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NUMERICAL METHODS

Unit I : Solution of equations and eigen value problems

Part A

1. If g(x) is continuous in [a,b] then under what condition the iterative method x = g(x) has
unique solution in [a,b].
1 3 
2. Find inverse of A =   by Gauss – Jordan method.
2 7
3. Why Gauss Seidel iteration is a method of successive corrections.
4. Compare Gauss Jacobi and Gauss Siedel methods for solving linear system of the form AX = B.
5. State the conditions for convergence of Gauss Siedel method for solving a system of
equations.
6. Compare Gaussian elimination method and Gauss-Jordan method.
7. What type of eigen value can be obtained using power method.
1 2 
8. Find the dominant eigen value of A =   by power method.
3 4
9. How is the numerically smallest eigen value of A obtained.
10. State two difference between direct and iterative methods for solving system of equations.

Part B

 2 −1 0 
 
1. Find all the eigen values of the matrix − 1 2 − 1 by power method (Apply only 3
 
 0 − 1 2 
iterations).

2. Use Newton’s backward difference formula to construct an interpolating polynomial of degree 3

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for the data:

f( - 0.75) = - 0.0718125, f( - 0.5) = - 0.02475, f( - 0.25) = - 0.3349375 and f(0) = 1.101. Hence find f
1
(- ).
3

6. Solve the system of equations using Gauss Seidel iterative methods.

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20x – y – 2z = 17, 3x + 20y – z = -18, 2x – 3y +20z = 25.

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7.Find the largest eigen values and its corresponding vector of the matrix

 1 3 − 1
 3 2 4  by power method.
 
− 1 4 10 

2 2 3

8. Using Gauss- Jordan obtain the inverse of the matrix 2 1 1

 
1 3 5

9. Using Gauss Seidel method solve the system of equations starting with the values x = 1 , y = -2
and z = 3,
x + 3y + 5z = 173.61, x – 27y + 2z = 71.31, 41x – 2y + 3z = 65.46

10. Solve the following equations by Jacobi’s iteration method

x + y + z = 9, 2x – 3y + 4z = 13, 3x + 4y + 5z = 40.
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Unit II : Interpolation and Approxiamtion

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Part A

1. Construct a linear interpolating polynomial given the points (x0,y0) and (x1,y1).
2. Obtain the interpolation quadratic polynomial for the given data by using Newton’s forward
difference formula.
X: 0 2 4 6

Y : -3 5 21 45

3. Obtain the divided difference table for the following data.

X : -1 0 2 3

Y : -8 3 1 12

4. Find the polynomial which takes the following values.

X : 0 1 2

Y : 1 2 1

5. Define forward, backward, central differences and divided differences.

6. Evaluate ∆10 (1-x) (1-2x) (1-3x)--------(1-10x), by taking h=1.

7. Show that the divided difference operator ∆ is linear.

8. State the order of convergence of cubic spline.

9. What are the natural or free conditions in cubic spline.

10. Find the cubic spline for the following data

X:0 2 4 6
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Y:1 9 21 41

11. State the properties of divided differences.

1 −1
12. Show that ∆3 ( ) = .
bcd a abcd

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13. Find the divided differences of f(x) = x3 + x + 2 for the arguments 1,3,6,11.
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14. State Newton’s forward and backward interpolating formula.

15. Using Lagranges find y at x = 2 for the following

X:0 1 3 4 5

Y:0 1 81 256 625

Part B

1. Using Lagranges interpolation formula find y(10) given that y(5) = 12, y(6) = 13,

y(9) = 14 and y(11) = 16.

2. Find the missing term in the following table

x:0 1 2 3 4

y:1 3 9 - 81

3. From the data given below find the number of students whose weight is between

60 to 70.

Wt (x) : 0-40 40-60 60-80 80-100 100-120

No of
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students : 250 120 100 70 50

4. From the following table find y(1.5) and y’(1) using cubic spline.

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X : 1 2 3
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Y : -8 -1 18

5. Given sin 450 = 0.7071, sin 500 = 0.7660, sin 550 = 0.8192, sin 600 = 0.8660, find

sin 520 using Newton’s forward interpolating formula.

6. Given log 10 654 = 2.8156, log 10 658 = 2.8182, log 10 659 = 2.8189, log 10 661 =

2.8202, find using Lagrange’s formula the value of log 10 656.

7. Fit a Lagrangian interpolating polynomial y = f(x) and find f(5)

x:1 3 4 6

y : -3 0 30 132

8. Find y(12) using Newton’ forward interpolation formula given

x : 10 20 30 40 50

y : 46 66 81 93 101

9. Obtain the root of f(x) = 0 by Lagrange’s inverse interpolation given that f(30) = -30,

f(34) = -13, f(38) = 3, f(42) = 18.

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10. Fit a natural cubic spline for the following data

x:0 1 2 3

y:1 4 0 -2

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11. Derive Newton’s divided difference formula.

12. The following data are taken from the steam table:

Temp0 c : 140 150 160 170 180

Pressure : 3.685 4.854 6.502 8.076 10.225

Find the pressure at temperature t = 1420 and at t = 1750

13. Find the sixth term of the sequence 8,12,19,29,42.

14. From the following table of half yearly premium for policies maturing at different ages, estimate
the premium for policies maturing at the age of 46.
Age x : 45 50 55 60 65

Premium y : 114.84 96.16 83.32 74.48 68.48

15. Form the divided difference table for the following data

x : -2 0 3 5 7 8

y : -792 108 -72 48 -144 -252

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Unit III - Differentiation and Integration

Part A

1. What the errors in Trapezoidal and Simpson’s rule.

2. Write Simpson’s 3/8 rule assuming 3n intervals.
1
dx
3. Evaluate ∫ 1+ x
−1
4
using Gaussian quadrature with two points.

4. In Numerical integration what should be the number of intervals to apply Trapezoidal, Simpson’s
1/3 and Simpson’s 3/8.
1
x 2 dx
5. Evaluate ∫ 1+ x 4 using Gaussian three point quadrature formula.
−1
1
6. State two point Gaussian quadratue formula to evaluate ∫ f ( x)dx .
−1
7. Using Newton backward difference write the formula for first and second order derivatives at the
end value x = x0 upto fourth order.
dy d2y
8. Write down the expression for and at x = x0 using Newtons forward difference formula.
dx dx 2
9. State Simpson’s 1/3 and Simpson’s 3/8 formula.
Π

∫
10. Using trapezoidal rule evaluate sin xdx by dividing into six equal parts.
0

Part B

1. Using Newton’s backward difference formula construct an interpolating polynomial

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of degree three and hence find f(-1/3) given f(-0.75) = - 0.07181250, f(-0.5) =

- 0.024750, f(-0.25) = 0.33493750, f(0) = 1.10100.

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dxdy
∫ ∫ 1+ x + y by Simpson’s 1/3 rule with ∆x = ∆y = 0.5 where 0<x,y<1.
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2. Evaluate

2 2
dx dy
3. Evaluate I = ∫∫ x+ y
1 1
by using Trapezoidal rule, rule taking h= 0.5 and h=0.25. Hence the value of

the above integration by Romberg’s method.

4. From the following data find y’(6)
X:0 2 3 4 7 9

Y: 4 26 58 112 466 922

2 2
dx dy
5. Evaluate ∫∫ x
1 1
2
+ y2
numerically with h= 0.2 along x-direction and k = 0.25 along y direction.

6. Find the value of sec (31) from the following data

θ (deg ree) : 31 32 33 34

Tan θ : 0.6008 0.6249 0.6494 0.6745

7. Find the value of x for which f(x) is maxima in the range of x given the following table, find also
maximum value of f(x).
X: 9 10 11 12 13 14

Y : 1330 1340 1320 1250 1120 930

8. The following data gives the velocity of a particle for 20 seconds at an interval of five seconds. Find
initial acceleration using the data given below
Time(secs) : 0 5 10 15 20

Velocity(m/sec): 0 3 14 69 228

7
dx
9. Evaluate ∫ 1+ x
3
2
using Gaussian quadrature with 3 points.

dy d2y
10. For a given data find and at x = 1.1
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dx dx 2
X : 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Y: 7.989 8.403 8.781 9.129 9.451 9.750 10.031

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UNIT – IV : INITIAL VALUE PROBLEMS FOR ORDINARY DIFFERENTIAL EQUATIONS

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PART – A

1. By Taylor series, find y(1.1) given y ′ = x + y, y(1) = 0.

2. Find the Taylor series upto x3 term satisfying 2 y ′ + y = x + 1, y (0) = 1 .
dy
3. Using Taylor series method find y at x = 0.1 if = x 2 y − 1, y (0) = 1 .
dx
4. State Adams – Bashforth predictor and corrector formula.
5. What is the condition to apply Adams – Bashforth method ?
dy
6. Using modified Euler’s method, find y (0.1) if = y 2 + x 2 , y ( 0) = 1 .
dx
7. Write down the formula to solve 2nd order differential equation using Runge-Kutta method of 4th
order.
8. In the derivation of fourth order Runge-Kutta formula, why is it called fourth order.
9. Compare R.K. method and Predictor methods for the solution of Initial value problems.
dy
10. Using Euler’s method find the solution of the IVP = log( x + y ), y (0) = 2 at x = 0.2 taking
dx
h = 0 .2 .

PART-B

dy
11. The differential equation = y − x2 is satisfied
dx
by y(0) = 1, y(0.2) = 1.12186, y(0.4) = 1.46820, y(0.6) = 1.7379 .Compute the value of y(0.8)
by Milne’s predictor - corrector formula.
12. By means of Taylor’s series expension, find y at x = 0.1,and x = 0.2 correct to three decimals
dy
places, given − 2 y = 3e x , y(0) = 0.
dx
13. Given y ′′ + xy ′ + y = 0, y (0) = 1, y ′(0) = 0, find the value of y(0.1) by using R.K.method of
fourth order.
dy
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14. Using Taylor;s series method find y at x = 0.1, if = x 2 y − 1 , y(0)=1.

dx
dy
15. Given = x 2 (1 + y ) , y(1) = 1, y(1.1) = 1.233, y(1.2) = 1.548, y(1.3)=1.979, evaluate y(1.4) by
dx
Adam’s- Bashforth method.
dy y 2 − x 2
16. Using Runge-Kutta method of 4th order, solve = with y(0)=1 at x=0.2.
dx y 2 + x 2
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17. Using Milne’s method to find y(1.4) given

that 5 xy ′ = y 2 − 2 = 0 given that
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y (4) = 1, y (4.1) = 1.0049, y (4.2) = 1.0097, y (4.3) = 1.0143 .

dy
18. Given = x 3 + y, y (0) = 2, y (0.2) = 2.443214, y (0.4) = 2.990578, y (0.6) = 3.823516 find
dx
y(0.8) by Milne’s predictor-corrector method taking h = 0.2.
dy
19. Using R.K.Method of order 4, find y for x = 0.1, 0.2, 0.3 given that = xy + y 2 , y (0) = 1 also
dx
find the solution at x = 0.4 using Milne’s method.
dy
20. Solve = y − x 2 , y(0) = 1.
dx
Find y(0.1) and y(0.2) by R.K.Method of order 4.

Find y(0.3) by Euler’s method.

Find y(0.4) by Milne’s predictor-corrector method.

21. Solve y ′′ − 0.1(1 − y 2 ) y ′ + y = 0 subject to y(0) = 0, y ′(0) = 1 using fourth order Runge-Kutta
Method.
Find y(0.2) and y ′(0.2) . Using step size ∆x = 0.2 .

xy
22. Using 4th order RK Method compute y for x = 0.1 given y ′ = given y(0) = 1 taking h=0.1.
1+ x2
dy
23. Determine the value of y(0.4) using Milne’s method given = xy + y 2 , y (0) = 1 , use Taylors
dx
series to get the value of y at x = 0.1, Euler’s method for y at x = 0.2 and RK 4th order method for
y at x=0.3.
dy
24. Consider the IVP = y − x 2 + 1, y (0) = 0.5
dx
(i) Using the modified Euler method, find y(0.2).
(ii) Using R.K.Method of order 4, find y(0.4) and y(0.6).
(iii) Using Adam- Bashforth predictor corrector method, find y(0.8).
25. Consider the second order IVP y ′′ − 2 y ′ + 2 y = e 2t S int, with y(0) = -0.4 and y’(0)=-0.6.
(i) Using Taylor series approximation, find y(0.1).
(ii) Using R.K.Method of order 4, find y(0.2).
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UNIT-5
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PART-A

1. Define the local truncation error.

2. Write down the standard five point formula used in solving laplace equation U xx + U yy = 0 at the
point ( i∆x, j∆y ).

3. Derive Crank-Niclson scheme.

4. State Bender Schmidt’s explicit formula for solving heat flow equations

5. Classify x 2 f xx + (1-y 2 ) f yy = 0

6. What is the truncation error of the central difference approximation of

y ' (x)?

7. What is the error for solving Laplace and Poissson’s equation by finite difference method.

d2y
8. Obtain the finite difference scheme fore the difference equations 2 + y = 5.
dx 2

9. Write dowm the implicit formula to solve the one dimensional heat equation.

10. Define the diagonal five point formula .

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PART-B
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1. Solve the equqtion U t = U xx subject to condition u(x,0) = sin πx ; 0 ≤ x ≤ 1 ,u(0,t) =

u(1,t) =0 using Crank- Nicholson method taking h = 1/3 k = 1/36(do on time step)

2. Solve U xx + U yy = 0 for the following square mesh with boundary values

1 2

u1 u2
1 4

2 u3 u4 5

4 5

3. Solve U xx = U tt with boundary condition u(0,t) = u(4,t) and the initial condition

u t (x,0) = 0 , u(x,0)=x(4-x) taking h =1, k = ½ (solve one period)

4. Solve xy II + y = 0 , y(1) =1,y(2) = 2, h = 0.25 by finite difference method.

5. Solve the boundary value problem xy II -2y + x = 0, subject to y(2) = 0 =y(3).Find

y(2.25),y(2.5),y(2.75).

∂y ∂2 y
6 . Solve the vibration problem = 4 2 subject to the boundary conditions
∂t ∂x
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1 1
y(0,t)=0,y(8,0)=0 and y(x,0)= x(8-x).Find y at x=0,2,4,6.Choosing ∆ x = 2, ∆ t = up
2 2

compute to 4 time steps.

7. Solve ∆2 u = -4(x + y) in the region given 0 ≤ x ≤ 4, 0 ≤ y ≤ 4. With all boundaries kept

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at 0 0 and choosing ∆ x = ∆ y = 1.Start with zero vector and do 4 Gauss- Seidal iteration.
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00 00 00 00 00

00

00

00

00 00 00 00 00

8. Solve u xx + u yy = 0 over the square mesh of sid e 4 units, satisfying the following

conditions .

u(x,0) =3x for 0 ≤ x ≤ 4

u(x, 4) = x 2 for 0 ≤ x ≤ 4

u(0,y) = 0, for 0 ≤ y ≤ 4

u(4,y) = 12+y for 0 ≤ y ≤ 4

∂ 2u ∂u
9. Solve −2 = 0, given that u(0,t)=0,u(4.t)=0.u(x,0)=x(4-x).Assume h=1.Find
∂x 2
∂t
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the values of u upto t =5.

10. Solve y tt = 4y xx subject to the condition y(0,t) =0, y(2,t)=o, y(x,o) = x(2-x),

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∂y
( x,0) = 0 . Do 4steps and find the values upto 2 decimal accuracy.
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∂t
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