ROEVER COLLEGE OF ENGINEERING & TECHNOLOGY
Elambalur, Perambalur – 621 212.
DEPARTMENT OF METHAMATICS
QUESTION BANK
MA 2264 – NUMERICAL METHODS
UNIT – I
SOLUTION OF EQUATIONS AND EIGEN VALUE PROBLEMS
PART – A
1. What is the iterative formula of Newton – Raphson Method?
2. What is the order of convergence and the convergence condition for Newton’s Raphson method?
3. Derive a formula to find the value of N where N > 0, using Newton – Raphson method.
4. Find an iterative formula to find the reciprocal of a given number N( N ≠ 0).
4. What are the merits of Newton’s method of Iteration?
5. When should we not use Newton Raphson?
6. State the sufficient condition for the existence and uniqueness of fixed point iteration.
7. Write down the order of convergence and the condition for convergence of fixed point iteration method.
9. What do you mean by the order of convergence of an iterative method for finding the root of the equation f(x) = 0.
8. Give two direct methods to solve a system of linear equations.
9. What are the advantages of iterative methods over direct methods for solving a system of linear equations?
10. Give two indirect methods to solve a system of linear equations.
11. State any two difference between direct and iterative methods for solving system of equations.
12. State the principle used in Gauss – Jordan method.
13. Compare Gaussian elimination and Gauss – Jordan Methods in solving the linear system
 A X   B .
14. Compare Gauss - elimination with Gauss seidal method.
16. Solve the equations x + 2y = 1 and 3x – 2y = 7 by Gauss – Elimination method.
17. using Gauss elimination method solve: 5x + 4y = 15, 3x + 7y = 12.
15. Compare Gauss – Jacobi & Gauss Seidal Methods.
16. Solve by Gauss–Jordan Method the following system of equations 2x1 + x2 = 3, x1 + 2x2 = 3.
17. Find inverse of A  

1 3  by Gauss Jordan Method.

2 7
18. Write the sufficient condition for Gauss – seidal Method to converge.
19. Define eigen value and eigen vector.
20. State the basic principle involved for finding A-1 by Gauss – Jordan Method?
21. How will you find the smallest eigenvalue of a square matrix A?
22. If the eigen values of A are 1, 3, 4 then the dominant eigenvalue of A is ------------.
23. The power Method will work satisfactorily only if A has a -------------eigen value.
25. What is the use fo Power method?
PART – B
Method – I– NewtoN’s RaphsoN:
1. Find the root of 4 x  e x  0 that lies between 2 and 3 by Newton’s Method.
2. *Find by Newton’s iterative formula for the reciprocal of a number N and hence find the value of 1/23,
correct to five decimal places.
3. (i)Solve by Newton – Raphson Method the Equation xsinx + cosx = 0, correct to four decimal
places. (ii) *Solve by N – R method x log10 x  1.2  0 .
4. Find by Newton’s Method, the real root of the equations (i)* 3x = cosx +1. And (ii) xe x  cos x .
5. Establish the formula to find the square root of N, using Newton – Raphson Method. Hence
find the square root of (i)15, (ii)11, (iii) 5 correct to 3 decimal places.
6. Prove the quadratic convergence of Newton Raphson Method. Find a +ve real root of
f(x) = x3 – 5x + 3 = 0, using this method.
7. Find the +ve real root of (i) x3 – 2x + 0.5 = 0 and (ii) x4 – x – 10 = 0using Newton – Raphson Method correct
to 3 decimal places.
8. Obtain the +ve of 2x3 – 3x - 6 = 0 that lies between 1 and 2 by using Newton – Raphson method.
9. Find by Newton’s method, the root of ex = 4x near x = 2, correct to four decimal places.
Method – II – FIXED POINT ITERATION :
1. Find a real root of the equation cosx = 3x – 1 correct to 5 decimal places by fixed point iteration method.
2. *Solve e x  3 x  0 by the method of fixed point iteration.
3. Find the –ve root of the equation x3 – 2x + 5 = 0.
4. Use the method of fixed point iteration to solve the equation 3x  log10 x  6 .
5. Solve the following by iteration method.
(i) 3x – cosx - 2 = 0.
(ii) 2 x  log10 x  7 .
Method – II – GAUSS – JORDAN :
1. Using Gauss – Jordan method, solve the following system of equations
(i) 2x – y + 3z = 8, x + 2y + z = 4 , 3x + y –z = 0 and (ii) 10x + y + z = 12, 2x + 10y + z = 13, x + y + 5z = 7.
2. Solve by Gauss – Jordan Method the following system
(i) 10x + y – z = 11.19
* (ii) 10x + y + z = 12
x + 10y + z = 20.08
x + 10y + z = 12
-x +y + 10z = 35.61.
x + y + 10z = 12.
3. Solve the following equations by Gauss Jordan method.
(i) x + y + z = 9, 2x – 3y + 4z = 13, 3x + 4y + 5z = 40.
(ii) 2x + y + 4z = 12, 8x – 3y + 2z = 20, 4x + 11y – z = 33.
(iii)5x – y = 9; - x + 5y – z = 4; - y + 5z = -6.
Method – III – GAUSS - ELIMINATION :
1. Solve the given system of equations by Gaussian elimination method
-x1 +x2 + 10x3 = 35.61
10x1 + x2 – x3 = 11.19
x1 + 10x2 + x3 = 20.08.
2. Solve the system of equations by Gauss elimination method.
(i) 10x – 2y + 3z = 23
(ii) 3.15x -1.96y + 3.85z = 12.95
2x + 10y – 5z = -33
2.13x + 5.12y – 2.89z = -8.61
3x – 4y + 10z = 41.
5.92x + 3.05y +2.15z = 6.88.
Method – IV – Inverse of the matrix:
2
 8 1
4 1
1 1 
 3
1. Find the inverse of (i) A   2 3 1 , (ii) * A   15 6 5  , (iii) A   5 1





 10 1
 1 2 2 
 5


2
2





Jordan method.
2. Using Gauss – Jordan Method find the inverse of the following matrices
1
3 
2
2 2
3 1 2 
 1
4 1
(iii)
(iv) *






A   2 6
A   2 3 1
 i  A   1 3 3   ii  A   2 3 1
1 2 1 
 4 8
 2 4 4 
 1 2 2 







3 

2  using Gauss 4 
6 
6 
8 
1 2 6 
0 1 2
 1 2 1
 1 1 1
1 1 2


 (ix)







.
,
(viii)
A   2 5 15 
A  1 2 3
(v) A   4 3 1  (vi ) A   1 2 3  (vii) A   4 1 0 
 6 15 46 
3 1 1
 2 1 3 
2 3 1
 3 5 3










Method – V – Gauss – Jacobi:
1. Solve by Gauss – Jacobi Method, the following equation
(i) 4x1 + x2 + x3 = 6
(ii) **8x -3y +2z = 20
x1 + 4x2 + x3 = 6
4x + 11y – z = 33
x1 + x2 + 4x3 = 6.
6x + 3y + 12z = 35
2. Solve by Gauss – Jacobi method the equations
20x + y – 2z = 17, 3x + 20y – z = -18, 2x – 3y + 20z = 25.
Method – VI – Gauss – seidal:
1. Solve the following system of equations by Gauss – Seidal method
(i) *4x + 2y +z = 14, x + 5y – z = 10, x + y + 8z = 20.
(ii) 2x + 10y + z = 13, 10x + y + z = 12, x + y + 5z = 7.
(iii)* 20x + y – 2z = 17, 3x + 20y – z = -18, 2x – 3y + 20z = 25.
(iv)* x + y + 54z = 110, 27x + 6y – z = 85, 6x + 15y + 2z = 72.
(v) 8x -3y +2z = 20, 4x + 11y – z = 33, 6x + 3y + 12z = 35.
(vi)9x – y + 2z = 9, x + 10y – 2z = 15, 2x – 2y - 13z = -17.
(vii) **10x + 2y + z = 9, x + 10y – z = -22, -2x + 3y + 10z = 22.
2. Solve by Gauss – Seidal method
(i) 6x – 3y + z = 11
(ii) 28x + 4y – z = 32 (iii) 6x -3y + z = 1 (iv) 28x + 4y – z = 32
x + 3y + 10z = 24
2x + y – 8z = -15
2x + y -8z = -15
x + 3y + 10z = 24
2x + 17y + 4z = 35
x – 7y + z = 10.
x – 7y + z = 10.
2x + 17y + 4z = 35
3. Using Gauss Seidal Method, solve the following system start with x = 1, y = -2, z = 3
x + 3y + 52z = 173.61
x – 27y + 2z = 71.81
41x – 2y + 3z = 65.46
4. Solve by Gauss – Seidal iteration the given system of equations starting with (0, 0, 0) as solution. Do 5
iteration only 4x – x2 –x3 = 2, - x1 + 4x2 – x4 = 2, -x1 + 4x3 – x4 = 1, - x2 –x3 + 4x4 = 1.
Method – VII – EIGEN VALUE OF A MATRIX BY Power method
5 0 1
by power method.
A   0 2 0 
1 0 5


2. Determine the Largest eigen value and the corresponding eigenvector of the matrix using the power
1. **Find the Largest eigenvalues of
method
 2 1 0 
.
A   1 2 1
 0 1 2 


1 6 1
2 0 

 0 0 3


3. Find the dominant Eigen value and the corresponding Eigen vector of A   1
1
Using power method with the initial Eigen vector X   1  .
0
 
1
 
 1 1 3
5 1  .

3 1 1


4. Solve by power method to find the dominant Eigen value for the matrix  1
5. Find the numerically largest eigenvalues and the corresponding Eigen vector using power method, given
4
3
5
 1 3 1
(i)
, (ii) *
. Starting vector is ( 1, 1, 1)T.


A   10 8
6
A   3 2 4 
 20 4 22 
 1 4 10 




6. Obtain by the power method, the dominant eigen value and the corresponding eigen vector Correct to two
decimal places for the matrix  2
2
2 
 1  as the initial approximation to the eigen vector.

 
 2 / 3 5 / 3 5 / 3  taking  0 
 1
0
5 / 2 11/ 2 

 
7. *Find the numerically largest eigenvalues and the corresponding Eigen vector using power method, given
2 
 25 1
. Starting vector is ( 1, 0, 0)T.

A 1
3
0 

 2
0 4 


8. Find the numerically largest eigenvalues and the corresponding Eigen vector using power method, given
(i)
1
A   0
0

2
4
0
3
 15 4 3 
, (ii)

2
A   10 12 6 
 20 4 2 
7 


Method – VIII – EIGEN VALUE OF A MATRIX BY JACOBI method
1. Find all the Eigen values and Eigen vectors of the following matrix by using Jacobi method.
 1
2 2 
 2 0 1
5 0 1
4 1 1

 , (ii)
(i)
(iii)
(iv)




A 0 2 0 
A   0 2 0 
A   1 1 2 
A 2 3
2
 1 0 2 
1 0 5
1 2 1


 2






2 1 

 6 2 2 
 3 1 1 
1 0 0 
(vi)
(vii)




A   2 3 1
A   1 5 1
A   0 3 1
 1 1 3 
 0 1 3 
 2 1 3 






(v)
UNIT – II
INTERPOLATION AND APPROXIMATIONS
PART – A
1. State Lagrange’s interpolation formula.
2. What is the assumption we make when Lagrange’s formuls is used?
3. Use Lagrange’s formula, to find the quadratic polynomial that takes these values.
X: 0 1 3
Y: 0 1 0
Then find y(2).
4. Explain briefly interpolation.
5. State the order of convergence of cubic spline.
6. State the properties of cubic spline.
7.* Derive Newton’s forward and backward difference formula. (OR) State Gregory – Newton forward
difference interpolation formula.
8. State Newton’s forward difference formula for equal intervals.
8. Obtain the divided difference table for the following data
x
f(x)
-1
-8
0
3
2
1
3
12
9. Fit a polynomial which takes the following values:
x
Y
0
1
1
2
2
1
x
Y
0
5
1
6
3
50
10. Using Lagrange’s formula, find the polynomial to the given data.
10. What do you understand by inverse interpolation?
11. What is the nature of nth divided differences of a polynomial of nth degree?
12. Find the second divided difference with arguments a, b, c if f(x) = 1/x.
13. Form the divided difference table for the data (0, 1), (1, 4), (3, 40) and (4, 85).
14. **Define a cubic spline S(x) which is commonly used for interpolation.
15. If y(x) = y, I = 0, 1, …, n write down the formula for the cubic spline polynomial y(x), valid in xi-1< x < xi.
16. What is meant by Natural cubic spline?
17. What in the error in Newton’s forward interpolation formula?
18. When to use Newton’s forward interpolation and when to use Newton’s backward interpolation?
19. Find the divided difference of f(x) = x2 + x + 2 for the arguments 1, 3, 6, 11.
20. Find the divided difference of f(x) = x3 – x2 + 3x + 8 for the arguments 0, 1, 4, 5.
21. Find the second divided difference with arguments a, b, c if f(x) = 1/x.
20. Show that

bcd
3
1
1
.
 
abcd
a
21. What is inverse interpolation?
PART – B
Method – I – LagRaNge’s iNteRpoLatioN foRmuLa:
1. Using Lagrange’s interpolation formula, find x corresponding to
y = 85 given
2. Given the values
x
f(x)
5
150
7
392
11
1452
13
2366
17
5202
x
y
2
94.8
5
87.9
8
81.3
14
68.7
Evaluate f(9), using (i) Lagrange’s formula (ii) Newton’s divided difference formula.
3. Find f(x) as a polynomial in x from the given data and find f(8) and f(6).
x
f(x)
3
168
7
120
9
72
10
63
4. **Find y(40) from the following data using Lagrange’s interpolation formula y(30) = 148, y(35) = 96,
y(45) = 68 and y(55) = 34.
x
1
3 4
6
5. Fit a Lagrange’s interpolating polynomial y = f(x) and find f(5).
y=f(x)
-3
0
30
132
6. Find the polynomial f(x) by using Lagrange’s formula and hence find f(3).
x
f(x)
0
2
1
3
2
12
3
147
7. ***Given the values Find f(27) by using Lagrange’s interpolation formula.
x
f(x)
14
68.5
17
64
31
44
35
39.5
8. Using Lagrange’s interpolation formula fit a polynomial to the following data: And hence find y at
x = 1.5
x
-1
0
2
3
y
-8
3
1
12
9. The following table gives certain corresponding values of x and log10x. Compute the value of
log10323.5, by using Lagrange’s interpolation formula.
x
Log10x
321.0
2.50651
322.8
2.50893
324.2
2.51081
325.0
2.51188
10. Use Lagrange’s interpolating formula to fit a polynomial to the given data f(-1) = -8, f(0) = 3,
f(2) = 1 and f(3) = 12. Hence find the value of f(1).
11. Use Lagrange’s method to find log10656, given that log10654 = 2.8156, log10658 = 2.8182, log10659 = 2.8189
and log10661 = 2.8202.
Method – II - NewtoN’s divided diffeReNce method:
x
4
5
7
10
11
13
1. **Using Newton’s divided difference formula, find f(8) for the given
y 48 100 294 900 1210 2028
data:
2. Determine f(x) as a polynomial in x for the following data, using Newton’s
x
-4
-1 0 2 5
divided difference formula. Also find f(2).
f(x) 1245 33 5 9 1335
3. Using Newton’s divided difference formula find the equation y = f(x) of
least degree and passing through the points (-1, -21), (1, 15), (2, 12), (3, 3). Find also y at x = 0.
4. From the following table find f(x), by Newton’s divided difference
x 0 2 3 4
6 7
interpolation formula.
y 0 8 0 -72 0 1008
5. Use Newton’s divided difference formula, find u(3) given u(1) = -26, u(2) = 12, u(4) = 256,
u(6) = 844.
x
5
7
11
13
17
6. Given the values
y
150
392
1452
2366
5202
Evaluate f(9) using Newton’s divided difference formula
7. If f(0) = f(1) = 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 interpolation formula. Hence, find f(6).
x
3
7
9
10
f(x)
8. Using divided difference, find f(x) as a polynomial in x from the given data:
9. Use Newton’s divided difference formula to find f(5) from the following data:
168
x
f(x)
0
4
120
2
26
72
3
58
63
4
112
7
466
10. *Use Newton’s divided difference formula to find f(x) from the following data and hence find f(4).
x
f(x)
0
2
1
3
2
12
5
147
11. Use Newton’s divided difference formula to find f(x) from the following data
x
y
1
1
2
5
7
5
8
4
Method – III – Cubic spline approximation:
x
1. ****Find the cubic spline approximation for the function y = f(x) from the data,
given that y0  y3  0 .
''
''
i
x
y
2. Fit the 4 points by the cubic splines, using the conditions y0 ''  y3''  0 .
0
1
1
1
2
5
-1
0
1
2
y -1
2
3
3
4
11 8
1
3
35
3. Find the natureal cubic spline to fit the data.
x
0
1 2
f(x)
-1
3 29
Hence find f(0.5) and f(1.5).
4. Fit a natural cubic spline for the following data
x
0
1
2
3
y
1
4
0
-2
'
5. Fit a cubic spline curve for the points (2, 11), (3, 49) and (4, 123). Hence find y(2.5) and y (3.5) , assume that
y '' (2)  0 and y '' (4)  0 .
6. Fit the cubic spline for the data: Hence evaluate y(1.5)
x
1
2
3
y
-6
-1
16
7. The following values of x and y are given: Find the cubic splines and evaluate
y(1.5) .
8. Fit the straight line for the data.
x
0
1
2
3
y
1
2
9
28
x
1
2
3
4
y
1
2
5
11
9. Using cubic spline, compute y(1.5) from the given data.
x
1
2
3
y
-8
-1
18
Method – IV – Newton Forward and backward difference:
1. Find y at x = 84 using backward difference for the given data.
x
y
40
184
50
204
60
226
70
250
2. The following are data from the steam table. Find the pressure at temperature t = 1750C and t = 1420C.
Using Newton’s Backward and Forward difference Formula.
80
276
90
304
Temp0C
Pressure Kgf/cm2
140
3.685
150
4.854
160
6.302
170
8.076
180
10.225
3. From the following table find the interpolating polynomial for y by Newton’s Forward and Backward
x 1
3
5
7
interpolation formulae.
y
4. Estimate the value of f(22) from the following using Newton forward
interpolation formula.
10
16
45
1.00000
46
1.03553
47
1.07237
48
1.11061
49
1.15037
30
291
60
6. Use Newton’s Backward difference formula to construct an interpolating polynomial of degree 3 for the data
f(-0.75) = -0.07181250, f(-0.5) = -0.024750, f(-0.25) = 0.33493750, f(0) = 1.10100. Hence find f(-1/3).
7. *Using Newton’s forward interpolation formula, find the polynomial f(x)
x
4
6
8
Satisfying the following data. Hence evaluate y at x = 5.
y
1
3
8
x
Tanx˚:
25
332
30
45
204
tan 45 15 by
20
354
12
40
231
5. From the following table, find the value of
Newton’s forward interpolation formula.
x
f
3
35
260
50
1.19175
8. Using Newton’s forward interpolation formula, find the polynomial f(x) satisfying the following Data. Hence
find f(2).
x
y
0
14
5
379
10
1444
15
3584
9. *Using Newton’s forward interpolation formula find the cubic Polynomial which takes places the
following values: Evaluate f(4) using Newton’s backward formula. Is it the same as obtained from
the cubic Polynomial found above.
x
y(x)
0
1
1
2
10. Find e-1.1 using Newton’s forward interpolation
Formula.
2
1
3
10
x
e-x
1.00
0.3679
1.25
0.2865
1.50
0.2231
1.75
0.1738
UNIT – III
NUMERICAL DIFFERENTION AND INTEGRATION
PART – A
1. Using Newton’s backward difference formula, write the formulae for the first and second order
derivatives at the end value x = xn upto the fourth order difference term.
2. State Newton’s formula to find
3.
f ' ( x) using the forward difference.
 d2y 
 dy 
State Newton’s forward difference formula to find 
 and  2 
 dx  x x0
 dx  x x
.
0
4. By differentiating Newton’s backward difference formula, find the first derivative of the function f(x).
5. Why is Trapezoidal rule so called?
1
6. Evaluate

1
2
dx
x
by Trapezoidal rule, dividing the range into 4 equal parts.
7. State Simpson’s rule. (Or) State Simpson’s one – third rule. (or) Write down the formula for Simpson’s
1/3 and 3/8 rule
8. When does Simpson’s rule give exact result?
2.00
0.1353
9. What is the order of errors in Trapezoidal rule and Simpson’s formula.
x6
10. Six sets of values of x and y are given write the formula to get
 ydy .
x1
11. What are the errors in Trapezoidal and Simpson’s rules of numerical integration.
xn
12. In order to evaluate
 ydy by Simpson’s 1/3 rule as well as by Simpson’s 3/8 rule what is the restriction
x0
on the number of intervals.

13. Using Trapezoidal rule evaluate
 sin x dx by dividing the range into 6 equal parts.
0
6
14. Write down the Trapezoidal rule to evaluate
 f  x  dx with h = 0.5.
1
15. Compare Trapezoidal rule and Simpson’s 1/3 rule for evaluating numerical integration.
1
16. State three point Gaussian quadrature formula. Evaluate
 f  x  dx .
1
6
17. State two point Gaussian quadrature formula to evaluate
 f  x  dx .
1
18. If the range is not (-1, 1) then what is the idea to solve the Gaussian quadrature problems.
1
19. Apply Gauss two point formula to evaluate
dx
 1 x
2
.
1
b d
20. State Trapezoidal rule for evaluating
  f ( x, y )dxdy .
a c
b d
21. State Simpson’s rule for evaluating
  f ( x, y )dxdy .
a c
PART – B
Method – I: DERIVATIVES FROM DIFFERENCE TABLES:
x
f(x)
1. Find f '  3 and f ''  3 for the following data
2. Determine y(x) as a polynomial in x for
the following data, using Newton’s
divided difference formula. Given that
dy
d2y
& 2
dx
dx
Find
x
y(x)
1.0
7.989
3.0
-14
1.1
8.403
3.2
-10.032
1.2
8.781
3.4
-5.296
1.3
9.129
3.6
-0.256
1.4
9.451
3.8
6.672
1.5
9.75
4.0
14
1.6
10.031
at x = 1.1.
3. Consider the following table of data
'
Find f  0.25 ) using Newton’s formula
'
difference approximation and f  0.95
using Newton’s Backward difference approximation.
x
y
0.2
0.9798652
0.4
0.917771
0.6
0.8080348
0.8
0.6386093
1.0
0.3843735
4. The following data gives the velocity of a particle for 20 seconds at an interval of 5 seconds. Find
the initial acceleration using the entire data
Time(sec)
0
5
10
15
20
Velocity(m/sec)
0
8
14
69
228
5. Find the maximum and minimum value of y tabulated below
x
y
-2
2
-1
-0.25
0
0
1
-0.25
2
2
3
15.75
4
56
x
y
6. Give the following data, find y '  6  , y '  5  and the maximum value
0
4
52
26
3
58
7. Find the first
of
 x
1
3
4
112
7
466
two derivatives
x
y =  x
at x =
for the given
1
3
50
3.6840
51
3.7084
52
3.7325
53
3.7563
54
3.7798
55
3.8030
56
3.8259
50 and x = 56,
table:
''
'
8. Find y and y at x = 1.25 for the data given
x
y
1.00
1.00000
1.05
1.02470
1.10
1.04881
1.15
1.07238
9. Compute f '  0 and f ''  4  from the data
1.20
1.09544
1.25
1.11803
x
y
0
1
1.30
1.14017
1
2.718
2
7.381
3
20.086
4
54.598
10. Find at x = 1.5 and x = 4.0 from the following data using Newton’s formulae for differentiation.
x
y
1.5
3.375
2.0
7.0
2.5
13.625
3.0
24.0
3.5
38.875
4.0
59.0
Method – II : tRapezoidaL aNd simpsoN’s 1/3 aNd 3/8 RuLes:
1
1. Using Trapezoidal rule, evaluate
5
2. Evaluate
dx
 1 x2
1
taking 8 intervals.
dx
 4 x  5 by Simpson’s one – third rule and hence find the value of log
e
5 (n = 10).
0

3. By dividing the range into ten equal parts, evaluate
 sin x dx by Trapezoidal and Simpson’s rule.
0
Verify your answer with integration
1
4. Using Simpson’s 1/3 rule evaluate
 xe dx taking 4 intervals. Compare your result with actual value.
x
0
6
5.
dx
0 1 x2
9
922
by (i) Trapezoidal rule (ii) Simpson’s rule. Also check up the results by actual integration.
1.4
6. Compute the value of
 (sin x  log
e
x  e x ) dx taking h = 0.2 and using Trapezoidal rule,
0.2
Simpson’s 1/3 rule. Also compare with exact result.
rd
5.2
7. By dividing the range into six equal parts, evaluate
 log
e
xdx
using Simpson’s rule.
4
1
8. Evaluate
dx
0 1 x2
using Trapezoidal rule with 10 subintervals. Hence approximate the value of π.
9. The velocity of a particle at a distance S from a point on
its path is given by the table below. Estimate the time
taken to travel 60 meters by using Simpson’s one – third
rule. Compare your answer with Simpson’s 3/8 rule and also
using Trapezoidal rule.
S in meter
V m/sec
0
47
10
58
20
64
30
65
40
61
50
52
60
38
 /2
10. Dividing the range into 10 equal parts, find the value of
 sin x dx by (i) Trapezoidal rule
0
(ii) Simpson’s 1/3 rule and (iii)* Simpson’s 3/8 rule(equal part is not mention).
Method – III : RombeRg’s method:
rd
2
1. Evaluate
x
dx
using Romberg’s method. Hence obtain an approximate value for π.
4
2
0
1
dx
 x  1 correct to 3 – decimal places using Romberg’s method. Hence find the value of loge 2.
2. Evaluate
0
1
3. Compute
x
0
dx
by using Trapezoidal rule, taking h = 0.5 and h = 0.25. hence find one value of the above
1
2
integration by Romberg’s method.
Method – IV: Gaussian quadrature formulas:
1
1. Using three – point Gaussian quadrature formulas, evaluate (i)
dx
1 x2  1 (ii)
1
t
0
dx
.
1
2
x2  2x  1
0 1  ( x  1)4 dx by Gaussian three point formula.
2
2. Evaluate
1
x 2 dx
3. Evaluate  4
by using three points Gauss quadratic.
x 1
1
1
4. Using three point Gaussian quadrature, evaluate
dx

x4  1
0
2
5. Evaluate
1
 1 x
2
.
dx using Gauss three point formula.
1
3
6. Using Gaussian Two and Three formula, evaluate
1
 1  xdx .
2
Method – V : doubLe iNtegRaLs usiNg tRapezoidaL aNd simpsoN’s RuLe:
2 2
1. Evaluate
  f ( x, y )dxdy by Trapezoidal rule for the following data:
0 0
Y
x
0
1
2
0
2
3
4
0.5
3
4
6
1
4
6
8
1.5
5
9
11
2
5
11
14
1 1
2. Using Simpson’s 1/3 rule evaluate
dxdy
  1  x  y taking h = k = 0.5.
0 0
1 2
3. Evaluate
  (1  x
0 1
2 xy
dxdy by Trapezoidal rule with h = k =0.25.
)(1  y 2 )
2
2.0 1.5
 I
4. Using Trapezoidal rule evaluate
n
( x  2 y )dxdy . Choosing Δx = 0.15 and Δy = 0.25.
1.4 1.0
1 1
5. Evaluate
e
x y
dxdy using Simpson’s and Trapezoidal rule.
0 0
1.4 2.4
1
  xy dxdy by Simpson’s rule taking h = 0.1 and k = 0.1. verify your result by actual integration.
6. Evaluate
1 2
5 4
7. Evaluate
1
  x  y dxdy by Trapezoidal rule in x – direction with h = 1 and Simpson’s 1/3 rule in y – direction
1 1
with k = 1.
2.4 4.4
8. Evaluate
  xydxdy using Simpson’s rule (h = k = 0.1).
2
4
2 1
9. *Evaluate
  4xydxdy by using Simpson’s rule taking h = ¼ and k = ½.
0 0
UNIT – IV
INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS
PART – A
1. Write the Taylor’s series formula at y(x0) = y0.
2. Write the merits and demerits of the Taylor method of solution.
3. Which is better Taylor’s method or R-K method?
4. Solve the differential equation
dy
 x  y  xy, y  0  1 by Taylor series method to get the value
dx
of y at x = h.
5. What is meant by initial value problem and give an example for it.
6. Find y(0.1) by Taylor’s series given
dy
1  y, y (0) 1 .
dx
7. Write down the fourth order Taylor Algorithm.
8. State Modified Euler algorithm to solve
y '  f  x, y  , y(x0) = y0, at x = x0 + h.
9. Write down Euler algorithm to the differential equation
10. Find y(0.2) when
11. Solve
12. Solve
dy
 f  x, y  .
dx
y '   2 xy 2 , y(0) = 1 and h = 0.2 by Euler’s method.
dy
1  y , y(0) = 0, for x = 0.1 by Euler’s method.
dx
dy
  y , y  0  1 to find y(0.01) using Euler method.
dx
13. Write the Runge – Kutta algorithm of second order for solving
y '  f  x, y  , y(x0) = y0.
dy
 f  x, y 
dx
14. Write down the R – K formula of 4th order to solve
with y(x0) = y0.
15. In the deviation of fourth order Runge – Kutta formula, why it is called fourth order.
16. ** Write down the formula to solve 2nd order differential equation using Runge – Kutta method of
4th Order.
 
17. What are the values of k1 and l1 to solve y  xy  y  0 , y(0) = 1, y 0  0 by Runge –
Kutta Method of 4th order?
18. Compare Taylor’s series and R – K method.
19. Write Milne’s predictor corrector formula.
20. What is the error term in Milne’s corrector formula?
21. How many prior values are required predict the next value in Milne’s method.
22. What is the error term in Milne’s Predictor formula?
23. Predictor corrector methods are ……………starting methods.
24. Write down Adams – Bashforth predictor formula.
25. How many prior values are required predict the next value in Adam’s method?
26. What is a predictor – corrector method of solving a differential equation?
27. What is the condition to apply Adams Bashforth method?
28. What do we mean by saying that a method is self – starting? Not self starting.
29. Compare the Milne’s predictor – corrector and Adam – Bashforth predictor – corrector methods
For solving ordinary differential equations.
PART - B
"
'
'
Method –I : Taylor series method:
1. ***Using Taylor series method find y(1.1) and y(1.2) correct to four decimal places given
dy
 xy1/3 and y(1) = 1.
dx
2. **Using Taylor series method, find y at x = 0.1 and x = 0.2 given
dy
 2 y  3e x , y(0) = 0.
dx
3. Using Taylor series method, compute the value of y(0.1), y(0.2), y(0.3) and y(0.4) correct to three decimal
dy
1  2 xy given that y(0) = 0.
dx
dy
 x  y (i) When y(0) = 1. Using Taylor’s series up to x = 0.1, 0.2 with h = 0.1.
**Solve numerically
dx
places from
4.
(ii)When y(1) = 0, find y(1.1).
5. Using Taylor series method, find y at x = 0.2 and x = 0.4 given
dy
 y 2  x 2 , y(0)=1
dx
dy 2
 x  y , y(0)=1,correct to 4 decimal places.
dx
dy 2
 x 1  y  , y(1)=1.
Using Taylor series method, find y at x = 1.1 and x = 1.2 given
dx
6. Using Taylor series method, find y at x = 0.1 given
7.
Method – II – euLeR aNd modified euLeR’s method:
1. Using Euler’s method find y(0.2) and y(0.4) from
dy
 x  y , y(0) = 1 with h = 0.2.
dx
2. Using modified Euler’s method, solve
dy
1  y , y(0) = 0, for x = 0.1, 0.2, and 0.3. compare your results with
dx
exact solutions.
3. Using modified Euler’s method find y at x = 0.1 and x = 0.2 given
4. Solve
dy
 log10  x  y  , y (0)  2
dx
dy
2x
 y  , y (0) 1.
dx
y
by Euler’s Modified method and find the values of y(0.2), y(0.4),and
y(0.6) take h = 0.2.
5. Using Modified Euler method find
a. y(0.1) and y(0.2) given
b. y(0.2) and y(0.4) given
y '  x 2  y 2 , y(0) = 1 with h = 0.1
y '  x 2  y 2 , y(0) = 1 with h = 0.2.
6. Find y(0.25) and y(0.5), using Modified Euler’s Method with h = 0.25 given that y
'
 3x 2  y ,
y(0) = 4. Compare the values with the exact solutions.
7.
**consider the initial value problem
dy
 y  x 2  1, y  0   0.5 using the modified Euler method,
dx
find y(0.2)
8. *Using modified Euler’s method, find y(4.1) and y(4.2) if
5x
dy
 y 2  2  0, y  4  1 .
dx
Method – III: 4th order R – K method:
1. Find y(0.2) using R – K method of order 4 from
y'  x  y
, y(0) = 1.
dy 2
 x y  x , y(0) = 1, using R – K method of 4th order find y at x = 0.1.
dx
dy y 2  x 2

Compute y(0.2) given
, y(0) = 1 by R – K method.
dx y 2  x 2
2. Given
3.
y '  y  x 2 , y(0.6) = 1.7379 by using R – K method of order 4 with h = 0.1
dy
 x  y 2 , y(0) = 1.
Using R.K. method of 4th order find y(0.1) given initial value problem
dx
'
3
Find y(0.2) and y(0.4) using R – K method of order 4 from y  x  y , y(0) = 2.
dy
1
Find y(0.1) using R – K method of order 4 from
, y(0) = 1.

dx x  y
dy
 log  x  y  , y(0) = 1 using R – K method of 4th order, taking h = 0.1.
Find y(0.2), given
dx
4. Find y(0.8) given that
5.
6.
7.
8.
Method – IV : 2ND ORDER DIFFERENTIAL EQUATION USING R – K METHOD OF 4TH ORDER:
1. **Using R – K method of order 4, solve
y"  2 y '  2 y  e 2 x sin x with y(0) = -0.4 y '  0    0.6 .
2. Use R – K 4th order method to find y(0.2) for the equation y
"
 xy '  y  0 given that y(0) = 1,
y '  0   0, takeh  0.2
3. **Given
y"  xy '  y  0 , y(0) = 1, y '  0   0 . Find the value of y(0.1) by R – K method of 4th order.
Method – V : miLNe’s pRedictoR aNd coRRectoR method:
1. * Using Milne’s method find y(2) if y(x) is the solution of
dy x  y

dx
2
given y(0) = 2, y(0.5) = 2.636,
y(1) = 3.8595 and y(1.5) = 4.968.
2. Using Milne’s method , find y (4.4) given
2 xy '  y 2  2  0 , y(4) = 1, y(4.1) = 1.0049,
y(4.2) = 1.0097 and y(4.3) = 1.0143.
3. ***Solve
dy
 xy  y 2 , y (0) 1
dx
, using Milne’s predictor – corrector formula and find y(0.4).
Using Taylor series method to find y(0,.1), y(0.2) and y(0.3).
dy
 y  x 2 , y (0) 1
dx
4. Solve
(i) Find y(0.1) and y(0.2) by R – K method of order 4.
(ii) Find y(0.3) by Euler’s method
(iii) Find y(0.4) by Milne’s predictor – corrector method.
dy 2
 x  y 2  2 , using Milne’s predictor – corrector method for x = 0.3,y(0) = 1.
dx
5. Solve
Evaluate
the values of y for x = - 0.1, 0.1 and 0.2 using Taylor’s series.
6.
**Using Milne’s method find y(4.4) given 5 xy
'
 y 2  2  0 given y(4) = 1, y(4.1)= 1.0049,
y(4.2) = 1.0097 and y(4.3) = 1.0143.
7. Given
2
2
dy y 1  x 
and y(0) = 1, y(0.1) = 1.06, y(0.2) = 1.12, y(0.3) = 1.21, evaluate y(0.4) by

dx
2
Milne’s predictor – corrector method.
8. *Given that
dy
1  y 2
dx
; y(0.6)= 0.6841, y(0.4) = 0.4228, y(0.2) = 0.2027, y(0) = 0, find y(-0.2) using Milne’s
method.
Method – VI – adam’s pRedictoR aNd coRRectoR methods:
1. Given
dy 2
 x 1  y 
dx
, y(1) = 1, y(1.1) = 1.233, y(1.2) = 1.548, y(1.3) = 1.979, evaluate y(1.4) by Adams –
Bashforth method.
2. Solve
dy
1  y
dx
with the initial condition x = 0, y = 0, using Euler’s algorithm and tabulate the solutions at x
= 0.1, 0.2, 0.3, 0.4. using these remits find y(0.5) using Adams – Bashforth predictor – corrector method.
3. Using Adams – Bashforth predictor – corrector formulae, evaluate y(1.4). if y satisfies
dy y 1
 
dx x x 2
and y(1)
= 1, y(1.1) = 0.996, y(1.2) = 0.986, y(1.3) = 0.972.
4. Find y(0.1), y(0.2,), y(0.3), from y
'
 x2  y
,y(0) = 1 using Taylor’s series method and hence obtain y(0.4)
using Adams – Bashforth method.
5. By using Adam’s pc method find y when x = 0.4, given
= 1.023.
dy xy

dx 2
, y(0) = 1, y(0.1) = 1.01, y(0.2) = 1.022, y(0.3)
Method – VII - Simultaneous differential equations:
1. Solve for y(0.1) from the simultaneous differential equations
dy
dz
 2 y  z ;  y  3 z ; y(0) = 0, z(0) =
dx
dx
0.5 using Runge kutta method of the fourth order.
UNIT – V
BOUNDARY VALUE PROBLEMS IN ORDINARY
AND PARTIAL DIFFERENTIAL EQUATIONS
PART – A
1. Write an explicit formula to solve numerically the heat equation(parabolic equation) uxx – aut = 0.
2. Write down the Crank – Nicolson formula to solve ut = uxx.
3. Write down the implicit formula to solve one dimensional heat flow equation
4. What is the central difference approximation for
y"
5. Write the difference scheme for solving the Poisson equation
6. Write down the finite difference formula for
 2 u  f ( x, y ) .
y ' and y " .state the finite difference scheme to solve the
ytt  a 2 y xx .
Classify the PDE y xx  xu yy  0 .
equation
7.
8. Write down Bender – Schmidt’s difference scheme in general form and using suitable value of
scheme in simplified form.
9. State Standard Five Point Formula with relevant diagram.
10. Define a difference quotient.
11. State the SFPF in solving Laplace equation.
12. State the implicit scheme to solve the dimensional heat equation numerically.
13. Write the difference scheme quotients of uxx and uyy.
14. Write the finite difference scheme for the second order differential equation
15. State the explicit finite difference scheme for one dimensional wave equation
 , write the
1
.
n
2
 2u
2  u
.

t 2
x 2
y"  f ,h 
16. Define SFPF and DFPF.
17. Define elliptic, parabolic and hyperbolic type of partial differential equations.
PART – B
METHOD – I – BENDER SCHMIDT METHOD:
1. Solve
 2u u
, subject to u(0, t) = u(1, t) = 0 and u(x, 0) = sinπx, 0 < x < 1 using Bender – Schmidt method.

x 2 t
2. Solve with the conditions u(0, t) = 0 = u(4, t), u(x, 0) = x(4 - x)
(1) Taking h = 1 employing Bender- Schmidt recurrence equation. Continue the solution through 10 time
steps
(2) Assume h = 1. Find the values of up to t = 5, using explicit method.
3. Solve with boundary conditions u(0, t) = 0, u(8, t) = 0, u(x, 0) = x(8 - x)/2up to t = ½ , taking h = 1,
k = 1/8.
4. Given
 2 f f

 0 , f(0, t) = f(5, t) = 0, f(x, 0) = x2(25 – x2), find f in the range taking h = 1 and upto 5
2
x
t
seconds.
Method – II:
1. Find the pivotal values of the equation
u(x, 0) = x(4 - x) and
 2u
 2u
 4 2 with given conditions u(0, t) = 0, u(4, t) = 0,
t 2
x
u ( x,0)
 0 by taking h = 1, for 4 time steps.
t
2. Evaluate the Pivotal values of the equation taking h = 1 and up to on half of the period of the oscillation 16uxx
– utt = 0 given u(0, t) = u(5, t) = ut(x, 0) = 0and u(x, 0) = x2(5 - x).
3. Solve 25uxx – utt = 0 for u at the Pivotal points given u(0, t) = u(5, t) = ut(x, 0) = 0 and u(x, 0) = 2x for
0
< x < 2.5 and = 10 – 2x for 2.5 < x <5 for one half period of vibrations.
4. Solve the equation
 2u  2u
, 0 < x < 1, t > 0 satisfying the conditions u(x, 0) = 0, u(0, t) = 0 and u(1, t) =

x 2 t 2
½sinπt. Compute u(x, t) for 4 time – steps by taking h = ¼.
Method – III:
y "  x  y with the boundary conditions y(0) = y(1) = 0.
Solve the finite difference method, the boundary value problem y "( x)  y ( x)  0 , where y(0) = 0 and y(1)
1. Solve the equation
2.
= 1, taking h = 0.25.
3. Using the finite difference method, compute y(0.5), given
y " 64 y  10  0 , x ∈ (0, 1), y(0) = y(1) = 0, sub
dividing the interval into (1) 4 equal parts (2) 2 equal parts.
4. Solve
y " y  0 with the boundary conditions y(0) = 0 and y(1) = 1.
Method – IV:
1. Solve the elliptic equation for the following square mesh with boundary values as shown
2. Solve
0
1000
2000
1000
500
0
500
1000
500
0
1000
2000
1000
1000
500
0
 2u  8 x 2 y 2 for square mesh given u = 0 on the boundaries dividing the square in to 16 sub squares
of length 1 unit.
3. *Solve uxx +uyy = -10(x2 +y2 + 10) over the square mesh with sides x = 0, y = 0, x = 3, y = 3 with u = 0, on the
boundary and mesh length 1 unit.
4. Solve  u  0 over the square mesh of side 4 units, satisfying the boundary conditions.
(a) u(0, y) = 0 for 0 < y < 4
(b) u(4, y) = 12 +y for 0 < y < 4
(c) u(x, 0) = 3x for 0 < x < 4
(d) u(x, 4) = x2 for 0 < x < 4.
5. Solve uxx +uyy = 0 in 0 ≤ x ≤ 4, 0 ≤ y ≤ 4. Given that u(0, y) = 0, u(4, y) = 8 + 2y; u(x, 0) = x2/2 and u(x, 4) = 2
taking h = k = 1. Obtain the result correct to one decimal.
2
6. Solve  u  0 in the square region bounded by x = 0, x= 4, y = 0, y = 4 and with boundary conditions u(0, y)
= y2/2, u(4, y) = y2, u(x, 0) = 0and u(x, 4) = 8 +2x taking h = k = 1. (Perform 4 iterations)
2