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MA2264-NUMERICAL METHODS

KINGS COLLEGE OF ENGINEERING-PUNALKULAM

1

KINGS COLLEGE OF ENGINEERING

DEPARTMENT OF MATHEMATICS ACADEMIC YEAR 2012-2013 (EVEN SEMESTER)

QUESTION BANK SUBJECT NAME: MA2264- NUMERICAL METHODS

YEAR/SEM: II / IV BRANCH: EEE&CIVIL

UNIT - I SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS

PART-A(2 Marks) 1. Write Newton’s formula to find the cube root of N. Let f(x) = x3 – N and f’(x) = 3x2 .By Newton method,

2

3

2

3

13

2

3 n

n

n

n

nnx

Nx

x

Nxxx

+⇒

−−=+

2. State the order of convergence of Newton’s Raphson method. The rate of convergence in Newton’s Raphson method is of order 2.

3. Find an iterative formula to find N by Newton’s method.

Taking x2 = N and using Newton Raphson formula

+=+

n

n

nx

Nxx

2

12

1

4. State the criterion for the convergence in Newton Raphson method.

Newton – Raphson method converges if [ ]2'' )(')()( xfxfxf < in the interval

considered. 5. State the fixed point iteration theorem.(or) If g(x) is continous in [a,b], then under what condition in [a,b]? Let f(x) = 0 be the given equation whose exact root is α . The equation f(x)=0 be

rewritten as x = ( )xφ . Let I be the interval containing the root x = α . If ( ) 1' <xφ for

all x in I, then the sequence of approximations nxxx ,........., 21 will converge to α , if

the initial starting value 0x is chosen in I.

6. What is the order of convergence for fixed point iteration? The convergence is linear and the convergence is of order 1. 7. Solve x+y = 2, 2x+3y = 5 by Gauss Elimination method. Given x+y = 2, 2x+3y = 5



2

[ ]

122 21

2

10

11

5

2

32

11,

5

2

32

11

RRR

BA

y

x

−

=

=

=

↔

y = 1 x+y = 2 x+1=2, ∴x=1. 8. When Gauss-Elimination method fails? This method fails if the element in the top of the first column is zero. We can rectify this by interchanging the rows of the matrix. 9. Solve the following system of equations by Gauss – Jordan method. 5x + 4y = 15, 3x + 7 y = 12.

=

12

15

73

45

y

x

The given system is equivalent to AX = B

[A, B] =

12

15`

73

45

Now , we will make the matrix A is a diagonal matrix.

=

15

15`

230

45

R2↔ 5R2 – 3R1

=

15

285`

230

0115

R1↔ 23R1 – 3R2

115x = 285

6522.0

23

15

1523

4783.223

57

==

=

==

y

y

x

10. Distinguish Gauss Elimination method and Gauss Jordan method.

Gauss Elimination method Gauss Jordan method Co – efficient matrix A of the system reduces into upper triangular matrix.

Co – efficient matrix A of the system reduces into diagonal of unit matrix

Back substitution process gives solution.

Solution obtained directly



3

11. Distinguish between direct and iterative (indirect) methods of solving simultaneous equations.

S.No. Direct method Iterative method 1. We get exact solution Approximate solution. 2. Simple, take less time Time consuming

laborious.

12. Write a sufficient condition for Gauss_seidal method to converge? The process of iteration by Gauss-Jacobi method will converge if in each equation of the system, the absolute value of the largest coefficient is greater than the sum of the absolute values of the remaining coefficients.

13. Compare Gauss-Jacobi and Gauss seidal methods. S.No. Gauss-Jacobi Gauss-seidal 1. Convergence rate is slow The rate of convergence

of Gauss-seidal method is roughly twice that of Gauss-jacobi

2. Indirect method Indirect method 3. Condition for convergence is

the coefficient matrix is diagonally dominant

Condition for convergence is the coefficient matrix is diagonally dominant

14. Find the inverse of the coefficient matrix by Gauss-Jordan method 5x – 2y = 10 , 3x + 4y = 12.

Coefficient matrix A=

−

43

25

[ ]

510

05

1

435

21

10

01

43

25,

11

RR

IA

↔

−=

−=

122 3

15

3

05

1

5

263

5

21

RRR −↔

−

−

=

26

5

26

5

26

3

05

1

135

21

22 ×↔

−

−= RR



4

−=∴

−==∴

+↔

−=

−

−

53

24

26

1

26

5

26

313

1

13

2

5

2

26

5

26

313

1

13

2

10

01

1

1

211

A

A

RRR

15. Find the inverse of A =

72

31by Gauss-Jordan method.

[ ]

−

−=∴

−

−

−=

−

−=

=

=

−

↔

↔

12

37

312

37

10

01

212

01

10

31

10

01

72

31,

72

31

1

211

122

A

RRR

RRR

IA

A

16. Write two iterative methods in solving a set of simultaneous equations. 1. Gauss Jacobi method 2. Gauss Seidel method 17.Check the conditions for convergence for the system of equations 10x-5y-2z=3 ,4x-10y+3z=-3 ,x+6y+10z=-3 The given system of equations are diagonally dominant. 18. Determine the largest eigenvalue and the corresponding eigen vector of the

matrix

11

11 correct to two decimal places using power method?

32

21

21

12

2

2

1

1

11

11

21

12

2

2

1

1

11

11

XAX

XAX

=

=

=

=

=

=

=

=

∴This shows that the largest eigenvalue = 2.



5

The corresponding eigen vector =

1

1

19. Find the dominant eigenvalue of A =

43

21 by power method?

Let

=

1

11X

43

32

21

38.51

46.0389.5

38.5

46.2

1

46.0

43

21

29.51

46.029.5

29.5

43.2

1

43.0

43

21

71

43.07

7

3

1

1

43

21

XAX

XAX

XAX

=

=

=

=

=

=

=

=

=

=

=

=

The dominant eigenvalue = 5.38 and the corresponding eigenvector =

1

46.0

. 20. How will you find the smallest eigen value of a square matrix A? By power method, the largest eigen value of A-1 can be found. Then smallest eigen value of A is the reciprocal of largest eigen value of A-1.

PART – B

1 .a. Find by Newton’s method , the root of xlog10 x=1.2 correct five decimal (8) places.

b. Find the iteration formula to find N where N is a positive integer by (8)

Newton’s method and hence find 11 2.a. Derive a Newton-Raphson iteration formula for finding the cube root of a (8)

positive number N. Hence find 3 12 b. Obtain an iteration formula, using N - R values to find the reciprocal of a

given number N and hence find 19

1 , correction of 4 decimal places (8)

3.a. Find the double root of x3 – x2 – x – 1 = 0 choosing with the initial value of 0.8 (8) b. Find a real root of the equation x3+x2-100 = 0. (8) 4.a. Find a real root of the equation cos x = 3x-1 correct to 5 decimal places by fixed point iteration method. (8) b.Solve the system of equations 10x-2y+3z=23; 2x+10y-5z=-33; 3x-4y+10z=41 using Gauss – elimination method (8) 5.a. Solve the following system of equation using Gauss – elimination method 2x + y + 4z = 12 , 8x -3y +2z = 20 , 4x +11y – z =33. (8) b.Using Gauss-Jordan method, solve the following system of equations 10x+y+z=12,2x+10y+z=13,x+y+5z=7 (8) 6. a.Using Gauss-Jordan method, solve the following system of equations

2x –y +2z = 8, -x +2y + z = 4, 3x + y -4z = 0 (8) b.Solve the following system of equations by Gauss-Jacobi Method 27x+6y-z=85;



6

x+y+54z=110; 6x+15y+2z=72. (8) 7. a Using Gauss – Seidel method, solve the following system. Start with x = 1, y = -2, z = 3, x+3y+52z = 173.61; x-27y+2z = 71.31; 41x-2y+3z = 65.46 (8) b.Solve the following system by Gauss – Seidel method 28x+4y-z=32, x+3y+10z=24, 2x+17y+4z=35 (8) 8.a. Using Gauss- elimination method , find the inverse of the matrix

−

−

221

132

214

(8)

b. Find the numerically largest eigen values of A =

− 402

031

2125

by power

method corresponding eigen vector (correct to 3 decimal places). Start

with initial eigen value

0

0

1

. (8)

9.a. Determine the largest eigen value and the corresponding eigen vector correct to 3

decimal places, using power method for the matrix A =

321

242

123

(8)

b.Determine the largest eigen value and the corresponding eigen vector of the matrix

−

−−

−

210

121

012

(8)

10. a.Using Jacobi method ,find the eigen values and eigen vectors of

A=

122

232

221

(8)

b. Using Jacobi method ,find the eigen values and eigen vectors of

A=

−

−

201

020

102

(8)



7

UNIT II INTERPOLATION AND APPROXIMATION

PART- A (2 Marks) 1. Write the Lagrange’s formula to find y if three sets of values ( )0,0 yx , ( )11 , yx and

( )2,2 yx are given.

2

1202

10

1

2101

20

0

2010

21

))((

))((

))((

))((

))((

))((y

xxxx

xxxxy

xxxx

xxxxy

xxxx

xxxxy

−−

−−+

−−

−−+

−−

−−=

2. Obtain the Lagrange’s interpolating polynomial for the observed data of points (1,1),(2,1) and (3,-2).

( )2)23)(13(

)2)(1(1

)32)(12(

)3)(1(1

)31)(21(

)3)(2(

))((

))((

))((

))((

))((

))((2

1202

101

2101

200

2010

21

−−−

−−+

−−

−−+

−−

−−=

−−

−−+

−−

−−+

−−

−−=

xxxxxx

yxxxx

xxxxy

xxxx

xxxxy

xxxx

xxxxy

= ( )4932

1 2 +− xx

3. Find the polynomial which takes the following values

X: 0 1 2 Y: 1 2 1

x y ∆y ∆2y 0 1 1 1 2 -2 -1 2 1

( )

0

2

002

1y

ppypyy ∆

∠

−+∆+=

= 1 + 2x + x2 4. Given f(0) = -1, f(1) = 1 and f(2) = 4. Find the Newton’s interpolating formula.

( ) ( )23

2

1

1

0

1,3,2,1

2

0

2

100

−+=⇒=−

=

=∆=∆=∆−=

xxxfxx

p

yyyy

5. State any two properties of divided differences. (i) The divided differences are symmetrical in all their arguments. (ii) The divided differences of sum or difference of two functions is equal to the sum or difference of the corresponding separate divided differences.



8

6. State Newton’s divided difference interpolation formula? ),....,()).....()((.....),,()(),()()()( 1011021001000 nn xxxfxxxxxxxxxfxxxxfxxxfxf −−−−++−+−+=

7. What are the nth divided differences of a polynomial of the nth degree. The nth divided differences of a polynomial of the nth degree are constant.

8. Show that the divided differences are symmetrical in their arguments.

( )( ) ( ) ( ) ( )

( )01

10

10

01

01

10 ,, xxfxx

xfxf

xx

xfxfxxf =

−

−=

−

−=

This shows divided differences are symmetrical in their arguments 9. Find the divided difference for the following data

X: 2 5 10 Y: 5 29 109

x y ∆y ∆2y 2 5

825

529=

−

−

5 29 1210

816=

−

−

16510

29109=

−

−

10 109 10. Find the divided differences of f(x) = x2+x+2 for the arguments 1,3,6,11. The corresponding f(x) values are 4,32,224,1344. The divided differences are f(1,3) = 14. f(3,6) = 64, f(6,11) = 224, f(1,3,6) = 10, f(3,6,11) = 20 and f(1,3,6,11) = 1 11. What is a cubic spline. A cubic polynomial which has continuous slope and curvature is called a cubic spline. 12. What is meant by natural cubic spline. A cubic spline fitted to the given data such that the end cubics approach linearity at their extremities is called a natural cubic spline. 13. State the conditions required for a natural cubic spline. S’’(x0) = M0 = 0, S’’(xn) = Mn = 0 are called the free conditions for a natural cubic spline. 14. Given f(0)=-2,f(1)=2 and f(2)=8.Find the root of the Newton’s forward interpolating polynomial equation f(x)=0.

x y ∆y ∆2y 0 -2 4 1 2 2 6 2 8



9

( ) .....!3

)2)(1(

!2

)1(

!10

3

0

2

00 +∆−−

+∆−

+∆+= yuuu

yuu

yu

yxy

where u = (x-x0)/h Here x0=0, h=1, u=x and Y(x) = x2+3x-2. The roots of the equation f(x)=0.

2

173 ±−=x .

15. State Newton – Gregory forward difference interpolation formula

.....!3

)2)(1(

!2

)1(

!11

)1(

)()()(

0

3

0

2

0

0

000

+∆−−

+∆−

+∆+=

∆+=

==+=

yuuu

yuu

yu

y

yExPEuhxPxP

u

u

n

u

nn

Where u = (x-x0)/h

16. Find the value of Y at x =21 using Newton’s forward difference formula from the following table:

X: 20 23 26 29 Y: 0.3420 0.3907 0.4384 0.4848

( ) .....!3

)2)(1(

!2

)1(

!10

3

0

2

00 +∆−−

+∆−

+∆+= yuuu

yuu

yu

yxy

x y ∆y ∆2y ∆3y 20 .3420 0.487 23 .3907 -0.0046 0.0105 0.441 26 .4348 0.0059

0.05

29 .4848

where u = (x-x0)/h = 3

1

3

2021=

−

y(x) = 0.3420+0.0162+0.0005+0.00013 = 0.3583 17.State Newton’s backward difference interpolation formula.

( ) ( )( ) ( )n

n

nn yn

nvvvvy

vvy

vyxy ∇

−+++++∇

++∇+=

!

1.......21.........

!2

)1(

!10

2

18. Find the polynomial for the following data by Newton’s backward difference formula:

X: 0 1 2 3 F(x) -3 2 9 18



10

x y ∆y ∆2y ∆3y 0 -3 5 1 2 2 0 7 2 9 2

3 18 9

( ) .....!2

)1(

!10

2 +∇+

+∇+= yvv

yv

yxy nn

1

3−=

−=

x

h

xxv n

y(x) = 2x2-x+3

19. Find f(2.5) from the data:

X: 1 2 3 F(x) 0 1 8

( ) ( ) ( )( )

75.362

5.05.115.105.2

5.1,6,7,1,0 0

2

100

=++=

==∆=∆=∆=

f

pyyyy

20. Find the sixth term in the sequence 8,12,19,29,42,......

X y ∆ ∆ 2 ∆ 3

0 1 2 3 4

8 12 19 29 42

4 7 10 13

3 3 3

0 0

The sixth term is

( ) ( ) 58....5511 0

2

210

5

0

5

5 =+∆+∆+=∆+== yCCyyEy

PART – B 1. a. Find the Lagrange’s polynomial of degree 3 to fit the data: (8)

Y(0)= - 12, y(1)=0,y(3)=6 and y(4) =12. Hence find y(2). b. Using Lagrange’s formula, fit a polynomial to the data (8)

X: -1 0 2 3

Y: -8 3 1 12 Hence find y at x=1.5 and x=1 2. a. Using Lagrange’s formula, fit a polynomial to the data (8)



11

X: 0 1 2 4 5 6 f(x): 1 14 15 5 6 19 Also find f(3).

b. Using Lagrange’s formula, find y at x=6 for the following data: (8) X: 2 5 7 10 12 Y: 18 180 448 1210 2028

3. a. Find the age corresponding to the annuity value 13.6 given the table (8) Age (x) 30 35 40 45 50 Annuity value (y)

15.9 14.9 14.1 13.3 12.5

b. If f(0) = 0 , f(1) = 0, f(2) = -12, f(4) =(0), f(5) =600 and f(7) = 7308,find a polynomial that satisfies this data using Newton’s divided difference (8) interpolation formula. Hence, find f(6) 4. a. Using Newton’s divided difference formula find f(x) and f(3) from the following data: (8)

X: 0 1 2 4 f(x) 1 14 15 5

b. Using Newton’s divided difference formula to find f(5) from the following data (8)

X: 0 2 3 4 7

f(x) 4 26 58 112 466 5. a. Using Newton’s divided difference formula find the cubic function of x. from the following data: (8)

X: 0 1 4 5 f(x) 8 11 68 123

b. Using Newton’s divided difference formula find the cubic function of x (8) from the following data:

6. Using cubic spline, find y(1.5) from the following data. (16)

x 1 2 3 y -8 -1 18

7. Find the cubic spline for the data:

x 1 2 3 4

f(x) 1 2 5 11

Assume that y’’(1)=0 and y’’(4)=0. (16) 8. a. Find the polynomial of degree two for the data by Newton’s forward (8) difference method:

X: 0 1 2 3 4 5 6 7 F(x) 1 2 4 7 11 16 22 29

b. From the following table of half-yearly premium for policies maturing at different ages, estimate the premium for policies maturing at age 46 and 63. (8)

X: 0 1 4 5 f(x) 2 3 12 147



12

Age X: 45 50 55 60 65 PremiumY: 114.84 96.16 83.32 74.48 68.48

9. a. From the following table,find the value of tan(0.12) (8)

x: 0.10 0.15 0.20 0.25 0.30

y=tanx 0.1003 0.1511 0.2027 0.2553 0.3093

b. Use Newton’s backward difference formula to construct an interpolating polynomial of degree 3 for the data (8)

f(-0.75) = -0.07181250 , f(-0.5) = -0.024750

f(-0.25) = 0.33493750 , f(0) = 1.10100 Hence find f(-3

1 ).

10. a. Given (8) Marks 0-40 40-60 60-80 80-100 100-120 No: of students

250 120 100 70 50

Estimate the number of students who obtained marks between 60 and70. b. The following data are taken from the steam table: (8)

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 at t=175.

UNIT – III NUMERICAL DIFFERENTIATION AND INTEGRATION

PART – A(2 Marks) 1. State Newton’s formula to find f’(x) using the forward differences.

−∆+∆−∆=

=

.....3

1

2

110

3

0

2

0

0

yyyhdx

dy

xx

2. Find dx

dy at x = 1 from the table

x : 1 2 3 4 y : 1 8 27 64

x y ∆y ∆2y ∆3y 1 1 7 2 8 12 19 6 3 27 18 37 4 64



13

−∆+∆−∆=

=

.....3

1

2

110

3

0

2

0

0

yyyhdx

dy

xx

= 1 , x0 =1 , ∆y0 = 7 , ∆2y0 = 12 , ∆3y0 = 6

=

= 01xdx

dy 3

3. Using Newton’s backward difference formula, write the formulae for the first and second order derivatives at the end values x = xn upto the fourth order difference term.

∇+∇=

+∇+∇+∇=

+∇+∇+∇=

=

=

=

nn

xx

nnn

xx

nnn

xx

yyhdx

yd

yyyhdx

yd

yyyhdx

dy

n

n

n

33

33

3

332

22

2

32

2

31

...12

111

.......3

1

2

11

4. Find y’(5) from the following table x : 0 1 2 3 4 5 y : 4 8 15 7 6 2 x y ∇ y ∇ 2y ∇ 3y ∇ 4y ∇ 5y

0 4 4 1 8 3 7 -18 2 15 -15 40 -8 22 -72

3 7 7 -32 -1 -10 4 6 -3 -4 5 2

+∇+∇+∇=

=

.......3

1

2

11 32

nnn

xx

yyyhdx

dy

n

= -4-1.5-3.3333-8-14.4 = -31.2333



14

5. State the formula for Trapezoidal rule of integration

∫nx

x

dxxf

0

.)( = 2

h[(y0+ yn) + 2(y1+y2+........+yn-1)]

6. A curve passes through (2,8), (3,27), (4,64) and (5,125) Find the area of the Curve between the x axis and the lines x = 2 and x = 5 by Trapezoidal rule.

∫5

2

ydx = 2

h[(y0 + yn) + 2(y1+y2+.... +yn-1)]

= 2

1[(8+125) + 2(27+64)]

= 157.5 sq.units.

7. Given f(0)=-1,f(1)=1 and f(2)=4, find ∫2

0

)( dxxf by Trapezoidal rule.

x 0 1 2 y -1 1 4 Here h =1. A = Sum of the first and the last ordinates = -1+4 = 3 B = sum of the remaining ordinates =1.

8. Evaluate ∫ +

1

01 x

dx with h = 0.5 using Trapezoidal rule.

t(s x 0 0.5 1

Y 1 0.6666 0.5

∫1

0

ydx = 2

h[(y0 + yn) + 2(y1+y2+.... +yn-1)]

= 2

5.0[(1+0.5) + 2(0.6666)]

= 0.7083 9. What is the order of trapezoidal rule ? Error in the Trapezoidal formula is of the order h2

10. Using Simpson’s rule find ∫4

0

.dxex given e0 = 1, e1 = 2.72, e2 = 7.39, e3 = 20.09

and e4 = 54.6.

∫4

0

e dxx =

3

h[(y0 + y4) + 2y2+4(y1+ y3)]

= 3

1[(1+54.6) + 2(7.39)+4(2.72+20.09)]

= 53.8733.



15

11. What is the local error term in Trapezoidal formula and in Simpson’s one third rule?

Principal part of the error in the interval (x1, x2) = ''

1

2

12y

h−

Where y1 is the value of y and y1’’ is the value of the second derivative of y at x =x1.

Principal part of the error in the interval (x1, x3) = ivy

h1

5

90−

Where y1 is the value of y and ivy1 is the value of the fourth derivative of y at x =x1.

12. Find ∫−

2

2

4 .dxx by Simpson’s rule taking h = 1. ..

t(sec) x -2 -1 0 1 2

Y 16 1 0 1 16

∫−

2

2

4 .dxx = 3

1[(16+16)+4(2)] = 13.3 sq units

13. Compare Trapezoidal rule and Simpson’s 1/3 rule for evaluating numerical integration. S.NO Trapezoidal Rule Simpson’s 1/3 1. Number of intervals may be Number of intervals is even even or odd 2. Order of error = h2 Order of error = h4 3 . Principal part of the error Principal part of the error

in the interval (x1,x2) = - y’’ in the interval (x1,x2) = - yiv

14. State Romberg’s method integration formula to find the value of I = ∫b

a

dxxf .)(

using h and h/2.

I = I2+

−

3

II 12

15. If I 1 = 0.775 , I 2 = 0.7828 find I using Romberg’s method.

I = I2+

−

3

II 12 , I = 0.7828+

−

3

775.07828.0 = 0.7802

16. If I = ∫−

1

0

2 .dxex then I 1 = 0.731 , I 2 = 0.7430 with h = 0.5 and h = 0.25 Find I

using Romberg’s method.

I = I2+

−

3

II 12

I = 0.743+

−

3

731.0743.0 = 0.747



16

17. Compute ∫−

−2

2

2 .dxe

x

using Gaussian two point formula.

Given the range is not (-1,1) so by using the formula to make them as (-1,1).

X = 22

abz

ab ++

− Here a = -2 , b =2

=2

22

2

22 −+

+z

x = 2z 2

xz =⇒

dx = 2dz

∫−

−2

2

2 .dxe

x

= ∫−

−1

1

).2( dzez = 2 ∫

−

−1

1

dzez = 2

+

−

3

1

3

1ff = 2(0.5614+1.7813) = 4.6854

18. State three point Gaussian quadrature formula. Three points Gaussian quadrature formula is,

)0(9

8

5

3

5

3

9

5)(

1

1

fffdxxf +

+

−=∫

−

This formula is exact for polynomials up to degree 5.

19. State Trapezoidal rule for evaluating ∫∫d

c

b

a

dxdyyxf .),(

I = 4

hk[ (Sum of values of f at the four corners ) + 2(Sum of values of f at the

remaining nodes on the boundary) + 4(Sum of the values of f at the interior nodes)]

20. State Simpson’s rule for evaluating ∫∫d

c

b

a

dxdyyxf .),(

I = 9

hk[(Sum of the values of f at the four corners) + 2(Sum of the values of f at the

odd positions on the boundary except the corners) + 4(Sum of the values of f at the even positions on the boundary) + {4(Sum of the values of f at odd positions) + 8(Sum of the values of f at even positions) on the odd row of the matrix except boundary rows} + {8(Sum of the values of f at the odd positions) + 16(Sum of the values of f at the even positions) on the even rows of the matrix)].



17

PART– B 1. a. Find the value of f ’(8) from the following table (8)

x : 6 7 9 12 )(xf : 1.556 1.690 1.908 2.158

b. Find the first and second derivative of the function tabulated below at x = 0.4 x : 0.4 0.5 0.6 0.7 0.8 y : 1.5836 1.7974 2.0442 2.3275 2.6511 (8)

2. a. Find dx

dy at x = 1.5 given

x : 1 2 3 4 5 y : 77 78 127 248 375 (8) b. Find the first and second derivatives of y w. r. to x at x = 10 x : 3 5 7 9 11 y : 31 43 57 41 27 (8)

3. a. Find dx

dyand

2

2

dx

yd at x=51 for the following data (8)

x: 50 60 70 80 90 y: 19.96 36.65 58.81 77.21 94.61 b. Find the maximum and minimum value of y = f(x) given the data

x : 0 1 2 3 4 5

f(x) : 0 4

1 0

4

9 16

4

225 (8)

4. a. A river is 80 mts wide.The depth ‘d’ in mts at a distance x mts from one bank is given by the following table. Calculate the area of cross section of the river using

Simpson’s 3

1rd rule

x : 0 10 20 30 40 50 60 70 80 d : 0 4 7 9 12 15 14 8 3 (8)

b. By dividing the range into 10 equal parts, evaluate ∫π

0

sin xdx by using

Simpson’s 3

1rd rule. (8)

5. a. Evaluate ∫ +

6

0

21 x

dxby Trapezoidal rule and simpson’s 3/8 rule. (8)

b. Calculate dxxex

∫−

7.0

5.0

taking 5 ordinates by trapezoidal rule and Simpson’s1/3 rule (8)

6. a. Evaluate dxex

∫−

4.1

1

2

by taking h=0.1using Simpson’s 3/8 rule. (8)



18

b. Evaluate ∫ +

1

0

21 x

dx using Romberg’s method. Hence obtain an approximate

value of Π . (8)

7. a. Evaluate ∫ +

2

0

2 4x

dx using Romberg’s method hence obtain an approximate

Value of Π (8)

b. Evaluate ∫5

1x

dxusing Gaussian Quadrature with three points. (8)

8. a. Evaluate I = ∫− +

1

1

21 t

dtby Three point Gaussian Quadrature formula. (8)

b. Evaluate ∫ +=

1

01 x

dxI using two point Gaussian Quadrature formula. (8)

9. a. Evaluate ∫ ∫6.2

2

4.4

4xy

dxdy by using Simpson’s rule (8)

b. Evaluate dxdyyx∫ ∫ +

2

1

2

1

22

1 taking h =0.2 k = 0.25 by both trapezoidal and

Simpson’s rule (8)

10.a. Evaluate ( )

dxdyxy

xy∫ ∫

+

5.0.

01

5.0

0 1

)sin( by using Trapezoidal rule h = 0.25 and k = 0.25 (8)

b. Evaluate ( )dxdyyx∫ ∫ +2/

0 2/

cos

π π

π

by using Trapezoidal rule, Simpson’s 1/3 rule and

taking h=k=4

π. (8)

UNIT – IV INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL

EQUATIONS PART – A(2 Marks) 1. State the disadvantage of Taylor series methods?

In the differential equation dx

dy= f(x, y), the function f(x, y) may have complicated

algebraical structure. Then the evaluation of higher of order derivatives may become tedious. This is the demerit of this method. 2. What is the major drawback of Taylor series method?

If f(x, y) involves some complicated algebraic structures then the calculation of higher derivatives becomes tedious and the method fails. This is the major drawback of this method.

3. State Taylor series algorithm for the first order differential equation? To find the numerical solution of with the condition y(x0) = y0. We expand y(x) at a



19

general point xm in a Taylor series, getting

iv

mmmmm yh

yh

yh

hyyym !4

4

!3!2

'''3

''2

'

1 ++++=+ +……

Here yrm denotes the rth derivatives of y w.r.to x at the point (xm, ym).

4. Which is better Taylor’s method or R.K. method? R.K methods do not require prior calculation of higher derivatives of y(x), as the Taylor method does. Since the differential equations using in applications are often complicated, the calculation of derivatives may be difficult. Also the R.K formula involve the computation of f(x,y) at various positions, instead of derivatives and this function occurs in the given equation.

5. Solve the differential equation 1)0(, =++= yxyyxdx

dy by Taylor series method to

get the value of y at x = h? Given y’ = x +y + xy, x0 = 0, y0 = 1 We know that the Taylor’s series formula for y1 is

.......!3!2!1

'''

0

3''

0

2'

001 ++++= yh

yh

yh

yy

y’ = x + y + xy y0’ = x0 +y0 + x0y0=0+1+0=1 y’’ = 1 + y’ + xy’ + y y0’’ = 1+y0’+x0y0’+y0=1+1+0+1=3 y’’’ = y ‘’ + xy’’ + y’ + y’ y0’’’ = y0’+x0y0’’ +2y0’= 1+0+2=3 . ‘ . y’’’ = y’’ + xy’’ + 2y’

. ‘ . .......)3(!3

)3(!2!1

132

1 ++++=hhh

y

6. Write down Euler method to the differential equation ),( yxfdx

dy= .

yn+1 = yn + h f(xn,yn) when n=0,1,2…. This is Euler method. It can also be written as Y(x+h) = y(x) + hf(x,y) 7. State modified Euler method to solve y ’ = f(x,y), y(x0)=y0 at x = x0 + h.

+++=+ )1 ,(

2,

2 nyxf

hy

hxhfyy nnnnn

+++= )000001 ,(

2,

2yxf

hy

hxhfyy

8. Using Modified Euler’s method, find y(0,1) if 1)0(,22 =+= yyxdx

dy

Given f(x,y) = x2+y2,x0=0,y0=1,h=0.1,x1=0.1 By modified Euler method



20

+++=+ )1 ,(

2,

2 nyxf

hy

hxhfyy nnnnn

……… (1)

+++= )000001 ,(

2,

2yxf

hy

hxhfyy ……… (2)

f(x0,y0) = x02+ y0

2 = 0 + 1 = 1

y1 = 1 + (0.1) f

++ )1(

2

1.01,

2

1.00

= 1 + (0.1)f[0.05, 1.05] = 1 + (0.1)[(0.05)2+(1.05)2] = 1 + (0.1)[0.0025+1.1025] = 1+ 0.1105 = 1.1105 . ‘ . y(0.1) = 1.1105 9. What is the error of Euler’s method?

Error at (x= 1x ) = ),(''2

11

2

yxyh

Error = 0(h2) = Error is of order h2

10. What are the limitations of Euler’s method? 1. The attainable accuracy is limited by length of step h. 2. The method is slow and has limited accuracy. 11. What is the Error in modified Euler’s method?

Error = *12

2h

− constant

Error = 0(h3), order of h3 12. Write down the Runge-Kutta formula of fourth order to solve dy/dx = f(x,y) with y(x0) = y0. Let h denote the interval between equidistant values of x. If the initial values are (x0,y0), the first increment in y is computed from the formulas. k1 = h f(x0,y0)

k2 = h f

++

2,

2

1

00

ky

hx

k3 = h f

++

2

2,

200

ky

hx

k4 = h f(x0+h,y0+k3) and ∆y= 1/6=(k1+2k2+2k3 + k4) then x1=x0 +h, y1 = y0 + ∆y The increment in y in the second interval is computed in a similar manner using the same four formulas, using the values x1,y1 in the place of (x0,y0) respectively.



21

13. State the special advantage of Runge-Kutta method over Taylor series method? (or) Compare Taylor’s series and R.K. method? R-K methods do not require prior calculation of higher derivatives of y(x) as the Taylor method does. Since the differential equations are using in applications Often complicated, the calculation of derivatives may be difficult. Also the R.K. formulas involve the computations of f(x,y) at various positions, Instead of derivatives and this function occurs in the given equation 14. Write Milne’s predictor corrector formula? .

Milne’s predictor formula is

( )nnnnn yyy

hyy '2''2

3

41231 +−+= −−−+

Milne’s corrector formula is

( )1111 ''4'3

+−−+ +++= nnnnn yyyh

yy

15. How many prior values are required to predict the next value in Milne’s method & Adam’s method? Four prior values. 16. What is the error term in Milne’s corrector formula?

The error term is -'

0

4

90y

h∆

17. What is the error term in Milne’s predictor formula?

The error term is'

0

4

45

14y

h∆

18. Write down Adams-Bashforth formula? Adam’s predictor corrector formulas are

( )321,1 '9'37'59'5524

−−−+ −+−+= kkkkkpk yyyyh

yy

( )211,1 ''5'19'924

−−++ +−++= kkkkkck yyyyh

yy

19. What is a Predictor-Collector method of solving a differential equation? Predictor-collector methods are methods which require the values of y at xn,xn-1, xn-2,……. For computing the value of y at xn+1. We first use a formula to find the value of y at xn+1 and this is known as a predictor formula. The value of y so got is improved or corrected by another formula known as corrector formula. 20. What is the Error of Adam Bashforth method?

predictor )(720

251 )(5 ξivfh

Corrector )(720

251 )(5 ξivfh



22

PART – B

1. a. Using Taylor series method find y at x = 0.1 if .1)0(,12 =−= yyxdx

dy (8)

b. Solve .1)0(,22 =+= yyxdx

dyUse Taylor series at x=0.2 and 0.4, Find x = 0.1. (8)

2. a. Solve the system of equations ,2xzdx

dy−= ,xy

dx

dz+= with y(0) = 1,z(0) = 1 by

Taking h = 0.1 , to get y(0.1) and z(0.1).Here y and z are dependent variables and x is independent. (8) b. Using Taylor series method, find y(1.1) and y(1.2) correct to four decimal places.

Given 3/1xydx

dy= and y(1) =1. (8)

3. a. By Taylor’s series method find y(0.1) given that y’’ = y + xy’ , y(0) =1 , y’(0) = 0 (8) b. Using Euler’s modified method find y(0.1) from ,1)0(,' =++= yxyyxy with h=0.05 (8)

4. a. Using Euler’s method find y(0.3)of y(x) satisfies the initial value problem.

,1114.1)2.0(,)1(2

1' 22 =+= yyxy with h=0.1. (8)

b. Using modified Euler’s method, compute y(0.1) with h=0.1 from

.1)0(,2

=−= yy

xy

dx

dy (8)

5. a. Solve ,0)0(,1' =−= yyy find y(0.1) ,y(0.2),y(0.3) by modified Euler’s method. (8)

b. Given .2)0(,3 =+= yyxdx

dyCompute y(0.2), y(0.4) and y(0.6) by Runge-Kutta

method of fourth order. (8)

6. a. Using R.K. method of 4th order, solve .1)0(,22

22

=+

−= y

xy

xy

dx

dyat x = 0.2. (8)

b. Using R.K. method of fourth order find y(0.1) for the initial value problem y’’ = xy’ - y ,y(0)=3, .0)0(' =y (8)

7. Determine the value of y(0.4) using Milne’s method given ;.1)0(,2 =+= yxyydx

dyuse

Taylor series to get the values of y(0.1), y(0.2) and y(0.3). (16) 8. Given y’ = 1-y and y(0) = 0 , find (i) y(0.1) by Euler method (ii) y(0.2) and y(0.3) by modified Euler method (iii) y(0.4) by Milne’s method (16)

9. a. Using Milne’s method find y(0.8) given .2)0(,1

=+

= yyxdx

dygiven y(0.2)=2.0933,

y(0.4.) =2.1755 y(0.6)=2.2493. (8)

b. Given ,979.1)3.1(,548.1)2.1(,233.1)1.1(,1)1(),1(2 ====+= yyyyyxdx

dyevaluate

y(1.4) by Adams – Bash forth method. (8)



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10. Consider the initial value problem .5.0)0(,12 =+−= yxydx

dy (16)

a. using Taylor series, find y(0.2) b. Using 4th order Runge-Kutta method, find y(0.4) and y(0.6)

c. Using Adam-Bashforth Predictor.

UNIT V BOUNDARY VALUE PROBLEMS IN ORDINARY AND PARTIAL

DIFFERENTIAL EQUATIONS PART - A(2 Marks) 1. State the conditions for the equation. Auxx + Buyy + Cuxy +Dux +Euy + Fu =G where A, B, C, D, E, F, G are function of x and y to be (i) elliptic (ii) parabolic (iii) hyperbolic

The given equation is said to be (i) Eliptic at a point (x,y) in the plane if B2-4AC<0 (ii) Parabolic if B2-4AC=0 (iii) Hyperbolic if B2-4AC>0

2. State the condition for the equation Auxx+2Buxy+Cuyy=f(ux,uy,x,y) to be (a)elliptic (b) parabolic(c) hyperbolic when A, B, C are functions of x and y. The equation is elliptic if (2B2)-4AC<0 i.e. B2-AC<0. It is parabolic if B2-4AC>0. 3. What is the classification of fx-fyy=0? Here A=0, B=0, C=-1 B2-4AC=0-(4)(0)(-1)=0 Therefore the equation is parabolic. 4. State Schmidt’s explicit formula for sloving heat flow equation?

[ ]

jijij

jijijiji

uuuif

uuuu

,1,11,1

,1,,11,

2

1,

2

1

)21(

−++

−+++

+==

+−+=

λ

λλλ

5. What is the classification of one dimensional heat flow equation?

The One dimensional heat flow equation is t

u

x

u

∂

∂=

∂

∂22

2 1

α here A=1, B=0, C=0.

B2-4AC = 0 Hence the one dimensional heat flow equation is parabolic. 6. Write down the Crank-Nicholson formula to solve ut=uxx?

(or) Write down the implicit formula to solve one dimensional heat flow equation?

)()1(2)1(2)(

)1(2

1

2

1)1(

2

1

2

1

,1,1,1,1,11,1

,1,11,1,11,1

jijijijijiji

ijjijijijiji

uuuuuu

or

uuuuuu

−+++−++

−+++−++

+−−=+−+

−+−−

=+−+

λλλλ

λλλλλλ



24

7. What type of equations can be solved by using Crank-Nicholson’s difference formula? Crank-Nicholoson’s difference formula is used solve parabolic equations of the Form uxx = aut

8. Write the Crank Nicholson difference scheme to solve uxx=aut with u(0,t) = T0, u(l,t)=T1 and the initial condition as u(x,0) = f(x)? The scheme is

ijjijijijiji uuuuuu )1(2

1

2

1)1(

2

1

2

1,1,11,1,11,1 −+−

−=+−+ −+++−++ λλλλλλ

9. For what purpose Bender-Schmidt recurrence relation is used To solve one dimensional heat equation 10. State the explicit scheme formula for the solution of the wave equation? The formula to solve numerically the wave equation a2uxx-utt=0 is

1,,1,1

22

,

22

1, )()1(2 −−++ −++−= jijijijiji uuuauau λλ

11. For what value of λλλλ, the explicit, method of solving the hyperbolic equation

2

2

22

2 1

t

u

cx

u

∂

∂=

∂

∂is stable, where λλλλ = C∆t/∆x?

The explicit method is stable only if λ=1/c 12. Write the diagonal five-point formula to solve the Laplace equation uxx+uyy=0?

[ ]1,11,11,11,1,4

1++−++−−− +++= jijijijiji uuuuu

13. Write down the standard five point formula to solve Laplace equation

02

2

2

2

=∂

∂+

∂

∂

y

u

x

u

[ ]1,1,1,,1,1,4

1+−−+− ++++= jijijijijiji uuuuuu

14. What is the purpose of Liebmann’s process? The purpose of Liebmann’s process is to find the solution of the Laplace equation uxx + uyy = 0 by iteration. 15. If u satisfies Laplace equation and u = 100 on the boundary of a square what will be the value of u at an interior grid point? Since u satisfies Laplace equation and u = 100 on the boundary of a square ui,j = 1/4(100+100+100+100)=100 16. State Liebmann’s iteration process formula?

[ ])1(

1,1,,1

)1(

,1

1

,4

1 ++−+

+−

+ +++= n

ji

n

ji

n

ji

n

ji

n

ji uuuuu

17. Write down the finite difference form of the equation ∇∇∇∇2u= f(x,y)

),(4 2

,1,1,,1,1 jhihfhuuuuu jijijijiji =−+++ +−+−

18. State the general form of Poisson’s equation in partial derivatives?

),(2

2

2

2

yxfy

u

x

u=

∂

∂+

∂

∂



25

19. What is Shooting method? In this method, the given boundary value problem is first converted into an equivalent initial value problem and then solved using any of the methods. This methods makes use of the techniques of solving initial value problems. 20. What is the procedure of shooting method? * Convert the problem into an initial value problem. * Initialise the variables including two guesses at the initial slope. * Solve the equations with these guesses using either a one-step or a multistep method. * Interpolate from these results to find an improved value of the slopes obtained. * Repeat the process until a specified accuracy in the final function value is obtained.

PART - B 1. a. Solve by finite difference method, the boundary value problem y’’(x) – y(x) = 0 where y(0) = 0 and y(1) = 1, taking h = 1/4 . (8) b. Using the finite difference method, find y(0.25),y(0.5) and y(0.75) satisfying the (8) differential equation y’’(x) + y = x subject to the boundary conditions y(0) = 0,y(1)= 2.

2. a. Solve t

u

x

u

∂

∂=

∂

∂2

2

2

given u(0,t)=0, u(4,t)=0,u(x,0)=x(4-x) taking ∆x= ∆ t=1. Find the

value of u upto t=3 using Bender-Schmidt explicit difference scheme. (8) b. Using Schmidt’s process solve 25 uxx=ut where 0<x<1 t>0 with boundary conditions

u(0,t)=0=u(10,t); u(x,0)=25

)10( xx − and choosing h=1 and k suitably. Find ui,j for

i=1,2,3….9 and j= 1,2,3,4 (8)

3. a. Solve2

2

x

u

t

u

∂

∂=

∂

∂, 0≤ x≤ 1, t≥ 0 with u(x,0)=x(1-x), 0<x<1 and u(0,t)=u(1,t) = 0, for all

t >0 using explicit method with ∆ x=0.2 for 3 time steps. (8)

b. Solve 2

2

x

u

t

u

∂

∂=

∂

∂ in 0<x<5, 0≥t given that u(x,0) =20, u(0,t)=0,u(5,t) =100.Compute

u for the time-step with h=1 by Cranck – Nicholson method (8)

4.a. Solve 2

2

x

u

∂

∂=

t

u

∂

∂,0<x<2,t>0,u(0,t)=u(2,t)=0,t>0 and u(x,0)=

2sin

xπ,0 ≤ x≤ 2, and

∆ t=0.25 and ∆ x=0.5 for two times steps by Crank-Nicholson implicit finite difference method. (8)

b. Solve ,2

2

2

2

x

y

t

y

∂

∂=

∂

∂0<x<1.t>0 given u(x,0)=0, )0,(x

t∂

∂=u(0,t)=0 and u(1,t)=100sinπt,

complete u(x,t) for 4 times steps with h=0.25. (8) 5. Solve the Laplace’s equation over the square mesh of side 4 units satisfying the boundary conditions: (16)

U(0,y)=0,0 4≤≤ y ;u(4,y)=12+y, 0 4≤≤ y

U(x,0)=3x, 0 4≤≤ x ;u(x,4)= 2x ,0 4≤≤ x



26

6. By iteration method , solve the laplace equation uxx+ uyy=0 , over the square region, satisfying the boundary condition. (16)

u(0, y) = 0 , 0 ≤ y ≤ 4 u(4, y) = 12+y, 0 ≤ y ≤ 4 u(x,0) = 3x , 0 ≤ x ≤ 4 u(x, 4) = x2 0≤ x ≤ 3

7. Solve uxx+uyy = 0 for the following square mesh with boundary values as shown in the figure below (16) 1000 1000 1000 1000 2000 500 2000 0

1000 500 0 0 8. Solve the Laplace equation at the interior points of the square region given below (16) 0 500 1000 500 0

1000 1000 2000 2000 1000 1000

0 500 1000 500 0

9. Solve )10(10 222 ++−=∇ yxu over the square mesh with sides x=0,y=0,x=3,y=3

With u=0 on the boundary and mesh length of 1 unit. (16)

10. Solve the Poisson equation 228 yxuu yyxx =+ in the square mesh given u=0 on the four

Boundaries dividing the square into 16 sub squares of length 1 unit . (16)

1u

2u

3u

4u

U1

U2

U3

U4

U5

U6

U7

U8

U9

MA 2264 NM EEE Q.B - Welcome to Kings College of ...kingsindia.net/question bank even 13-14/EEE/II...

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Transcript of MA 2264 NM EEE Q.B - Welcome to Kings College of ...kingsindia.net/question bank even 13-14/EEE/II...