NUMERICAL METHODS

LECTURE NOTES
SUBJECT NAME: NUMERICAL METHODS
SUBJECT CODE:MA1251
PREPARED BY
A.NALAYINI DEVI M.SC., M.PHIL.,
LECTURER, DEPARTMENT OF MATHEMATICS,
N.P.R.COLLEGE OF ENGINEERING AND TECHNOLOGY,
NATHAM-624 401

UNIT I
SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS
Solution of equation
Fixed point iteration: x=g(x) method
Newton’s method
Solution of linear system by Gaussian elimination
Gauss – Jordon method
Gauss – Jacobi method
Gauss – Seidel method
Inverse of a matrix by Gauss Jordon method
Eigenvalue of a matrix by power method
Jacobi method for symmetric matrix.

Solution of Algebraic and Transcendental Equations
The equation of the form
( ) ....... 0 1
2
2
1
0 1 n n
n n n f x a x a x a x a x a ------------(A) is called rational integral
equation
Here, 0 0 a , n is a positive integer, a0, a1, a2, ....., an are constants.
The rational integral equation is classified into two parts
1. Algebraic Equation
2. Transcendental Equation
Algebraic Equation
In this equation, f(x) is a polynomial purely in x as in (A)
Example: 3 1 0, 2 3 2 1 0 3 4 3 2 x x x x x x
Transcendental Equation
In this equation, f(x) contains some other functions such as trigonometric,
logarithmic or exponential etc.,
Example: 3x-cosx-1=0, xlog
10
x-1.2=0.
Properties
1. If f(a) and f(b) have opposite signs then one root of f(x)=0 lies between a
and b
2. To find an equation whose roots are with opposites signs to those of the
given equ. change x to (–x)
3. To find an equation whose roots are reciprocals of the roots of of the given
equ. change x to
x
1
4. Every equation of an odd degree has atleast one real root whose sign is
opposite to that of its last term.
5. Every equation of an even degree with last term negative have atleast a pair
of real roots one positive and other negative.
Methods for solving Algebraic and Transcendental Equations
Fixed point iteration: x = g(x) method (or) Method of successive
approximation.
Newton’s method (or) Newton’s Raphson method
Fixed point iteration: x = g(x) method (or) Method of successive
approximation
Let f(x) =0 be the given equation whose roots are to be determined.
Steps for this method
1. Use the first property find ‘a’ and ‘b’ where the roots lies between.

2. Write the given equation in the form x = φ(x) with the condition '(x) 1
3. Let the initial approximation be x0 which is lies in the interval (a, b)
4. Continue the process using xn = φ(xn-1)
5. If the difference between the two consecutive values of xn is very small then
we stop the process and that value is the root of the equation.
Convergence of iteration method
The iteration process converges quickly if '(x) 1where x= (x) is the
given equation. If '(x) 1, n x will become infinitely large and hence this
process will not converge. The convergence is linear.
Example: Consider the equation f(x) = x3+x-1 = 0
we can write x= (x) in three types
1. x=1-x3
2.
2 1
1
x
x
3. 1 3 x 1 x
but we take the type which has the convergence property '(x) 1
f(x) = x3+x-1
f(0)=-ve and f(1)=+ve
hence the root lies between 0 and 1
Now consider The equation (1.) x=1-x3
Here (x)= 1-x3 , ’(x)=-3x2
at x=0.9 ’(x)=-3(0.9)2
’(x)=-3(0.81)=-2.81
'(x) 1
this equation x=1-x3 will not converge
so the iteration will not work if we consider this equation
Now consider
the equation
2 1
1
x
x
Here
2 2 (1 )
2
' ( )
x
x
x
at x=0.9
0.5494
3.2761
1.8
(1 0.9 )
2 0.9
' ( )
2 2
x
'(x) 1
the equation is converge
we use this equation.
no need to consider the third type.
Problems based on Fixed point iteration
1. Find a real root of the equation x3+x2-1=0 by iteration method.

Solution:
Let f(x)= x3+x2-1
f(0)= -ve and f(1)=+ve
Hence a real root lies between 0 and 1.
Now can be written as ( )
1
1
x
x
x
In this type only '(x) 1in (0, 1)
Let the initial approximation be x0=0.5
0.75488
0.75485 1
1
( )
0.75485
0.75500 1
1
( )
0.75500
0.75427 1
1
( )
0.75427
0.75767 1
1
( )
0.75767
0.74196 1
1
( )
0.74196
0.81649 1
1
( )
0.81649
0.5 1
1
( )
7 6
6 5
5 4
4 3
3 2
2 1
1 0
x x
x x
x x
x x
x x
x x
x x
Here the difference between x6 and x7 is very small.
therefore the root of the equation is 0.75488
--------------------------------------------------------------------------------------------
2. Find the real root of the equation cosx = 3x-1, using iteration method.
Solution:
Let f(x)= cosx - 3x-1
f(0)= +ve and f(1)=-ve
A root lies between 0 and π/2
The given equation can be written as
3
sin
'( )
(1 cos ) ( )
3
1
x
x
x x x
Clearly, '(x) 1in (0, π/2)
Let the initial approximation be x0=0

(1 cos0.60710) 0.60710
3
1
( )
(1 cos0.60709) 0.60710
3
1
( )
(1 cos0.60718) 0.60709
3
1
( )
(1 cos0.60668) 0.60718
3
1
( )
(1 cos0.60933) 0.60668
3
1
( )
(1 cos0.59529) 0.60933
3
1
( )
(1 cos0.66667) 0.59529
3
1
( )
(1 cos0) 0.66667
3
1
( )
8 7
7 6
6 5
5 4
4 3
3 2
2 1
1 0
x x
x x
x x
x x
x x
x x
x x
x x
Since the values of x7 and x8 are equal, the required root is 0.60710
--------------------------------------------------------------------------------------------3.
Find the negative root of the equation x3-2x+5=0
Solution:
The given equation is x3-2x+5=0 ------------------------------------------------(1)
we know that if α, β, γ are the roots of the equation (1), then the equation whose
roots are -α, -β, - γ is x3+(-1) 0 x2+(-1)2(-2x)+(-1)35=0------------------------(2)
The negative root of the equation (1) is same as the positive root of the equation
(2)
Let f(x)= x3-2x+5
Now f(2)=-ve and f(3) =+ve
Hence the root lies between 2 and 3. Equation (2) can be written as
(2 5) 3 ( )
1
x x x
where '(x) 1 in (2, 3)
Since the values of x6 and x7 are equal, the root is 2.09455
Therefore the negative root of the given equation is -2.09455
--------------------------------------------------------------------------------------------------
Newton’s method (or) Newton’s Raphson method (Method of tangents)
Let f(x)=0 be the given equ. whose roots are to be determined.
FORMULA:
' ( )
( )
1
n
f x
n
f x
n
x
n
x

Steps for this method
i. Use the first property find ‘a’ and ‘b’ where the roots lies between.
ii. The initial approximation x0 is ‘a’ if f (a) f (b) ; the initial approximation x0
is ‘b’ if f (b) f (a) in the interval (a, b).
iii. Use the formula and continue the process
iv. If the difference between the two consecutive values of xn+1 is very small
then we stop the process and that value is the root of the equ.
Note
 The process will evidently fail if f’(x) = 0 in the neighbourhood of the root.
In such cases Regula-Falsi method should be used.
 If we choose initial approximation x0 close to the root then we get the root of
the equ. very quickly.
 The order of convergence is two
Condition for convergence of Newton’s Raphson method
Problems based on Newton’s Method
1. Compute the real root of xlog10x=1.2 correct to three decimal places using
Newton’s Raphson Method
Solution:
Let f(x)= xlog10x-1.2
Now f(2)=-ve and f(3) =+ve
Hence the root lies between 2 and 3.
f(x)= xlog10x-1.2
log 0.4343 log 0.4343
log
1
(log )
dx
d
log
1
'( ) log .
10 10
10 10
x e
e
x
e x
x
f x x x a a


2
f (x). f "(x) f '(x)

iteration value of
' ( )
( )
1
n
n
n n f x
f x
x x
initial iteration x0 3
1 2.746
log 0.4343
log 1.2
3
10 0
0 10 0
1 x
x x
x
2 2.741
log 0.4343
log 1.2
2.746
10 1
1 10 1
2 x
x x
x
3 2.741
log 0.4343
log 1.2
2.741
10 2
2 10 2
3 x
x x
x
Hence the real root of f(x)=0, correct to three decimal places is 2.741
-------------------------------------------------------------------------------------------2.
Evaluate 12 to four decimal places by Newton’s Raphson Method
Solution:
Let x= 12 x2=12 x2-12 =0
Let f(x)= x2-12 and f’(x)=2x
Now f(3)=-ve and f(4) =+ve.Hence the root lies between 3 and 4. Here
f (3) f (4) the root is nearer to 3. Therefore the initial approximation is x0=3
iteration value of
' ( )
( )
1
n
n
n n f x
f x
x x
initial iteration x0 3
1
3.5
' (3)
(3)
3 1 f
f
x
2
3.4642
' (3.5)
(3.5)
3.5 2 f
f
x
3 3.4641
' (3.4642 )
(3.4642 )
3.4642 3 f
f
x
4 3.4641
' (3.4641)
(3.4641)
3.4641 4 f
f
x
Hence the value of 12 is 3.4641
---------------------------------------------------------------------------------

Solving Simultaneous Equations with two variables using Newton’s
Method
Let the simultaneous equations with two variables be f(x, y) =0 and g(x, y) =0
x1=x0+h and y1=y0+k
x 0 0
x 0 0
2
0 y 0
0 y 0
1
x 0 y 0
x 0 y 0
1 2
g g
f f
g g
f f
g g
f f
D
D
D
and k
D
D
where
D
D
h
Problems
1. Find the solution of the equation 4x2+2xy+y2=30 and 2x2+3xy+y2=3
correct to 3 places of decimals, using Newton’s Raphson method, given
that x0=-3 and y0=2.
Solution:
Let f(x, y) = 4x2+2xy+y2-30 and g(x, y) = 2x2+3xy+y2-3
fx=8x +2y, fy=2x+2y, gx=4x+3y, gy=3x+2y
x0 y0 x 0 f
y 0 f x 0 g y 0 g 0 f 0 g
-3 2 -20 -2 -6 -5 -2 1
0.3636
88
32
0.1364
88
12
32
- 6 1
- 20 - 2
g g
f f
12
1 -5
- 2 - 2
g g
f f
88
- 6 -5
- 20 - 2
g g
f f
D
2
1
x 0 0
x 0 0
2
0 y 0
0 y 0
1
x 0 y 0
x 0 y 0
D
D
k
D
D
h
D
D
x1=x0+h and y1=y0+k

x1= -3.1364 and y1=2.364
x1 y1 x 1 f
y 1 f x 1 g
y 1 g 1 f 1 g
-3.1364 2.364 -20.360 -1.544 -5.452 -4.680 0.0995 0.0169
0.0023
0.0051
0.1984
g g
f f
0.4395
g g
f f
86.8669
g g
f f
D
2
1
x 1 1
x 1 1
2
1 y 1
1 y 1
1
x 1 y 1
x 1 y 1
D
D
k
D
D
h
D
D
x2= -3.131 and y2=2.362
x2 y2 x 2 f y 2 f x 2 g y 2 g 2 f 2 g
-3.131 2.362 -20.324 -1.538 -5.438 -4.669 0.0008 -0.0009
0.0003
0.0001
0.0226
g g
f f
0.0051
g g
f f
86.5291
g g
f f
D
2
1
x 2 2
x 2 2
2
2 y 2
2 y 2
1
x 2 y 2
x 2 y 2
D
D
k
D
D
h
D
D
x3= -3.1309 and y3=2.3617.Since the two consecutive values of x2, x3 and y2, y3
are approximately equal, the correct solution can be taken as x= -3.1309 and
y=2.3617.
---------------------------------------------------------------------------------------------------------------------

Method of False Position (or) Regula Falsi Method (or) Method of
Chords
FORMULA:
( ) ( )
( ) ( )
1 f b f a
af b bf a
x
 This is the first approximation to the actual root.
 Now if f(x1) and f(a) are of opposite signs, then the actual root lies between
x1 and a.
 Replacing b by x1 and keeping a as it is we get the next approx. x2 to the
actual root.
 Continuing this manner we get the real root.
Note:
 The convergence of the root in this method is slower than Newton’s
Raphson Method
Problems based on Regula Falsi Method
1. Sove the equation xtanx=-1 by Regula-Falsi method starting with x0=2.5 and
x1 = 3.0 correct to 3 decimal places.
Solution:
Let f(x) = xtanx+1
f(2.5)=-ve and f(3)=+ve
Let us take a=3 and b=2.5
iteration a b
( ) ( )
( ) ( )
f b f a
af b bf a
x
1 3 2.5
2.8012
(2.5) (3)
3 (2.5) 2.5 (3)
1 f f
f f
x f(2.8012)=+ve
2 2.8012 2.5
2.7984
(2.5) (2.8012 )
2.8012 (2.5) 2.5 (2.8012 )
2 f f
f f
x f(2.7984)=+ve
3 2.7984 2.5
2.7984
(2.5) (2.7984 )
2.7984 (2.5) 2.5 (2.7984 )
3 f f
f f
x
Since the two consecutive values of x2 and x3 are approximately equal, the
required root of the equation f(x)=0 is 2.7984
--------------------------------------------------------------------------------------------
2. Find the root of xex = 3 by Regula-Falsi method correct to 3 decimal places.
Solution:
Let f(x)= xex – 3
f(1)= +ve and f(1.5) = -ve
The root lies between 1 1nd 1.5
Take a = 1 and b = 1.5

iteration a b
( ) ( )
( ) ( )
f b f a
af b bf a
x
1 1 1.5
1.035
(1.5) (1)
1 (1.5) 1.5 (1)
1 f f
f f
x f(1.035)=-
ve
2 1.035 1.5
1.045
(1.5) (1.035)
1.035 (1.5) 1.5 (1.035)
2 f f
f f
x f(1.045)=-
ve
3 1.045 1.5
1.048
(1.5) (1.045)
1.045 (1.5) 1.5 (1.045)
3 f f
f f
x f(1.045)=-
ve
4 1.048 1.5
1.048
(1.5) (1.048)
1.048 (1.5) 1.5 (1.048)
4 f f
f f
x
Since the two consecutive values of x3 and x4 are equal, the required root of the
equation f(x)=0 is 1.048
--------------------------------------------------------------------------------------------
Solutions of linear algebraic equations
A system of m linear equations (or a set of m simultaneous linear equations)
in ‘n’ unknowns n x , x ,..., x 1 2 is a set of equations of the form,
m m mn n m
n n
n n
a x a x a x b
a x a x a x b
a x a x a x b
...
:::::::::: :::::::::: :::::::::: :::::::::: ::::
......
......
1 1 2 2
21 1 22 2 2 2
11 1 12 2 1 1
(1)
Where the coefficients of n x , x ,..., x 1 2 and m b ,b ,..., b 1 2 are constants.
The left hand side members of (1) may be specified by the square array of
the coefficients, known as the coefficient matrix.
m m mn
n
n
a a a
a a a
a a a
A
...... .
:::::::::: :::::::::: :
........ .
........ .
1 2
21 22 2
11 12 1
Whereas the complete set may be specified by the rectangular array
m m mn m
n
n
a a a b
a a a b
a a a b
M
...... .
:::::::::: :::::::::: :
........ .
........ .
1 2
21 22 2 2
11 12 1 1
is known as the augmented matrix.
There are two methods to solve such a system by numerical methods.
 Direct methods
 Iterative or indirect methods.

Gaussian elimination method, Gauss-Jordan method, belongs to Direct
methods,
Gauss-Seidel iterative method and relaxation method belongs to iterative
methods.
Back Substitution
Let A be a given square matrix of order ‘n’, b a given n-vector. We wish to
solve the linear system.
Ax b
For the unknown n-vector x. The solution vector x can be obtained without
difficulty in case A is upper-triangular with all diagonal entries are non-zero. In
that case the system has the form
nn n n
n n n n n n n n n
n n n n n n n
n n n n
n n n n
a x b
a x a x a x b
a x a x b
a x a x a x b
a x a x a x a x b
:::::::::: :::::::::: :::::::::: :::::::::: :::::
...
...
1, 1 1 1, 2 1
2, 2 2 2, 1 1 2
22 2 2, 1 1 2 2
11 1 12 2 1, 1 1 1 1
(1)
In particular, the last equation involves only n x ; hence, since 0 nn a , we must have
nn
n
n a
b
x
Since we now know, the second last equation
n 1,n 1 n 1 n 1,n n n 1 a x a x b
Involves only one unknown, namely, n 1 x .
As 0 n 1,n 1 a , it follows that
1, 1
1 1,
1
n n
n n n n
n a
b a x
x
With n x and n 1 x now determined, the third from last equation
n 2,n 2 n 2 n 2,n 1 n 1 n 2,n n n 2 a x a x a x b
Contains only one true unknown, namely, n 2 x . Once again, wince 0 n 2,n 2 a , we
can solve for n 2 x .
2, 2
2 2, 1 1 1 2,
2
n n
n n n n n n n n
n a
b a x b a x
x and so on.
This process of determining the solution of (1) is called Back substitution.

Gauss elimination method
Basically the most effective direct solution techniques, currently being used
are applications of Gauss elimination, method which Gauss proposed over a
century ago. In this method, the given system is transformed into an equivalent
system with upper-triangular coefficient matrix i.e., a matrix in which all elements
below the diagonal elements are zero which can be solved by back substitution.
Note
 This method fails if the element in the top of the first column is zero.
Therefore in this case we can interchange the rows so as to get the pivot
element in the top of the first column.
 If we are not interested in the elimination of x, y, z in a particular order, then
we can choose at each stage the numerically largest coefficient of the entire
coefficient matrix. This requires an interchange of equations and also an
interchange of the position of the variables.
Problems based on Gauss Elimination Method
1. Solve 2x+y+4z=12; 8x-3y+2z=20; 4x+11y-z=33 by gauss elimination method
Solution:
The given equations are,
2x+y+4z=12
8x-3y+2z=20
4x+11y-z=33
above equations can be written as
33
20
12
4 11 1
8 3 2
2 1 4
z
y
x
A X B
where, A-co efficient matrix; B-constants; X- unknown variables
The augmented matrix can be written as

3 1 3
3 1 3
2 1 2
9 7
0 0 189 189
0 7 14 28
2 1 4 12
~
2
0 9 9 9
0 7 14 28
2 1 4 12
~
4
4 11 1 33
0 7 14 28
2 1 4 12
~
4 11 1 33
8 3 2 20
2 1 4 12
,
R R R
R R R
R R R
A B
solutions are obtained from above matrix by back substitution method as
2x+y +4z = 12 (1)
7y +14z = 28 (2)
189z = 189 (3)
from the above equations we get z=1, y=2, x=3
thus the solution of the equations are x=3 ; y= 2 ; z=1
------------------------------------------------------------------------------------------------------
2. Solve 3x+4y+5z=18; 2x-y+8z=13; 5x-2y+7z=20 by gauss elimination method
Solution:
3x+4y+5z=18
2x-y+8z=13
5x-2y+7z=20
20
13
18
5 2 7
2 1 8
3 4 5
z
y
x
A X B

26 11
0 0 408 408
0 11 14 3
3 4 115 18
~
2 3 , 5 3
0 26 4 30
0 11 14 3
3 4 5 18
~
5 2 7 20
2 1 8 13
3 4 5 18
,
3 2 3
2 1 2 3 1 3
R R R
R R R R R R
A B
solutions are obtained from above matrix by back substitution method as
3x+4y +5z = 18 (1)
11y -14z = -3 (2)
-408z = -408 (3)
from the above equations we get z=1, y=1, x=3
thus the solution of the equations are x=3 ; y=1 ; z=1
------------------------------------------------------------------------------------------------------
Gauss – Jordan Method
This method is a modified from Gaussian elimination method. In this
method, the coefficient matrix is reduced to a diagonal matrix (or even a unit
matrix) rather than a triangular matrix as in the Gaussian method. Here the
elimination of the unknowns is done not only in the equations below, but also in
the equations above the leading diagonal. Here we get the solution without using
the back substitution method since after completion of the Gauss – Jordan method
the equations become
n n a
a
a
x
x
x
...
...
...
...
0 0 0 ... 1
... ... ... ... ...
... ... ... ... ...
0 1 0 ... ...
1 0 0 ... 0
2
1
2
1
Note:
 This method involves more computation than in the Gaussian method.
 In this method we can find the values of n x , x ,..., x 1 2 immediately without
using back substitution.
 Iteration method is self-correcting method, since the error made in any
computation is corrected in the subsequent iterations.

Problems based on Gauss Jordan Method
1. Solve 3x+4y+5z=18; 2x-y+8z=13; 5x-2y+7z=20 by gauss elimination method
Solution:
3x+4y+5z=18
2x-y+8z=13
5x-2y+7z=20
20
13
18
5 2 7
2 1 8
3 4 5
z
y
x
A X B
408
,
4488
,
13464
R
R
0 0 1 1
0 1 0 1
1 0 0 3
~
R 408R 111 , 408 14
0 0 408 408
0 4488 0 4488
13464 0 0 40392
~
R 11R 4 , 26 11
0 0 408 408
0 11 14 3
33 0 111 210
~
2 3 , 5 3
0 26 4 30
0 11 14 3
3 4 5 18
~
5 2 7 20
2 1 8 13
3 4 5 18
,
3
3
2
2
1
1
1 1 3 2 1
1 1 2 3 2 3
2 1 2 3 1 3
R
R
R
R
R R R R
R R R R
R R R R R R
A B
without back substitution method we get z=1, y=1, x=3
Thus the solution of the equations are x=3 ; y=1 ; z=1
--------------------------------------------------------------------------------------------
2. Solve 10x+y+z=12; 2x+10y+z=13; 2x+2y+10z=14 by gauss Jordan method
Solution:

10x+y+z=12
2x+10y+z=13
2x+2y+10z=14
14
13
12
2 2 10
2 10 1
10 1 1
z
y
x
A X B
A-co efficient matrix
B-constants
X- unknown variables
A, B
2 2 10 14
2 10 1 13
10 1 1 12
2365
R
,R
-115885
R
,R
1158850
R
R
0 0 1 1
0 1 0 1
1 0 0 1
~
R 2365R 45R ,R 2365R 4R
0 0 2365 2365
0 115885 0 115885
1158850 0 0 1158850
~
R 49R R ,R 9R - 49R
0 0 2365 2365
0 49 4 53
490 0 45 535
~
R R -5R ,R R -5R
0 9 49 58
0 49 4 53
10 1 1 12
~
3
3
2
2
1
1
1 1 3 2 2 3
1 1 2 3 2 3
2 1 2 3 1 3
witho
ut the back substitution x=1 ; y= 1 ; z=1
-------------------------------------------------------------------------------------------------------
Jacobi’s (or Gauss – Jacobi’s) iteration method
Let the system of simultaneous equations be
3 3 3 3
2 2 2 2
1 1 1 1
a x b y c z d
a x b y c z d
a x b y c z d
(1)

This system of equations can also be written as
c d a x b y
z
b d a x c z
y
a d b y c z
x
3 3 3 3
2 2 2 2
1 1 1 1
1
1
1
(2)
Let the first approximation be 0 0 x , y and 0 z . Substituting 0 0 x , y and 0 z in (2) we get,
3 3 0 3 0
3
1
2 2 0 2 0
2
1
1 1 0 1 0
1
1
1
1
1
d a x b y
c
z
d a x c z
b
y
d b y c z
a
x
Substitution the values of 1 1 x , y and 1 z in (2) we get the second approximations
2 2 x , y and 2 z as given below
3 3 1 3 1
3
1
2 2 1 2 1
2
1
1 1 1 1 1
1
2
1
1
1
d a x b y
c
z
d a x c z
b
y
d b y c z
a
x
Substituting the values of 2 2 x , y and 2 z in (2) we get the third approximations 3 3 x , y
and 3 z .
This process may be repeated till the difference between two consecutive
approximations is negligible.
Problems based on Gauss Jacobi Method
1. Solve the following equations by Gauss Jacobi’s iteration method, 20x+y-2z=17,
3x+20y-z=-18, 2x-3y+20z=25.
Solution:
The given equations are
20x+y-2z=17,
3x+20y-z=-18,
2x-3y+20z=25.
The equations can be written as,

z x y
y x z
x y z
25 2 3
20
1
18 3
20
1
17 2
20
1
Iterati
on x 17 y 2z
20
1
y 18 3x z
20
1
z 25 2x 3y
20
1
Initial x0=0 y0=0 z0=0
1
17 0.85
20
1
1 x 18 0.9
20
1
1 y 25 1.25
20
1
1 z
2
1.02
17 0.9 2.5
20
1
2 x
0.965
18 3(0.85) 1.25
20
1
2 y
1.03
25 1.7 2.7
20
1
2 z
3
1.00125
17 0.965 2(1.3)
20
1
3 x
1.0015
18 3(1.02) 1.03
20
1
3 y
1.00325
25 2(1.02) 3(0.965)
20
1
3 z
4
1.0004
17 1.0015 2(1.00325)
20
1
4 x
1.000025
18 3(1.00125) 1.00325
20
1
4 y
0.99965
25 2(1.00125) 3( 1.0015
20
1
4 z
x3~x4; y3~y4 ; z3~z4
Therefore the solution is x=1; y= -1; z=1
----------------------------------------------------------------------------------------------------------
2. Solve the following equations by Gauss Jacobi’s method 9x+2y+4z=20;
x+10y+4z=6; 2x-4y+10z=-15
Solution:
The given system of equations is
9x+2y+4z=20;
x+10y+4z=6;
2x-4y+10z=-15
The equations can be written as
z x y
y x z
x y z
15 2 4
10
1
6 4
10
1
20 2 4
9
1
Let the initial values be x0=y0=z0=0

Iterati
on x 20 2y 4z
9
1
y 6 x 4z
10
1
z 15 2x 4y
10
1
1
2.222
9
20
1 x 0.6
10
6
1 y 1.5
10
15
1 z
2 x2=2.7556 y2=0.9778 z2=-1.7044
3 x3=2.762 y3=1.0062 z3=-1.66
x2~x3; y2~y3 ; z2~z3
Therefore the solution is x=2.8; y= 1 ; z=-1.7
----------------------------------------------------------------------------------------------------------
Gauss – Seidal Iterative Method
Let the given system of equations be
n n n nn n n
n n
n n
n n
a x a x a x a x C
a x a x a x a x C
a x a x a x a x C
a x a x a x a x C
......
..............................................................
..............................................................
......
......
......
1 1 2 2 3 3
31 1 32 2 33 3 3 3
21 1 22 2 23 3 2 2
11 1 12 2 13 3 1 1
Such system is often amenable to an iterative process in which the system is first
rewritten in the form
n n C a x a x a x
a
x 1 12 2 13 3 1
11
1 ......
1
(1)
n n C a x a x a x
a
x 2 22 2 23 3 2
22
2 ......
1
(2)
n n C a x a x a x
a
x 3 32 2 33 3 3
33
3 ......
1
(3)
………………………………………….
n n n nn n
nn
n C a x a x a x
a
x ......
1
2 2 3 3 (4)
First let us assume that ..... 0 2 3 n x x x in (1) and find 1 x . Let it be *
1 x . Putting *
1 x
for 1 x and ..... 0 3 4 n x x x in (2) we get the value for 2 x and let it be *
2 x . Putting
*
1 x for 1 x and *
2 x for 2 x and ..... 0 3 4 n x x x in (3) we get the value for 3 x and let
it be *
3 x . In this way we can find the first approximate values for n x , x ,..., x 1 2 .
Similarly we can find the better approximate value of n x , x ,..., x 1 2 by using the
relation

n n C a x a x a x
a
x 1 12 2 13 3 1
11
*
1 ......
1
n n C a x a x a x
a
x 23 3 2
*
2 21 1
22
*
2 ......
1
n n C a x a x a x
a
x 3
*
32 3
*
3 31 1
33
*
3 ......
1
*
, 1 1
*
2 3
*
1 1
* ......
1
n n n n n n
nn
n C a x a x a x
a
x
Note:
 This method is very useful with less work for the given systems of equation whose
augmented matrix have a large number of zero elements.
 We say a matrix is diagonally dominant if the numerical value of the leading
diagonal element in each row is greater than or equal to the sum of the numerical
values of the other elements in that row.
 For the Gauss – Seidal method to coverage quickly, the coefficient matrix must be
diagonally dominant. If it is not so, we have to rearrange the equations in such a
way that the coefficient matrix is diagonally dominant and then only we can apply
Gauss – Seidal method.
Problems based on Gauss Seidal Method
1. Solve x+y+54z=110, 27x+6y-z=85, 6x+15y+2z=72, by using Gauss Seidal
method.
Solution:
The system of equations is
x+y+54z=110,
27x+6y-z=85,
6x+15y+2z=72,
The co-efficient matrix is
6 15 2
27 6 1
1 1 54
2 6 15
6 27 1
1 1 54
6 15 2
1 1 54
27 6 1
2 6 15
1 1 54
27 6 1
Here R1 R2

1 1 54
6 15 2
27 6 1
54 1 1
15 6 2
27 6 1
Here R2 R3
Here the matrix is diagonally dominant
The diagonally dominant matrix is
1 1 54
6 15 2
27 6 1
Thus the matrix is diagonally dominant, now the system of equations is
27x+6y-z=85,
6x+15y+2z=72,
x+y+54z=110.
x=
27
1 (85-6y+z)
y=
15
1 (72-6x-2z)
z=
54
1 (110-x-y)
The initial values be x0=y0=z0=0.
Iteration x=
27
1 (85-6y+z) y=
15
1 (72-6x-2z)
z=
54
1 (110-x-y)
1 x1=85/27=3.148 y1=
15
1 (72-18.888)
=3.5408
z1=
54
1 (110-3.148-
3.541)
=1.9132
2 x2=
27
1 (85-1.2448+1.9132)
=2.432
y2=
15
1 (72-14.592-3.8264)
=3.572
z2=
54
1 (110-2.432-
3.572)
=1.9259
3 x3=
27
1 (85-21.432+1.9258)
=2.4257
y3=
15
1 (72-14.5542-3.8516)
=3.5729
z3=
54
1 (110-2.4257-
3.5729)
=1.9259
4 x4=
27
1 (85-
21.4374+1.9259)
=2.4255
y4=
15
1 (72-14.553-3.8518)
=3.5730
z4=
54
1 (110-2.4255-
3.5730)
=1.9259
5 x5=
27
1 (85-21.438+1.9259)
=2.4255
y5=
15
1 (72-14.553-3.8518)
=3.5730
z5=
54
1 (110-2.4255-
3.5730)
=1.9259

x4~x5 ; y4~y5 ; z4~z5, thus the solution is x=2.4255, y=3.5730, z=1.9259.
------------------------------------------------------------------------------------------------------------
2. Solve 8x-3y+2z=20; 6x+3y+12z=35; 4x+11y-z=33 by Gauss Seidal method
Solution:
8x-3y+2z=20;
6x+3y+12z=35;
4x+11y-z=33
The co-efficient matrix is
4 11 1
6 3 12
8 3 2
1 4 11
3 6 12
8 3 2
Thus the matrix is not diagonally dominant
6 3 12
4 11 1
8 3 2
12 6 3
11 4 1
8 3 2
Now the matrix is diagonally dominant
8x-3y+2z=20
4x+11y-z=33
6x+3y+12z=35
x=
8
1
(20+3y-2z)
y=
11
1
(33-4x+z)
z=
12
1
(35-6x-3y)
Let the initial values be x0=y0=zo=0

Iteration
x=
8
1
(20+3y-2z)
y=
11
1
(33-4x+z) z=
12
1
(35-6x-3y)
Initial xo=0 yo=0 zo=0
1 x1=20/8=2.5
y1=
11
1
(33-
10)=2.0909
z1=
12
1
(35-15-
6)=11.439
2
x2=
8
1
(20+6-2.334)
=2.9583
y2=
11
1
(33-
11.833+1.607)
=2.0758
z2=
12
1 (35-17.7498-
6.2274)
=0.91875
3
x3=
8
1
(20+6.2274-
1.8371)
=3.0260
y3=
11
1
(33-
12.104+0.91857)
=1.9825
z3=
12
1
(35-18.156-
5.9475)
=0.9077
4
x4=
8
1
(20-5.9475-
1.8154)
=3.0165
y4=
11
1
(33-
12.066+0.9077)
=1.9856
z4=
12
1 (35-18.0996-
5.9568)
=0.9120
5
x5=
8
1
(20+5.9568-
1.824)
=3.0166
y5=
11
1
(33-
12.0664+0.9120)
=1.9859
z5=
12
1 (35-18.0996-
5.9577)
=0.9120
x4~x5 ; y4~y5 ; z4~z5,
Therefore the solution x=3.0165; y=1.9856; z=0.9120
--------------------------------------------------------------------------------------------------------
Inverse of a Matrix
Gauss Jordan Method
Let A=
3 3 3
2 2 2
1 1 1
a b c
a b c
a b c
be the given matrix
Step 1:
Write the augmented matrix
0 0 1
0 1 0
1 0 0
/
3 3 3
2 2 2
1 1 1
a b c
a b c
a b c
A I
Step 2:

Use either row or column operations make the augmented matrix [A/I] as [I /A-1].
Here A-1 is the required inverse of the given matrix.
Problems based on Inverse of a Matrix
1. Point the inverse of a matrix
1 4 9
3 2 3
2 1 1
by using Gauss-Jordan Method.
Solution:
5 3.5 0.5
12 8.5 1.5
3 2.5 0.5
4
,
4
,
4
0 0 1 5 3.5 0.5
0 1 0 12 8.5 1.5
1 0 0 3 2.5 0.5
~
2 , 3
0 0 4 20 14 2
0 4 0 48 34 6
4 0 0 12 10 2
~
, 7
0 0 4 20 14 2
0 1 3 3 2 0
2 0 2 4 2 0
~
3 2 , 2
0 7 17 1 0 2
0 1 3 3 2 0
2 1 1 1 0 0
~
1 4 9 0 0 1
3 2 3 0 1 0
2 1 1 1 0 0
/
1
3
3
2
2
1
1
1 1 3 2 2 3
1 1 2 3 2 3
2 1 2 3 1 3
A
R
R
R
R
R
R
R R R R R R
R R R R R R
R R R R R R
A I
--------------------------------------------------------------------------------------------
2. Point the inverse of a matrix
0 4 8
4 8 4
8 4 0
by using Gauss-Jordan Method.
Solution:

16
3
8
1
16
1
8
1
4
1
8
1
16
1
8
1
16
3
16
,
24
,
48
16
3
8
1
16
1
0 0 1
8
1
4
1
8
1
0 1 0
16
1
8
1
16
3
1 0 0
~
2 , 2
0 0 16 1 2 3
0 24 0 3 6 3
48 0 0 9 6 3
~
3 , 3
0 0 16 1 2 3
0 12 8 1 2 0
24 0 8 4 2 0
~
2
0 4 8 0 0 1
0 12 8 1 2 0
8 4 0 1 0 0
~
0 4 8 0 0 1
4 8 4 0 1 0
8 4 0 1 0 0
/
1
3
3
2
2
1
1
1 1 3 2 2 3
1 1 2 3 2 3
2 1 2
A
R
R
R
R
R
R
R R R R R R
R R R R R R
R R R
A I
--------------------------------------------------------------------------------------------
Eigen values and Eigenvectors
Let A be any square matrix of order n. then for any scalar , we can form a
matrix A I where I is the nth order unit matrix. The determinant of this matrix
equated to zero is called the characteristic equation of A. i.e., the characteristic
equation of the matrix A is 0 1 A . Clearly this a polynomial of degree n in
having n roots for , say n , ,..., 1 2 . These values are called eigenvalues of the
given matrix A.

For each of these eigenvalues, the system of equations A I X 0 has a non-trivial
solution for the vector
n x
x
x
X
...
2
1
. This solution
n x
x
x
X
...
2
1
is called a latent
vector or eigen vector corresponding to the eigenvalue .
If A is of order n, then its characteristic equation is of nth degree. If n is large, it is
very difficult to find the exact roots of the characteristic equation and hence the
eigenvalues are difficult to find. But there are numerical methods available for
such cases. We list below two such methods called
 Power method
 Jacobi’s method
The second method can be applied only for symmetric matrices.
Power method
This method can be applied to find numerically the greatest eigenvalue of a square
matrix (also called the dominant eigenvalue). The method is explained below.
Let n , ,..., 1 2 be the eigenvalues of A and let 1 be the dominant eigenvalue.
i.e., n ....... 1 2 3
if the corresponding eigenvectors are n x , x , x ,.... x 0 1 2 , then any arbitrary vector y can
be written as n n y a x a x ... a x 0 0 1 1 , since the eigenvectors are linearly
independent. Now
n n
k k A y A a x ... a x 0 0
a x a x a x A X X k k
n
k
n n
k k ...  0 1 0 1 2 1
...... 1
1
2
1 0 0 1 a x a x
k
k
But i n
j
i 1. 2,..., . Hence 1 0 0 A y a x k k and 0 0
1
1
1 A y a x k k .
Hence, if k is large,
A y
A y
k
k 1
1 where the division is carried out in the
corresponding components.
Here y is quite arbitrary. But generally we choose it as the vector having all its
components ones.

Note:
 If n , ,..., 1 2 are the eigenvalues of A, then the eigenvalue 1 is dominant if
1 i for i 2,3,...,n .
 The eigenvector
n x
x
x
X
...
2
1
corresponding to the eigenvalue 1 is called the
dominant eigenvector.
 If the eigenvalues of A are -3,1,2, then -3 is dominant.
 If the eigen values of A are -4,1,4 then A has no dominant eigenvalue since
4 4 .
 The power method will work satisfactorily only if A has a dominant
eigenvalue.
 Eigen vector may be
1
0
0
( )
0
0
1
( )
1
1
1
or or for 3 x 3 matrix.
Problems based on Eigen value of a matrix
1. Using power method to find a dominant eigen value of a given matrix
0 1 2
1 2 1
2 1 0
Solution:
Let A=
0 1 2
1 2 1
2 1 0
and x0=
0
0
1

0.609
1
0.805
3.416
2.082
3.416
2.75
0.541
1
0.875
0 1 2
1 2 1
2 1 0
0.541
1
0.875
0.541
1
0.875
3.428
1.856
3.428
3
0.428
1
1
0 1 2
1 2 1
2 1 0
0.428
1
1
0.428
1
1
2.8
1.2
2.8
2.8
0.2
0.8
1
0 1 2
1 2 1
2 1 0
0.2
0.8
1
0.2
0.8
1
2.5
0.5
2
2.5
0
1 2
1
0 1 2
1 2 1
2 1 0
0
1 2
1
0
1 2
1
2
0
1
2
0
0
1
0 1 2
1 2 1
2 1 0
4
4
3
3
2
2
1
1
0
Ax
Here x
Ax
Here x
Ax
Here x
Ax
Here x
Ax

0.68
1
0.726
3.414
2.348
3.414
2.48
0.674
1
0.74
0 1 2
1 2 1
2 1 0
0.674
1
0.74
0.674
1
0.74
3.414
2.3
3.414
2.528
0.65
1
0.764
0 1 2
1 2 1
2 1 0
0.65
1
0.764
0.65
1
0.764
3.414
2.3
3.414
2.61
0.609
1
0.805
0 1 2
1 2 1
2 1 0
0.609
1
0.805
7
7
6
6
5
5
Ax
Here x
Ax
Here x
Ax
Here x
Here x7 =x8 approximately.
=3.414
Eigen value=3.414
Eigen vector=
0.68
1
0.72
--------------------------------------------------------------------------------------------

Steps for finding the smallest eigen value
 First obtained the largest eigen value 1 of the given matrix.
 Let B=A- 1I. Let be the largest eigen value of the matrix B then the
numerically smallest eigen value of A is + 1.
2. Find the largest eigen value and the corresponding eigen vector of the
matrix
0 0 7
0 4 2
1 2 3
and hence find the remaining eigen values.
Solution:
Let A=
0 0 7
0 4 2
1 2 3
Step 1
To find the largest Eigen value of A
Let x0=
1
0
0
A x0=
0 0 7
0 4 2
1 2 3
1
0
0
=
7
2
3
=7
1
0.29
0.43
1 x
1
0.29
0.43
A 1 x
0 0 7
0 4 2
1 2 3
1
0.29
0.43
=
7
0.84
4.01
=7
1
0.12
0.57
2 x
1
0.12
0.57
A 2 x
0 0 7
0 4 2
1 2 3
1
0.12
0.57
=7
1
0.22
0.54
3 x
1
0.22
0.54

A 3 x
0 0 7
0 4 2
1 2 3
1
0.22
0.54
=7
1
0.16
0.56
4 x
1
0.16
0.56
A 4 x
0 0 7
0 4 2
1 2 3
1
0.16
0.56
=7
1
0.19
0.556
5 x
1
0.19
0.556
A 5 x
0 0 7
0 4 2
1 2 3
1
0.19
0.556
=7
1
0.18
0.56
6 x
1
0.18
0.56
5 x x6 approximately
Eigen value = =7
Eigen vector =
1
0.18
0.56
Step 2
To find the largest Eigen value of B
B=A- I=
0 0 0
0 11 2
6 2 3
Let y0=
1
0
0
B y0=
0 0 0
0 11 2
6 2 3
1
0
0
=
0
0
0
.

Here the we get eigen vector is zero . But the eigen vector should be non- zero. So
we consider th value of y0 be
0
0
1
B y0=
0 0 0
0 11 2
6 2 3
0
0
1
=
0
0
6
=-6
0
0
1
1 y
0
0
1
Here 1 y 0 y
0
0
1
The numerically largest eigen value of B = -6 =
Numerically smallest eigen value of A = + 1 = -6 + 7 =1= 2
1+ 2 + 3 = sum of the main diagonals of A = 1-4+7 =4
ie) 7 +1+ 3 = 4
3 =-4
The eigen values are 1, -4, 7.
--------------------------------------------------------------------------------------------

UNIT II
INTERPOLATION AND APPROXIMATION
Lagrangian Polynomials
Divided differences
Interpolating with a cubic spline
Newton’s forward difference formula
Newton’s backward difference formulA

LAGRANGIAN POLYNOMIALS
Formula
5
5 0 5 1 5 2 5 3 5 4
0 1 2 3 4
4
4 0 4 1 4 2 4 3 4 5
0 1 2 3 5
3
3 0 3 1 3 2 3 4 3 5
0 1 2 4 5
2
2 0 2 1 2 3 2 4 2 5
0 1 3 4 5
1
1 0 1 2 1 3 1 4 1 5
0 2 3 4 5
0
0 1 0 2 0 3 0 4 0 5
1 2 3 4 5
( )
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y x
Problems based on Lagrange’s Method
1. Using Lagrange’s formula to calculate f(3) from the following table (A.U. N/D. 2007)
Solution:
x x0=0 x1=1 x2=2 x3=4 x4=5 x5=6
f(x) y0=1 y1=14 y2=15 y3=5 y4=6 y5=19
We know that Lagrange’s formula is
5
5 0 5 1 5 2 5 3 5 4
0 1 2 3 4
4
4 0 4 1 4 2 4 3 4 5
0 1 2 3 5
3
3 0 3 1 3 2 3 4 3 5
0 1 2 4 5
2
2 0 2 1 2 3 2 4 2 5
0 1 3 4 5
1
1 0 1 2 1 3 1 4 1 5
0 2 3 4 5
0
0 1 0 2 0 3 0 4 0 5
1 2 3 4 5
( )
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y x
(19)
6 0 6 1 6 2 6 4 6 5
0 1 2 4 5
(6)
5 0 5 1 5 2 5 4 5 6
0 1 2 4 6
(5)
4 0 4 1 4 2 4 5 4 6
0 1 2 5 6
(15)
2 0 2 1 2 4 2 5 2 6
0 1 4 5 6
(14)
1 0 1 2 1 4 1 5 1 6
0 2 4 5 6
(1)
0 1 0 2 0 4 0 5 0 6
1 2 4 5 6
( )
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
y x
x 0 1 2 4 5 6
f(x) 1 14 15 5 6 19

(19)
6 5 4 2 1
1 2 4 5
(6)
5 4 3 1 1
1 2 4 6
(5)
4 3 2 1 2
1 2 5 6
(15)
2 1 2 3 4
1 4 5 6
(14)
1 1 3 4 5
2 4 5 6
1 2 4 5 6
1 2 4 5 6
( )
x x x x x x x x x x
x x x x x x x x x x
x x x x x x x x x x
y x
put x=3 we get
(19)
240
3(3 1)(3 2)(3 4)(3 5)(3 6)
(6)
60
3(3 1)(3 2)(3 4)(3 6)
(5)
48
3(3 1)(3 2)(3 5)(3 6)
(15)
48
3(3 1)(3 4)(3 5)(3 6)
(14)
60
3(3 2)(3 4)(3 5)(3 6)
240
(3 1)(3 2)(3 4)(3 5)(3 6)
y(3)
(19)
240
3(2)(1)( 1)( 2)
(6)
60
3(2)(1)( 1)( 3)
(5)
48
3(2)(1)( 2)( 3)
(15)
48
2(1)( 1)( 2)( 3)
(14)
60
3(1)( 1)( 2)( 3)
240
2(1)( 1)( 2)( 3)
10
240
2400
240
12 1008 2700 900 432 228
240
228
60
108
48
180
48
540
60
252
240
12
Answer: y(3) = 10.
------------------------------------------------------------------------------------------------------------
2. Using Lagrange’s formula fit a polynomial to the data (A.U. N/D. 2006)
Solution:
x x0=0 x1=1 x2=3 x3=4
f(x) y0= -12 y1=0 y2=6 y3=12
( )
3
3 0 3 1 3 2 3 4 3 5
0 1 2 4 5
2
2 0 2 1 2 3 2 4 2 5
0 1 3 4 5
1
1 0 1 2 1 3 1 4 1 5
0 2 3 4 5
0
0 1 0 2 0 3 0 4 0 5
1 2 3 4 5
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y x
x 0 1 3 4
f(x) -12 0 6 12

( ) 7 18 12 (1)
6 12 6 12
( 1)[ 6 12]
( 1)[ 3 4 12 4 3 ]
( ) ( 1)( 3)( 4) ( 1)( 4) ( )( 1)( 3)
(12)
(4)(3)(1)
( )( 1)( 3)
(6)
(3)(2)( 1)
( )( 1)( 4)
( 12)
( 1)( 3)( 4)
( 1)( 3)( 4)
( )
(12)
(4 0)(4 1)(4 3)
( 0)( 1)( 3)
(6)
(3 0)(3 1)(3 4)
( 0)( 1)( 4)
(0)
(1 0)(1 3)(1 4)
( 0)( 1)( 3)
( 12)
(0 1)(0 3)(0 4)
( 1)( 3)( 4)
( )
3 2
3 2 2
2
2 2 2
y x x x x
x x x x x
x x x
x x x x x x x x
y x x x x x x x x x x
x x x x x x x x x
y x
x x x x x x
x x x x x x
y x
Substituting x=2 in (1), we get
y(2) = 23-7(22)+18(2)-12 = 8 – 28 +36 -12 = 44 -40 =4
Answer: y(2) = 4.
------------------------------------------------------------------------------------------------------------
3. Using Lagrange’s interpolation find the polynomial through (0, 0), (1, 1) and (2, 2)
(A.U. M/J. 2007)
x 0 1 2
f(x) 0 1 2
Solution:
x x0=0 x1=1 x2=2
f(x) y0= 0 y1=1 y2=2
The required polynomial is y x
( )
2
( 2) ( 1)
2
2 1
1
1
1 1
2
( )
2
2 0 2 1
0 1
1
1 0 1 2
0 2
0
0 1 0 2
1 2
( )
( )
2 2
2
2 0 2 1
0 1
1
1 0 1 2
0 2
0
0 1 0 2
1 2
y x x
x x x x
x x x x
x x x x
y x
x x x x x x
y x
y
x x x x
x x x x
y
x x x x
x x x x
y
x x x x
x x x x
y x
------------------------------------------------------------------------------------------------------------

4. The following table gives certain corresponding values of x and log
10
x. Compute the value of
log
10
323.5, by using Lagrange’s formula
Solution:
x x0=321.0 x1=322.8 x2=324.2 x3=325.0
f(x) y0= 2.50651 y1= 2.50893 y2= 2.51081 y3= 2.51188
( )
3
3 0 3 1 3 2 3 4 3 5
0 1 2 4 5
2
2 0 2 1 2 3 2 4 2 5
0 1 3 4 5
1
1 0 1 2 1 3 1 4 1 5
0 2 3 4 5
0
0 1 0 2 0 3 0 4 0 5
1 2 3 4 5
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y
x x x x x x x x x x
x x x x x x x x x x
y x
Substituting these values in Lagrange’s interpolation formula, we get,
(323.5) 2.50987
0.07996 1.18794 1.83897 0.43708
(2.51188)
(325 321)(325 322.8)(325 324.2)
(323.5 321)(323.5 322.8)(323.5 324.2)
(2.51081)
(324.2 321)(324.2 322.8)(324.2 325)
(323.5 321)(323.5 322.8)(323.5 325)
(2.50893)
(322.8 321)(322.8 324.2)(322.8 325)
(323.5 321)(323.5 324.2)(323.5 325)
(2.5061)
(321 322.8)(321 324.2)(321 325)
(323.5 322.8)(323.5 324.2)(323.5 325)
(323.5)
f
f
------------------------------------------------------------------------------------------------------------
Inverse Interpolation
The process of finding a value of x for the corresponding value of y is called inverse
interpolation.
x 321.0 322.8 324.2 325.0
f(x) 2.50651 2.50893 2.51081 2.51188

n
n n n n
n
n
n
n
n
x
y y y y y
y y y y y y
x
y y y y y y
y y y y y y
x
y y y y y y
y y y y y y
x
1 1
0 1 1
1
1 0 1 2 1
0 2
0
0 1 0 2 0
1 2
0 ...
...
...
...
...
...
...
Problems based on Inverse Interpolation
1. Find the value of x when y=85, using Lagrange’s formula from the following table.
Solution:
3
3 0 3 1 3 2
0 1 2
2
2 0 2 1 2 3
0 1 3
1
1 0 1 2 1 3
0 2 3
0
0 1 0 2 0 3
1 2 3
x
y y y y y y
y y y x y y
x
y y x y y y
y y y y y y
x
y y y y y y
y y y y y y
x
y y y y y y
y y y y y y
x
Substituting the above values, we get,
(14)
(68.7 94.8)(68.7 87.9)(68.7 81.3)
(85 94.8)(85 87.9)(85 81.3)
(8)
(81.3 94.8)(81.3 87.9)(81.3 68.7)
(85 94.8)(85 87.9)(85 68.7)
(5)
(87.9 94.8)(87.9 81.3)(87.9 68.7)
(85 94.8)(85 81.3)(85 68.7)
(2)
(94.8 87.9)(94.8 81.3)(94.8 68.7)
(85 87.9)(85 81.3)(85 68.7)
x
x = 0.1438778+3.3798011+3.3010599-0.2331532 = 6.3038
Therefore the value of x when y = 6.3038
---------------------------------------------------------------------------------------------------
2. The following table gives the value of the elliptic integral
0 2 sin
2
1
1
( )
d
y
for
certain values of θ. Find if y( ) = 0.3887
Solution:
x 2 5 8 14
y 94.8 87.9 81.3 68.7
x x0=2 x1=5 x2=8 x3=14
y y0=94.8 y1=87.9 y2=81.3 y3=68.7
21° 23° 25°
y( ) 0.3706 0.4068 0.4433

3
3 0 3 1 3 2
0 1 2
2
2 0 2 1 2 3
0 1 3
1
1 0 1 2 1 3
0 2 3
0
0 1 0 2 0 3
1 2 3
y y y y y y
y y y x y y
y y x y y y
y y y y y y
y y y y y y
y y y y y y
y y y y y y
y y y y y y
we have
21.999
(25)
(0.3706 0.3706)(0.3706 0.4068)
(0.3887 0.3706)(0.3887 0.4068)
(23)
(0.4068 0.3706)(0.4068 0.4433)
(0.3887 0.3706)(0.3887 0.4433)
(21)
(0.3706 0.4068)(0.3706 0.4433)
(0.3887 0.4068)(0.3887 0.4433)
Therefore the value of θ such that y( ) = 0.3887 is θ=21.999°
θ1=21° θ2=23° θ3=25°
y( ) y1=0.3706 y2=0.4068 y3=0.4433

DIVIDED DIFFERENCES
Problems based on Newton’s Divided Difference Formula
2.2.1 Let the function y f x take the values n f x , f x , f x ,... f x 0 1 2 corresponding to
the values n x , x x ,...x 0 1, 2 for the argument x where 1 0 2 1 1 , ,..., n n x x x x x x need not be
necessarily equal.
The first divided of f x for the argument 0 1 x , x is defined as
`
1 0
1 0
0 1 ,
x x
f x f x
f x x ...(i)
Similarly
2 1
2 1
1 2 ,
x x
f x f x
f x x
3 2
3 2
2 3 ,
x x
f x f x
f x x and so on.
Thus, for defining a first divided difference, we need the functional values
corresponding to two arguments.
The second divided difference of f(x) for three arguments
x0, x1, x2 is defined as
2 0
1 2 0 1 , ,
x x
f x x f x x
... (ii)
Similarly
3 1
2 3 1 2
1 2 3
, ,
, ,
x x
f x x f x x
f x x x
The third divided difference of f(x) for the four arguments
x0, x1, x2,x3 is defined as
3 0
1 2 3 0 1 2
0 1 2 3
, , , ,
, , ,
x x
f x x x f x x x
f x x x x ... (iii)
The quantities in (i), (ii) and (iii) are called divided differences of orders 1, 2, 3
respectively.
1. Using Newton’s Divided Difference formula, find the value of f(8) and f(5) given
the following data.
x 4 5 7 10 11 13
f(x) 48 100 294 900 1210 2028

Solution:
x x0=4 x1=5 x2=7 x3=10 x4=11 x5=13
f(x) f(x0)=48 f(x1)=100 f(x2)=294 f(x3)=900 f(x4)=1210 f(x5)=2028
The divided difference table for the given data is given below.
Newton’s divided difference formula,
f(x) = f(x0) + (x-x0)f(x0,x1) + (x-x0) (x-x1) f(x0,x1,x2) + (x-x0) (x-x1) (x-x2) f(x0,x1,x2,x3)
+ ……………..
x f(x) Δf(x) Δ2f(x) Δ3f(x) Δ4f(x)
4 48
52
5 4
100 48
5 100 15
7 4
97 52
97
7 5
294 100
52
5 4
100 48
7 294 21
10 5
292 97
0
202
10 7
900 294
52
5 4
100 48
10 900 27
11 7
310 202
0
310
11 10
1210 900
52
5 4
100 48
11 1210 33
13 10
409 310
409
13 11
2028 1210
13 2028
Using divided differences and the given data in (1),
f(x) = 48 + 52(x-4) + 15(x-4)(x-5) +(x-4)(x-5)(x-7)
when x=8,
f(8) = 48 + 208 +180 +12 =448
Therefore f(8) =448
when x=15,
f(15) = 48 + 572 +1650 +880 =3150
Therefore f(15) =3150
------------------------------------------------------------------------------------------------------------
2. Use Newton’s divided difference formula, to fit a polynomial to the data

and find y when x=1
Solution:
x x0=-1 x1=0 x2=2 x3=3
y y0=8 y1=3 y2=1 y3=12
The divided difference table for the given data as follows.
x y Δy Δ2y Δ3f(x)
-1 -8
11
0 1
3 8
0 3 4
2 1
1 11
1
2 0
1 3
2
3 1
4 4
2 1 4
2 1
11 1
11
3 2
12 1
3 12
By Newton’s divided difference formula,
f(x) = f(x0) + (x-x0)f(x0,x1) + (x-x0) (x-x1) f(x0,x1,x2) + (x-x0) (x-x1) (x-x2) f(x0,x1,x2,x3)
+ ……………..
Using these we get,
f(x) = -8+(x+1)11+(x+1)x(-4)+(x+1)x(x-2)2
= -8+11x+11-4x2-4x+2x3-2x2-4x
= 2x3-6x2+3x+3
y = 2x3-6x2+3x+3
y(1) = 2-6+3+3 =2
Answer: y(1)=2
------------------------------------------------------------------------------------------------------------
x -1 0 2 3
y 8 3 1 12

Interpolating with a Cubic Spline
Definition: Cubic Spline
A cubic polynomial approximating the curve in every subinterval is called is called
satisfying the following properties.
1. F(xi)=fi for i = 0, 1, 2, ………, n.
2. On each interval [xi-1, xi], 1 ≤ i ≤ n, F(x) is a third degree polynomial.
3. F(x), F’(x) and F”(x) are continuous on the interval [x0, xn].
The second derivatives at the end points of the given range are denoted as M0 and Mn
respectively.
Natural cubic spline
A cubic spline F(x) with end conditions M0=0 and Mn=0 where f”(xi)=Mi in the interval
[x0, xn]
i.e., f”(x0) = 0 and f”(xn) = 0 is called a natural cubic spline.
Fitting a natural cubic spline for the given data:
In case of natural cubic spline the derivatives f”(x0) = 0 and f”(xn) = 0. i.e., M0=0 and
Mn=0
Suppose that the values of x are equally spaced with a spacing h.
The cubic spline approximation in the subinterval (xi-1, xi) is given by
i i
i
i i
i
i
i
i
i
i M
h
y
h
x x
M
h
y
h
x x
M
x x
M
x x
h
F x
6 6 6
( )
6
1 ( )
( )
2
1
1
2
1
3
1
1
3
Mi-1+4Mi+Mi+1 = 2
6
h
[yi-1+yi+1-2yi] where i = 1, 2, ….., n- 1 with M0=0 and Mn=0.
Problems based on Cubic spline
1. Obtain the cubic spline approximation for the function y = f(x) from the following data,
given that y0”=y3”=0.
x -1 0 1 2
y -1 1 3 35
Solution:
x x0=-1 x1=0 x2=1 x3=2
y y0=-1 y1=1 y2=3 y3=35
The values of x are equally spaced with h =1.
Therefore we have
Mi-1+4Mi+Mi+1 =6 [yi-1+yi+1-2yi] where i = 1, 2, ….., n- 1
Further M0=0 and M3=0
M0+4M1+M2 =6 (y0+y2-2y1)
4M1+M2 =0 ----------------------------------------------------------------------------(1)
and

M1+4M2+M3 = 6 (y1+y3-2y2)
= 6(1-6+35) = 180
M1+4M2 = 180 -----------------------------------------------------------------------(2)
Solving (1) and (2), M1=-12 M2 = 48
The cubic spline in (xi-1, xi) is given by
i i
i
i i
i
i
i
i
i y M
x x
y M
x x
M
x x
M
x x
y
6
1
6 1
1
6 1
( )
6
( ) 1
1 1
3
1
1
3
---------------(3)
where i = 1, 2, ….., n- 1.
In the interval -1≤ x≤ 0, i.e., x0≤ x≤ x1(i=1) the cubic spline is given by
1 1
0
0 0
1
1
3
0
0
3
1
6
1
6 1
1
6 1
( )
6
( )
y M
x x
y M
x x
M
x x
M
x x
y
[( 1) ( 12)] ( )( 1) ( 1)(1 2)
6
1 3 y x x x
= (-2)(x3+3x2+3x+1)+x+3x+3
y = -2x3-6x2-2x+1
In the interval 0≤ x≤1, i.e., x1≤ x≤ x2(i=2) the cubic spline is given by
2 2
1
1 1
2
2
3
1
1
3
2
6
1
6 1
1
6 1
( )
6
( )
y M
x x
y M
x x
M
x x
M
x x
y
[(1 ) ( 12) (48)] (1 )(1 2) ( 0)(3 4)
6
1 3 3 y x x x x
= (-2)(1-x)3+8x3+3-3x-x
y = 2x3-6x2+6x-2+8x3+3-4x
y = 10x3-6x2+2x+1
3 3
2
2 2
3
3
3
2
2
3
3
6
1
6 1
1
6 1
( )
6
( )
y M
x x
y M
x x
M
x x
M
x x
y
[(2 ) (48)] (2 )(3 8) ( 1)(35)
6
1 3 y x x x
= 8(8-12x+6x2-x3)+5x-10+35x-35
y = -8x3+48x2-56x+19
y = -8x3+48x2-56x+19
Hence the required cubic spline approximation for the given function is
8x 48x 56x 19 for 1 x 2
10x 6x 2x 1 f or 0 x 1
2x 6x 2x 1 f or 1 x 0
3 2
3 2
3 2
y
2. Obtain the natural cubic spline which agrees with y(x) at the set of data points given
below:
Hence find y(2.5)
x 2 3 4
y 11 49 123

Solution:
The values of x are equally spaced with h =1.
Therefore we have
Mi-1+4Mi+Mi+1 =6 [yi-1+yi+1-2yi] where i = 1, 2, ….., n- 1
Further M0=0 and M2=0
M0+4M1+M2 =6 (y0+y2-2y1)
4M1=6(11-98+123)
M1= 54
The cubic spline in (xi-1, xi) is given by
i i
i
i i
i
i
i
i
i y M
x x
y M
x x
M
x x
M
x x
y
6
1
6 1
1
6 1
( )
6
( ) 1
1 1
3
1
1
3
---------------(3)
where i = 1, 2, ….., n- 1.
In the interval 2≤ x≤ 3, i.e., x0≤ x≤ x1(i=1) the cubic spline is given by
1 1
0
0 0
1
1
3
0
0
3
1
6
1
6 1
1
6 1
( )
6
( )
y M
x x
y M
x x
M
x x
M
x x
y
[( 2) (54)] (3 )(11) ( 2)(49 9)
6
1 3 y x x x
= 9(x3-6x2+12x-8)+33-11x+40x-80
y = 9x3-54x2+137x-119
2 2
1
1 1
2
2
3
1
1
3
2
6
1
6 1
1
6 1
( )
6
( )
y M
x x
y M
x x
M
x x
M
x x
y
[(4 ) (54)] (4 )(40) ( 3)(123)
6
1 3 y x x x
= 9(64-48x+12x2-x3)+160-40x+123x-369
y = -9x3+108x2+349x+367
Hence the required cubic spline approximation for the given function is
-9x 108x 349x 367 f or 3 x 4
9x 54x 137x -119 for 2 x 3
3 2
3 2
y
------------------------------------------------------------------------------------------------------------
Newton’s Forward and Backward Difference Formulas
Introduction:
If a function y=f(x) is not known explicitly the value of y can obtained when a set of
values of (xi, yi) i = 1, 2, 3, ….n are known by using the methods based on the principles of finite
differences, provided the function y=f(x) is continuous. Here the values of x being equally
spaced, i.e., xn = x0 +nh, n = 0, 1, 2, …., n
Forward Differences
x 2 3 4
y 11 49 123

If y0, y1, y2,…, yn denote the set of values of y, then the first forward differences of y =
f(x) are defined by
Δy0 = y1-y0; Δy1=y2-y1; …….; Δyn-1=yn-yn-1
where Δ is called the forward difference operator.
Forward Difference table
Formula
.......
3!
( 1)( 2)
2!
( 1)
( ) 0
3
0
2
0 0 0 y
n n n
y
n n
y x nh y n y
x y Δ Δ2 Δ3 Δ4 Δ5
x0 y0
Δy0
x1 y1 Δ2y0
Δy1 Δ3y0
x2 y2 Δ2y1 Δ4y0
Δy2 Δ3y1 Δ5y0
x3 y3 Δ2y2 Δ4y1
Δy3 Δ3y2
x4 y4 Δ2y3
Δy4
x5 y5

Problems based on Newton’s forward interpolation formula
1. Using Newton’s Forward interpolation formula, find f(1.5) from the following data
x 0 1 2 3 4
f(x) 858.3 869.6 880.9 829.3 903.6
Solution:
Difference Table
To
find
y for
x =
1.5
0.5
1
1.5 1.0 0 n
h
x x
n
By
Newt
on’s
forw
ard interpolation formula,
(1.5) 875.7375
(1.5) 869.6 5.65 0.01875 0.46875
6
(0.28125)
2
(0.0375)
(1.5) 869.6 5.65
(0.3)
6
(0.5)(0.5)(1.5)(2.5)
(0.1)
2
(0.5)(0.5)(1.5)
(1.5) 869.6 (0.5)(11.3)
( 0.3)
6
(0.5)(0.5 1)(0.5 2)(0.5 3)
(0.1)
2
(0.5)(0.5 1)(0.5 2)
(0)
2
(0.5)(0.5 1)(0.5 2)
(1.5) 869.6 (0.5)(11.3)
.......
3!
( 1)( 2)
2!
( 1)
( ) 0
3
0
2
0 0 0
y
y
y
y
y
y
n n n
y
n n
y x nh y n y
------------------------------------------------------------------------------------------------------------
2. Using Newton’s forward interpolation, find the value of 10 log ,given log3.141 =
0.4970679364 log3.142 = 0.4972061807 log3.143 = 0.4973443810
log3.144 = 0.49748253704 log3.145 = 0.4974825374
Solution
x x0=0 x1=1 x2=2 x4=3 x5=4
f(x) 858.3 869.6 880.9 829.3 903.6
x y Δy Δ2y Δ3y Δ4y
0 858.3
869.6-858.3=11.3
1 869.6 11.3-11.3=0
880.9-869.6=11.3 0.1-0=0.1
2 880.9 11.4-11.3=0.1 -0.2-0.1=-0.3
892.3-880.9=11.4 -0.1-0.1=-0.2
3 892.3 11.3-11.4=-0.1
903.6-892.3=11.3
4 903.6
x y=logx Δy Δ2y Δ3y Δ4y

Here x0=3.141, h = 0.001, y0 = 0.4970679364
The Newton’s forward interpolation formula is
x0+nh=π=3.1415926536
n=
0.001
3.1415926536-3.141
=0. 5926536
( ) 0.4971498727
( ) 0.4970679364 0.0000819310 0.0000000053
2
0.5926536( 0.4073464)( 0.440 10 )
( ) 0.4970679364 (0.5926536)(0.000138244)
.......
3!
( 1)( 2)
2!
( 1)
( )
7
0
3
0
2
0 0 0
y
y
y
y
n n n
y
n n
y x nh y n y
----------------------------------------------------------------------------------------------------------
3. From the following data, estimate the no. of persons earning weekly wages between 60
and 70 rupees.
Wages(in Rs.) Below 40 40-60 60-80 80-100 100-120
No.of person(in thousands) 250 120 100 70 50
Solution
Here x0=40, h = 20, y0 = 250
The Newton’s forward interpolation formula is
x0+nh=70
n=
20
70 - 40
=1.5
0 0.4970679364
0.1382443x10-3
1 0.4972061807 11.3-11.3=0
0.1382003x10-3 0.1-0=0.1
2 0.4973443810 11.4-11.3=0.1 -0.2-0.1=-0.3
0.1381564x10-3 -0.1-0.1=-0.2
3 0.49748253704 11.3-11.4=-0.1
0.1381124x10-3
4 0.4974825374
x x0=40 x1=60 x2=80 x4=100 x5=120
y 250 250+120=370 370+100=470 470+70=540 540+50=590
wages
x
Frequency(y) Δy Δ2y Δ3y Δ4y
Below 250

(70) 423.5937
(70) 250 180 7.5 0.625 0.46875
(20)
24
(1.5)(1.5 1)(1.5 2)(1.5 3)
( 10)
6
(1.5)(1.5 1)(1.5 2)
( 20)
2
(1.5)(1.5 1)
(70) 250 (1.5)(120)
.......
3!
( 1)( 2)
2!
( 1)
( ) 0
3
0
2
0 0 0
y
y
y
y
n n n
y
n n
y x nh y n y
No
. of
per
so
ns
wh
ose
we
ekly wages below 70 =423.5937
No. of persons whose weekly wages below 60 = 370
No. of persons whose weekly wages No. of persons whose weekly wages below 70-
between 60 and 70 = No. of persons whose weekly wages below 60
= 423.5937-370
---------------------------------------------------------------------------------------------------------
40
120
Below
60
370 -20
100 -10
Below
80
470 -30 20
70 10
Below
100
540 -20
50
Below
120
590

Newton’s Backward Interpolation formula
The formula is used mainly to interpolate the values of y near the end of a set of values of
y a short distance ahead (to the right) of y.
Formula:
.......
3!
( 1)( 2)
2!
( 1)
( ) 2 3
n n n n n y
n n n
y
n n
y x nh y n y
Problems based on Newton’s backward difference formula
1. The following data are taken from the steam table:
Temp:°C 140 150 160 170 180
Pressure kgf/cm2 3.685 4.854 6.302 8.076 10.225
Find the pressure at temperature t = 142°C and t = 175°C
Solution: We form the difference table:
0.2
5
1
10
142 140 0
h
t t
u
10
5
1
142 140 4 4 P P
...
2
1
0
2
0 0 P
u u
P u P
0.279
2
0.2 0.8
3.685 0.2 1.169
0.002
24
0.2 0.8 1.8 2.8
0.047
6
0.2 0.8 1.8
=3.685+0.2338-0.02332+0.002256-0.0000672
=3.897668
=3.898
0.5
10
175 180
10 ,
2
1
175 180 4 4 P t P where v o
...
2
1 2
n n n P P P
0.375
2
0.5 0.5
10.225 0.5 2.149
+ 0.002
24
0.5 0.5 1.5 2.5
0.049
6
0.5 0.5 1.5
=10.225-1.0745-0.0046875-0.0030625-0.000078125
=9.10048438=9.100
t p Δp Δ2p Δ3p Δ4p
140 3.685
1.169
150 4.854 0.279
1.448 0.047
160 6.302 0.326 0.002
1.774 0.049
170 8.076 0.375
2.149
180 10.225

UNIT III
NUMERICAL DIFFERENTIATION AND INTEGRATION
Differentiation using interpolation formulae
Numerical integration by trapezoidal rule
Simpson’s 1/3 and 3/8 rules
Romberg’s method
Two and Three point Gaussian quadrature formulas
Double integrals using trapezoidal and simpsons’s rules.

Numerical Differentiation
 Differentiation using Forward Interpolation formula(for equal interval)
 Differentiation using Backward Interpolation formula(for equal interval)
 Differentiation using Stirling’s(Central Difference) Formula(for equal interval)
 Maximum and Minimum
 Differentiation using divided difference(for unequal interval)
Forward difference formula to compute the derivative
Newton’s forward interpolation formula is
...
4!
1 2 3
3!
1 2
2!
1
0
4
0
3
0
2
0 0 0 y
r r r r
y
r r r
y
r r
f x rh y y
(Here using of r for n is only for convenience)
Differentiating w.r.t. we get,
...
12
2 9 11 3
6
3 6 2
2
2 1
' 0
4
3 2
0
3
2
0
2
0 0 y
r r r
y
r r
y
r
hf x rh y ---
(1)
...
12
6 18 11
" 1 0
4
2
0
3
0
2
0
2 y
r r
h f x rh y r y
... (2)
...
2
2 3
"' 3 0
4
0 0
3 y
r
h f x rh y
...(3)
Similarly we can find the remaining derivatives.
If we want to find the derivatives at a point , 0 x x then o x rh x 0
i.e., r=0.
Hence on substituting this value of r=0 in the above formula (1), (2) and (3), we get
...
4
1
3
1
2
1 1
' 0
4
0
3
0
2
0 0 y y y y
h
f x
...
12
1 11
' ' 0
4
0
3
0
2
0 2 y y y
h
f x
... and so on.
2
3
3
1
' ' ' 0
4
0 3 0 y y
h
f x
Note: If the x value is nearer to the starting of the given table we use Forward
Interpolation formula
Backward difference formula to compute the derivatives
Newton’s backward difference formula is
...
4!
1 2 3
3!
1 2
2!
1 2 3 4
n n n n n n y
r r r r
y
r r r
y
r r
f x rh y y
...
12
2 6 11 3
6
2 6 2
2
1 2 1 4
3 2
3
2
' 2
n n n n n y
r r r
y
r r
y
r
y
h
f x rh

...
12
6 18 11
'' 1 4
2
2 2 3
n n n n y
r r
h f x rh y r y
...
2
2 3
''' 3 3 4
n n n y
r
h f x rh y
At the point , n x x i.e., , n n x rh x we have r=0.
...
4
1
3
1
2
1 1
' 2 3 4
n n n n n y y y y
h
f x
...
12
1 11
'' 2 3 4
n 2 n n n y y y
h
f x
... and so on.
2
1 3
''' 3 4
n 3 n n y y
h
f x
Note: If the x value is nearer to the end of the given table we use backward Interpolation formula
Central difference formula for computing the derivatives
We know that Stirling’s central difference formula is
...
5!
1 4
2
4!
1
2
1 2
3!
1
1
2 2!
1
3
5
2
5
2 2
4
2 2
2 3 3
2
2
0
0 0
y y
r r r
y
r r
r r y y
y
y y r
f x rh y r
(Here using of r for n is only for convenience)
i.e.,
...
5!
5 4
2
2 4!
1 2
3!
1
2 2!
1
3
5
2
5
5 3
4
3 3 3 4 2
2
2
0
0 0
y y
r r r
y
r r y y r r
y
y y r
f x rh y r
Differentiating (1) w.r.t. ‘r’ we get,
...
5!
5 15 4
12
2
1 2
12
3 1
1
2
1
'
3
5
2
5
4
2
4
3
3 3
2
0 2
0
y y
r r
y
r r
y y
r
y
y y
hy x rh
2
3
1
3
2
0 2
0 12
3 1
1
2
1 1
' y y
r
y
y y
h
y x rh
...
5!
5 15 4
12
2
3
5
2
5
4 2
2
4
3
y y
r r
y
r r
...(2)
Differentiating (2) w.r.t. ‘r’, we get,

...
5!
20 30
12
6 1
2
1
''
3
5
2
5
3
2
4
2
2
3
1
3
1
2
0 2
y y
r r
y
r
y y
r
y
h
y x rh
-----------(3)
If we want to find the derivative at a point , 0 x x then 0 0 x rh x
i.e., r=0
Substituting r=0 in (2) and (3) we get,
...
12
1 1
''
...
30
1
12
1
2
1 1
'
2
4
1
2
2
0
3
5
2
5
2
3
1
0 3
0
y y
h
y x
y y y y
y y
h
y x
Note:
 If the x value is middle of the given table we use central Difference formula
Maximum and Minimum:
Steps
 Write the Newton’s Forward difference formula y(x),
dx
dy
 Write the forward difference table
 Find
dx
dy
 Put
dx
dy
=0 and find the value of x
 Find 2
2
dx
d y
 For every value of x find 2
2
dx
d y
 If 2
2
dx
d y
<0 , y is maximum at that x(maximum point)
 If 2
2
dx
d y
>0 , y is minimum at that x(minimum point)
 To find the maximum and minimum value substitute the maximum and minimum
points in y(x) formula respectively.

Problems based on Differentiation using Interpolation formula
1. Find the first, second and third derivatives of the function tabulated below at the point
x=1.5
x 1.5 2.0 2.5 3.0 3.5 4.0
f(x) 3.375 7.0 13.625 24.0 38.875 59.0
Solution
The difference table is as follows:
x y=f(x) y y 2 y 3 y 4
1.5x0 3.375y0
3.625 0 y
2.0 7.0 3.0 0
2 y
6.625 0.75 0
3 y
2.5 13.625 3.75 0
10.375 0.75
3.0 24.0 4.50 0
14.875 0.75
3.5 38.875 5.25
20.125
4.0 59.0
Here we have to find the derivative at the point x=1.5 which is the initial value of the table.
Therefore by Newton’s forward difference formula for derivatives at x=x0, we have
...
3
1
2
1 1
' 0
3
0
2
0 0 y y y
h
f x
Here x0=1.5, h=0.5
0.75 ...
3
1
3.0
2
1
3.625
0.5
1
f ' 1.5
f ' 1.5 4.75

At the point , 0 x x
...
12
1 11
'' 0
4
0
3
0
2
0 2 y y y
h
f x
Here x0=1.5, h=0.5
3.0 0.75
0.5
1
'' 1.5 2 f
f '' 1.5 9.0
At the point , 0 x x
0.75 6.0
0.5
1
''' 1.5 3 f
f ''' 1.5 6.0
------------------------------------------------------------------------------------------------------------2.
Compute f’(0) and f’’(4) from the data
x 0 1 2 3 4
y 1 2.718 7.381 20.086 54.598
Solution:
x y=f(x) y y 2 y 3 y 4
0x0 1y0
1.718 0 y
1
2.718
2.945 0
2 y
4.663 5.097 0
3 y
2
7.381
8.042 8.668 0
4 y
12.705 13.765 n y 3 n y 4
3
20.086
21.807 n y 2
34.512 n y
4
54.598
xn yn
Here we have to find f’(0) .ie.x=0 which is the starting of the given table. So we use the
forward interpolation formula.
0
4
0
3
0 3 2
1 3
' ' ' y y
h
f x

...
4
1
3
1
2
1 1
' 0
4
0
3
0
2
0 0 y y y y
h
f x
8.668 0.2225
4
1
5.097
3
1
2.945
2
1
1.718
1
1
f ' 0
Here we have to find f”(4) .ie.x=4 which is the end of the given table. So we use the
backward interpolation formula.
...
12
1 11
'' 2 3 4
n 2 n n n y y y
h
f x
8.668 43.5177
12
11
21.807 13.765
1
1
'' 4 2 f
------------------------------------------------------------------------------------------------------------

2. Find 2
2
dx
d y
and
dx
dy
at x = 51 from the following data.
x 50 60 70 80 90
y 19.96 36.65 58.81 77.21 94.61
Solution:
Here h=10. To find the derivatives of y at x=51 we use Forward difference formula
taking the origin at x0=50.
We have 0.1
10
51 50 0
h
x x
r
at x=51, r=0.1
x 51 r 0.1 dx
dy
dx
dy
= ...
12
2 9 11 3
6
3 6 2
2
1 2 1
0
4
3 2
0
3
2
0
2
0 y
r r r
y
r r
y
r
y
h
The difference table is given by
x y=f(x) y y 2 y 3 y 4
50x
0
19.96y
0
16.69 0 y
60 36.65 5.47
0
2 y
22.1
6
-9.23
0
3 y
70 58.81 -3.76 11.09
0
4 y
18.4
0
2.76
80 77.21 -1.00
17.4
0

90 94.61
x 51 r 0.1 dx
dy
dx
dy
= 11.99
12
2(0.1) 9(0.1) 11(0.1) 3
9.23
6
3(0.1) 6(0.1) 2
(5.47)
2
0.2 1
16.69
10
1 2 3 2
= 16.69 2.188 2.1998 1.9863 1.0316
10
1
0.1
2
2
51
2
2
x r dx
d y
dx
d y
...
12
6 18 11
1
1
0
4
2
0
3
0
2
2 y
r r
y r y
h
0.1
2
2
51
2
2
x r dx
d y
dx
d y
(11.99)
12
6(0.1) 18(0.1) 11
5.47 0.1 1 ( 9.23)
100
1 2
= 5.47 8.307 9.2523 0.2303
100
1
------------------------------------------------------------------------------------------------------------
3. Find the maximum and minimum value of y tabulated below.
x -2 -1 0 1 2 3 4
y 2 -
0.25
0 -
0.25
2 15.75 56
Solution:
...
12
2 9 11 3
6
3 6 2
2
1 2 1
0
4
3 2
0
3
2
0
2
0 y
r r r
y
r r
y
r
y
dx h
dy
x y=f(x) y y 2 y 3 y 4 y 5
-2 2
-2.25
-1 -0.25 2.5
0.25 -3
0 0 -0.5 6
-0.25 3 0
1 -0.25 2.5 6
2.25 9 0
2 2 11.5 6
13.75 15
3 15.75 26.5
40.25
4 56
Choosing x0=0, r = x x
1
0
(6)
12
2 9 11 3
(9)
6
3 6 2
(2.5)
2
2 1
0.25
1
1 2 3 2 x x x x x x
dx
dy

0.25 2.5 1.25 4.5 9 3 4.5 5.5 1.5
1
1 2 3 2 x x x x x x
x x
dx
dy 3
Now 0 0 3 x x
dx
dy
x = 0, x = 1, x = -1.
3 1 2
2
2
x
dx
d y
at x=0 ve
dx
d y
2
2
at x=1 ve
dx
d y
2
2
at x=-1 ve
dx
d y
2
2
y is maximum at x=0, minimum at x=1 and -1
...
2!
( 1)
( ) 0
2
0 0 y
x x
y x y x y
Maximum value =y(0) =0, Minimum value =y(1)=-0.25.
------------------------------------------------------------------------------------------------------------
4. Consider the following table of data
x 0.2 0.4 0.6 0.8 1.0
f(x) 0.9798652 0.9177710 0.8080348 0.6386093 0.3843735
Find f’(0.25),f’(0.6) and f’(0.95).
Solution:
Here h=0.2
 0.25 is nearer to the starting of the given table. So we use Newton’s forward
interpolation formula to evaluate f’(0.25)
 0.95 is nearer to the ending of the given table. So we use Newton’s backward
interpolation formula to evaluate f’(0.95)
 0.6 is middle point of the given table. So we use Central Difference formula to
evaluate f’(0.6)
The difference table
x y=f(x) y y 2 y 3 y 4
0.2x0 0.9798652y0
-0.0620942
0 y
0.4 0.9177710 0.047642 0
2 y
-0.1097362 -0.0120473 0
3 y
0.6 0.8080348 -0.0596893
0
4 y
-
0.01310985
n y 4
-0.1694255 -

0.02515715 n y 3
0.8 0.6386093 -
0.08484645 n y 2
-
0.25427195 n y
1.0 0.3843735
xn yn
To find f’(0.25)
Newton’s forward interpolation formula for derivative
...
12
2 9 11 3
6
3 6 2
2
2 1
' 0
4
3 2
0
3
2
0
2
0 0 y
r r r
y
r r
y
r
hf x rh y
( 0.01310985)
12
2(0.25) 9(0.25) 11(0.25) 3
0.0120473
6
3(0.25) 6(0.25) 2
( 0.047642)
2
2(0.25) 1
- 0.0620942
0.2
1
' 0.25
3 2
2
f
=-0.2536(correct to four decimal places)
To find f’(0.95)
Newton’s backward interpolation formula for derivative
x y=f(x) y y 2 y 3 y 4
0.2x0 0.9798652(y-2)
-0.0620942 2 y
0.4 0.9177710 (y-1)
0.047642 2
2 y
-0.1097362 1 y
-0.0120473 2
3 y
0.6 0.8080348 (y0)
-0.0596893 1
2 y
2
4 y
-0.01310985
-0.1694255 0 y
-0.02515715 1
3 y
0.8 0.6386093(y1)
-0.08484645 0
2 y
-0.25427195 1 y
1.0 0.3843735(y2)

...
12
2 6 11 3
6
2 6 2
2
1 2 1 4
3 2
3
2
' 2
n n n n n y
r r r
y
r r
y
r
y
h
f x rh
0.25
0.2
0.95 1
h
x x
r n
( 0.01310985)
12
2( 0.25) 9( 0.25) 11( 0.25) 3
0.02515715
6
3( 0.25) 6( 0.25) 2
( 0.08484645)
2
2( 0.25) 1
- 0.25427195
0.2
1
' 0.95
3 2
2
f
f’(0.95) =-1.71604
To find f’(0.6)
Central Differ e nce formula (Stirling’s Formula)
...
5!
5 15 4
12
2
12
3 1
2
1
' 3
5
2
5
4 2
2
4
3
2
3
1
3
2
1
0 1 2
0 y y
r r
y
r r
y y
r
r y
y y
h
f x rh
( 0.01310985)...
12
2(0.2) 0.2
0.02515715 0.0120473
12
3(0.2) 1
(0.2)( 0.00596893)
2
0.1694255 0.1097362
0.2
1
' 0.6
2 3
f
f’(0.6)=-0.74295(correct to 5 decimal places)
------------------------------------------------------------------------------------------------------------
5. Given the following data, find y’(6), y’(5) and the maximum value of y
x 0 2 3 4 7 9
y 4 26 58 112 466 922
Solution:
Since the intervals are , we will use Newton’s divided difference formula.
Divided Difference Table
x y=f(x) f (x) ( ) 2 f x ( ) 3 f x ( ) 4 f x
0 4
11=f(x0,x1)
2 26 7= f(x0,x1,x2)
32 1= f(x0,x1,x2,x3)
3 58 11 0= f(x0,x1,x2,x3, x4)
54 1
4 112 16 0
118 1
7 466 22
228
9 922
By Newton’s Divided Difference formula,
y =f(x)=f(x0)+(x-x0)f(x0, x1)+ (x-x0) (x-x1)f(x0, x1,x2)+……….
=4+(x-0)11+(x-0)(x-2)7+(x-0)(x-2)(x-3)1
=x3+2x2+3x+4

Therefore, f’(x) = 3x2+4x+3
f'(6)=135
f’(5)=98.
------------------------------------------------------------------------------------------------------------------
Numerical Integration
 Single integral
 Trapezoidal
 Simpson’s one-third rule
 Simpson’s three-eighth rule
 Romberg method
 Two and Three point Gaussian Quadrature Formulas
 Double integral
 Trapezoidal rule
 Simpson’s Rule
Single Integral
Trapezoidal Rule
sumof the f irst and last ordinates sumof the remaining ordinates
h
f x dx
xn
x
2
2
( )
0
ie, 0 1 2 1 2 ...
2
( )
0
n n
x
x
y y y y y
h
f x dx
n
Simpson’s one third rule
sumof the remaining ordinates
h sumof the f irst and last ordinates sumof the remaining ordinates
f x dx
xn
x 4 even
2 odd
3
( )
0
ie, 0 1 3 1 2 4 2 2 ... 4 ...
2
( )
0
n n n
x
x
y y y y y y y y
h
f x dx
n
Simpson’s three eight rule
2 3
3 3
8
3
( )
0 sumof the remaining ordinateswhich are divisibleby
sum of the remaining ordinateswhich are not divisibleby
sum of the f irst and last ordinates
h
f x dx
xn
x
0 1 2 4 5 1 3 6 3 3 ... 2 ...
2
( )
0
n n n
x
x
y y y y y y y y y y
h
f x dx
n

Rule Degree of y(x) No.of intervals Error Order
Trapezoidal Rule One Any
12
( ) 2b a h M
E
h2
Simpson’s one third rule Two Even
180
( ) 4b a h M
E
h4
Simpson’s three eight
rule
three Multiple of 3
8
3 5 h
E
h5
Two and Three point Gaussian Quadrature Formulas
Gaussian Two point formula
 If the Limit of the integral is -1 to 1 then we apply
3
1
3
1
( )
1
1
f x dx f f . This
formula is exact for polynomials upto degree 3.
 If
b
a
f (x)dx then x=
2 2
b a
t
b a
and dx =
2
b a
using these conditions convert
b
a
f (x)dx into
1
1
f (t)dt and then we apply the formula
3
1
3
1
( )
1
1
f t dt f f
Gaussian Three point formula
 If the Limit of the integral is -1 to 1 then we apply
(0)
9
8
5
3
5
3
9
5
( )
1
1
f x dx f f f . This formula is exact for polynomials upto
degree 5.
 Otherwise
b
a
f (x)dx then x=
2 2
b a
t
b a
and dx =
2
b a
using these conditions
convert
b
a
f (x)dx into
1
1
f (t)dt and then we apply the formula
(0)
9
8
5
3
5
3
9
5
( )
1
1
f t dt f f f
Problems based on single integrals
1. Using Trapezoidal rule evaluate
2
0
sin xdx

Solution
x 0 15 30 45 60 75 90
y 0 0.5087 0.7071 0.8408 0.9306 0.9828 1
Trapezoidal rule
0 1 2 1 2 ...
2
( )
0
n n
x
x
y y y y y
h
f x dx
n
(0 1) 2(0.5087 0.7071 0.8408 0.9306 0.9828 )
2
sin 12
2
0
xdx
= 1.17024
------------------------------------------------------------------------------------------------------------
2. By dividing the range into ten equal parts, evaluate
0
sin xdx by Trapezoidal rule and
Simpson’s Rule. Verify your answer with integration
Solution:
Range = π-0=π. Hence h=
10
we tabulate below the values of y at different x’s
x 0
10
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
sinx 0 .309 .5878 .809 .9511 1 .9511 .809 .5878 .309 0
Trapezoidal rule
0 1 2 1 2 ...
2
( )
0
n n
x
x
y y y y y
h
f x dx
n
(0 0) 2(0.3090 0.5878 0.8090 0.9306 0.9511 1 0.9511 0.9306 0.8090 0.3090 )
2
sin 10
2
0
xdx
= 1.9843 nearly
Simpson’s one third Rule:
we use Simpson’s one third rule only when the no.of intervals is even
here the no of intervals =10(even)
0 1 3 1 2 4 2 2 ... 4 ...
2
( )
0
n n n
x
x
y y y y y y y y
h
f x dx
n

(0 0) 2(0.9511 0.5878 0.9306 0.9511 0.9306 ) 4(0.3090 0.8090 1 0.8090 0.3090 )
2
sin 10
2
0
xdx
= 2.00091 nearly
We use Simpson’s three eight rule only when the no.of intervals divisible by 3
Here the no of intervals =10 which is not divisible by 3.
So we cannot use this method.
By actual integration,
sin cos 2 0
0
xdx x
Hence, Simpson’s one third rule is more accurate than the Trapezoidal rule.
-------------------------------------------------------------------------------------------------------------------
-
3. Evaluate (i)
1
1
2 4 (3x 5x )dx and (ii)
1
0
2 4 (3x 5x )dx by Gaussian two and three point
formulas
Solution:
(i)
1
1
2 4 (3x 5x )dx
Gaussian two point formula
Given interval is -1 and 1.
Hence we can apply
3
1
3
1
( )
1
1
f x dx f f
1 2 4 2 4
1
2 4
3
1
5
3
1
3
3
1
5
3
1
(3x 5x )dx 3 =3.112
Given interval is -1 and 1.
Hence we can apply
(0)
9
8
5
3
5
3
9
5
( )
1
1
f x dx f f f
(0)
9
8
5
3
5
5
3
3
5
3
5
5
3
3
9
5
(3 5 )
2 4 2 4
1
1
2 4 x x dx
=
25
9
5
5
3
3
25
9
5
5
3
3
9
5
1
1
2 4 (3x 5x )dx = 4

(ii)
1
0
2 4 (3x 5x )dx
Here the interval is 0 to 1. So we use the formula
x=
2 2
b a
t
b a
and dx =
2
b a
dt
i.e,
2
t 1
x and dx=
2
dt
1.556
2
1
5
2
1
3
1
1
2 4
dt
t t
Here the interval is 0 to 1. So we use the formula
x=
2 2
b a
t
b a
and dx =
2
b a
dt i.e,
2
t 1
x and dx=
2
dt
1
1
1 2 4
1
2 4
2
1
5
2
1
3
2
1
2 2
1
5
2
1
3 dt
t t dt t t
2 4
2
1
5
2
1
( ) 3
t t
f t
(0)
9
8
5
3
5
3
9
5
( )
1
1
f t dt f f f
5
3
f 0.038138
2
0.7745 1
5
2
0.7745 1
3
2 4
5
3
f 4.28
2
0.7745 1
5
2
0.7745 1
3
2 4
f (0) 1.0625
2
0 1
5
2
0 1
3
2 4
(1.0625 )
9
8
0.038138 4.28
9
5
2
1
5
2
1
3
1
1
2 4
dt
t t
= 4(approximately)
2
2
4
2
1
5
2
1
3
2
1 1
1
2 4
dt
t t
------------------------------------------------------------------------------------------------------------

Double Integral
Trapezoidal Rule
Evaluate
d
c
b
a
f (x, y)dxdywhere a, b, c, d are constants.
D K L C
J M N H
I O P G
A E F B
I=
4
hk
[sum of values in]+2(sum of values in +4[sum of remaining values]
Simpson’s Rule
on the even row f of the matrix }
16(sum of the values of f at the even positions)
{8(sum of the values of f at the odd positions)
on the odd row f of the matrix except boundary rows}
8(sum of the values of f at the even positions)
{4(sum of the values of f at the odd positions)
4(sum of the values of f at the even positions on the boundary)
boundaryexcept the corners)
2(sum of the values of f at the odd positions on the
sum of the values of f at four corners
9
hk
I
Problems based on Double integrals
1. Evaluate dxdy
xy
1.4
1 22.4
1
using Trapezoidal and Simpson’s rule. Verify your result by actual
integration.
Solution:
Divide the range of x and y into 4 equal parts
0.1
4
1.4 1
0.1
4
2.4 2
k
h
Get the values of f(x, y) =
xy
1
at nodal points

yx 2 2.1 2.2 2.3 2.4
1 0.5 0.4762 0.4545 0.4348 0.4167
1.1 0.4545 0.4329 0.4132 0.3953 0.3788
1.2 0.4167 0.3968 0.3788 0.3623 0.3472
1.3 0.3846 0.3663 0.3497 0.3344 0.3205
1.4 0.3571 0.3401 0.3247 0.3106 0.2976
(i) Trapezoidal Rule
4 0.4329 0.4132 0.3953 0.3968 0.3788 0.3623 0.3663 0.3497 0.3344
0.3247 0.3401
0.4348 0.3788 0.3472 0.3205 0.3106
0.4545 0.4167 0.3846 0.4762 0.4545
0.5 0.4167 0.3571 0.2976 2
4
(0.1)(0.1)
I
I=0.0614
(ii) Simpson’s Rule:
16(0.3663 0.3344 0.4329 0.3953)
0.3401 0.3788) 8(0.3968 0.3623 0.3497 0.4132)
4(0.3846 0.4545 0.4762 0.4348 0.3788 0.3205 0.3106
0.5 0.4167 0.3571 0.2976 2(0.4167 0.4545 0.3472 0.3247)
9
(0.1)(0.1)
I
I=
9
0.01(55.2116)
I=0.0613
------------------------------------------------------------------------------------------------------------

UNIT IV
INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS
Taylor series method
Euler methods
Runge-Kutta method for solving first and second order equations
Milne’spredictor and corrector method
Adam’s predictor and corrector method.

Introduction
An ordinary differential equation of order n is a relation of the form
, , ', ",..., 0 n F x y y y y where y y x and r
r
r
dx
d y
y . The solution of this differential
equation involves n constants and these constants are determined with the help of n conditions
1 , ', ",..., n y y y y are prescribed at 0 x x , by
1
0 0
' 1
0 0 0 0 , ' ,... n n y x y y x y y x y
These conditions are called the initial conditions because they depend only on 0 x .
The differential equation together with the initial conditions is called an initial value problem.
Taylor’s Series
Point wise solution
If y x is the solution of (1), then by Tailor series,
...
1! 2! 3!
"'
0
3
" 0
0
2
' 0
0
0
0 y
x x
y
x x
y
x x
y x y
Put x x h 1 0 where h is the step-size, we have
...
1! 2! 3!
'"
0
3
"
0
2
'
1 1 0 0 y
h
y
h
y
h
y x y y (3)
once 1 y has been calculated form (1), " '
1
"
1
'
1 y , y , y can be calculated from
y' f x, y
Expanding y x in a Taylor series about 1 x x , we get
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
Where 2 2 y y x and x x h 2 1
The Tailor algorithm is given as follows
...
1! 2! 3!
" '
3
"
2
'
m 1 m m m m y
h
y
h
y
h
y y
Where r
r
r
m dx
d y
y at the point m m x , y where m 0,1,2,...
Problems based on Taylor’s Series
Solve ' ; 0 1 2 y y x y using Taylor series method and compute y 0.1 and y 0.2
Solution
Here 0, 1 0 0 x y .
Given y y x 2 ' ; 1 '
0 y
y" 2yy' 1 ; 3 "
0 y
'2 y"' 2yy" 2y ; 8 "'
0 y
y 6y" 2yy"' iv ; 34 0
iv y etc.,
To find

Here h 0.1
By Taylor algorithm
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
34
24
0.001
8
6
0.001
3
2
0.01
1 0.1 1 1 y
y 0.1 1.1164
To find y 0.2
x x h 2 1 where 0.2 2 x
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
0.1 1.1164 1.3463 ' 2
1 y
1 2 1.1164 1.3463 4.006 "
1 y
"' 2
1 y 2 1.1164 4.006 2 1.3463
= 12.5696
12.5696
6
0.0001
4.006
2
0.01
y 0.2 1.1164 0.1 1.3463
y 0.2 1.2732
----------------------------------------------------------------------------------------------------------
2. Evaluate y 0.1 and y 0.2 , correct to four decimal places by Taylor series method, if y x
satisfies y' xy 1, y 0 1
Solution
Here 0, 1 0 0 x y
y' xy 1 ; 1 '
0 y
y" xy' y ; 1 "
0 y
' y"' xy" 2y ; 2 "
0 y
y xy'" 3y" iv ; 3 0
iv y
To find y 0.1
By Taylor algorithm
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
Let x x h 1 0 where h 0.1
0.1 1 x
3
24
0.0001
2
3
0.001
1
2
0.01
1 0.1 1 y
= 1.1057
To find y 0.2
Let 2 2 y y x where x x h 2 1

0.2 2 x
By Taylor algorithm,
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
1 0.1 1.1057 1.11057 '
1 y
0.1 1.11057 1.1057 1.216757 "
1 y
0.1 1.216757 2 1.11057 2.3428157 "'
1 y
2.342815
6
0.0001
1.216757
2
0.01
1.1057 0.1 111057 2 y
1.2178 2 y
Hence y 0.1 1.1057 and y 0.2 1.2178 .
--------------------------------------------------------------------------------------------------------
3.Solve by Taylor series method, ' , 0 1 2 y xy y y at x 0.1and 0.2, , correct to four decimal
places.
Solution
Given 0, 1 0 0 x y
2 y' xy y ; 1 '
0 y
y" y xy' 2yy' ; 3 "
0 y
"' 2 ' " 2 2 "
' 2 y y xy y yy ; 10 "
0 y
y 3y" xy" 6y' y" 2yy"' iv ; 47 0
iv y
To find 0.1 1 y y
Let x x h 1 0 . Here h 0.1
0.1 1 0 1 x x h x
By Taylor algorithm
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
47
24
0.0001
10
6
0.001
3
2
0.01
1 0.11 1 y
1.1 0.015 0.0017 0.0002
y 0.1 1.1169
To find y 0.2
2 2 y y x where x x h 2 1
Here 0.1, 0.1, 0.02 1 2 x h x
By Taylor algorithm,
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
0.1, 1.1169 1 1 x y
1.35925 '
1
2
1 1 1
'
1 y x y y y

2 1.35925 '
1
'
1 1
'
1 1 1
"
1 y y x y y y y
1.1169 0.1 1.35925 2 1.1169 1.35925
4.2891
"
1 1
' 2
1
"
1 1
'
1
"'
1 y 2y x y 2 y 2y y
2.7185 0.42891 3.6951 6.5810
16.4235
"'
1 1
"
1 1
"'
1 1
"
1 1 y 3y x y 6y y 2y y iv
212.8673 1.64235 34.9797 36.6868
86.17615
86.17615
24
0.0001
16.4235
6
0.001
4.2891
2
0.01
1.1169 0.1 1.35925 2 y
1.1169 0.135925 0.02144 0.00274 0.00036
y 0.2 1.2774
Hence y 0.1 1.1169 and y 0.2 1.2774 .
------------------------------------------------------------------------------------------------------------
Taylor series method for simultaneous first order differential equation
We can solve the equations of the form g x y z
dx
dy
f x y z
dx
dy
, , ; , , with initial
conditions 0 0 y x y and 0 0 z x z .
The values of y and z at x x h 1 0 are given by Taylor algorithm,
...
1! 2! 3!
'"
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
...
1! 2! 3!
'"
0
3
"
0
2
'
1 0 0 z
h
z
h
z
h
z z
The derivatives on R.H.S of the above expressions are found at 0 x x using the given
equations.
Similarly 2 y and 2 z corresponding to x x h 2 1 are calculated by Taylor series method.
Problems
1.Evaluate x 0.1 and y 0.1 given tx
dt
dy
ty
dt
dx
1 ; given x 0, y 1 at t 0 by Taylor
series method.
Solution
Given 0, 0, 1 0 0 0 t x y
x' 1 ty y' tx
x" y ty' y" x tx'
x'" 2y' ty" y"' 2x' tx"
x 3y" ty"' iv y 3x" tx"' iv
Then
1 '
0 x 0 '
0 y

1 "
0 x 0 "
0 y
"' 0
0 x 2 "'
0 y
0 0
iv x 3 0
iv y
By Taylor algorithm, we have
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
3
24
0.0001
2
6
0.001
0
2
0.01
y 0.1 1 0.1 0
y 0.1 0.9997
...
1! 2! 3!
0.1 " '
0
3
"
0
2
'
1 0 0 x
h
x
h
x
h
x x x
1 .....
2
0.01
0 0.1 1
0.105
------------------------------------------------------------------------------------------------------------
2. Find x 0.2 and y 0.2 using Taylor series method given that ty x
dt
dy
xy t
dt
dx
2 ; 2 , with
initial conditions x 1, y 1 at t 0.
Solution
Given 0, 1, 1 0 0 0 t x y
Taking 0.2, 0.2 1 0 1 h t t h t
x' xy 2t y' 2ty x
x" xy' x' y 2 y" 2ty' 2y x'
x'" xy" 2x' y x" y y"' 4y' 2ty" x"
1 '
0 x 1 '
0 y
4 "
0 x 3 "
0 y
9 "'
0 x 8 '
0 y
By Taylor algorithm, we have
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 x
h
x
h
x
h
x x
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
0.2 ...
2
3
0.2 1 0.2 1 2 0.2 2 3 x
x 0.2 0.796
0.008
3
4
0.2 1 0.2 1.5 0.04 1 y y
0.8493
Hence x 0.2 0.796 and y 0.2 0.8493
------------------------------------------------------------------------------------------------------------

Taylor series method for second order differential equations
Consider the differential equation
dx
dy
f x y
dx
d y
, , 2
2
with the initial conditions 0 0 y x y
and '
0 0 y' x y , where '
0 0 y , y are known values.
This equation can be reduced into a set of simultaneous equations, by putting y' p
we have 0 0 y' p, y x y (1)
And '
0 0 0 p' f x, y, p , p x p y (2)
Successively differentiating y", the expression for iv y"' , y etc., are known.
By Taylor series method, we find
...
1! 2! 3!
'"
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
...
2! 3!
" '
0
3
"
0
2
0 0 p
h
p
h
y hp
Also by Taylor series method, we have
...
1! 2! 3!
" '
0
3
"
0
2
'
0 0
'
1 1 p
h
p
h
p
h
p y p
Then the values iv y y y1
"'
1
"
1 , , are found from "'
1
"
1 y , y etc
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
Thus we calculate , ,... 1 2 y y
Problems
1.Find y 0.2 and 0.4 given y" xy if y 0 1, y' 0 1 by Taylor series method.
Solution
Given " , 0, 1 0 0 y xy x y and y' 0 1
Then we have
y"' xy' y
y xy" 2y' iv
y 3y" xy"' v etc
0, 1, 2, 0 0 0
"'
0
"
0
iv v y y y y
By Taylor Algorithm,
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
Taking, 0.2, 0.2 1 0 1 h x x h x
2
24
0.0016
1
6
0.008
0
2
0.04
1 0.2 1 1 1 y y x
y 0.2 1.2014
To find '
1 y

Set p y' . Then p' xy
p" xy' y; p"' xy" 2y'; p 3y" xy"' iv
1, 0, 1, 2, 0 0
'"
0
"
0
'
0 0
iv p p p p p
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 p
h
p
h
p
h
p p
2
6
0.08
1
2
0.04
1 0.2 0 1 p
1 0.02 0.0027 '
1 y
1.0227
Let x x h 2 1 . Since 0.2, 0.2 0.4 1 2 h x x
By Taylor series method,
...
1! 2! 3!
0.4 " '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y x y y
" 0.2 1.2014 0.24028 1 1
"
1 y xy y x y
"' ' 0.2 1.0227 1.2014 1.40594 " '
1 y xy y y
2 ' " 0.2 0.024028 2 1.0227 0.00048 2.0454 2.0459 1
iv iv y y xy y
2.0459
24
0.0016
1.40594
6
0.008
0.024028
2
0.04
1.2014 0.2 1.0227 2 y
y 0.4 1.4084
Hence y 0.2 1.2014 and y 0.4 1.4084
------------------------------------------------------------------------------------------------------------
2.Find y at x 1.1,1.2 given " ', 1 3 2 y x y y y and y' 1 1, correct to four decimal places
using Taylor series method.
Solution
Given 1, 1; 1 '
0 0 0 x y y
" ' 0 "
0
3 2 y x y y y
"' 3 2 ' " 1 "
0
2 2 2 y x y y y y y
6 2 4 ' " 2 ' " "' 3 2 y x y yy y yy y y y iv
6 2 0 0 1 3 0
iv y
0, 0, 1, 3 0
"'
0
"
0
'
0
iv y y y y
By Taylor series method,
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 y
h
y
h
y
h
y y
Taking, 0.1, 1, 0.1 0 1 0 h x x x h
We have
3
24
0.0001
1
6
0.001
0
2
0.01
y 1.1 1 0.11
y 1.1 1.1002

To find '
1 y
Set p y' . Then ' ' , 1 '
0 0
3 2 p x y y p y
1 1 0, 1, 3 "'
0
"
0
1
0 p p p etc.,
...
1! 2! 3!
" '
0
3
"
0
2
'
1 0 0 p
h
p
h
p
h
p p
2
6
0.01
1
2
0.01
1 0.1 0 1 p
1.0053 1.0053 '
1 1 p y
" ' 1.1 1.1002 1.0053 " 3 2
1
3 2 y x y y y
0.11415
"' 3 2 ' " 2 2 2 y xy y y y y 3 1.1 2 1.1002 1.0053 1.1002 0.11415 "' 3 2 2
1 y
1.268
...
1! 2! 3!
" '
1
3
"
1
2
'
2 1 1 y
h
y
h
y
h
y y
Where , 1.2 2 2 2 y y x x
1.268
6
0.01
0.11415
2
0.01
1.1002 0.1 1.0053
y 1.2 1.2015
Hence y 1.1 1.1002 and y 1.2 1.2015
------------------------------------------------------------------------------------------------------------
Euler method
Let i y y x 1 , where x x h 1 0
Then y y x h 1 0 . Then by Taylor’s series,
" .....
2!
'
1! 0
2
1 0 0 y x
h
y x
h
y y x (1)
Neglecting the terms with 2 h and higher powers of h, we get from (1),
1 0 0 0 y y hf x , y (2)
Expression (2) gives an approximate value of y at x x h 1 0 .
Similarly, we get 2 1 1 1 y y hf x , y for x x h 2 1 .
for any m, m m m m y y hf x , y 1 ,m 0,1,2,... (3)
In Euler’s method, we use (3) to compute successively , ,... 1 2 y y etc., with an Error 2 0 h
Modified Euler method
The algorithm presented already in Modified Euler method in unit IV is sometimes
referred as Improved Euler Method.
Therefore a different algorithm for Modified Euler method to solve
0 0 f x, y , y x y
dx
dy
is explained with illustrations.

Explanation: Modified Euler Method
f x y
h
y
h
y x h y x j f x ,
2
,
2
(or) f x y
h
y
h
y y h f x n n n n ,
2
,
2 1
Where y y x h n 1 n and h is the step – size
Problems based on Euler’s Method
1,Using Euler’s method, compute y in the range 0 x 0.5, if y satisfies 3 , 0 1 2 x y y
dx
dy
.
Solution
Here , 3 , 0, 1 0 0
2 f x y x y x y
By Euler’s, method
, , 0,1,2,... 1 y y hf x y n n n n n
Choosing h 0.1, we compute the values of y using (1)
, , 0,1,2,... 1 y y hf x y n n n n n
2
1 0 0 0 y 0.1 y y hf x , y 1 0.1 3 0 1
1.1
2
2 1 1 1 y 0.2 y y hf x , y 1.1 0.1 0.3 1.1
1.251
2
3 2 2 2 y 0.3 y y hf x , y 1.251 0.1 0.6 1.251
1.4675
2
4 3 3 3 y 0.4 y y hf x , y 1.4675 0.1 0.9 1.4675
1.7728
2
5 4 4 4 y 0.5 y y hf x , y 1.7728 0.1 1.2 1.7728
2.2071
------------------------------------------------------------------------------------------------------------
2.Given
y x
y x
dx
dy
, with y 1 for x 0 . Find y approximately for x 0.1 by Euler’s method in
five steps.
Solution
Given 1, 0 0 0 y x , choosing h 0.002,
, 1,2,3,4,5 0 x x ih i i
To find 1 2 3 4 y , y , y , y and 5 y where i i y y x
Here
y x
y x
f x, y
Using , , 0,1,2,... 1 y y hf x y n n n n n
We get

1.02
1 0
1 0
, 1 0.02 1 0 0 0 y y hf x y
1.0392
1.02 0.02
1.02 0.02
, 1.02 0.02 2 1 1 1 y y hf x y
1.0577
1.0392 0.04
1.0392 0.04
, 1.0392 0.02 3 2 2 2 y y hf x y
1.0756
1.0577 0.06
1.0577 0.06
, 1.0577 0.02 4 3 3 3 y y hf x y
1.0928
1.0756 0.08
1.0756 0.08
, 1.0756 0.02 5 4 4 4 y y hf x y
Hence y 1.0928 and x 0.1
----------------------------------------------------------------------------------------------
3.Compute y at x 0.25 by Modified Euler method given y' 2xy, y 0 1
Solution
Here , 2 , 0, 1 0 0 f x y xy x y
Choose 0.25, 0.25 1 0 h x x h
By Modified Euler method
1 0 0 0 0 0 ,
2
,
2
f x y
h
y
h
y y h f x
, 2 0 1 0 0 0 f x y
, 0.125,1 0.25
2
,
2 0 0 0 0 f x y f
h
y
h
f x
1 0.25 0.25 1.0625 1 y
Hence y 0.25 1.0625 .
--------------------------------------------------------------------------------------------------
4.Using Modified Euler method, find y 0.1 and y 0.2 given , 0 1 2 2 x y y
dx
dy
.
Solution
Here 0, 1, 0.1 0 0 x y h
0 1 1.05
2
0.1
, 1
2
, , 0 0 0
2 2 f x y
h
f x y x y y
, 1 0.1 0.05,1.05
2
,
2 1 0 0 0 0 0 x y f
h
y
h
y y hf x

1 0.1 0.05 1.05 1.1105 2 2
y 0.1 1.1105
, 0.1,1.1105 0.1 1.1105 1.24321 2 2
1 1 f x y f
, 1.1105 0.05 1.24321 1.17266
2 1 1 1 f x y
h
y
2 1 1 1 1 1 ,
2
,
2
f x y
h
y
h
y y hf x
1.1105 0.1 f 0.15,1.17266
2 2 1.1105 0.1 0.15 1.17266
0.2 1.2503 2 y y
y 0.2 1.2503
-----------------------------------------------------------------------------------------------------
Fourth-order Range-Kutta method
This method is commonly used for solving the initial value problem
0 0 f x, y , y x y
dx
dy
Working Rule
The value of 1 1 y y x where x x h 1 0 where h is the step-size is obtained as follows.
We calculate successively.
1 0 0 k hf x , y
2
,
2
1
2 0 0
k
y
h
k hf x
2
,
2
2
3 0 0
k
y
h
k hf x
4 0 0 3 k hf x h, y k
Finally compute the increment
1 2 3 4 2 2
6
1
y k k k k
The approximate value of 1 y is given by
1 0 1 0 1 2 3 4 2 2
6
1
y y y y y k k k k
Error in R-K fourth order method 5 0 h
In general, the algorithm can be written as
1 1 2 3 4 2 2
6
1
y y k k k k m m where
m m k hf x , y 1
2
,
2
1
2
k
y
h
k hf x m m

2
,
2
2
3
k
y
h
k hf x m m
4 3 k hf x h, y k m m where m 0,1,2,...
Runge-Kutta method for second order differential equations
Consider the second order differential equation
dx
dy
f x y
dx
d y
, , 2
2
with initial conditions
0 0 y x y and '
0 0 y' x y . This can be reduced to a system of simultaneous linear first order
equations, by putting
dx
dy
z . Then we have,
z
dx
dy
with 0 0 y x y
f x y z
dx
dz
, , , with '
0 0 0 z x y' x y
i.e., g x y z
dx
dy
, , , where g x, y, z z
and f x y z
dx
dz
, , with initial conditions 0 0 y x y and
0 0 z x z where '
0 0 z y
Now, starting from 0 0 0 x , y , z , the increments y and z in y and z are given by (h-step size)
1 0 0 0 k hg x , y , z 1 0 0 0 l hf x , y , z
2
,
2
,
2
1
0
1
2 0 0
l
z
k
y
h
k hg x
2
,
2
,
2
1
0
1
2 0 0
l
z
k
y
h
l hf x
2
,
2
,
2
2
0
2
3 0 0
l
z
k
y
h
k hg x
2
,
2
,
2
2
0
2
3 0 0
l
z
k
y
h
l hf x
4 0 0 3 0 3 k hg x h, y k , z l 4 0 0 3 0 3 l hf x h, y k , z l
1 2 3 4 2 2
6
1
y k k k k 1 2 3 4 2 2
6
1
z l l l l
Then for x x h 1 0 , the values of y and z are y y y 1 0 and z z z 1 0 respectively.
By repeating the above algorithm the value of y at x x h 2 1 can be found.
Problems based on RK Method
1.The value of y at x 0.2 if y satisfies , 0 1 2 x y x y
dx
dy
taking h 0.1 using Runge-Kutta
method of fourth order.
Solution
Here , , 0, 1. 0 0
2 f x y x y x x y
Let x x h 1 0 , choosing 0.1, 0.1 1 h x .
Then by R-K fourth order method,
1 0 1 2 3 4 2 2
6
1
y y k k k k

, 0.1 0 0 0 1 0 0 k hf x y
0.05,1 0.00525
2
,
2
1
2 0 0 hf
k
y
h
k hf x
0.05,1.0026 0.00525
2
,
2
2
3 0 0 hf
k
y
h
k hf x
, 0.1,1.00525 0.0110050 4 0 0 3 k hf x h y k hf
0 0.00525 0.00525 2 0.011005 1.0053
6
1
1 1 y
y 0.1 1.0053
To find 2 2 y y x where x x h 2 1 . Then 0.2 2 x
2 1 1 2 3 4 2
6
1
y y k k k k
, 0.1 0.1 0.01 1.0053 0.0110 1 1 1 k hf x y
0.1 0.15 0.15 1.0108 0.01727
2
,
2
1 2
2 1 1
k
y
h
k hf x
0.1 0.15 0.15 1.013935 0.01728
2
,
2
2 2
3 2 2
k
y
h
k hf x
, 0.1 0.2 0.2 1.02258 0.02409 2
4 1 1 3 k hf x h y k
0.0110 2 0.01727 0.01728 0.02409 1.0227
6
1
1.0053 2 y
y 0.2 1.0227
Hence y 0.2 1.0227 .
---------------------------------------------------------------------------------------------------
2. Apply Runge-Kutta method to find an approximate value of y for x = 0.2 in steps of 0.1 if
, 0 1 2 x y y
dx
dy
, correct to four decimal places.
Solution
Here , , 0, 1 0 0
2 f x y x y x y
We choose h 0.1
1 1 2 3 4 2 2
6
1
y y k k k k o
, 0.1 0 1 0.1 2
1 0 0 k hf x y
0.1 0.05 1.05 0.1152
2
,
2
1 2
2 0 0
k
y
h
k hf x
0.1 0.05 1.0576 0.1168
2
,
2
2 2
3 0 0
k
y
h
k hf x
, 0.1 0.1 1.1168 0.1347 2
4 0 0 3 k hf x h y k

0.1 2 0.1152 0.1168 0.1347
6
1
1 1 y
1.1165 1 y
Hence y 0.1 1.1165 .
---------------------------------------------------------------------------------------------------
3.Use Runge-Kutta method to find y when x=1.2 in steps of 0.1, given that 2 2 x y
dx
dy
and
y 1 1.5.
Solution
Given , , 1, 1.5 0 0
2 2 f x y x y x y
Let x x h 1 0 , we choose h 0.1
1 1 2 3 4 2 2
6
1
y y k k k k o
, 0.1 1 1.5 0.325 2
1 0 0 k hf x y
0.1 1.05 1.6625 0.3866
2
,
2
1 2 2
2 0 0
k
y
h
k hf x
0.1 1.05 2.8673 0.397
2
,
2
2 2
3 0 0
k
y
h
k hf x
, 0.1 1.1 1.897 0.4809 2 2
4 0 0 3 k hf x h y k
0.325 2 0.3866 0.397 0.4809 1.8955
6
1
1.5 1 y
1.8955 1 y
To compute y 1.2 :
2 2 y y x where 1.2 2 1 x x h , since h 0.1
2 1 1 2 3 4 2
6
1
y y k k k k
, 0.1 1.1 1.8955 0.4803 2 2
1 1 1 k hf x y
0.1 1.15 2.1356 0.5883
2
,
2
1 2 2
2 1 1
k
y
h
k hf x
0.1 1.3225 2.1897 0.6117
2
,
2
2 2
3 2 2
k
y
h
k hf x
, 0.1 1.44 6.286 0.7726 4 1 1 3 k hf x h y k
0.4803 2 0.5883 2 0.6117 0.7726 2.5043
6
1
1.8955 2 y
Hence y 1.2 2.5043 .
---------------------------------------------------------------------------------------------------
4.Solve the equation xy
dx
dz
xz
dx
dy
1, for x = 0.3 and 0.6. Given that y 0, z 1 when x 0

Solution
Here , , 1 , 0, 0, 0 1 0 0 0 f x y z xz x y z and h 0.3
To find y 0.3 and z 0.3
, 0.3 0 1 0.3 1 0 0 k hf x y
0.3 0.15 1 0.3450
2
,
2
1
2 0 0
k
y
h
k hf x
0.3 0.15 0.9966 1 0.3448
2
,
2
2
3 0 0
k
y
h
k hf x
, 0.3 0.3 0.99224 1 0.3893 4 0 0 3 k hf x h y k
, , 0.3 0 0 1 0 0 0 l hf x y z
0.3 0.15 0.15 0.00675
2
,
2
,
2
1
0
1
2 0 0
l
z
k
y
h
l hf x
0.3 0.15 0.1725 0.00776
2
,
2
,
2
2
0
2
3 0 0
l
z
k
y
h
l hf x
, , 0.3 0.3 0.3448 0.031036 4 0 0 3 0 3 l hf x h y k z l
1 0 1 2 3 4 2 2
6
1
y 0.3 y y k k k k
0.3 2 0.3450 0.3448 0.3893
6
1
0
y 0.3 0.3448
1 0 1 2 3 4 2 2
6
1
z 0.3 z z l l l l
0 2 0.00675 0.00776 0.031036
6
1
z 0.3 1
z 0.3 0.9899
To find y at x = 0.6, the starting values are 0.3, 0.3448, 0.9899 1 1 1 x y z and h 0.3
, 0.3 1 0.3 0.9899 0.3891 1 0 0 k hf x y
0.3 1 0.45 0.9744 0.4315
2
,
2
1
2 0 0
k
y
h
k hf x
0.3 1 0.45 0.9535 0.4287
2
,
2
2
3 0 0
k
y
h
k hf x
, 0.3 1 0.6 0.9142 0.4645 4 0 0 3 k hf x h y k
, , 0.3 0.3 0.3448 0.0310 1 0 0 0 l hf x y z
0.3 0.45 0.53935 0.0728
2
,
2
,
2
1
0
1
2 0 0
l
z
k
y
h
l hf x
0.3 0.45 0.56055 0.0757
2
,
2
,
2
2
0
2
3 0 0
l
z
k
y
h
l hf x
, , 0.3 0.6 0.7735 0.1392 4 0 0 3 0 3 l hf x h y k z l

2 1 1 2 3 4 2 2
6
1
y y k k k k
0.3891 2 0.4315 0.4287 0.4645
6
1
0.3448
y 0.6 0.7738
2 1 1 2 3 4 2 2
6
1
z z l l l l
0.0310 2 0.0728 0.0757 0.1392
6
1
0.9899
z 0.3 0.9210
---------------------------------------------------------------------------------------------------
Using R-K method of fourth order solve 2 2 y" x y' y for x=0.2, given that y=1 and
y' 0 when x = 0.
Solution
Let y' z then y" z'
Hence the given equation reduces to the form,
z
dx
dy
and 2 2 xz y
dx
dz
Given 0, 1, 0 0 0 0 x y z and h 0.2
Take 2 2
1 2 f x, y, z z, f x, y, z xz y
, 0.2 0 0 1 0 0 k hf x y
0.2 0.1 0.02
2
,
2
1
2 0 0
k
y
h
k hf x
0.2 0.999 0.01998
2
,
2
2
3 0 0
k
y
h
k hf x
, 0.3 0.1958 0.03916 4 0 0 3 k hf x h y k
, , 0.2 0 1 0.2 2
1 0 0 0 l hf x y z
0.2 0.1 0.1 0.99 0.1998
2
,
2
,
2
1 2 2
0
1
2 0 0
l
z
k
y
h
l hf x
0.2 0.1 0.0999 0.99 0.1958
2
,
2
,
2
2 2 2
0
2
3 0 0
l
z
k
y
h
l hf x
, , 0.2 0.2 0.1958 0.98 0.1905 2 2
4 0 0 3 0 3 l hf x h y k z l
1 0 1 2 3 4 2 2
6
1
y y k k k k
0.2 2 0.02 0.01998 0.03916
6
1
1
y 0.2 0.9801
Also 0.2 2 0.1998 0.1958 0.1905 0.1969
6
1
z 0.2 0
------------------------------------------------------------------------------------------------------------------

Multi-Step Methods (Predictor-Corrector Methods)
Introduction
Predictor-corrector methods are methods which require function values at
1 2 3 , , , n n n n x x x x for the computation of the function value at n 1 x . A predictor is used to find the
value of y at n 1 x and then the corrector formula is used to improve the value of n 1 y .
The following two methods are discussed in this chapter
(1) Milne’s Method (2) Adam’s Method
Milne’s Predictor-corrector method
Consider the initial value problem 0 0 f x, y , y x y
dx
dy
. Assume that
, , , 0 0 1 1 2 2 y y x y y x y y x and 3 3 y y x where , 0,1,2,3 1 x x h i i i are known, these
are the starting values.
Milne’s predictor formula
'
3
'
2
'
4, 0 1 2 2
3
4
y y y
h
y y p and
Milne’s corrector formula
'
4
'
3
'
4, 2 2 4
3
y y y
h
y y c where p y f x y 4 4,
'
4 ,
---------------------------------------------------------------------------------------------------
Problems based on Predictor-Corrector method
1.Using Milne’s method, compute y(0.8) given that 1 , 0 1, 2 y y
dx
dy
y 0.2 0.2027 ,
y 0.4 0.4228 and y 0.6 0.6841
Solution
we have the following table of values
x y 2 y' 1 y
0
0.2
0.4
0.6
0
0.2027
0.4228
0.6481
1.0
1.0411
1.1787
1.4681
0, 0.2, 0.4, 0.6 0 1 2 3 x x x x
0, 0.2027 , 0.4228 , 0.6841 0 1 2 3 y y y y
1, 1.0411, 1.1787 , 1.4681 '
3
'
2
'
1
'
0 y y y y
To find y (0.8):
0.8 4 x . Here h = 0.2
By Milne’s predictor formula,
'
3
'
2
'
4, 0 1 2 2
3
4
y y y
h
y y p
2 1.0411 1.1787 2 1.4681
3
0.8
0
1.0239 4, p y

2
4 4
'
4 y f x , y 1 1.0239
= 2.0480
By Milne’s corrector formula,
'
4
'
3
'
4, 2 2 4
3
y y y
h
y y c
1.1787 4 1.4681 2.0480
3
0.2
0.4228
y 0.8 1.0294
---------------------------------------------------------------------------------------------------
2.Given ' , 0 1, 0.1 0.9052 , 0.2 0.8213 2 y x y y y y , find y(0.3) by Taylor series
method. Also fine y(0.4) by Milne’s method
Solution
Given 0 0 x , 1 0 y
0.1 1 x , 0.9052 1 y
0.2 2 x , 0.8213 2 y
0.3 3 x , 3 3 y y x
By Taylor algorithm
...
2!
"
2
2
'
3 2 2 y
h
y y hy
y x y 2 ' y" 2x y'
y"' 2 y", y y"' iv etc
8187
2 1.1813 0.8187
2 0.2 0.7813 1.1813
0.2 0.8213 0.7813
2
'"
2
"
2
' 2
2
iv y
y
y
y
0.8187
24
0.0001
0.8187
6
0.001
1.1813
2
0.01
0.8213 0.1 0.7813 3 y
y 0.3 0.7492
For 0.3 0.7492 3 3 x y and 0.09 0.7492 0.6592 '
3 y
Also 1, 0.01 0.905 0.8952 '
1
'
0 y y and 0.7813 '
2 y
By Milne’s method
0.6897
2 0.8952 0.7813 2 0.6952
3
0.4
1
2 2
3
4
4,
4,
'
3
'
2
'
4, 0 1
p
p
p
y
y
y y y
h
y y
0.16 0.6897 0.5297 '
4 y
By Correctors formula,

0.6897 0.4 0.6897
0.7813 4 0.6592 0.5297
3
0.1
0.8213
4
3
4,
4,
'
4
'
3
'
4, 2 2
y y
y
y y y
h
y y
c
c
c
---------------------------------------------------------------------------------------------------
Adam-Bash Forth Predictor-Corrector Method
Using Newton’s backward difference interpolation formula, we derive a set of predictor
and corrector formulae. This method also requires past four values to estimate the fifth value.
Adam’s predictor formula
'
3
'
2
'
1
'
1, 55 59 37 9
24 n p n n n n n y y y
h
y y
Adam’s corrector formula
'
2
'
1
' '
1, 1 59 19 5
24 n c n n n n n y y y
h
y y
The errors in these formulae are
iv h f 5
720
251
and iv h f 5
720
19
respectively.
In particular,
'
0
'
1
'
2
'
4, 3 3 55 59 37 9
24
y y y y
h
y y p
And
'
1
'
2
'
3
'
4, 3 4 9 19 5
24
y y y y
h
y y c
---------------------------------------------------------------------------------------------------
Given ' 1 , 0 0, 0.2 0.2027 , 0.4 0.4228 , 0.6 0.6841 2 y y y y y y , estimate
y 0.8 using Adam’s method.
Solution
Form the given data
0 0 x , 0 0 y , 1 '
0 y
0.2 1 x 0.2027 1 y 1.0411 '
1 y
0.4 2 x 0.4228 2 y 1.1786 '
2 y
0.6 3 x 0.6841 3 y 1.4680 '
3 y
To fine 4 y for 0.8 4 x . Here h =0.2
1.0235
55 1.4680 59 1.1786 37 1.0411 9
24
0.2
0.6841
55 59 37 9
24
4,
4,
'
0
'
1
'
2
'
4, 3 3
p
p
p
y
y
y y y y
h
y y

0.8 1.0297
1.0297
9 2.0475 19 1.4680 5 1.1786 1.0411
24
0.2
0.6481
1 1.0235 2.0475
9 19 5
24
4,
4,
' 2
4
'
1
'
2
'
3
'
4, 3 4
y
y
y
y
y y y y
h
y y
c
c
c
------------------------------------------------------------------------------------------------------------

UNIT V
BOUNDARY VALUE PROBLEMS IN ORDINARY AND PARTIALDIFFERENTIAL EQUATIONS
Finite difference solution of second order ordinary differential equation
Finite difference solution of one dimensional heat equation by explicit and implicit methods
One dimensional wave equation
Two dimensional Laplace equation
Poisson equation

 Solution of Boundary Value Problems in ODE
 Solution of One Dimensional Heat Equation
 Solution of One Dimensional Wave Equation
 Solution of Laplace Equation
 Solution of Poisson Equation
There are number of methods for solving second order boundary value problems.
 Finite Difference Method
 Shooting Method
Finite difference solution of second order ordinary differential equation
Let us consider
y”(x)+f(x)y’(x)+g(x)y(x)=0
with boundary conditions y(x0)=a and y(xn)=b
Formula
2
1 1 2
"( )
h
y y y
y x i i i and
h
y y
y x i i
2
' ( ) 1 1
Steps for solving the second order ODE
 Put y”(x), y’(x), y(x) by i i i y ", y ' , y respectively
 substitute the above formula
 Form an equation
 Substitute the boundary conditions
 Solve the equations by putting i=1, 2, 3
 Write the solution of the intermediate values of boundary conditions
Problems based on Finite difference method for ODE
1. Solve by finite difference method, the boundary value problem y”(x) – y(x)=2 where
y(0)=0 and y(1)=1, taking h=
4
1
Solution:
Given y”(x) – y(x)=2
Step1:
Put y”(x), y(x) by i i y ", y respectively
i.e, " 2 i i y y -------------------(1)
Step2:

substitute the formula 2
1 1 2
"( )
h
y y y
y x i i i
(1) becomes 2
2
2
1 1
i
i i i y
h
y y y
16
1
(2 ) 2 1
2
i 1 i i y h y y
8
1
(2 ) 1
2
i 1 i i y h y y
8
1
16
33
i 1 i i 1 y y y
2.0625 0.125 i 1 i i 1 y y y ----------------------------(2)
Step3:
The boundary conditions are y0=0, y4=1
h=
4
1
gives
Step4
Put i=1, 2, 3 we get the following equations
2.0625 0.125 1 21 y y ----------------------------(3)
2.0625 0.125 1 2 3 y y y ----------------------------(4)
2.0625 0.875 2 3 y y ----------------------------(5)
Solving these three equations we get
y1=0.0451 y2=0.2183 y3=0.5301
---------------------------------------------------------------------------------------------------------
2. Solve the equation y”(x) –x y(x)=0 where xi=0,
3
2
,
3
1
,given that y(0)+y’(0)=1 and y(1)=1.
Solution:
Step1:
Put y”(x), y(x),x by i i y ", y , xi respectively
i.e, " 0 i i i y x y -------------------(1)
Step2:
substitute the formula 2
1 1 2
"( )
h
y y y
y x i i i
(1) becomes 0
2
2
1 1
i i
i i i x y
h
y y y
x0=0 x1=
4
1 x2=
2
1 x3=
4
3 x4=1
y0=0 y1=? y2=? y3=? y4=1

0
9
1
2 i 1 i i i 1 y x y y ---------------------------(2)
Step3:
The conditions are y(0)+y’(0)=1 and y(1)=1
x0=0 x1=
3
1 x2=
3
2 x3=1
y0=? y1=? y2=? y3=1
Putting i=0, 1, 2 we get
2 0 1 0 1 y y y -------------------------------(3)
0
27
55
0 1 2 y y y -------------------------------(4)
0
27
56
1 2 3 y y y -------------------------------(5)
Step4:
To solve the above equations we need the values of y-1 and y3
we have y3=1
So now we have to find the value of y-1 by using the condition y(0)+y’(0)=1
y(0)+y’(0)=1 1 '
0 0 y y
1
2
1 1
0 h
y y
y [put i=0 in
h
y y
y i i
i 2
' 1 1 ]
put h=
3
1
we get
3
2 3 2 0 1
1
y y
y -----------------------------------------(6)
substitute (6) in (3),
2 3 1 0 1 y y -----------------------------------------(7)
0
27
55
0 1 2 y y y -----------------------------------------(8)
1 0
27
56
1 2 y y -------------------------------------------(9)
Solving the above three equations we get
y0=-0.9880 y1=-0.3253 y2=0.3253
-----------------------------------------------------------------------------------------------------------
Classification of second order PDE
The most general linear pde of second order can be written as
0 2
2 2
2
2
Fu
y
u
E
x
u
D
y
u
C
x y
u
B
x
u
A ------------------------(1)
where A, B, C, D, E, F are in general functions of x and y.
Equation (1) is said to be
 Elliptic if B2-4AC<0

 hyperbolic if B2-4AC>0
 Parabolic if B2-4AC=0
Note:
 The same differential equation may be elliptic in one region, parabolic in another region
and hyperbolic in some other region.
Examples
Elliptic Parabolic Hyperbolic
1. uxx+uyy=0 (Laplace
Equation)
2. uxx+uyy=f(x, y) (Poisson
Equation)
uxx= 2
1
ut (One dimensional
heat equation)
uxx= 2
1
utt (One dimensional
wave equation)
Problems based on the classification of 2nd order PDE
1. Classify uxx+4uxy+(x2+4y2)+uyy=sin(x+y)
Solution:
A = 1, B = 4, C = x2+4y2
B2-4AC=4[4-x2-4y2]
The equation is elliptic if 4-x2-4y2<0 1
4 1
2 2 x y
(outside the ellipse 1
4 1
2 2 x y
)
Parabolic 4-x2-4y2=0 1
4 1
2 2 x y
(on the ellipse 1
4 1
2 2 x y
)
Hyperbolic 4-x2-4y2>0 1
4 1
2 2 x y
(inside the ellipse 1
4 1
2 2 x y
)
------------------------------------------------------------------------------------------------------------
2. Classify (x+1)uxx-2(x+2)uxy+(x+3)uyy=0
Solution:
A =(x+1), B=-2(x+2), C=(x+3)
B2-4AC=4>0
hyperbolic.
------------------------------------------------------------------------------------------------------------
One Dimensional Heat Equation
One dimensional heat equation is
2
2
2
x
u
t
u
Classification: Parabolic
There are two types for finding the Finite difference solution of one dimensional heat
flow equation
(i) Bender-Schmidt Method(Explicit Method valid for
2
1
0 )
(ii) Crank-Nicholoson Method(Implicit Method)

Bender-Schmidt Method(Explicit Method)
Coefficient of
t
u
=a, xi=x0+ih and tj=t0+jk
Case 1: (If h is given and k is not given)
Assume that λ = ½
Find k, using the formula
Using the given (i), (ii) and (iii) conditions fill the first row, first and last column of the table
Next using the Bender-Schmidt recurrence relation,
i j i j i j u u u , 1 1, 1, 2
1
i.e,
ui-1,j ui+1,j
ui,j+1
2
a b
c
Case 2: (If h and k are given)
Find , using the formula
Using the given (i), (ii) and (iii) conditions fill the first row, first and last column of the table
Next using the Bender-Schmidt recurrence relation,
i j i j i j i j u u u u , 1 1, 1, , (1 2 )
Using the formula and tabulate the remaining rows.
Examples
1. Solve uxx=2ut when u(0, t) = u(4,t)=0 and with initial condition u(x, 0)=x(4-x) upto t=5
sec, assuming Δx=h=1.
Solution:
Given: a=Coefficient of
t
u
a=2 ; h=1 and k is not given.
Assume that λ=½ k= λah2 = 2 1
2
1 =1
By the given conditions x=0, 1, 2, 3, 4 and t=0, 1, 2, 3, 4, 5
k= λah2
a
b
C
k= λah2

Condition(i) u(0, t)=0 u(0, 0)=0, u(0, 1)=0, u(0, 2)=0, u(0, 3)=0, u(0, 4)=0, u(0, 5)=0
Condition(ii) u(4, t)=0 u(4, 0)=0, u(4, 1)=0, u(4, 2)=0, u(4, 3)=0, u(4, 4)=0, u(4, 5)=0
Condition(iii) u(x, 0)=x(4-x) u(1,0)=1(4-1)=3,
u(2,0)=2(4-2)=4,
u(3,0)=3(4-3)=3,
Bender-Schmidt recurrence relation,
i j i j i j u u u , 1 1, 1, 2
1
i.e,
ui-1,j ui+1,j
ui,j+1
2
a b
c
xi=x0+ih and tj=t0+jk
------------------------------------------------------------------------------------------------------------
2. Find the values of the function satisfying the pde 4uxx=ut and the boundary conditions
u(0, t) = u(8,t)=0 and u(x, 0)=4x -
2
2 x for points x=0,1, 2, 3, 4, 5, 6, 7, 8, t=
8
j , j=0,1, 2, 3,
4, 5
Solution:
Given: a=Coefficient of t
u
a=¼ ; Here h and k is not given. Take h=1
Assume that λ=½ k= λah2 =
8
1
1
4
1
2
1
By the given conditions x=0,1, 2, 3, 4, 5, 6, 7, 8 and
8
5
,
8
4
,
8
3
,
8
2
,
8
1
t 0,
Condition(i) u(0, t)=0
8
5
,
8
4
,
8
3
,
8
2
,
8
1
t 0,
Condition(ii) u(8, t)=0
8
5
,
8
4
,
8
3
,
8
2
,
8
1
t 0,
Condition(iii) u(x, 0)=4x -
2
2 x u(1,0)= 4 -
2
1 =3.5, u(2,0)= 8 -
2
4 =6, u(3,0)= 12 -
2
9 =7.5,
i
j
0 1 2 3 4
x=0 x=1 x=2 x=3 x=4
0 t=0 0 3 4 3 0
1 t=1 0 2 3 2 0
2 t=2 0 1.5 2 1.5 0
3 t=3 0 1 1.5 1 0
4 t=4 0 0.75 1 0.75 0
5 t=5 0 0.5 0.75 0.5 0
a b
C

u(4,0)= 16 -
2
16 =8, u(5,0)= 20 -
2
25 =7.5, u(6,0)= 24 -
2
36 =6, u(7,0)= 28 -
2
49 =3.5.
Bender-Schmidt recurrence relation,
i j i j i j u u u , 1 1, 1, 2
1
i.e,
ui-1,j ui+1,j
ui,j+1
2
a b
c
xi=x0+ih and tj=t0+jk
------------------------------------------------------------------------------------------------------------
3. Solve uxx=ut when u(0, t) =0, u(4, t) =0 with initial boundary condition u(x, 0) = x(4-x),
h= 1, k=
4
1
upto t=1 sec.
Solution:
Given: a=Coefficient of t
u
a=1 ; Here h and k are given. Take h=1, k=
4
1
=
4
1 2
a
kh
This is in the interval
2
1
0
So we use Bender-Schmidt method
Formula:
i j i j i j i j u u u u , 1 1, 1, , (1 2 )
i
j
0 1 2 3 4 5 6 7 8
x=0 x=1 x=2 x=3 x=4 x=5 x=6 x=7 x=8
0 t=0 0 3.5 6 7.5 8 7.5 6 3.5 0
1 t=
8
1 0 3 5.5 7 7.5 7 5.5 3 0
2 t=
8
2 0 2.75 5 6.5 7 6.5 5 2.75 0
3 t=
8
3 0 2.5 4.625 6 6.5 6 4.625 2.5 0
4 t=
8
4 0 2.3125 4.25 5.5625 6 5.5625 4.25 2.3125 0
5 t=
8
5 0 2.125 3.9875 5.125 5.5625 5.125 3.9875 2.125 0
a
i j i j i j i j u u u u , 1 1, 1, , (1 2 )
i j i j i j i j u u u u , 1 1, 1, , (1 2 )
b
C

put =
4
1
i j i j i j i j u u u u , 1 1, 1, , 2
1
4
1
ui-1,j ui,j ui+1,j
ui,j+1 a b d c
2
1
4
1
boundary condition
boundary condition
boundary condition boundary condition
i
j
0 1 2 3 4
x=0 x=1 x=2 x=3 x=4
0 t=0 0 3 4 3 0
1
t=
4
1
0 2.5 3.5 2.5 0
2
t=
2
1
0 2.125 3 2.125 0
3
t=
4
3
0 1.8125 2.5625 1.8125 0
4 t=1 0 1.5468 2.1875 1.5468 0
b c d
a

Crank-Nicholson’s Method(Implicit Method)
Formula
Case(1) (h is given and k is not given)
Assume =1 where 2 ah
k
find k using this formula
i j i j i j i j i j u u u u u , 1 1, 1 1, 1 1, 1, 4
1
ui-1,j ui+1,j
ui-1,j+1 ui,j+1 ui+1,j+1
a= ( )
4
1
b c d e
using the formula and tabulate the values
Case(2) (h and k are given)
Find the value of using the formula 2 ah
k
Formula:
i j i j i j i j i j i j u u u u u u , 1 1, 1 1, 1 1, 1, , (2 2) (2 2 )
Using the formula and tabulate the values.
Problems based on Crank- Nicholson Method
1. Solve uxx=2ut when u(0, t) = u(4,t)=0 and with initial condition u(x, 0)=x(4-x) assuming
Δx=h=1 and compute u for one time step.
Solution:
Here h is given and k is not given.
=1. k= ah2=2
d e
b a c

Formula
i j i j i j i j i j u u u u u , 1 1, 1 1, 1 1, 1, 4
1
we have to find u11, u21, u31
put j=0,
,1 1,1 1,1 1,0 1,0 4
1
i i i i i u u u u u
put i=1, 2, 3 we get the following equations
4u11-4u21=4; -u11+4u21-u31=6; 0u11-u21+4u31=4 respectively.
We solve these equations we get the solution
u11=1.571, u21=2.2857, u31=1.571
One dimensional wave equation
One dimensional wave equation is uxx= 2
1
utt
There are two types
Type1:
Subject to the boundary conditions u(0, t) =0, u(l, t) =0 and the initial condition u(x,0)=f(x) and
ut(x, 0)=0.
Type2:
Subject to the boundary conditions u(0, t) =0, u(l, t) =0 and the initial condition u(x,0)=0 and ut(x, 0)=
f(x)
Formula
1, 1, , 1
2 2
,
2 2
, 1 21 i j i j i j i j i j u a u a u u u -------------------------(1)
where
h
k
.
This formula is the explicit scheme for the wave equation.
If
a
h
k the equation (1) takes the form
i , j 1 i 1, j i 1, j i , j 1 u u u u
i
j
0 1 2 3 4
x=0 x=1 x=2 x=3 x=4
0 t=0 0 3 4 3 0
1 t=2 0 1.571 2.2857 1.571 0

ui,j-1
ui-1,j ui,j ui+1,j
ui,j+1 a=b+c-d
and ut(x, 0)=0 ,1 1,0 1,0 2
1
i i i u u u which gives the values of u at the first time step in terms
of the values of u at time t =0.
This is used to solve the type 1
Suppose the problem is given in type2
We write ut(x, 0)=f(x) as ut(x, 0)=f(ih) and substitute in the formula
( )
2
, 1 , 1 f ih
k
u u i j i j for j =0
And then find the value of u11, u21, u31.
The values of Remaining rows are same as type1
-----------------------------------------------------------------------------------------------
Problems based on one dimensional wave equation
1. Find the nodal values of the equation 16uxx=utt taking Δx=1 given that u(0, t) =0,
u(5,t)=0, ut(x, 0)=0 and u(x, 0)=x2(5-x) and upto one half of the period of vibration.
Solution:
One dimensional wave equation is uxx= 2
1
utt a=4.
u(5, t)=0 l=5
Period of vibration is =
2
2 5
a
l
=2.5
Half period of vibration =1.25
Δx=1 h=1. Since k is not given we choose
a
h
k
4
1
k
b
d
c
a

Formula:
i , j 1 i 1, j i 1, j i , j 1 u u u u ----------------------------(1)
Conditions
u(0,t)=0 for t=0,
4
1
,
2
1
,
4
3
,1,
4
5
u(5,t)=0 for t=0,
4
1
,
2
1
,
4
3
,1,
4
5
u(x,0) = x2(5-x) u(1,0)=4, u(2,0)=12, u(3,0)=18,u(4, 0)=16
ut(x, 0) =0 0
2
, 1 , 1
k
u u i j i j for j=0
i ,1 i , 1 u u we find the values of the second using this formula
ui-1,j ui+1,j
ui, j+1
2
b c
a
The remaining rows obtained by the following formula
ui,j-1
ui-1,j ui,j ui+1,j
ui,j+1 a=b+c-d
The values of u at the time steps
i
j
0 1 2 3 4 5
x=0 x=1 x=2 x=3 x=4 x=5
0 t=0 0 4 12 18 16 0
1 t=
4
1 0 6 11 14 9 0
2 t=
2
1 0 7 8 2 -2 0
3 t=
4
3 0 2 -2 -8 -7 0
4 t=1 0 -9 -14 -11 -10 0
5 t=
4
5 0 -16 -22 -32 -18 0
b
a
c
b
d
c
a

2. Solve uxx=utt,0<x<2, t>0 subject u(x,0)=0, ut(x, 0)=100(2x-x2), u(0, t) =0, u(2,t)=0.
Choosing h=½, compute u for 4 time steps.
Solution
Here h=½ and a = 1. We choose
a
h
k
2
1
k
The simplest explicit scheme is given by
i , j 1 i 1, j i 1, j i , j 1 u u u u --------------------------------(1)
The boundary conditions are
u(0,t)=0 for t=0,
2
1
, 1,
2
3
, 2
u(5,t)=0 for t=0,
2
1
, 1,
2
3
, 2
u(x,0) = 0 for x =0,
2
1
, 1,
2
3
, 2
ut(x, 0)=100(2x-x2) 100(2 )
2
, 1 , 1 2 2 ih i h
k
u u i j i j for j =0
Here h=k=½
4
100
2
,1 , 1
i
u u i i i for i=1, 2, 3.
4
100
2
, 1 ,1
i
u u i i i for i=1, 2, 3. ---------------------------(2)
From (1) we get i ,1 i 1,0 i 1,0 i , 1 u u u u
Sub (2) in (1) we get
4
50
2
2
1,0 1,0
1 ,
i
i
u u
u i i
i for i =1, 2, 3.
But 0 i ,0 u , i
4
50
2
1 ,
i
u i i for i =1, 2, 3
So we have u11=37.5, u21=50, u31=37.5
i
j
0 1 2 3 4
x=0 x=
2
1 x=1 x=
2
3 x=2
0 t=0 0 0 0 0 0
1 t=
2
1 0 37.5 50 37.5 0
2 t=1 0 50 75 50 0
3 t=
2
3 0 37.5 50 37.5 0
4 t=2 0 0 0 0 0

Two dimensional Laplace Equation
Laplace equation is 2u=uxx+uyy=0
Standard Five Point Formula(SFPF)
, 1, 1, , 1 , 1 4
1
i j i j i j i j i j u u u u u
ui,j-1
ui-1,j ui,j ui+1,j
ui,j+1
Diagonal Five Point Formula(DFPF)
, 1, 1, , 1 , 1 4
1
i j i j i j i j i j u u u u u
ui-1,j+1 ui+1,j+1
ui,j
ui-1,j-1
ui+1,j-1
Liebmann’s Iteration Process
u1 u2 u3
u4 u5 u6
u7 u8 u9
We compute the initial values of u1, u2, …,u9 by using SFPF and DFPF. First we compute u5 by
SFPF.
b13 b12
b14
b11 b10 b9
b8
b15 b7
b16 b6
b1 b2 b3 b4 b5

5 7 15 11 3 4
1
u b b b b
We compute u1, u3, u7, u9 by using DFPF
1 1 5 3 15 4
1
u b u b b 3 5 5 3 11 4
1
u u b b b
7 13 5 15 11 4
1
u b u b b 9 7 11 5 9 4
1
u b b u b
Finally we compute u2,u4,u6,u8 using SFPF
2 3 5 1 3 4
1
u b u u u 4 1 7 15 5 4
1
u u u b u
6 3 9 5 7 4
1
u u u u b 8 11 5 7 9 4
1
u b u u u
These initial values are called rough values.
After that we use only SFPF method
Once all the values are computed their accuracy can be improved by Gauss-Seidal method.
The Gauss-Seidal formula is given by
( 1)
1, , 1 , 1
( 1)
1,
( 1)
, 4
1 n
i j
n
i j
n
i j
n
i j
n
i j u u u u u

NUMERICAL METHODS

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NUMERICAL METHODS