4. 26 Numerical Methods
4 0 0 3 0 3, ,k h f x h y k z
(0.1) 0.1,0.9951, 0.0995f
(0.1)( 0.0995) 0.00995
4 0 0 3 0 3, ,h g x h y k z
(0.1) 0.1,0.9951, 0.0995g
(0.1) (0.1)( 0.0995) 0.9951 0.0985
1 2 3 41
2 26
y k k k k
10 2( 0.005) 2( 0.0049) ( 0.00995) 0.00496
6
The solution is 0 0( )y x h y y
(0.1) 1 0.00496 0.995y
Do Yourself:
Solve dy
xydx
given that (0.1) 2y , compute (1.2)y and
(1.4)y by using R-K method of fourth order. (A/M 2018)
Predictor and Corrector Methods
1, 3 2 14
2 23n p n n n nh
y y y y y
Corrector Formula:
1, 1 1 143n c n n n nh
y y y y y
4. 27 Initial Value Problems for ODE
Problem 4.14:
Given 2dyx y
dx, (0) 0, (0.2) 0.02, (0.4) 0.0795y y y
and (0.6) 0.1762y . Compute (0.8)y
(N/D 2016)
Solution:
Given 2dyx y
dx, (0) 0, (0.2) 0.02, (0.4) 0.0795y y y
and (0.6) 0.1762y .
2y x y and 0.2h .
0 0x 0 0y
1 0.2x 1 0.02y
2 0.4x 2 0.0795y
3 0.6x 3 0.1762y
4 0.8x 4 ?y Predictor Formula:
4, 0 1 2 34
2 23ph
y y y y y
2 21 1 1 0.2 (0.02) 0.1996y x y
2 22 2 2 0.4 (0.0795) 0.3937y x y
2 23 3 3 0.6 (0.1762) 0.569y x y
Substituting in (1), we get
4,4(0.2)
0 2(0.1996) (0.3937) 2(0.569)3py
4, 0 0.3049 0.3049py
4. 28 Numerical Methods
Corrector Formula:
4, 2 2 3 443ch
y y y y y
2 24 4 4 0.8 (0.3049) 0.707y x y
Substituting in (2), we get
4,0.2
0.0795 0.3937 4(0.569) 0.7073cy
4, 0.0795 0.2251 0.3046cy
(0.8) 0.3046y
Problem 4.15:
corrector formula to find (0.4)y , given 2 2(1 )
2
dy x y
dx, (0) 1, (0.1) 1.06, (0.2) 1.12y y y and
(0.3) 1.21y . (A/M 2010)
Or
Given that 2 21(1 ) ;
2
dyx y
dx (0) 1; (0.1) 1.06;y y
(0.2) 1.12y and (0.3) 1.21y . Evaluate (0.4)y
predictor corrector method. (N/D 2011),(N/D 2018)
Solution:
Given 2 21(1 )
2
dyx y
dx, (0) 1, (0.1) 1.06, (0.2) 1.12y y y
and (0.3) 1.21y .
2 21(1 )
2y x y and 0.1h .
4. 29 Initial Value Problems for ODE
0 0x 0 2y
1 0.1x 1 2.073y
2 0.2x 2 2.452y
3 0.3x 3 3.023y
4 0.4x 4 ?y
4, 0 1 2 34
2 23ph
y y y y y
2 2 2 21 1 1
1 11 1 (0.1) (1.06) 0.5674
2 2y x y
2 2 2 22 2 2
1 11 1 (0.2) (1.12) 0.6523
2 2y x y
2 2 2 23 3 3
1 11 1 (0.3) (1.21) 0.7979
2 2y x y
Substituting in (1), we get
4,4(0.1)
1 2(0.5674) (0.6523) 2(0.7979)3py
4, 1 0.2771 1.2771py
Corrector Formula:
4, 2 2 3 443ch
y y y y y
2 2 2 24 4 4
1 11 1 (0.4) (1.2771) 0.9460
2 2y x y
Substituting in (2), we get
4,0.1
1.12 0.6523 4(0.7979) 0.94603cy
4. 30 Numerical Methods
4, 1.12 0.1597 1.2797cy
(0.4) 1.2797y
Problem 4.16:
Given 25 2 0xy y , (4) 1,y (4.1) 1.0049,y
(4.2) 1.0097y and (4.3) 1.0143y . Compute (4.4)y using
(N/D 2014),(A/M 2015)
Solution:
Given 25 2 0xy y , (4) 1,y (4.1) 1.0049,y
(4.2) 1.0097y and (4.3) 1.0143y .
22
5
yy
x and 0.1h .
0 4x 0 1y
1 4.1x 1 1.0049y
2 4.2x 2 1.0097y
3 4.3x 3 1.0143y
4 4.4x 4 ?y
4, 0 1 2 34
2 23ph
y y y y y
2 21
11
2 2 (1.0049)0.0483
5 5(4.1)
yy
x
2 22
22
2 2 (1.0097)0.0467
5 5(4.2)
yy
x
2 23
33
2 2 (1.0143)0.0452
5 5(4.3)
yy
x
4. 31 Initial Value Problems for ODE
Substituting in (1), we get
4,4(0.1)
1 2(0.0483) (0.0467) 2(0.0452)3py
4, 1 0.0187 1.0187py
Corrector Formula:
4, 2 2 3 443ch
y y y y y
2 24
44
2 2 (1.0187)0.0437
5 5(4.4)
yy
x
Substituting in (2), we get
4,0.1
1.0097 0.0467 4(0.0452) 0.04373cy
4, 1.0097 0.009 1.0187cy
(4.4) 1.0187y
Problem 4.17:
Given 2dyxy y
dx, (0) 1y , (0.1) 1.1169y and
(0.2) 1.2773y , find (i) (0.3)y by R-K method of fourth order
and (ii) (0.4)y
(N/D 2017),(A/M 2018)
Solution:
Given 2dyxy y
dx, (0) 1, (0.1) 1.1169, (0.2) 1.2773y y y .
2y xy y and 0.1h .
4. 32 Numerical Methods
i) R-K Method
(0.3) 1.5023y
ii)
0 0x 0 1y
1 0.1x 1 1.1169y
2 0.2x 2 1.2773y
3 0.3x 3 1.5023y
4 0.4x 4 ?y
4, 0 1 2 34
2 23ph
y y y y y
2 21 1 1 1 (0.1)(1.1169) (1.1169) 1.3587y x y y
2 22 2 2 2 (0.2)(1.2773) (1.2773) 1.8853y x y y
2 23 3 3 3 (0.3)(1.5023) (1.5023) 2.7076y x y y
Substituting in (1), we get
4,4(0.1)
1 2(1.3587) (1.8853) 2(2.7076)3py
4, 1.8330py
Corrector Formula:
4, 2 2 3 443ch
y y y y y
2 24 4 4 4 (0.4)(1.8330) (1.8330) 4.0930y x y y
4. 33 Initial Value Problems for ODE
Substituting in (2), we get
4,0.1
1.2773 1.8853 4(2.7076) 4.09303cy
4, 1.8370cy
(0.4) 1.8370y
Problem 4.18:
Using Runge Kutta method of fourth order, find the value of y
at 0.2, 0.4, 0.6x given 3 , (0) 2dy
x y ydx
. Also find the
value of y at 0.8x
method. (M/J 2014)
Solution:
Given 3dyx y
dx, (0) 2y .
3( , )f x y x y and (0) 2y .
0 00, 2x y and 0.2h .
i) R-K Method First Formula:
1 0 0( , ) 0.2 (0,2)k h f x y f
30.2 (0) 2
0.4
12 0 0,
2 2
khk h f x y
0.2 0.40.2 0 , 2
2 2f
0.2 (0.1, 2.2)f
4. 34 Numerical Methods
30.2 (0.1) 2.2 0.4402
23 0 0,
2 2
khk h f x y
0.2 0.44020.2 0 , 2
2 2f
0.2 (0.1, 2.2201)f
30.2 (0.1) 2.2201 0.4442
4 0 0 3,k h f x h y k
0.2 0 0.2, 2 0.4442f
0.2 (0.2, 2.4442)f
30.2 (0.2) 2.4442 0.4904
1 2 3 41
2 26
y k k k k
10.4 2(0.4402) 2(0.4442) 0.4904
6
12.6592 0.4432
6
1 0(0.2)y y y y 2 0.4432
(0.2) 2.4432y
1 10.2, 2.4432x y
ii) R-K Method Second Formula:
1 1 1( , ) 0.2 (0.2, 2.4432)k h f x y f
4. 35 Initial Value Problems for ODE
30.2 (0.2) 2.4432
0.4902
12 1 1,
2 2
khk h f x y
0.2 0.49020.2 0.2 , 2.4432
2 2f
0.2 (0.3, 2.6883)f
30.2 (0.3) 2.6883 0.5431
23 1 1,
2 2
khk h f x y
0.2 0.54310.2 0.2 , 2.4432
2 2f
0.2 (0.3, 2.7148)f
30.2 (0.3) 2.7148 0.5484
4 1 1 3,k h f x h y k
0.2 0.2 0.2, 2.4432 0.5484f
0.2 (0.4, 2.9916)f
30.2 (0.4) 2.9916 0.6111
1 2 3 41
2 26
y k k k k
10.4902 2(0.5431) 2(0.5484) 0.6111
6
13.2843 0.5474
6
4. 36 Numerical Methods
2 1(0.4)y y y y 2.4432 0.5474
(0.4) 2.9906y
2 20.4, 2.9906x y
iii) R-K Method Third Formula:
1 2 2( , ) 0.2 (0.4, 2.9906)k h f x y f
30.2 (0.4) 2.9906
0.6109
12 2 2,
2 2
khk h f x y
0.2 0.61090.2 0.4 , 2.9906
2 2f
0.2 (0.5, 3.2961)f
30.2 (0.5) 3.2961 0.6842
23 2 2,
2 2
khk h f x y
0.2 0.68420.2 0.4 , 2.9906
2 2f
0.2 (0.5, 3.3327)f
30.2 (0.5) 3.3327 0.6915
4 2 2 3,k h f x h y k
0.2 0.4 0.2, 2.9906 0.6915f
0.2 (0.6, 3.6821)f
4. 37 Initial Value Problems for ODE
30.2 (0.6) 3.6821 0.7796
1 2 3 41
2 26
y k k k k
10.6109 2(0.6842) 2(0.6915) 0.7796
6
14.1419 0.6903
6
3 2(0.6)y y y y 2.9906 0.6903
(0.6) 3.6809y
3 30.6, 3.6809x y
iv)
0 0x 0 2y
1 0.2x 1 2.4432y
2 0.4x 2 2.9906y
3 0.6x 3 3.6809y
4 0.8x 4 ?y
4, 0 1 2 34
2 23ph
y y y y y
3 31 1 1 (0.2) 2.4432 2.4512y x y
3 32 2 2 (0.4) 2.9906 3.0546y x y
3 33 3 3 (0.6) 3.6809 3.8969y x y
Substituting in (1), we get
4. 38 Numerical Methods
4,4(0.2)
2 2(2.4512) (3.0546) 2(3.8969)3py
4, 2 2.5711 4.5711py
Corrector Formula:
4, 2 2 3 443ch
y y y y y 2)
3 34 4 4 (0.8) 4.5711 5.0831y x y
Substituting in (2), we get
4,0.2
2.9906 3.0546 4(3.8969) 5.08313cy
4, 2.9906 1.5817 4.5723cy
(0.8) 4.5723y
Do yourself:
1) Given 1
,yx y
(0) 2,y (0.2) 2.0933,y
(0.4) 2.1755,y (0.6) 2.2493y find (0.8)y using
(M/J 2012)
2) Given that 21dy
ydx
; (0.6) 0.6841,y
(0.4) 0.4228,y
(0.2) 0.2027,y (0) 0y , find ( 0.2)y
method. (N/D 2012)
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