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LAKIREDDY BALI REDDY COLLEGE OF ENGINEERING (AUTONOMOUS)

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B. Tech I Year (II-Semester) May/ June 2014 Subject: T 264- Numerical Methods

ECE – B Section

UNIT – V Numerical Solutions of Faculty Name: N V Nagendram

Ordinary Differential Equations

Introduction – Solution by Taylor’s series – Picard’s method of successive approximations Euler’s method and modified Euler’s method – Runge - Kutta Methods – Predictor – Corrector methods – Adams – Moulton method – Milne’s method – Curve fitting – Fitting a

straight line – 2nd degree curves – exponential Curve by method of least

squares.




B.Tech I Year (II-Semester) May/ June 2014 T 264- Numerical Methods

ECE - B Section


Ordinary Differential Equations Planned Topics Lectures

5.1 Introduction

5.2 Solution of a differential equation

5.3 Methods of numerical solutions of ordinary differential equations(ODEs)

(i) Taylor’s series method (ii) Euler’s method and Modified Euler’s method

(iii) Picard’s method of successive approximation (iv) Runge – Kutta method

(v) Predictor – Corrector methods : Adams Moulton method

5.4 Initial boundary conditions

Tutorial-1

5.5 Taylor – series method – merits and de merits of Taylor series

Tutorial-2

5.6 Taylor series method for simultaneous Ist order differential equations

Tutorial-3

5.7 Taylor series method for simultaneous 2nd order differential equations

Tutorial-4

5.8 Picard’s Method of successive approximations

Tutorial-5

5.9 Euler’s method

Tutorial-6

5.10 Runge – Kutta methods

Merits and de-merits of runge Kutta Method

I st order Runge – Kutta method 2 nd order Runge – Kutta Method

3 rd order Runge – Kutta Method 4 th order Runge – Kutta Method

Advantages of Runge Kutta Method over Taylor series

Tutorial-7

5.11 Predictor – Corrector Methods

Tutorial-8

5.12 Milne’s Predictor – Corrector Formulae

Tutorial -9

5.13 Curve fitting – Fitting a straight line – 2nd degree curves – exponential

Curve by method of least squares.




B. Tech I Year (II-Semester) May/ June 2014 T 264- Numerical Methods

ECE - B Section


Ordinary Differential Equations Lecture-1

Introduction:

There exists large number of ordinary differential equations, whose solution

cannot be obtained by the known analytical methods.

In such cases, we use numerical methods to get an approximate solution of

a given differential equation with given initial condition.

Consider an ordinary differential equation of first order and first degree of

the form ),( yxfdx

dy ............................................................................. (i)

With the initial condition y(x0) = y0 which is called initial value problem.

Solution of a differential Equation:

To find the solution of initial value problem of the form (1) by numerical

methods, we divide the interval (a, b) on which the solution is derived in

finite number of sub-intervals by the points.

a = x0 < x1 < x2 ....... < xn = b

The points are called mesh points. Let the points are equally spaced by xn =

x0 + nh.

The solution of an initial value problem which exists uniquely in [x0 , b depends on the theorem due to Lipschitz which states that

(i) If f(x, y) is a real function defined and continuous in (x0 , b), y (-, +)

where x0 and b are finite.

(ii) there exists a constant L is called Lipschitz ‘s constant such that for any two values y = y1, and y = y2 | f(x, y1) – f(x, y2)| < |y1 – y2| where x [x0 , b then for any y(x0) = y0 the initial value problem(1) , has unique solution for

x [x0 , b.

Methods of numerical solutions of ordinary differential Equations(ODE)

(i) Taylor’s series method:

Consider the first order differential Equation (O. D. En) ),(= yxfdx

dy....... (1)

Where y(x0) = y0. Let y = f(x) be a solution of equation (1).

Expanding y(x) i.e., f(x) in a Taylor’s series about the point x0, we get

y(x0 + h) = y(x0) + hy( x0) + !2

2h

y"( x0) + (2)

The derivatives in this expansion can be obtained by taking the total

derivatives of f(x, y) which is an implicit function of y.

Thus, we get y( x ) = f(x, y)

fffyffx

y

y

f

x

fyxyx +=+=+ = ) x (y" 1

fffffffff yxyyyyxxx

22 +++2+ = ) x (y etc.

Equation (2) can be written as

y1 = y(x0 + h) = y0 + !1

hy0 +

!2

2h

y"0 + !3

3h

y0 + (3)

Similarly, we obtain

yn+1 = yn + !1

hyn +

!2

2h

y"n + !3

3h

yn + (4)

Equation (4) may be written as

yn+1 = yn + !1

hyn +

!2

2h

y"n + O(h3) (5)

where O(h3) means that all the succeeding terms containing the 3 rd and

higher powers of h.

If the terms containing the third and higher powers of h are neglected then

the local truncation error in the solution is kh3 where k is a constant.

For better approximation terms containing higher powers of h are

considered.

Note: Taylors series method can be applied only when the various

derivatives of f(x, y) exist and the value of (x – x0) in the expansion of

y = f(x) near x0 must be very small so that the series will be convergent.





ECE - B Section


Ordinary Differential Equations Tutorial-01

Problems on Taylor’s series

Problem #01 Form the Taylor’s series for y(x) , find y(0.1) correct to 4

decimal places if y(x) satisfies 1=)0(,1+= yxydx

dy. [Ans. 1.1053

Problem #02 Solve 0=)1(,+= yyxdx

dy and find y(1.1), y(1.2) by Taylor’s

series method. Compare the solution with exact solution.

[Ans.0.11033847; 0.2461077

Problem #03 Using Taylor’s series, find y at x = 0.1, 0.2 correct to 3

decimals given 0=)0(,3=2 yeydx

dy x . [Ans. y(0.1)= 0.349; y(0.2)= 0.811

Problem #04 Applying the Taylor’s series method, find the value of y(1.1)

and y(1.2) correct to three decimal places given that 1=)1(,= 3

1

yyxdx

dy

correct to 4 decimal places. [Ans. y(1.1)= 1.10681; y(1.2)= 1.22772

Problem #05 Find by Taylor’s series method, the value of y at x = 0.1,x = 0.2

to five decimal places given that 1=)0(,1= 2yyx

dx

dy .

[Ans. y(0.1)= 0.90033; y(0.2)= 0.80256





ECE - B Section


Ordinary Differential Equations Tutorial-01 Problems on Taylor’s series

Problem #06 Find y(0.2), y(0.4) correct to 4 decimals 0=)0(,21= yxydx

dy .

[Ans. y(0.2)= 0.19475; y(0.4)= 0.35989

Problem #07 Solve 0=)0(,2+3= yyedx

dy xand find y at x = 0.1, 0.2 correct

to 3 decimal places. [Ans. y(0.1)= 0.3486955; y(0.2)= 0.8112658

Problem #08 Solve 1=)0(,+

=23

ye

yxx

dx

dyx

and find y(0.1),y(0.2),y(0.3)?

[Ans. y(0.1)= 1.0047;y(0.2)= 1.081812; y(0.3)= 1.03995

Problem #09 Solve 1=)1(,+= 3yxy

dx

dy find y(1.1), y(1.2), y(1.3) ?

[Ans. y(1.1)= 1.225;y(1.2)= 1.512; y(1.3)= 1.874

Problem #10 Solve 0=)8.1(,10

1+= =y 2

yyxdx

dy find y(2).[Ans.y(2)= 0.3809

Problem #11 Solve 4=)4(,+

1=

2y

yxdx

dyto obtain y(4.1), y(4.2)?

[Ans. y(4.1)= 4.005;y(4.2)= 4.0098

Problem #12 Solve 1=)0(,+= =y 2yyx

dx

dy and compute y(0.1) , y(0.2) ?

[Ans. y(0.1)= 1.1164;y(0.2)= 1.2725

Problem #13 Solve y(0.2) given 0=)0(,3+2= yeydx

dy xcompare the

numerical solution obtained with exact solution?

[Ans.y(0.2)=numerical 0.811;y(0.2)=exact 0.8112 Problem #14 Find y to 5 decimal places when x = 1.02 given that

2=)1(,1= yxydx

dy ? [Ans. y(1.02) = 2.02061

Problem #15 Find y(0.2), y(0.4) given 1=)0(,= 2yyx

dx

dy

[Ans. y(0.2) = 0.8511043;y(0.4)= 0.7750643





ECE - B Section


Ordinary Differential Equations Lecture-2 Euler’s method and Modified Euler’s method:

Consider the differential Equation ),(= yxfdx

dy ....... (1)

Assume that the solution of (1) is y = (x) .............. (2)

Given y(x0) = y0.

Exact value Error

Pn Approx value(y)

P P1 Q1P2 Rn

R1 R2

0 A M1 M2 B

The curve of solution (2) through P(x0, y0) is shown line above in the fig.

Now, we have to find the ordinate of any point Q on the curve.

Let PA = y0, QB = yn.

Divide the interval AB into n sub-intervals each of width h at M1, M2, ..... so

that h is very small.

In the interval AM1, we approximate the curve by the tangent at P. If the

ordinate through M1, meets this tangent in P1 (x0 + h, y1) then

y1 = P1M1 = P1R1 + R1M1 = PA + P1R1 = y0 + PR1 Tan = y0 + h P

dx

dy

= y0 + h ),( 00 yxf

y1 = y0 + h ),( 00 yxf

Let P1, Q1 be the curve of solution (1) through P1 and let its tangent at P1

meets the ordinate through L2 in P2 (x0 +2h, y2).

Then y2 = y1 + h )y h, +x( 1 0f

On repeating this process n times, we finally reach on an approximation

BPn of BQ given by

yn = yn-1 + h )y 1)h,-(n +x( 1-n 0f

This is Euler’s method to find an approximate solution of (1).

Modified Euler’s Method Working Rule:

01. Modified Euler’s formula )1(

)1+(1

)(

1 ,(),(2

i

kkkkk

i

k yxfyxfh

yy

02. when i = 1, 0

)1+(ky can be calculated from Euler’s method.

03. k = 0,1,2,3,............... gives number of iterations.

i = 1,2,3,............... gives number of times, a particular iteration k is

repeated when i = 1.

)0(

)1+()1+(

1

)1+( ,(+),(2

kkkkkk yxfyxfh

yy

Note: In Euler’s method, we approximate the curve of solution by the

tangent in each interval i.e., by a sequence of short lines. Unless h is

small, there will be large error in yn. the sequence of lines may also

deviate considerably from the curve of solution. The process is very

slow and the value of h must be smaller to obtain accuracy reasonably.





ECE - B Section



Problems on EULER’S METHOD

Problem #01 Using Euler’s method, find an appropriate value of y

corresponding to x = 1 given that )+(= yxdx

dy and y(0) = 1?

Problem #02 Using Euler’s method, find an appropriate value of y

corresponding to x =0.3 given that )+(= yxdx

dy and y(0) = 1?

Problem #03 Using Euler’s method, find an appropriate value of y(0.02) = 1

given that )+(= 2yx

dx

dy and y(0) = 1?

Problem #04 Solve by Euler’s method, the differential equation )+(

)(=

xy

xy

dx

dy

with the initial condition y(0) = 1 and find an appropriate value of y

corresponding to x = 0.1 correct to four decimal places?

Problem #05 Given ydx

dy= and y(0) = 1 determine the values of y at x =

(0.01) (0.01) (0.04) by Euler’s method. Compare the values with exact values?

Problem #06 Solve )1(= ydx

dy given y(0) = 0 using modified Euler’s method

and tabulate the solutions at x = 0.1, 0.2 and 0.3 Compare your results with exact solutions?





ECE - B Section


Ordinary Differential Equations Lecture-3 Picard’s method of successive approximation:

Consider the differential Equation ),(= yxfdx

dy, y(x0) = y0.

Integrating the differential Equation between x0 and x, x

x

x

x

dxyxfdy

00

),(=

y(x) – y(x0) x

x

dxyxf

0

),(= .................................................................... (1)

Equation (1) is known as an integral equation since the dependent variable y occurs in the function f(x, y) on the right hand side under the sign of integration. The first approximation y1 is obtained by replacing y by y0 in f(x, y) in equation (1)

x

x

dxyxfyy

0

),(+= 001 .................................................................... (2)

The second approximation y2 is obtained by replacing y by y1 in f(x, y) in equation (1)

x

x

dxyxfyy

0

),(+= 102 .................................................................... (3)

The successive approximations of y are given by

x

x

dxyxfyy

0

),(+= 203

-------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------

x

x

nn dxyxfyy

0

),(+= 10 The process of iteration is stopped when any two

values of iteration are approximately the same. If x is large the convergence

is slow.





ECE - B Section



Problems on PICARD’S METHOD

Problem #01 Solve the equation )1+(= xydx

dy, y(0) = 1 by Picard’s method of

successive approximation and hence find y when x = 0.1?

Problem #02 Use Picard’s method to obtain y when x = 0.2 correct to five

decimal places given )(= yxdx

dy , y(0) = 1. Check your answer by finding the

exact particular solution?

Problem #03 Use Picard’s method of approximation to find y when x = 0.1, x

= 0.2 given )+(= 2yx

dx

dy, y(0) = 0?

Problem #04 Use Picard’s process of successive approximations obtain a

solution of the equation )x+(= 22y

dx

dy, y(0) = 1?

Problem #04 Solve ydx

dy= , y(0) = 1 by Picard’s method and compare the

solution with exact solution?





ECE - B Section



Runge – Kutta Method

The basic disadvantage of Taylor’s series method lies in the calculation of higher order total derivatives of Y. Euler’s method requires the smallness of

h for attaining reasonable accuracy. In order to overcomes these

disadvantages, the Runge – Kutta methods are designed to obtain greater

accuracy and at the same time to avoid the need for calculating higher order

derivatives. The advantage of these methods is that the functional values

only required at some selected points on the sub-interval.

I st order Runge – Kutta Method:

Consider the differential equation ),(= yxfdx

dy, y(x0) = y0

y1 = y(x0 + h)

Expanding by Taylor’s series

y1 = y(x0 + h) = y0 + !1

hy0 +

!2

2h

y"0 + !3

3h

y0 +

Also by Euler’s method

y1 = y0 + h f(x0 , y0) = y0 + hy0

It follows that the Euler’s method agrees with the Taylor’s series solution up

to the term in h.

Hence, Euler’s method is the Runge – Kutta method of the Ist order.

2nd order Runge – Kutta Method:

By Euler’s method y1 = y0 + h f(x0 , y0) ....................................... (1)

The modified Euler’s

method gives y1 = y0 + 2

h[ f(x0 , y0) + f(x0 +h, y1) ................. (2)

Substituting (1) in (2) we get,

y1 = y0 + 2

h[ f0 + f0 (x0+ h , y0+ hf0)................ (3)

where f0 = f0 (x0 , y0)

Expanding y1 = y(x0 + h) by Taylor’s series

y1 = y(x0 + h) = y0 + !1

hy0 +

!2

2h

y"0 + !3

3h

y0 +

Also expanding f(x0 +h, y0+hf0) by Taylor’s series for a function of two

variables, (3) gives

y1 = y0 + 2

h[ f0 + { f(x0 , y0) + h

0

x

f+h f0

0

y

f+ O(h2)

where O(h2) gives the terms containing 2nd and higher order powers of h

= y0 + 2

1[ hf0 + hf0 + h2

0

x

f+

0

y

f+ O(h3))

= y0 + hf0 + !2

2h

f0 + O(h3))

= y0 + !1

hy0 +

!2

2h

y"0 + O(h3)

y

ff

x

f

x

f+ ............................... (4)

Comparing (3) and (4) it follows that Euler’s series modified method agrees with the Taylor’s series solution upto the term in h2.

Hence the modified Euler’s method is the Runge – Kutta method of the 2 nd

order.

The second order Range – Kutta Formula is y1 = y0 + )+(2

121 kk

Where k1 = h. f(x0, y0), k2 = h. f(x0+ h , y0+ k1)

3 rd order Runge – Kutta Method

Proceeding as explained in 2 nd order Runge – Kutta method, it can be seen

that 3 rd order Range – Kutta method agrees with the Taylor’s series solution upto in term in h3.

The 3 rd order Runge - Kutta formula is

y1 = y0 + )+4+(6

1321 kkk where k1 = h. f(x0, y0), k2 = h. f(x0+

2

1 h , y0+

2

1 k1)

k3 = h. f(x0+ h , y0+ k1) where k1 = h. f(x0+ h , y0+ k1)

4 th order Runge – Kutta Method:

This method is most commonly used and is often referred to as unge – Kutta

method only.

Proceeding as mentioned above with local discretisation error in this method

O(h5), the increment k of y corresponding to an increment h of x by Runge –

Kutta method from dx

dy= f(x, y), y(x0) = y0 is given by

k1 = h. f(x0, y0), k2 = h. f(x0+

2

1 h , y0+

2

1 k1)

k3 = h. f(x0+

2

1 h , y0+

2

1 k2) , k4 = h. f(x0+ h , y0+ k3)

and finally computing

k = )+2+2+(6

14321 kkkk

which gives the required approximate value as y1 = y0 + k

Note: k = )1(4

1432 kkkk





ECE - B Section


Ordinary Differential Equations Tutorial-04 Problems on RUNGE – KUTTA METHOD

Problem #01 Find y(0.2) given that )+(= yxdx

dy and y(0) = 1 Using Runge –

Kutta Method?

Problem #02 Find an approximate value for x = 0.2 given that )+(= 2yx

dx

dy

and y(0) = 1?

Problem #03 Find y(0.3) given that 0=++ 2xyy

dx

dy and y(0) = 1 by taking h

= 0.1 Using Runge – Kutta Method?

Problem #04 Solve 22

22

+=

xy

xy

dx

dy To find y(0) = 1 at x=0.2,0.4 by Using Runge

– Kutta Method?

Problem #05 Evaluate y(1.1), Y(1.2) and y(1.3) using Runge – Kutta method

of order four for the initial value problem )+(= 22yx

dx

dy,y(1)=0?





ECE - B Section



Predictor – Corrector Methods:

We have discussed so far the methods in which a differential equation over

an interval can be solved if the value of y only is known at the beginning of

the interval. But, in the Predictor-Corrector method, four prior values are

needed for finding the value of y at xk.

The advantageous of these methods is to estimate error from successive

approximations to yk.

If xk and xk + 1 be two consecutive points, such that xk + 1 = xk + h , then

Euler’s method is yk + 1 = yk + h.f(x0 + kh, yk), k = 0,1,2,3,..... ............. (1)

and modified Euler’s method is

yk + 1 = yk +

2

h[f(xk , yk) + f(xk + 1 , yk + 1) , k = 0,1,2,3,..... ................... (2)

First we estimate yk + 1 by (1) and then this value of yk + 1 is substituted in (2)

to get a better approximation of yk + 1.

This procedure is repeated till two consecutive iterated values of yk + 1 are

approximately equal. This technique of refining an initially crude estimate of

yk + 1 by means of a more accurate formula is known as Predictor – Corrector method.

Thereofore, the equation (1) is called the Predictor while the equation (2)

serves as a Corrector of yk + 1. Two such methods, namely, Adams-Moulton

method and Milne’s method are discussed.





ECE - B Section



Adams - Moulton Method:


dy, y(x0) = y0

On integration we get y = y0 + x

x

dxyxf

0

),(

Let y1 = y0 + 1

0

),(

x

x

dxyxf for all x0 x x1 .................................... (1)

By Newton’s Backward Difference Formula, we have

....+ )y,f(x!3

)2+)(1+(+ )y,f(x

!2

)1+(+ )y,f(x p + )y,f(x = y),f(x 00

3

00

2

0000 ppppp

Where p = h

xx 0 ------------------ (2)

Substituting (2) in (1) we get

y1 = y0 + dxfpp

fpf

x

x

1

0

+!2

)1+(++ 0

2

00 where )y,f(x =f 000

y1 = y0 + h dpfpp

fpf

1

0

0

2

00 +!2

)1+(++ where p= dx,

x =x0,p=0,

x =x1,p=1,

y1 = y0 + h 0

432 +720

251+

8

3+

12

5+

2

1+1 f

Since y0 ,y-1, y-2, ...... etc are known and so y1 can be calculated.

y1p = y0 + h 0

432 +720

251+

8

3+

12

5+

2

1+1 f

------------------- (3)

this is called Adam-Bashforth formula and is used as Predictor formula

where P stands for Predictor.

A Corrector formulae can be derived in a similar manner by using Newton’s backward Formula at f1 viz.

....+f!3

)2+)(1+(+f

!2

)1+(+ f p + f = y),f(x 1

3

1

2

11 ppppp

------------------------ (4)

Again using (4) in (1) we get,

y1 = y0 + dxfpp

fpf

x

x

1

0

+!2

)1+(++ 1

2

11

y1 = y0 + h dpfpp

fpf

0

1

1

2

11 +!2

)1+(++

y1 = y0 + h 1

432 +720

19 -

24

1 -

12

1-

2

1-1 f

-------------------- (5)

the right hand side of (5) depends on y1,y0,y-1,y-2,...........

where y1p is used for y1. The new value of y1 is given by

y1c = y0 + h

pf1

432

720

251-

8

3-

12

5-

2

1-1

-------------------------------- (6)

where p

f1 = f1 ( x1, y1p)

The relation (6) is called Adams-Moulton Correct Formula.

Neglecting 4 th and higher order difference’s and expressing the remaining difference operators in their function values, the formulae (3) and (6)

become.

y1(p) = y0 +

24

h 3210 9-37+59 -55 ffff --------------------------------- (7)

y1(c) = y0 +

24

h 2101 +519 +55 ffff

p ---------------------------------- (8)

respectively.

The corrector formula (8) is repeatedly applied to get an improved value of f1

at each stage to find a better value of y1 unless it becomes stable and then

we proceed to calculate y2.

The error in formulae (7) and (8) are approximately )4(

0

2

720

251fh and )4(

0

5

720

19fh

respectively.

Such formula expressed in ordinate form, are often called explicit

predictor-corrector formula.

To apply Adams-Moulton method, we require four starting values of y which

are to be calculated by means of Picard’s method or Taylor’s series method or Euler’s method or Runge-Kutta methods. It is practice, Adams-Moulton

formulas have been found to be most useful.

Formulae

Adams – Bash forth Predictor formula

yn+1 (p) = yn +

24

h 321 9-37+59 -55 nnnn ffff

yn+1 (p) = yn +

24

h 1

3

1

2

1

1

1 9-37+59 -55 nnnn yyyy

Since y = f(x, y)

Adams – Bash forth Corrector formula

yn+1(c) = yn +

24

h 211+ +519 +9 nnnn ffff

or

yn+1(c) = yn +

24

h 1

2

1

1

1)(1

1+ +519 +9 nnn

p

n yyyy

Error Estimates

EAB = )(720

251 5 vyh and EAM = )(

720

19 5 vyh





ECE - B Section


Ordinary Differential Equations Tutorial-06 Problems on ADAMS - MOULTON’S METHOD

Problem #01 Given 22 )1(2

1yx

dx

dy and y[0 = 1, y(0.1) = 1.06, y(0.2)= 1.12,

y(0.3)= 1.21. Evaluate by Milne’s Predictor – Corrector method?

Problem #02 Using Milne’s Predictor – Corrector method to find y(0.8) from

given )+1( 2y

dx

dy and y[0 = 0, y(0.21) =0.02027, y(0.4)= 0.4228, y(0.6)=

0.6841?

Problem #03 Find the numerical solution of given )( 2yx

dx

dy and y[0 = 0,

0 x 1 by Milne’s Predictor – Corrector method?

Problem #04 Find the numerical solution of given )+( yxdx

dy and y[0 = 1,

from 0.20 x 0.30 by Milne’s Predictor – Corrector method?

Problem #05 Find the solution of initial value problem given )+( 22yxx

dx

dy

and y[1 = 1, 0.00 x 1.30 (interval 0.1). Evaluate by any numerical

method at x = 1.4 by Adam – Moulton corrector formula method?

Problem #06 Find the value of y(1.4) given )+( 22yx

dx

dy and y[1 = 0. Using

Adams – Moulton Predictor – Corrector method and Milne’s Predictor –

Corrector method?

Problem #07 Using Adams – Moulton Predictor – Corrector formulae find

y(0.4) given )(2

1xy

dx

dy and y[0 = 1, y(0.1) = 1.01, y(0.2)= 1.022, y(0.3)=

1.023?

Problem #08 Using Adams – Moulton ,find y(0.8) given )+1( 2y

dx

dy and

y[0 = 0?





ECE - B Section



MILNE’S METHOD:


dy, f(0) = y(x0) = y0 = 0

On integration we get y = y0 + x

x

dxyxf

0

),(

Let y1 = y0 + 4

0

),(

x

x

dxyxf for all x0 x x4 .................................... (1)

By Newton’s Forward Difference Formula is given by

....+f!3

)2-)(1-(+f

!2

)1-(+ f p + f = y),f(x 0

3

0

2

00 ppppp

.......................... (2)

Substituting (2) in (1), we get

y4 = y0 +

4

0

........+f!3

)2-)(1-(+f

!2

)1-(+ f p + f 0

3

0

2

00

x

x

dxnnnnn

y4 = y0 + h

4

0

0

3

0

2

00 ........+f!3

)2-)(1-(+f

!2

)1-(+ f p + f dp

nnnnn

y4 = y0 + h

........+f

3

8+f

3

20+ f 8 + f4 0

3

0

2

00

y4 = y0 + 3

4h 321 f2+ f - f2 ...................................................... (3)

After neglecting four and higher order differences and expressing the

differences f0, 2 f0, and 3 f0 in terms of the function value. Formula (3) is

called Predictor.

The value of y4 can be determined if those of y0, y1, y2 and y3 are known and

we obtain a first approximation as

y4 = y2 + 2

h 432 f+ f 4+ f ............................................................ (4)

Formula (4) is called a Corrector.

An improved value of f4 is computed and again the corrector is applied to

find a still a better value of y4 and f4 to desired degree of accuracy,

y5 = (x0 + 5h ) can be determined from the Predictor as

y5 = y2 + h3

4 432 f2+ f - f2 and then f5 = f(x0 + 5h, y5 ) is calculated.

Then, a better approximation to the value of y5 can be obtained from the

corrector as y5 = y3 + h3

1 543 f+ f4 + f

We repeat this step till the value of y5 can be obtained to the desired

accuracy. If the corrected value of y differs from the predicted value, we then

compute the value )(

1+1+29

1 p

nn yy

If the value of is very small, we proceed to the next interval. If not, the

value is made small.

IMPORTANT FORMULAE:

Milne’s Predictor Formula: (p)

1+ny = yn - 3 + 3

4h 2-n1-nn f2+ f - f2 or (p)

1+ny = yn - 3 + 3

4h 2-n1-nn y2+ y - y2

Milne’s Corrector Formula:

(C)

1+ny = yn - 1 + 3

1h 1-nn

(p)

1+n f+ f 4+ f or (C)

1+ny = yn - 1 + 3

1h 1-nn1+n

(p) y+ y 4+ y

Error Estimates:

E(p) = )(29

281

5 vyh and E(C) = )(

90

12

5 vyh





ECE – B Section

UNIT – I Important Questions Faculty Name: N V Nagendram

UNIT-I

1. Find the Eigen values and Eigen vectors of the following Matrix

3 2 2

2 3 2

2 2 3

A

2. Determine the Rank of the following Matrix

3 3 0

1 4 5

4 4 8

A

3. Verify Cayley Hamilton theorem for the Matrix

2 1 1

1 2 1

1 1 2

A

and find its inverse

4. Find the Eigen values and Eigen vectors of the following Matrix

5 2 0

1 2 3

1 2 4

A

5. Find the Echelon form and how reduced Echelon form of the Matrix

1 2 3 1

2 1 2 2

3 1 2 3

A

Hence find the Rank of A


1 2 3

2 1 1

3 1 2

A


7. Define the Rank of a Matrix. Find the non-singular Matrices P&Q such that PAQ is the

normal form for the Matrix

1 2 3

2 1 0

3 1 2

A

8. Solve the following system of linear equations by Gauss -Seidal method


3 2 2

2 3 2

2 2 3

A


10. Reduce the matrix

5114

1352

35110

2131

to the Echelon form hence find its Rank

11. Solve the following system of linear equations by Gauss elimination method

12. Show that the sum of Eigen Values of a matrix is the sum of the elements of the

principle diagonal of that matrix

13. Verify Cayley Hamilton theorem for the Matrix A=

211

121

212


14. Solve the following system of linear equations by Gauss -Seidal method

15. Reduce the Matrix in to Normal form and find its Rank

2 1 3 6

2 3 1 2

1 1 1 2

A

16 Verify Cayley Hamilton theorem for the Matrix A=

211

121

212

and find A4





ECE – B Section

UNIT – II Important Questions Faculty Name: N V Nagendram

UNIT-II

1. Determine the real root of 3 4 9 0x x which lies b/w 2 &3corrcet to three decimal

places by using bisection method

2. Find the root of the equation 102 log 7xx which lies b/w3.5&4.0 by the method of

False position

3. Determine the real root of 10log 1.2xx which lies b/w 2 &3corrcet to three decimal

places by using bisection method

4. Find the root of the equation 4 10 0x x using Newton -Raphson method

5. Find the Orthogonal transformation which transforms the Quadratic form

Q= 2 2 2

1 2 3 2 33 3 2x x x x x to the canonical form. Hence find its Index, Rank and

Signature

6. Find the root of the equation 3 2 5 0x x using Regula-false position

7. Identify the Nature, Index and Signature of the Quadratic form 2 2 2

1 2 3 1 2 1 3 2 34 4 2 4x x x x x x x x x

8. Find the root of the equation 1 0xxe using Newton -Raphson method

9. Reduce the Quadratic form Q= 2 2 2

1 2 3 2 3 3 1 1 22 3 8 4 12x x x x x x x x x to the canonical

form. Hence find its Index, Rank and Signature

10. Find the root of the equation 2

10log 12xx usingRegula-false position

11. Find the positive root of the equation 2sin 0x x

12. Find the solution of 2( ) 25 0f x x using method of False position and Newton -

Raphson method

13. Find a positive root of 13 xx correct to two decimal places by bisection method

14. Find the root of the equation 2 3 5 0x x using Newton -Raphson method

15. Apply bisection method to find a root of the equation 0cos xx

16. Reduce the Quadratic form Q= 323121

2

3

2

2

2

1 222353 xxxxxxxxx to the canonical

form. Hence find its Index, Rank and Signature

17. Find the solution of 2( ) 25 0f x x using method of False position

18. Find the root of the equation 4 10 0x x using Newton -Raphson method





ECE – B Section

UNIT – III Important Questions Faculty Name: N V Nagendram

UNIT-III 1. With the usual notation, show

i) 3 3

2 5y y ii) 6 6

2 8y y iii) 1/2(1 )

2. The values of x and y are given as below

x 1 2 3 4 5

y 4 6 8 10 12

Find the value of y when x=4.5 use Lagrange’s interpolation formulae

3. With the usual notation, show that

i) 11 E ii) iii) hD

E e where D is the differential operator

4. Derive the Lagrange’s interpolation formulae using this formulae ,find the value of

f(9), given the values

x 5 7 11 13 17

f(x) 150 392 1452 2366 5202

5. Evaluate (i) xxx 312113 (ii) 53 xne

6. Derive the Newtons backward interpolation formula.using this for estimate the

population of the city in the year 1995

Year (x) 1961 1971 1981 1991 2001

Population in lakhs (y) 46 66 81 93 101

7. With the usual notation, show that (i) 2/11

(ii)2 (iii)

8. Derive the Lagrange’s interpolation formula. Using this formula, find y(1) given that

9. With the usual notation, show that (i)22 E (ii)

2 (iii) 11

10. Derive the Lagrange’s interpolation formula. Using this formula, find y(2) given that

11. The population of a certain town is shown in the following table

Year 1990 1995 2000 2005 2010

Population in thousands 40 60 79 102 130

Find the population in 2007

x -1 0 2 3

f(x) -8 3 1 2

x 0 1 3 4

f(x) -12 0 12 24

12. The values of x and y are given as below

x 1 2 3 4 5

y 4 6 8 10 12

Find the value of y when x=4.5 use Lagrange’s interpolation formulae

13. The following gives the values of Tan for interval 0.10≤ ≤0.30

0.10 0.15 0.20 0.25 0.30

Tan 0.1003 0.1511 0.2027 0.2553 0.3093

Calculate the value of Tan (0.12)

14. Given the following table of values

x 0.4 0.5 0.7 0.8

f(x) -0.916 -0.693 -0.357 -0.223

Estimate the value of f (0.6) using Lagrange’s interpolation

15. With the usual notation, show that (i) 2/11

ii) 6 6

2 8y y

(iii) 11

16. Construct a polynomial function in x for y specified in the following table and hence

find y(0.5)

x 0 1 2 3 4 5

f(x) 5.2 8.0 10.4 12.4 14.0 15.2

17. Using Gauss backward formula

Year x 1931 1941 1951 1961 1971 1981

Sales in lakhs y 12 15 20 27 39 52

Find the sales for the year 1996

18. By using Lagrange’s interpolation formula, fit a polynomial to the data

x 0 1 3 4

F(x) -12 0 6 12

19. With the usual notation, show that i) 11 E (ii)

22 E (iii)2

20. Obtain the Newton’s divided difference interpolating polynomial for the following

data and hence find f(3)

x 0 1 2 4 5 6

f(x) 1 14 15 5 6 19





ECE – B Section

UNIT – IV Important Questions Faculty Name: N V Nagendram

UNIT-IV

1. Find the value of dy/dx when x=0.5 from the following data

x 0 1 2 3 4

y 768 913 1039 1143 1225

2. Using Simpson’s rule, evaluate /2

0

sin xdx

by dividing the range in to six equal parts

3. Evaluate

6

2

01

dx

x by using (i) Trapezoidal rule (ii) Simpson’s rules compare your results with the

actual value.

4. Find dy/dx and d2y/dx

2 at x=1.2 from the following data

x 1 1.2 1.4 1.6 1.8 2

y 2.72 3.32 4.06 4.95 6.05 7.39

5. Find an approximate value of5loge

by calculating to 4 decimal places by Simpson’s rule 4 5

dx

x

dividing range in to 10 equal parts.

6. The following table of values of x and y is given

x 0 1 2 3 4 5 6

y 6.989

7

7.403

6

7.781

5

8.129

1

8.451

0

8.750

6

9.030

9

Find dy/dx when (i)x=1 (ii)x=3 (iii)x=6 also find d2y/dx

2 when x=3



x 1 1.2 1.4 1.6 1.8 2

y 2.72 3.32 4.06 4.95 6.05 7.39

8. Given that

x 4 4.2 4.4 4.6 4.8 5.0 5.2

logx 1.3863 1.4351 1.4816 1.5216 1.5686 1.6094 1.6484

Evaluate

5.2

4

log xdx by Simpson’s rule


2 from the data near x=2 using Backward Difference Formula

x 1 1.2 1.4 1.6 1.8 2

y 2.72 3.32 4.06 4.95 6.05 7.39

10. Using Simpson’s rule, evaluate 3

0

sin xdx

from the following data

x 0 / 4 / 2 3 / 4

y 0.0 0.7071 1.0 0.7071 0



x 1 1.2 1.4 1.6 1.8 2

y 3 6.26 11.07 17.84 26.99 39

12. Using trapezoidal rule, approximately calculate the value of

6

2

0

dx

a x and compare with actual value.

13. A rod is rotating in plane .The following table gives the angle θ in the radians through which the rod has rotated for time t seconds, calculate the angular velocity of the rod when t=0.6 second

14. Evaluate 7loge

by Simpson’s 3/8 rule

15. Compute f1(x) and f

11(x) at x=15 using the following data

16. Compare your results with exact values. evaluate I= ,1

1

0

x

dxwith n=10 and h=0.1using Simpson’s

1/3 rule



x 1 1.2 1.4 1.6 1.8 2

y 3 6.26 11.07 17.84 26.99 39

18. Given that

x 4 4.2 4.4 4.6 4.8 5.0 5.2

Log

x

1.386

3

1.435

1

1.481

6

1.521

6

1.568

6

1.609

4

1.648

4

Evaluate

5.2

4

log xdx by (a) trapezoidal rule (b) Simpson’s 1/3 rule (c) Simpson’s 3/8 rule

t 0 0.2 0.4 0.6 0.8 1.0 1.2

θ 0 0.12 0.49 1.12 2.02 3.20 4.67

x 15 17 19 21 23 25

f(x) 3.873 4.123 4.359 4.583 4.796 5.8





ECE – B Section

UNIT – V Important Questios Faculty Name: N V Nagendram

UNIT-V 1. Solve the initial value problem , (0) 2 (0.6)y y x y y using Euler’s method to

find (0.6)y with h=0.2.

2. Use Runge - Kutta fourth order method to obtain solution to differential equation 2

2 2

2

d yxy y

dx with y(0)=1 and 0

dy

dx for x=0.2.

3. Given that 2 2 , (0) 0y y x y Determine the first three non-zero terms in

Taylor’s series and hence obtain y(1). 4. Fit a least Square curve of the form

by ax for the following data where a and b are

constants

x 61 26 7 2.6

y 350 400 500 600

5. Using Euler’s method find the value of y for x=0.2 when log( )y x y with (0) 1y

6. Use Runge - Kutta fourth order method to solve the initial value problem

, (0) 2dy

y x ydx

and find y(0.1) and y(0.2) correct to four decimal places.

7. Given that2 , (0) 1y x y y . Determine the first five non-zero derivative terms in

the Taylor’s series and hence obtain y(0.1) correct to four decimal places.

8. Fit a parabola by the method of least squares to the following data

x 1 2 3 4 5

y 5 12 26 60 100

9. Solve , (0) 1y y x y by using Picard’s method and hence find y(0.1) and y(0.2) up to third approximation

10. Using simple Runge - Kutta fourth order method to obtain solution to differential

equation 2 , (1) 2dy

xy ydx

at the points x=1.2, 1.4 taking h=0.2.

11. Solve the initial value problem 21 , (0) 0y y y and compute y(0.8) using the

Milne’s predictor corrector method

12. Fit a second order method to the following data by the method of least squares

x 1.0 1.5 2.0 2.5 3.0 3.5 4.0

y 1.1 1.3 1.6 2.0 2.7 3.4 4.1

13. Solve the differential equation xydx

dy1 subject to y(0)=1 by Taylor series method

and hence find y(0.2).

14. Apply Runge - Kutta fourth order method to solve the initial value problem 2 2

2 2, (0) 1

dy y xy

dx y x

with h= 0.2, x=0, 0.2, 0.4.

15. Using modified Euler’s method solve numerically the equation xydx

dy 2 with

y(1)=1 to find y(1.2)

16. Estimate the chlorine residual parts/million(Y) in swimming pool five hours after its

chemical treatment by fitting an exponential curve of the form Y=ABX

to the

following data

Number of hours(X) 2 4 6 8 10 12

Chlorine residual parts

per million 1.8 1.5 1.4 1.1 1.1 0.9

17. Solve by Taylor series method the initial value problem21 , (0) 0y y y to find the

vales of y at x=0. 0.2, 0.6

18. Use Runge - Kutta fourth order method to solve the initial value problem xyy 1with

y(0)=1 find y(0.4)

19. Using Euler’s method find the value of y for x=0.4 when 2 2 , (0) 0y y x y with

h=0.1

20. Fit a least Square curve of the form y= aebx

for the following data where a and b are

constants

x 1 2 3 4 5

y 2.6 3.3 4.2 5.4 6.9

END OF UNIT V AND IMPORTANT QUESTIONS ON UNIT I TO UNIT V

L.B.R.C.E MYLAVARAM

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