Mathematics
10CS41
UNIT -1
NUMERICAL METHODS-1
1. Use the Picard’s method to obtain the fourth approximation to the solution of
dy
= x + y , y(0) = 1 and hence find y at x = 0.1, 0.2
dx
2. Use Picard’s method to find y at x = 0.25, 0.5, 0.75 given
dy
x2
, y (0) =1 by taking two
 2
dx y  1
approximations.
3. Find the values of y for x =0(0.1)0.5 given 𝑦 ′ = 1 + xy, which passes through (0.1) from Picard’s
third approximation.
4. Solve
dy
 y  2 x by Picard’s method given that y = 3 at x =1 initially. Also compute y (1.1) from
dx
the third approximation.
5. Solve,
dy
 x 2 y  1 with y (0) =1 using Taylor’s series method and find y(0.1) consider up to
dx
fourth degree terms .
6. Solve,
[June/July 08, Dec2012, Dec08/Jan.09, June/July 2011].
dy
 2 y  3e x , y (0) = 0. Using Taylor’s series method and find y (0.1), y (0.2).
dx
[June-July2009, 2013]
7. Employ Taylor’s series method to find an approximate solution correct to fourth decimal places for
dy
the following initial value problem at x = 0.1,
[Dec. 09/Jan.10]
 x  y 2 y (0) = 1 .
dx
8. Find the y(0.1) correct to 6 decimal places by Taylor’s series method when
y (0) =1.0
(consider up to 4th degree term).
dy
 xy  1 ,
dx
[May/June2010]
9. Using the Taylor’s method, find the third order approximate solution at x = 0.4 of the problem
dy
= x2y + 1, with y(0) = 0. Consider terms up to fourth degree.
dx
Dept Of ISE
EWIT
[June 2012]
Page 1
Mathematics
10CS41
dy
 y  x 2  0 , y(0) =1, y (0.1) = 0.9052, y (0.2)=0.8213. Find correct to four decimal places
dx
y(0.3) and y (0.4) using modified Euler’s method .
[Dec.2010].
10. Given
11. Given ,
dy
1

 2 y 2 , y (0) = 0, Find y (0.5) in two steps, using the Modified Euler’s method.
2
dx 1  x
[Dec.2011]
12. Using Modified Euler’s method to find y(0.1) given
perform two iterations in each step.
13. Using Modified Euler’s method to solve
h = 0.1
14. Solve the differential equation
dy
 x 2  y , y (0) =1 by taking h =0.05,
dx
[Dec. 09/Jan.10]
dy
 x  y y(0) = 1 at x = 0.1 for three iterations taking
dx
[June-July2013]
dy
= -xy2 underthe initial condition y(0) = 2, by using the Modified
dx
Euler’s method, at the points x = 0.1 and x = 0.2. Take the step size h = 0.1 and carry out two
modifications at each step.
[June-July2012]
15. Using Runge- Kutta 4th order method to solve
0.2 and 0.4 take h = 0.2
dy y 2  x 2

with y (0) = 1 and find y for x =
dx y 2  x 2
[June-July2008, Dec. 2012]
16. Use Runge- Kutta method of fourth order to solve
h = 0.2 .
dy
 x  y , y (0) =1 at x = 0.2 with step length
dx
[June-July2009, June-July2013 ]
17. Using Runge- Kutta method of order 4, compute y(0.2) for the equation
2x
𝑦′ = y [May/June 2010]
y (0) = 1.0 (take h = 0.2)
y
18. Apply Runge- Kutta method of order 4, to compute y (2.0) given, 10
dy
 x 2  y 2 , y (0) =1
dx
taking h = 0.1
[Dec.2010].
19. Using Runge- Kutta method of order 4, find y (0.2) for the equation
h =0.1
Dept Of ISE
dy y  x
, y (0) = 1 taking

dx y  x
[Dec.2011].
EWIT
Page 2
Mathematics
20. Given,
10CS41
dy
 xy  y 2 y(0) = 1, y(0.1) = 1.1169, y(0.2) = 1.2773, y (0.3) = 1.5049, find y (0.4) accurate
dx
up to 3 decimal Places using Milne’s Predictor-Corrector method.
[June/July08, Dec12].
21..Use Milne’s Predictor-Corrector method to find y at x = 0.8, given
X
Y
0
0
0.2 0.4
0.02 0.0795
dy
 x  y 2 with,
dx
0.6
0.1762
Apply corrector once.
[June-July2009]
22. Using Milne’s Predictor-Corrector method to find y(0.3) correct to three decimals given
x
y
-0.1
0
0.1
0.2
0.908783 1.0000 1.11145 1.25253
[June-July2013]
23. If
dy
 2e x  y , y (0) = 2, y(0.1) = 2.010, y(0.2) = 2.04 and y(0.3) = 2.09 find y (0.4) correct to four
dx
decimal places. By using Milne’s Predictor-Corrector method. (Use corrector formula twice).
[Dec.09/Jan.10]
24. Given 2
dy
 (1  x 2 ) y 2 and y(0) = 1, y (0.1) = 1.06, y (0.2) = 1.12, y (0.3) = 1.21 evaluate
dx
y (0.4) by Milne’s method.
[Dec.2011]
25. Given that 𝑦 ′ = x 2 (1  y ) and y (1) = 1.0, y (1.1) = 1.233, y (1.2) = 1.548 and y (1.3) = 1.979
compute y (1.4) by Adm’s Bashforth method (Apply corrector formula twice).
[May/June2010, Dec.08/Jan.09, Dec.08/Jan.14].
Dept Of ISE
EWIT
Page 3
Mathematics
10CS41
UNIT-II
NUMERICAL METHODS – 2
1. Obtain second approximation to the solution of
𝑑𝑦
=
𝑑𝑡
2𝑡
3𝑥
𝑑𝑥
𝑑𝑡
1
= (𝑦 + 1),
2
2
𝑥 = , 𝑦 = 3 𝑎𝑡 𝑡 = 1 using Picard’s method.
,
3
2. Employing the Picard’s method, obtain the second order approximate solution of the
following problem at x = 0.2
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑦𝑧,
𝑑𝑧
𝑑𝑥
= 𝑦 + 𝑧𝑥, 𝑦(0) = 1, 𝑧(0) = −1
[DEC.2012, JUNE2012]
3. Use Picard’s method to find 𝑦(0.1) 𝑎𝑛𝑑 𝑧(0.1) given that
𝑑𝑧
𝑑𝑦
𝑑𝑥
= 𝑥 + 𝑧,
= 𝑥 − 𝑦 2 and (0) = 2, 𝑧(0) = 1 , carry out two approximations.
𝑑𝑥
4. Find the second approximation for the solution of the following system of equations by
applying Picard’s method given that
𝑑𝑥
𝑑𝑡
= (𝑥 + 𝑦)𝑡,
𝑑𝑦
𝑑𝑡
= (𝑥 − 𝑡 )𝑦 and 𝑥 = 0, 𝑦 =
1 𝑎𝑡 𝑡 = 0 carry out two iterations.
5. Obtain the Picard’s third approximation to the solution of the system of equations
2𝑥 − 6𝑦 ,
𝑑𝑦
𝑑𝑡
= 𝑥 + 2𝑦 ;
𝑑𝑥
𝑑𝑡
=
𝑡 = 0, 𝑥 = 0 , 𝑦 = 2 .
hence find x and y at t = 0.1
6. Approximate y and z at x = 0.2 using Picard’s method for the solution of
dz
= x3(y + z) with y(0) =1 , z(0) = ½ Perform two steps (y1, y2, z1, z2)
dx
𝑑𝑦
𝑑𝑥
= z,
[JUNE/JULY 2013]
7. Use Picard’s second approximation to compute y(0.1) and z (0.1) by solving the system of
equations
dy
dz
= x + z,
= x – y2 ; y(0) = 2, z(0) = 1
dx
dx
8. Use fourth order Runge-Kutta method to solve the system of equations
𝑑𝑥
𝑑𝑡
Dept Of ISE
= 𝑦 − 𝑡,
𝑑𝑦
𝑑𝑡
= 𝑥 + 𝑡, 𝑥 = 1, 𝑦 = 1 𝑎𝑡 𝑡 = 0. find x(0.1) and y(0.1).
EWIT
Page 4
Mathematics
10CS41
9. Use the fourth order Runge-Kutta method to obtain the
solution of the following system of equations at x = 0.3
𝑑𝑦
= 1 + 𝑥𝑧,
𝑑𝑥
𝑑𝑧
+ 𝑥𝑦 = 0;
𝑑𝑥
𝑦(0) = 0 ,
[DEC/JAN 2014]
𝑧(0) = 1.
10. Solve the system of equations to find the values of x and y at t=0.2 given that
𝑑𝑦
2𝑥 + 𝑦,
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
𝑎𝑛𝑑 (𝑡0 , 𝑥0 , 𝑦0 ) = (0, 0, 0.5) using Runge-Kutta method of
= 𝑥 − 3𝑦
order 4.
11. When a pendulum swings in a resisting medium, its equation of motion is of the
form
𝑑2𝜃
𝑑𝑡 2
+𝑎
𝑑𝜃
𝑑𝑡
+ 𝑏𝑠𝑖𝑛𝜃 = 0 where a and b are constants. Assuming a = 0.2, b = 10
find the second approximation to the solution of the equation with the initial conditions
𝜃 = 0.3 radian and
𝑑𝜃
𝑑𝑡
= 0 𝑤ℎ𝑒𝑛 𝑡 = 0 by Picard’s method. Also find
𝜃 𝑎𝑡 𝑡 = 0.01(0.01)0.03
12. Given
𝑑2 𝑦
𝑑𝑥 2
− 𝑥2
𝑑𝑦
𝑑𝑥
− 2𝑥𝑦 = 1, 𝑦(0) = 1, 𝑦 ′ (0) = 0.evaluate y(0.1)
using Runge-Kutta method of order 4.
13. Use Picard’s method to obtain the third approximation to the solution of
𝑑2𝑦
𝑑𝑥 2
+ 3𝑥
𝑑𝑦
𝑑𝑥
− 6𝑦 = 0 given that when x = 0 , y = 1,
14. Solve 𝑦 ′′ = 1 + 2𝑦𝑦 ′ ;
𝑑𝑦
𝑑𝑥
= 0.1.
𝑦(0) = 1 = 𝑦 ′ (0) by applying Picard’s method.
Find y(0.2).
15. Apply Picard’s method to find y(1.1) from the second approximation to the solution
of the equation 𝑦 ′′ + 𝑦 2 𝑦 ′ = 𝑥 3 given that y and 𝑦′ have the value 1 at
x = 1.
Dept Of ISE
[DEC/JAN 2014]
EWIT
Page 5
Mathematics
10CS41
16. Use Picard’s method to obtain the third approximation to the solution of
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
+ 2𝑥
𝑑𝑥
+ 𝑦 = 0 given that when x = 0 , y = 0.5,
𝑑𝑦
𝑑𝑥
= 0.1. also tabulate the value
of y(0.1) from all the three approximations.
17. Given
𝑑2 𝑦
𝑑𝑥 2
− 𝑥2
𝑑𝑦
𝑑𝑥
− 2𝑥𝑦 = 1, 𝑦(0) = 1, 𝑦 ′ (0) = 0. Evaluate y(0.1) using Runge-
Kutta method of order 4.
18. Given 𝑦 ′′ − 𝑥𝑦 ′ − 𝑦 = 0 with the initial conditions y(0)=1 , 𝑦 ′ (0) = 0, compute
y(0.2) and 𝑦 ′ (0.2) using fourth order Runge-Kutta method.
19. By Runge-Kutta method solve
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦 2
= 𝑥 ( ) − 𝑦 2 for x = 0. 2 correct to four
𝑑𝑥
decimal places, using the initial conditions 𝑦 = 1 𝑎𝑛𝑑 𝑦 ′ = 0 𝑤ℎ𝑒𝑛 𝑥 = 0.
[JUNE/JULY 2013]
20. Using the Runge - Kutte method, solve the following differential equation at x = 0.1
under the given condition:
𝑑2 𝑦
𝑑𝑥 2
= 𝑥 3 (𝑦 +
𝑑𝑦
𝑑𝑥
), y(0) =1, y1 (0) = 0.5
21. Use fourth order Runge kutta method to solve the equation
y = 1 and
[JUNE 2012]
dy
d2 y
=x
+ y given that
2
dx
dx
dy
dy
= 0 when x = 0. Compute y and
at x = 0.2
dx
dx
22. Solve y11 + 4y = xy given that y(0) = 3 and y1 (0) = 0 Compute y (0.1) using
Runge kutta of order 4
23. Apply Milne’s method to solve
𝑑2𝑦
𝑑𝑥 2
=1+
𝑑𝑦
𝑑𝑥
given 𝑦(0) = 1 = 𝑦 ′ (0). find y(0.4)
by generating the initial values from Picard’s method. Apply y(0.4) theoretically.
24. Obtain the solution of the equation 2
𝑑2 𝑦
𝑑𝑥 2
= 4𝑥 +
𝑑𝑦
𝑑𝑥
by computing the value of the
dependent variable corresponding to the value 1.4 of the independent variable by
Dept Of ISE
EWIT
Page 6
Mathematics
10CS41
applying Milne’s method using the following data
X
1
1.1
1.2
1.3
𝑦
2
2.2156
2.4649
2.7514
𝑦′
2
2.3178
2.6725
3.0657
25. Apply Milne’s method to find y(0.8) given that
𝑑2𝑦
𝑑𝑥 2
= 1 − 2𝑦
table of initial values.
X
𝑑𝑦
𝑑𝑥
and the following
[DEC/JAN 2014]
0
0.2
0.4
0.6
𝑦
0
0.02
0.0795
0.1762
𝑦′
0
0.1996
0.3937
0.5689
Apply the corrector formula twice in presenting the value of y at x = 0.8
26. Apply Milne’s method to find y(0.4) given the equation 𝑦 ′′ + 𝑦 ′ = 2 𝑒 𝑥 and the
following table of initial values
X
0
0.1
0.2
0.3
Y
2
2.01
2.04
2.09
0
0.2
0.4
0.6
𝑦′
27. Apply Milne’s method to find y(0.4) given the equation 𝑦 ′′ = 1 + 𝑦′ and the
following table of initial values
X
0
0.1
0.2
0.3
Y
1
1.1103
1.2427
1.399
1
1.2103
1.4427
1.699
𝑦′
Dept Of ISE
EWIT
Page 7
Mathematics
10CS41
28. Apply Milne’s method to find y(0.4) given the equation 𝑦 ′′ + 𝑥𝑦 ′ + 𝑦 = 0 and the
following table of initial values
X
0
0.1
0.2
0.3
Y
1
0.995
0.9801
0.956
0
-0.0995
-0.196
-0.2867
𝑦′
29. Given 𝑦 ′′ + 𝑥𝑦 ′ + 𝑦 = 0 y(0) = 1, 𝑦 ′ (0) = 0 find the values of y(0.1),
𝑦 ′ (0.1),y(0.2), 𝑦 ′ (0.2), y(0.3) and 𝑦 ′ (0.3) by using the Picard’s method with third
order approximation. Thereafter, find the value of y at x = 0.4 by the Milne’s method.
30. Using the Milne’s method, obtain an approximate solution at the point x = 0.4 of the problem
𝑑2𝑦
𝑑𝑥 2
+ 3𝑥
𝑑𝑦
𝑑𝑥
− 6𝑦 = 0 , y(0) = 1, 𝑦 ′ (0) = 0.1 given that y(0.1) = 1.03995,
y(0.2) = 1.138036, y(0.3) = 1.29865, 𝑦 ′ (0.1) = 0.6955, 𝑦 ′ (0.3) = 1.873, 𝑦 ′ (0.2) = 1.258
[DEC.2012, JUNE 2012]
Dept Of ISE
EWIT
Page 8