x cosnx dx
=
0
for n even
4
n 2
for n odd
0
4) find half range fourie sine series for f(x)=x (  -x) in in [0 ,  ]
solution:
forier cosine expansion is given by f(x)=
Where

bn sinnx
bn=
2


 x(  - x)
sinnx dx =
0
for n even
0
8
n 3
for n odd
SOME EXAMPLES :
1. Obtain the Fourier expansion of x sin x as cosine series in (0,  ) and show that
1
1
1
1
 2



 ......... 
1.3 1.5 5.7 7.9
4
2. Find the Fourier series corresponding to the following f(x) defined in (-2, 2) as follows
2,  2  x  0
f ( x)  
of f ( x) in (-2, 2)
 x, 0  x  2
3. Find the finite Fourier cosine transform of f(x) =x 2 in (0, 1)
4. Expand f(x)=x sin x , 0<x<2  as a Fourier Series
5. Define a periodic function. Find the Fourier Expansion for the function f(x)=x-x2, 1<x<1
6. State and prove Fourier integral theorem

x2
7. Find the finite cosine transform of f ( x)   x 
3
2
8. Find a Fourier series for
f(x) if f(x) is defined in
  ,    x  0
f ( x)  
0 x 
 x,
Deduce that
2
8

1
1
1


 ...........
12 32 52
  x  
as
UNIT VI
DEFINITION:
PARTIAL DIFF. EQUATIONS
PROBLEMS:
1) Eliminate the arbitrary constants from the following equation
Z=ax+by+a 2 b 2
Solution:
Z=ax+by+a 2 b 2 ………………..(1)
Diff . (1) w,r,t, x
We get p=a
Diff. (1) w.r.t. y
We get q=b
There fore z=px+qy+p 2  q 2 is req. diff . equation.
2) Eliminate the arbitrary function from the following equation
Xyz=f(x 2  y 2  z 2 )
Solution:
Xyz=f(x 2  y 2  z 2 )………………..(1)
Diff . (1) w,r,t, x
We get yz+xyp= f ' (2x+p)………….(2)
Diff. (1) w.r.t. y
We get xz+xyq= f ' (2y+q)………….(3)
Dividing (1) and (2)
yz  xyp
f (2x + p)
=
xz  xyq
f (2y + q)
There fore 2z (x 2  y 2 )+z(px-qy)-2xy(py-qx)=0 is req. diff . equation.
PROBLEMS:
1) Solve px+qy=z
Solution:
The subsidry equations are
=>
dx dy dz


x
y
z
dx dy

x
y
Integratig this we get,
x
 c1
y
Now
dy dz

y
z
integrating this, we get
y
 c2
z
x y
hence the general solution is f( ,
y z
)=0
2) solve pq=1
Solution:
Let the complete solution is z=ax+by+c
Diff.this we get
p=a ,
q=b
from given p.d.e. ab=1
 b=1/a
 z=ax+(1/a)y+c is required solution.
3) solve p+q= sinx+siny
solution:
p-sinx=siny-q=a(say)
let
dz=pdx+qdy
dz=(a+sinx)dx+(siny-a)dy
integrating this we get,
z=-(cosx+cosy)+a(x-y)+c is required solution.
CLAIRUATS FORM :
4) solve (p+q)(z-px-qy)=1
Solution :
We can write z-px-qy=1/(p+q)
=> z=px+qy+1/(p+q)
The required solution is z= ax+by+1/(a+b)
SOME EXAMPLES:
1. Form the Partial differential equation by eliminating the arbitrary constants from
( x  a) 2  ( y  b) 2  z 2  r 2
2. Solve the Partial differential equation z ( p  q )  x  y
3. Form the Partial differential equation by eliminating the arbitrary function
2
2Z 

2
2
2
2

x  a  y b b
4. Form the Partial differential equation by eliminating the arbitrary function from
xyz  f ( x 2  y 2  z 2 )
5. Solve the Partial differential equation x( y  z ) p  y ( z  x)q  z ( x  y )
6. Solve the Partial differential equation
p
q
 2 z
2
x
y
7. Solve the Partial differential equation
p2 x  q2 y  z
8. For the partial differential equation by eliminating arbitrary constants
px 2  qy 2
10. Solve the partial differential equation ( y  z ) p  ( z  x)q  x  y
9. Solve the partial differential equation z(x-y)=
z  ax 3  by 3
UNIT-VII
SECOND ORDER P.D.E AND ITS APPLICATIONS
PROBLEMS
1) Solve by the method of separation of variables 2xz x -3yz y =0 ……….(1)
Solution : given equation is 2xz x -3yz y =0
Let Z= X(x)Y(y) be the solution of (1)…………………….(2)
Diff (2) independently w.r.t. x and y
Z ' =X ' Y
,
Z ' =XY '
(1)=> 2x X ' Y -3y XY ' =0
Separating the var’s and euquating this to an constant  we get two ordinary
Diff. equqtions and solving them we get
X=Ax  / 2 and Y=B y  / 3
(2)=>
Z= Ax  / 2 B y  / 3
=>Z= C x  / 2 y  / 3 is the required solution.
ONE-DIMENTIONAL HEAT EUATION:
2)
2
u
2  u
is one-dimentional heat equation.
c
t
x 2
Solution:
2
u
2  u
………….(1)
c
t
x 2
Let u(x,t)=X(x)T(t) be a required solution………….(2)
Diff (2) seperatly w.r.t . t and x
u '  XT ' and u '  X ''T
(1)=> XT ' = '  c 2 X ''T ………………..(3)
Separating the var’s and euquating this to an constant  we get two ordinary
Diff. equqtions and solving them we get
X=Ae px  Be  px
T=Ce c
22 2
p t
(2)=> u(x,t)= (Ae px  Be  px ) Ce c
22 2
p t
is the req. solution
ONE-DIMENTIONAL WAVE EUATION:
3)
2
 2u
2  u

c
is one-dimentional wave equation.
t 2
x 2
Solution: similarly as above
Answer: u(t,x)= (Ae px  Be  px )( De cpt  Ee  cpt )
TWO-DIMENTIONAL LAPLACE EQUATION:
4)
 2u  2u

 0 is the two dimensional laplace equation
x 2 y 2
Solution:
 2u  2u

 0 ……………………………..(1)
x 2 y 2
Let u(x,t)=X(x)Y(y) be a required solution………….(2)
Diff (2) seperatly w.r.t . t and x
u ''  X ''Y
and u ' '  X Y ''
(1)=> X ''Y  XY ''  0 ………………..(3)
Separating the var’s and euquating this to an constant  we get two ordinary
Diff. equqtions and solving them we get
X=Ae px  Be  px
Y=C cospy+D sinpy
(2)=> u(x,y)= (Ae px  Be  px ) C cospy+D sinpy) is the req. solution
UNIT VIII
FORIER INTEGRALS
FOURIER INTEGRAL THEOREM :
FOURIER SINE AND COSINE INTEGRALS :
FOURIER TRANSFORMS AND INVERSE TRANSE FORMS:
FINITE FORIER TRANSFORMS :
PROBLEMS:
1) Find the fourier transform of f(x)= { 1 for x <a
0 for x  a
Solution:

We have F[f(x)]=
e
ipx
f ( x)dx

By sub. The above function we get
F[f(x)]=2sin pa/p
2) find the fourier sine and cosine transform of the function x
Solution:

F s ( p)=  f ( x) sin pxdx
0
By sub. The above function and integratin we get
F s ( p)  0
F c ( p) 
1
p2
SOME EXAMPLES :
e i k x , a  x  b
f ( x)  
0 , x  a and x  b
f ( x)  cos x 0  x  
2 Find a Fourier cosine transform of
0
xa
1 Find the Fourier transform of
3 Using Fourier integral show that
4 Using Fourier integral show that


0
1  cos  

sin x d 
e ax  e bx 
2(b 2  a 2 )

 
,0  x  

 2

x 
 0, if

 (
0
2
 sin x
d
 a 2 )(2  b 2 )