Langranges Interpolation Method

UNIT V NUMERICAL METHODS

Interpolation for equal and unequal integrals: Lagrange’s methods – Newton's forward and backward difference formulae - Divided difference method. ODE: Taylor series – Euler– Runge-Kulta methods

Prepared by Dr. A.R. VIJAYALAKSHMI

Interpolation for equal intervals Newton’s forward interpolation formula is

u(u 1) 2 u(u 1)(u  2) 3 y  y0  u  y0   y0   y0  .................. 2! 3! where

x  x0 u h Newton’s backward interpolation formula is

u(u 1) 2 u(u 1)(u  2) 3 y  yn  uyn   yn   yn  .................. 2! 3! where

x  xn u h

1.Obtain the interpolation quadratic polynomial for the given data by using Newton forward difference formula X : 0 2 4 6 Y : -3 5 21 45 Sol. The difference table is X

0

y f (X)

y

2 y

3 y

-3 8

2

5

8 16

4

21

8 24

6

45

0

Newton Forward Interpolation formula is

u (u  1) 2 u (u  1)(u  2) 3 y  y 0  uy 0   y0   y 0  .......... .. 2! 3!

x  x0 x0 x u    h 2 2

where

( x / 2 )( x / 2  1 ) y   3  ( x / 2 )( 8 )  (8 )  0 2! y  3  4 x  x ( x  2)

y  x  2x  3 2

2. Using Newton’s Forward Interpolation formula find the polynomial f(x) Satisfying the following data. Hence find f(2). x : 0 5 10 15 f(x) : 14 379 1444 3584 Sol. The difference table is

X

0

y f (X)  y

2 y

3 y

14 365

5

379

700 1065

10

1444

1075 2140

15

3584

375

Newton Forward Interpolation formula is u( u  1) 2 u( u  1)(u  2) 3 y  y0  u y 0   y0   y0  .... 2! 3!

where

x  x0 x  0 x u   h 5 5

x ( x / 5)( x / 5  1) ( x / 5)( x / 5  1)( x / 5  2) y  14  (365)  (700)  (375) 5 2! 3! 1  14  73 x  x ( x  5)(14)  x ( x  5)( x  10). 2

1 ( i . e .) f ( x )   x  13 x  56 x  28 2 3

2

1  f ( 2)  2  13( 2)  56 ( 2)  28  100 2 3

2



3.Construct Newton’s forward interpolation polynomial for the following data x: 4 6 8 10 y: 1 3 8 16 Use it to find the value of y for x = 5. Sol. The difference table is

X

y  f (X )

4

1

y

3 y

2

y

2 6

3

3 5

8

8

3 8

10

16

0

Newton Forward Interpolation formula is u ( u  1) 2 u ( u  1 )( u  2 ) 3 y  y0  u y0   y0   y 0  ... 2! 3! x  x0 x  4 u  h 2 x 4 1  x  4  x  4  y 1 (2)    1  (3 )  0  2 2!  2   2 

where

3  1  ( x  4)  ( x  4)( x  6) 8 8  8 x  32  3( x  10 x  24)  8 2

y (5) 



1 3(5)  22(5)  48  13  1.625 8 8 2

1 3 x  22 x  48 8 2

4.Given sin45  0.7071, sin50  0.7660, sin55  0.8192, sin60  0.8660 . Find sin 52 by Newton’ s formula 0

0

0

0

0

Sol.

To find sin 52 , we use Newton’s forward formula. Let y  sin x 0

The difference table is

x

y= sinx0

y

2 y

0.0589

-0.0057

3 y

45 0.7071 50 0.7660

-0.0007 0.0532

55 0.8192

-0.0064 0.0468

60 0.8660

0

Newton Forward Interpolation formula is u ( u  1) 2 u ( u  1)( u  2 ) 3 y  y 0  u y 0   y0   y 0  .... 2! 3!

where u  x  x0  52  45  7  1.4 h

5

5

y  0.7071  (1.4)(0.0589)  

(1.4)(1.4  1) (0.0057) 2!

(1 . 4 )(1 . 4  1)(1 . 4  2 ) (  0 . 0007 ) 3! y = 0.7880

(i.e ) sin 52  0.7880 0

5. The following data are taken from the steam table Temp 0 c : 140 150 160 170 180 Pressure kg f/cm2 : 3.685 4.854 6.302 8.076 10.22 Find the pressure at temperature t = 1750 Sol. To find the pressure f(t) at temperature t = 1750 , we use Newton’s Backward formula. The difference table is

t 140

y = f(t)

f (t )

 2 f (t )

 3 f (t )

 4 f (t )

3.685 1.169

150

4.854

0.279 1.448

160

6.302

0.047 0.326

1.774 170

8.076

180

10.225

0.049 0.375

2.149

0.002

Newton Backward Interpolation formula is

u(u  1) 2 u(u  1)(u  2) 3 y  yn  uyn   yn   yn  .... 2! 3! where u  x  xn  175  180  0.5 h 10 y  10.225  (0.5)(2.149) 

(  0 . 5 )(  0 . 5  1 ) ( 0 . 375 ) 2!

(0.5)(0.5  1)(0.5  2) (0.5)(0.5 1)(0.5  2)(0.5  3)  (0.049)  (0.002) 3! 4! y = 9.1005

6.From the following data, estimate the number of persons earning weekly Wages between 60 and 70 rupees. Wage Below 40 40 – 60 60 – 80 80 – 100 100 – 120 (in Rs.) No. of person 250 120 100 70 50 (in thousands ) Sol. The difference table is

Wage x Below 40

No. of persons y

y

2 y

3 y

4

 y

250 120

Below 60

370

-20 100

Below 80

470

-10 -30

70 Below 100

540

Below 120

590

10 -20

50

20

Let us calculate the number of persons whose weekly wages below 70. So we will use Newton’s forward formula. Newton Forward Interpolation formula is

u(u  1) u(u  1)(u  2) y  y  u y  y   y  .... 2! 3! 2

0

where

0

3

0

0

x  x 0 70  40 u   1 .5 h 20

(1.5)(1.5  1) y  250  (1.5)(120)  ( 20) 2! (1.5)(1.5  1)(1.5  2) (1.5)(1.5  1)(1.5  2)(1.5  3)  (10)  (20) 3! 4!



y  423 .59  424

Number of person whose weekly wages below 70 = 424

Number of person whose weekly wages below 60 = 370

 Number of persons whose weekly    424  370  54 thousands . wages between Rs.60 and Rs.70 

Interpolation for unequal intervals Lagrange’s interpolation formula y  f ( x) 



( x  x 1 )( x  x 2 )( x  x 3 ).......( x  x n ) y0 ( x 0  x 1 )( x 0  x 2 )( x 0  x 3 ).......( x 0  x n ) ( x  x 0 )( x  x 2 )( x  x 3 ).......( x  x n ) y1 ( x 1  x 0 )( x 1  x 2 )( x 1  x 3 ).......( x 1  x n )

( x  x 0 )( x  x1 )( x  x 3 ).......( x  x n )  y2 ( x 2  x 0 )( x 2  x1 )( x 2  x 3 ).......( x 2  x n ) +……………………………….+

( x  x0 )( x  x1 )( x  x 2 )( x  x3 ).......(x  xn1 )  yn ( xn  x0 )( xn  x1 )( xn  x 2 )( x n  x3 ).......(x n  xn1 )

7.Find the quadratic polynomial that fits y(x) = x4 at x = 0,1,2 Sol.

The following data is

x :

0

1

2

y=x4 :

0

1

16

By Lagrange’s formula

y  f ( x) 

y  f ( x) 

( x  x1 )( x  x2 ) ( x  x0 )( x  x2 ) ( x  x0 )( x  x1 ) y0  y1  y2 ( x0  x1 )( x0  x2 ) ( x1  x0 )( x1  x2 ) ( x2  x0 )( x2  x1 )

( x 1)(x  2) ( x  0)(x  2) ( x  0)(x 1) (0)  (1)  (16) (0 1)(0  2) (1 0)(1 2) (2  0)(2 1)

y   x ( x  2)  8 x ( x  1)

y( x)  7 x  6 x 2

8.Using Lagrange’s interpolation formula calculate the profit in the year 2000 from the following data

Year : 1997 Profit in lakhs   : 43 of Rs. 

1999

2001

2002

65

159

248

Sol. Lagrange’s interpolation formula is

y  f ( x) 

( x  x1 )( x  x2 )( x  x3 ) ( x  x0 )( x  x2 )( x  x3 ) y0  y1 ( x0  x1 )( x0  x2 )( x0  x3 ) ( x1  x0 )( x1  x2 )( x1  x3 ) 

( x  x0 )( x  x1 )( x  x3 ) ( x  x0 )( x  x1 )( x  x2 ) y2  y3 ( x2  x0 )( x2  x1 )( x 2  x3 ) ( x3  x0 )( x3  x1 )( x3  x2 )

Here x = 2000

(2000  1999)(2000  2001)(2000  2002)  y  f ( x)  43 (1997  1999)(1997  2001)(1997  2002)

( 2000  1997 )( 2000  2001 )( 2000  2002 )  65 ( 2000  1997 )( 2000  2001 )( 2000  2002 ) (2000  1997)(2000  1999)(2000  2002)  159 (2001  1997)(2001  1999)(2001  2002) 

y  f ( x) 

( 2000 ( 2002

 1997 )( 2000  1997 )( 2002

 1999 )( 2000  1999 )( 2002

(1)(1)(2) (3)(1)(2) 43  65 (2)(4)(5) (2)(2)(3) 

( 3 )( 1)(  2 ) ( 3 )( 1)(  1) 159  248 ( 4 )( 2 )(  1) ( 5 )( 3 )( 1 )

y = – 2.15 + 32.5 + 119.25 – 49.6 , y = 100.

 2001 ) 248  2001 )

9. Using Lagrange’s interpolation formula fit a polynomial to following data x : –1 0 2 3 y : –8 3 1 12 and hence find y at x = 1.5 Sol. Lagrange’s interpolation formula is

y  f ( x) 



( x  x1 )(x  x2 )(x  x3 ) ( x  x0 )(x  x2 )(x  x3 ) y0  y1 ( x0  x1)(x0  x2 )(x0  x3 ) ( x1  x0 )(x1  x2 )(x1  x3 )

( x  x0 )(x  x1 )(x  x3 ) ( x  x0 )(x  x1 )(x  x2 ) y2  y3 ( x2  x0 )(x2  x1 )(x2  x3 ) ( x3  x0 )(x3  x1 )(x3  x2 )

y  f ( x) 

( x  0)( x  2)( x  3) ( x  1)( x  2)( x  3) (8)  (3) (1  0)(1  2)(1  3) (0  1)(0  2)(0  3) 

( x  1)( x  0)( x  3) ( x  1)( x  0)( x  2) (1)  (12) (2  1)(2  0)(2  3) (3  1)(3  0)(3  2)

2 1 1 2 2 y  x( x  5x  6)  ( x 1)(x  5x  6)  x( x2  2x  3)  x( x2  x  2) 3 2 6

1 y  [(4 x  20 x  24 x )  (3 x  12 x  3 x  18) 6 3

2

3

2

 ( x  2x  3x)  (6x  6x 12x)] 3

2

3

2

1 y  (12 x  36 x  18 x  18 ) , 6 3

2

y  2 x  6 x  3x  3 3

2

y(1.5) = 2(1.5)3 – 6(1.5)2 + 3(1.5) + 3 = 0.75

10.Find the polynomial f(x) by using Lagrange’s formula and hence find f(3) for x: 0 1 2 5 f(x) : 2 3 12 147 Sol. Lagrange’s interpolation formula is

y  f ( x) 



( x  x1 )( x  x2 )( x  x3 ) ( x  x0 )( x  x2 )( x  x3 ) y0  y1 ( x0  x1 )( x0  x2 )( x0  x3 ) ( x1  x0 )( x1  x2 )( x1  x3 )

( x  x0 )( x  x1 )( x  x3 ) ( x  x0 )( x  x1 )( x  x2 ) y2  y3 ( x2  x0 )( x2  x1 )( x2  x3 ) ( x3  x0 )( x3  x1 )( x3  x2 )

( x  1)( x  2)( x  5) ( x  0)( x  2)( x  5) y  f ( x)  ( 2)  (3) (0  1)(0  2)(0  5) (1  0)(1  2)(1  5)



( x  0 )( x  1)( x  5 ) ( x  0 )( x  1)( x  2 ) (12 )  (147 ) ( 2  0 )( 2  1)( 2  5 ) ( 5  0 )( 5  1)( 5  2 )

y

1 3 49 (x 1)(x2  7x  10)  x(x 2  7x 10)  2x(x 2  6x  5)  x( x 2  3x  2) 5 4 20

11.Given log 654  2 . 8156 , log 10

log10 661 2.8202 . Find The following data is x :

Sol.

log 10 656 654

658

log10 x : 2.8156

2.8182

Let y  log x 10

10

658  2 . 8182 , log10 659  2.8189

by using Lagrange’s formula 659

661

2.8189

2.8202

Here x = 656

Lagrange’s interpolation formula is

( x  x1)(x  x2 )(x  x3 ) ( x  x0 )(x  x2 )(x  x3 ) y  f ( x)  y0  y1 ( x0  x1)(x0  x2 )(x0  x3 ) ( x1  x0 )(x1  x2 )(x1  x3 ) 

( x  x0 )( x  x1 )( x  x3 ) ( x  x0 )( x  x1 )( x  x2 ) y2  y3 ( x2  x0 )( x2  x1 )( x2  x3 ) ( x3  x0 )( x3  x1 )( x3  x2 )

( 656 ( 654

y  .



 658 , )( 656  658 )( 654

 659 )( 656 ,  659 )( 654

 661 ) 2, . 8156  661 )

( 656  654 )( 656  659 )( 656  661 ) 2 . 8182 ( 658  654 )( 658  659 )( 658  661 )

( 656  654 )( 656  658 )( 656  661 )  2 . 8189 ( 659  654 )( 659  658 )( 659  661 ) ( 656  654 )( 656  658 )( 656  659 )  2 . 8202 ( 661  654 )( 661  658 )( 661  659 )

y

(  2 )(  3)(  5 ) ( 2 )(  3)(  5 ) 2 .8156  2 . 8182 (  4 )(  5 )(  7 ) ( 4 )(  1)(  3) ( 2)( 2)( 5) ( 2)( 2)( 3)  2.8189  2.8202 (5)(1)( 2) (7 )(3)( 2)

y = 0.6033 + 7.0455 – 5.6378 + 0.8058 , y = 2.8168 (i.e.)

log

10

656

 2.8168

Lagrange’s formula for inverse interpolation.

( y  y1 )( y  y 2 )( y  y 3 ).......( y  y n ) x  f ( y)  x0 ( y 0  y1 )( y 0  y 2 )( y 0  y 3 ).......( y 0  y n ) ( y  y )( y  y )( y  y ).......( y  y )  x  .... ( y  y )( y  y )( y  y ).......( y  y ) 0

2

3

n

1

1

0

1

2

1

3

1

n

Newton divided difference formula 3  ( x  x )( x  x )( x  x )  y0 y  y0  ( x  x0 )y0  ( x  x0 )( x  x1 ) y0 0 1 2 2

+………..

Divided differences are symmetrical in their arguments. f ( x1 )  f ( x0 ) f ( x0 )  f ( x1 ) f ( x0 , x1 )    f ( x1 , x0 ). x1  x0 x0  x1

1.Find the second divided differences with arguments a,b,c if f(x) = 1/x. Sol. The divided difference table is y = 1/x x a

1/a

b

1/b

c

1/c

y

2 y

–1/ab –1/bc

1/abc

2.If f(x) = 1/x2, find f(a,b) and f(a,b,c) by using divided differences. Sol. The divided difference table is

x

y = 1/x2

a

1/a2

y

2 y

– (a+b)/a2b2 b

1/b2 (ab + bc + ca) / a2b2c2 – (b+c)/b2c2

c

1/c2

12.Using Newton divided difference formula find u(3) given u(1) = –26, u(2) = 12, u(4) = 256, u(6) = 844. Sol. The divided difference table is

x

y = u(x)

1

–26

y

3 y

2 y

38 2

12

28 122

4

256

43 294

6

844

3

Newton divided difference formula is

y  y0  ( x  x0 )y0  ( x  x0 )( x  x1 )2 y0 (x  x0)(x  x1)(x  x2)3 y0 . Here x = 3

y = –26 + (3 –1).(38) + (3 –1)(3 – 2)(28) + (3 –1)(3 – 2)(3 – 4)(3) = 132 – 32 = 100 (i.e.) u(3) = 100.

13.Given u   4 , u   2 , u  220 , u  546 , u  1148 0

Find u & u 2

1

4

5

6

3

Sol. The divided difference table is x 0

y u

y

2 y

3 y

x

4 y

–4 2

1

–2

18 74 9

4

220

63

1

326 15 5

546

138 602

6

1148

Newton divided difference formula is

y  y0  ( x  x0 )y0  ( x  x0 )( x  x1 )2 y0  (x  x0 )(x  x1 )(x  x2 )3 y0  (x  x0 )(x  x1 )(x  x2 )(x  x3 )4 y0  ................ y   4  ( x  0 )( 2 )  ( x  0 )( x  1)(18 )  ( x  0)( x 1)( x  4)(9)  ( x  0)( x 1)( x  4)( x  5)(1) u   4  ( 2 )( 2 )  ( 2 )( 1 )( 18 )  ( 2 )( 1 )(  2 )( 9 )  ( 2 )( 1 )(  2 )(  3 )( 1 ) 2

= – 4 + 4 + 36 – 36 + 12= 12. = – 4 +(3)(2) + (3)(2)(18) + (3)(2)( –1)(9) + (3)(2)( –1)( –2)(1) = – 4 + 6 + 108 – 54 + 12= 68.

u  – 4  (3)(2)  (3)(2)(18)  (3)(2)( – 1)(9)  (3)(2)( – 1)( – 2)(1) 3

 – 4  6  108 – 54  12  68.

14.From the following table, find the value of tan 45 015’ by Newton’s Forward Interpolation formula. x o : 45 46 47 48 49 50 tan x o : 1 1.03553 1.07237 1.11061 1.15037 1.19175 Sol.

y xo

y = tan xo

45

1

2 y

3 y

4 y

5 y

0.03553 46

1.03553

0.00131 0.03684

47

1.07237

0.00009 0.00140

0.03824 48

1.11061

0.00012 0.00152

0.03976 49

1.15037 1.19175

-0.00005 -0.00002

0.00010 0.00162

0.04138 50

0.00003

Newton Forward Interpolation formula is

y  y 0  uy 0 

where

u ( u  1) 2 u ( u  1)( u  2 ) 3  y0   y 0  .......... .. 2! 3!

x  x0 4515  45 u   0.25  h 1

( 0 . 25 )( 0 . 25  1) y  1  ( 0 . 25 )( 0 .03553 )  ( 0 . 00131 ) 2! ( 0 .25 )( 0 .25  1)( 0 .25  2 )  ( 0.00009 ) 3! 

(0.25)(0.25  1)(0.25  2)(0.25  3) (0.00003) 4! (0.25)(0.25  1)(0.25  2)(0.25  3)(0.25  4)  (0.00005) 5!

y  1  0.00888  0.00012  0.0000049 ............... (i.e.) tan 45 015’ = 1.00876

Numerical Solution of Ordinary Differential Equations Taylor series

( x  x0 ) 2 ( x  x0 ) 3 y ( x )  y 0  ( x  x 0 ) y 0  y 0  y 0 2! 3! ( x  x0 ) 4 iv  y  ....... 4!

dy 1 • Using Taylor’ s series find y at x  0.1 if  x y  1, dx y(0)  1. 2

Sol. Given

y   x 2 y  1, x 0  0 , y 0  1

y   x 2 y  1, y   x 2 y   y . 2 x y  x 2 y  y.2x  2x. y  2 y.1

2  y0  x0 y0 1  0 1  1

2

 y 0  x 0 y 0  2 x 0 y 0  0  0  0 2  y0  x0 y0  4x0 y0  2 y0  0  0  2  2

yiv  x2 y  y.2x  4x.y  4y.1 2y  y0 iv  x0 2 y0  2x0 y0  4x0 y0  6 y0  y0

iv

 0  0  0  ( 6)  6

Taylor’ s series about x  x is given by 0

(x  x ) (x  x ) (x  x )       y( x)  y  ( x  x ) y  y  y  y  ....... 2! 3! 4! 2

3

0

0

0

0

4

0

0

iv

0

0

0

( x  0) ( x  0) ( x  0) y ( x )  1  ( x  0 )(  1 )  (0)  (2)  (  6 )  ....... 2! 3! 4! 2

3

4

x3 x4 1 x    .......... ... 3 4

x3 x4 y( x)  1  x  0  (2)  (  6 )  .......... ... 6 24 ( 0 .1) 3 ( 0 .1) 4 y ( 0 .1)  1  ( 0 .1)    .......... ... 3 4

dy 2.Use Taylor series solution to solve numericall y  xy dx , y(1)  1. Tabulate y for x  1.1, 1.2 Sol. Given

1 3

x 0  1, y 0  1

1 3

1 3

y   xy  y 0  x 0 y 0  1(1)  1 2 3

1 3

1 1 4 1 1 2   1  y  x y . y  y  y0  x0 y0 3 . y0  y0 3 3 3 3 3 

1 2 / 3 1 2 / 3 1 2 / 3   2  5 / 3         y  xy   y y  xy y  y y .1  y y  3 3 3  9   y 0

2 4 1 1 8     9 9 3 3 9

Taylor’ s series about x  x is given by 0

(x  x ) (x  x ) (x  x )       y( x)  y  ( x  x ) y  y  y  y  ....... 2! 3! 4! 2

3

0

0

0

0

0

0

3 4 ( x  1 )     3!  3

(1 .1  1) 2 y (1 .1)  1  (1 .1  1)(1)  2!

iv

0

0

( x  1) 2 y ( x )  1  ( x  1)(1)  2!

4

0

8    ..... 9

3  4  (1 .1  1)   3! 3

8    ..... 9

2 ( 0 .1 ) 4 ( 0 .1 )  1  0.1    ........ 3 27 2

3

= 1 + 0.1 + 0.0067 + 0.00014 = 1.1068

(1 .2  1) 2  4  (1 . 2  1) 3 y (1 . 2 )  1  (1 .2  1)(1)    2! 3!  3

2 ( 0 .2 ) 2 4 ( 0 .2 ) 3  1  0.2    ........ 3 27

8    ..... 9

= 1 + 0.2 + 0.0267 + 0.0012 = 1.2279

d y dy 3.Find the value of y(1.1) and y(1.2) from  y  x , dx dx y ( 1 )  1 , y  ( 1 )  1 by using Taylor’ s series method. 2

2

2

Sol. Given

y  y y  x  (1) 2

3

Put y  z  (2) , then y  z  (3)

Sub (2) and (3) in (1), we get

z  y2 z  x3 (ie ).z   x  y z  (4) 3

2

The initial conditions are y(1)  1, y  (1)  1

( i .e ) y ( 1 )  1 , z ( 1 )  1 (sin ce y   z ) ( i .e ) x  1 , y  1 , z  1 0

0

0

Now to solve (1), it is enough if we solve the two first order differential equations (2) and (4).

3

y  z  y 0  z 0  1 y   z   y 0   z 0  0

y   z   y0  z 0  1

y iv  z  iv  y 0  z 0  3

z  x3  y 2 z 3

2

 z 0  x 0  y 0 z 0  1  1  0 z   3 x 2  y 2 z   z .2 y . y   z 0   3 (1 )  0  2 (1 )( 1 )( 1 )  1

z  6x  y 2 z  z.2 y. y  2[ y z. y  y y.z  yz.y]  z0  6(1)  (1)(1)  0  2[0  0  1]  6 1  2  3

Taylor’ s series about x  x is given by 0

(x  x ) (x  x ) (x  x )       y( x)  y  ( x  x ) y  y  y  y  ....... 2! 3! 4! 2

3

0

0

0

0

4

0

0

iv

0

0

0

( x  1) ( x  1) ( x  1) y ( x )  1  ( x  1 )( 1 )  (0)  (1 )  ( 3 )  .......... . 2! 3! 4! 2

3

4

(1.1  1) 3 (1.1  1) 4 y(1.1)  1  (1.1  1)(1)   (3)  ..... 6 24 (0.1)3 (0.1)4  1  0.1    ........ 6 8 = 1 + 0.1 + 0.00017 + 0.0000125 = 1.1002

(1.2  1) 3 (1.2  1) 4 y (1.2)  1  (1.2  1)(1)   (3)  ..... 6 24

(0.2)3 (0.2)4 1 0.2  ........ 6 8 = 1 + 0.2 + 0.0013+ 0.0002 = 1.2015

Euler’s method To solve

use

y' 

dy  f ( x , y ) with y ( x 0 )  y 0 dx

yn1  yn  h f ( xn , yn ) , n  0,1,2,... It’s order is

h2

dy y  x 1.Use Euler’ s method to approximate y when x  0.1 given that  dx y  x with y  1 for x  0. Sol.

We break up the interval 0.1 into five subintervals, we get the answer in more accurate form. So take h = 0.02

yx Given f ( x , y )  , x 0  0, y0  1 and h  0.02 yx x1  x 0  h = 0 + 0.02 = 0.02

y1  y0  hf ( x0 , y0 )  y 0  x0  y1  y 0  h   y  x 0   0 1 0 = 1.02 1 (0.02)  1 0

y 2  y1  hf ( x1 , y1 )  y  x1  y 2  y1  h  1  y  x  1 1 

 1 . 02  0 . 02   1.02  (0.02)   1 . 02  0 . 02  = 1.0392

(i.e.) y(0.02) = 1.02

(i.e.) y(0.04) = 1.0392

x2  x1  h = 0.02 + 0.02

= 0.04

y3  y2  hf(x2 , y2 )  y 2  x2  y3  y 2  h   y  x  2 2  1.0392 0.04 = 1.0392+ (0.02) 1.0392 0.04

(i.e.) y(0.06) = 1.0577

= 1.0577

x3  x2  h

= 0.04+ 0.02= 0.06

y 4  y3  hf ( x3 , y3 )  y3  x3  1.0577 0.06 y4  y3  h  =1.0577+ (0.02)1.0577 0.06  y3  x3 

= 1.0756

(i.e.) y(0.08) = 1.0756

x 4  x3  h

= 0.06 + 0.02= 0.08

y5  y4  hf (x4 , y4 )  y4  x4  y5  y 4  h  y  x 4   4 1 .0756  0 .08  = 1.0756 + (0.02)  1 .0756  0 .08 

= 1.0928

(i.e.) y(0.1) = 1.0928

•Solve

dy  1 y dx

with the initial condition x = 0, y = 0.Using modified Euler’s

.Method tabulate the solutions at x = 0.1, 0.2, 0.3, 0.4. Sol.

Given

Also given

f ( x, y)  1  y

x0 0, y0 0

and h = 0.1

h y1  y0   f ( x0 , y0 )  f [ x0  h, y0  hf ( x0 , y0 )] 2

f ( x0 , y0 )  1  y0 =1–0=1

y1  0 

0 .1 1  f [0  0.1,0  ( 0.1)(1)] 2

 0  0

0 .1 1  f [ 0 . 1, 0 . 1 ] 2

0.1 1  (1  0.1) = 0.095 2

.

(i.e.) y(0.1) = 0.095

x1  x0  h

= 0 + 0.1= 0.1

h y 2  y 1   f ( x 1 , y 1 )  f [ x 1  h , y 1  hf ( x 1 , y 1 )]  2 f (x , y )  1  y 1

1

1

= 1 – 0.095 = 0.905

0 .1 y 2  0 . 095  0 .905  f [ 0 .1  0 .1,0 .095  ( 0 .1)( 0 .905 )] 2 y2

0 .1  0 . 095  0 . 905  f [ 0 . 2 , 0 . 1855 ] 2

 0 . 095 

0 .1 0 .905  (1  0 .1855 ) 2

 0.095 

0.1 0.905  0.8145 2

= 0.18098

(i.e.) y(0.2) = 0.18098

x 2  x1  h y3  y2 

= 0.1+ 0.1= 0.2

h  f ( x 2 , y 2 )  f [ x 2  h , y 2  hf ( x 2 , y 2 )]  2

f ( x2 , y 2 )  1  y 2 = 1 – 0.18098= 0.81902 y 3  0 . 18098  y 3  0.18098 

 0.18098 

0 .1 0 .81902  f [ 0 .2  0 .1,0 .18098  ( 0 .1)( 0 .81902 )] 2 0. 1 0.81902  f [0.3,0.2629] 2

0 .1 0.81902  (1  0.2629) 2

 0 . 18098 

0 .1 0 . 81902  0 .7371  2

= 0.2588 (i.e.) y(0.3) = 0.2588

x3  x 2  h

= 0.2+ 0.1= 0.3

y4  y3 

h  f ( x3 , y3 )  f [ x3  h, y3  hf ( x3 , y3 )] 2

f ( x3 , y 3 )  1  y3 y 4  0 . 2588  y 4  0 .2588 

= 1 – 0.2588 = 0.7412

0 .1 0 .7412  f [ 0 .3  0 . 1,0 .2588  ( 0 .1)( 0 .2588 )] 2

0.1 0.7412  f [ 0.4,0 .3329 ] y 4  0.2588  0.1 0.7412  (1  0.3329) 2 2

y 4  0.2588 

0 .1 0.7412  0.6671 2

= 0.3292 (i.e.) y(0.4) = 0.3292

•Using R-K method of fourth order, solve

y  3 x 

1 y 2

with y(0) = 1 at x = 0.2 taking h = 0.1

Sol. Given

1 f ( x, y)  3x  y 2

Also given

x0  0, y0  1

Take h = 0.1 To find y(0.1)

k1  hf ( x0 , y0 ) y0    (0.1)3x0   2  1   (0.1)  3 ( 0 )    0 . 05 . 2 

h k1   k2  hf  x0  , y0   2 2  0.05   0.1  (0.1) f  0  , 1  2 2   1 . 025     (0.1) f ( 0 . 05 , 1 . 025 )  0.1  3 ( 0 . 05 )  2   = 0.0663

k  h  k 3  hf  x 0  , y 0  2  2 2 

0.0663  0.1  (0.1)f  0  , 1  2 2  

 (0.1) f (0.05,1.0332) 1.0332   0.13(0.05)   2   = 0.0667

= 0.0666

1  [0.05  2(0.0663)  2(0.0667)  0.0833] 6

y1  y0  y (i.e.) y(0.1) = 1.0666

k 4  hf  x 0  h , y 0  k 3   (0.1) f (0  0.1,1 0.0667) 1.0667    (0.1) 3(0.1)   2  

1 y  [k1  2k2  2k3  k4 ] 6

= 0.0833