unit1 vrs

Taylor’s and Picard’s methods 2

Dr. V. Ramachandra Murthy

Numerical Methods

Unit-I: Numerical Methods-I

Numerical solution of ordinary differential equations of first order and first degree: Picard’s method, Taylor’s series method, Modified Euler’s method, Runge-Kutta method of fourth order. Milne’s and Adams-Bashforth predictor and corrector methods [ No derivation of formulae] Unit-II: Numerical Methods-II

Numerical solution of simultaneous first order differential equations: Picard’s method, Runge-Kutta method of fourth order. Numerical solution of second order ordinary differential equations: Picard’s method, Runge-kutta method and Milne’s method. Numerical Solution of Ordinary Differential Equations(ODE)

The most general form of an ODE of nth order is given by -------- (1) A general solution of Eqn (1) is of the form ------- (2) If particular values are given to the constants then the resulting solution is called a particular solution.

To obtain a particular solution from the general solution (2), we must be given n conditions so that the constants can be determined. If all the n conditions are specified at the same value of x then the problem is termed as initial value problem. If the conditions are specified at more than one value of x, then the problem is termed as boundary value problem.

0dx

yd....,,.........

dx

yd,

dx

yd,

dx

dyy,x,φ

n

n

3

3

2

2

=

( ) 0c....,,.........c,c,cy,x,ψ n321 =



Though there are many analytical methods for finding the solution of the equation of the form (1), there exist large number of ODE’s whose solution cannot be obtained by the known analytical methods. In such cases, we use numerical methods to get an approximate solution of a given differential equation under the prescribed conditions. Numerical solution of a Differential Equation

Consider the first order differential equation Let be the solution values at the points We wish to find the approximate values to these solution values. Let the initial condition be . Let the exact solution y(x) of the given differential equation be represented by a continuous curve. Divide the interval on which the solution is derived into a finite number of equispaced subintervals.

0x 1x

2x 1-mx

mx

For each , the approximate values of the dependent variable y(x) are calculated using a suitable recursive formula. These values are and these are shown by points. Computation of these approximate values is known as Numerical solution of the Differential equation. Numerical solution of ODE’s of first order and first degree Single step Methods:

• Taylor’s series method

y)f(x,dx

dy=

)x........y( ), y(x),y(x m10 mx,...,1 x,0x

m10 y........, , y,y

00 y)y(x =

[ ]m x,0x

Approximate

solution

Exact solution

ix

m10 y......, , y,y

00 y)y(x,y)f(x,dx

dy==



• Picard’s method • Modified Euler’s method • Runge-Kutta method of fourth order

Taylor’s Series method Let y = f(x) be a solution of the equation Expanding it by Taylor’s series about we get

This may be written as

Putting , we get

Similarly

In general,

Where Problem (1): Solve numerically up to x=1.2 with h=0.1 by Taylor’s

x0 =1 x1 =1.1 x2 =1.2

00 y)y(x,y)f(x,dx

dy==

0xx =

( ) ( ) ( ).....)(xf

3!

xx)(xf

2!

xx)(xf

1!

xx)f(xf(x) 0

///

3

00

//

2

00

/00 +

−+

−+

−+=

( ) ( ) ( ).....y

3!

xxy

2!

xxy

1!

xxyy(x)

///

0

3

0//

0

2

0/

00

0 +−

+−

+−

+=

hxxx 01 +==

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

.....y3!

hy

2!

hy

1!

hy)f(xy

///

1

3//

1

2/

1122 ++++==

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

),........(xfy),(xfy),f(xy n

////

nn

//

nnn ===

0y(1) y,xdx

dy=+=



series method correct to

four decimal places. Soln: Given data: , h=0.1 From the Taylor’s series, we have ---------- (1) Where n=0, 1, 2,……. Put n=0 in Eqn (1) ----------- (2) Substituting all these values in Eqn(2) we get ∴

y0 =0 y1 =? y2 =?

yxy / +=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

yxy / +=/// y1y += ///// y1y +=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

101yxy 00

/

0 =+=+=

211y1y/

0

//

0 =+=+=

2yy//

0

///

0 ==

( ) ( ).....(2)

3!

0.1(2)

2!

0.1(1)

1!

0.10f(1.1)y

32

1 ++++==

0.1103y(1.1)y1 ==



Put n=1 in Eqn(1) ---------(3) Substituting all these values in Eqn(3) we get ∴ _________________________________________________________ Problem (2): Apply Taylor’s series method to find the value of y(1.1) and y(1.2) correct to 4 decimal places given that ; y(1)=1 taking the first four terms of the Taylor’s series expansion. Soln: Given data: , h=0.1 From the Taylor’s series, we have ------------(1)

x0 =1 x1 =1.1 x2 =1.2

y0 =1 y1 =? y2 =?

.....y3!

hy

2!

hy

1!

hy)f(xy

///

1

3//

1

2/

1122 ++++==

1.21030.11031.1yxy 11

/

1 =+=+=

2.21031.21031y1y/

1

//

1 =+=+=

2.2103yy//

1

///

1 ==

( ) ( )...(2.2103)

3!

0.1(2.2103)

2!

0.1(1.2103)

1!

0.10.1103y

32

2 ++++=

0.2427y(1.2)y2 ==

31

/ xyy =

.....///ny

3!

3h//ny

2!

2h/ny

1!

hny)1nf(x1ny ++++=+=+

31

xydx

dy=

31

/ xyy = /3

2

31

// .y.y3

1x.yy

−

+= ( )

−++=

−−−

2/35

//32

/3

2

/// yy3

2yy

3

xyy

3

2y



Put n=0 in Eqn (1) ----------- (2) Substituting all these values in Eqn(2) we get ∴

Put n=1 in Eqn(1) ---------(3) Substituting all these values in Eqn(3) we get ∴

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

.....y3!

hy

2!

hy

1!

hy)f(xy

///

1

3//

1

2/

1122 ++++==

( )( ) 111yxy 31

31

00/0 ===

1.3333(1)3

1(1)(1).y.y

3

1.xyy

3

2

31

/0

3

2

003

1

0//0 =

+=+=

−−

( ) 8888.0yy3

2yy

3

xyy

3

2y

2/0

35

0//0

32

00/

03

2

0///0 =

−+=

−−−

( ) ( ).....(0.8888)

3!

0.1(1.3333)

2!

0.1(1)

1!

0.11y

32

1 ++++=

1.1068y(1.1)y1 ==

( ) ( ) ( ) 4242.11.13781.1068(1.1)3

11.1068.y.yx

3

1yy 3

2-3

1/1

3

2-

113

1

1//1 =+=+=

( ) 1.13781.1068(1.1)yxy 31

31

11/1 ===

( ) 8438.0yy3

2yy

3

xyy

3

2y

2/1

35

1//1

32

11/

13

2

1///1 =

−+=

−−−

( ) ( )(0.8438)

3!

0.1(1.4242)

2!

0.1(1.1378)

1!

0.11.1068y

32

2 +++=

1.2278y(1.2)y2 ==



Problem (3): Use Taylor’s series method to approximate y when x=0.1 correct to 4 decimal places given that and y=1 when x=0 by taking the first five terms of the Taylor’s series expansion. Soln: Given data: , h=0.1 From the Taylor’s series, we have ----------(1) Where n=0, 1, 2,……. Put n=0 in Eqn(1) -----------(2)

121020030 =+=+=′ yxy

Substituting all these values in Eqn(2) we get y(0.1)=1.1272

x0 =0 x1 =0.1

y0 =1 y1 =?

2/ yx3y +=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

2/ yx3y +=//// 2yy3y +=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

2y3xdx

dy+=

( )( )2////// yyy2y +=

( ) ( )///////////////IV y3yyy2y2yyyyy2y +=++=

5(2)(1)(1)3y2y3y /00

//0 =+=+= ( )( ) ( )( ) 121(1)(5)2yyy2y

22/

0

//

00

///

0 =+=+=

( ) ( ) 54(3)(1)(5)(1)(12)2y3yyy2y //

0

/

0

///

00

IV

0 =+=+=

( ) ( ) ( )1.1272(54)

4!

0.1(12)

3!

0.1(5)

2!

0.1(1)

1!

0.11y

432

1 =++++=



Problem (4): Given with the initial condition y=1 when x=0. Compute y(0.2) correct to 4 decimal places by using Taylor’s series method. Soln: Given data: , h=0.2 From the Taylor’s series, we have ------------(1) Put n=0 in Eqn(1) -----------(2) Substituting all these values in Eqn(2) we get ∴

x0 =0 x1 =0.2

y0 =1 y1 =?

xy1y /+=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

xy1dx

dy+=

xy1y /+= yxyy ///

+=////////// 2yxyyyxyy +=++=

////////////IV 3yxy2yyxyy +=++=

11(0)(1)yyxy 0

/

00

//

0 =+=+=

2(2)(1)(0)(1)2yyxy /

0

//

00

///

0 =+=+=

1(0)(1)10y0x1/0

y =+=+=

3(3)(1)(0)(2)3yyxy //

0

///

00

IV

0 =+=+=

( ) ( ) ( )........(3)

4!

40.2(2)

3!

30.2(1)

2!

20.2(1)

1!

0.211y +++++=

1.2228y(0.2)y1 ==



Problem (5): Use Taylor’s series method to find the value of y at x=0.1 and x=0.2 correct to 5 decimal places from , y(0)=1. Soln: Given data: , h=0.1 From the Taylor’s series, we have ----------(1) Where n=0, 1, 2,……. Put n=0 in Eqn(1) -----------(2) Substituting all these values in Eqn(2) we get

x0 =0 x1 =0.1 x2 =0.2

y0 =1 y1 =? y2 =?

1-yxy 2/ =

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

1-yxdx

dy 2=

/2// yx2xyy +=

//2///2///// yx4xy2yyx2xy2xy2yy ++=+++=

///2//////2//////IV yx6xy6yyx2xy4xy4y2yy ++=++++=

1-yxy 2/ =

0(0)(-1)2(0)(1)yxy2xy /0

2000

//0 =+=+=

2(0)(0)4(0)(-1)2(1)yxy4x2yy //

0

2

0

/

000

///

0 =++=++=

11(0)(1)1-yxy 020

/0 −=−==

-6(0)(2)6(0)(0)6(-1)yxy6x6yy ///0

20

//00

/0

IV0 =++=++=

( ) ( ) ( )0.90030...(-6)

4!

40.1(2)

3!

30.1(0)

2!

20.1(-1)

1!

0.111y =+++++=



Put n=1 in Eqn(1) ---------(3) Substituting all these values in Eqn(3) we get ∴ _________________________________________________________ Problem (6): Using Taylor’s series method find y to five decimal places when x=1.02

given that and y=2 when x=1

x0 =1 x1 =1.02

y0 =2 y1 =?

.....y3!

hy

2!

hy

1!

hy)f(xy

///

1

3//

1

2/

1122 ++++==

1.40590=

0.990991(0.90030)1(0.1)-yxy 21

21

/1 −=−=

( )( ) ( ) ( ) 17015.099099.01.09003.01.02yxy2xy2/

12111

//1 =−+=+=

(0.17015)(0.1)99099)4(0.1)(-0.2(0.9003)yxy4x2yy 2//1

21

/111

///1 ++=++=

( ) ( )

( ) (-5.82979)

4!

0.1

(1.40590)3!

0.1(0.17015)

2!

0.1(-0.99099)

1!

0.10.9003y

4

32

2

+

+++=

0.80226y(0.2)y2 ==

-5.82979

(1.40590)(0.1)7015)6(0.1)(0.1)6(-0.99099

yxy6x6yy

2

///

1

2

1

//

11

/

1

IV

1

=

++=

++=

1)dx-(xydy =



Soln: Given data: , h=0.02 From the Taylor’s series, we have ------------(1) Put n=0 in Eqn(1) -----------(2) Substituting all these values in Eqn(2) we get ∴

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

1-xydx

dy=

1-xyy /= yxyy ///

+=

////////// 2yxyyyxyy +=++=

////////////IV 3yxyy2yxyy +=++=

32(1)(1)yyxy 0

/

00

//

0 =+=+=

52(1)(1)(3)2yyxy /

0

//

00

///

0 =+=+=

11(1)(2)1-yxy 00

/

0 =−==

14(3)(3)(1)(5)3yyxy //

0

///

00

IV

0 =+=+=

( ) ( ) ( )....(14)

4!

40.02(5)

3!

30.02(3)

2!

20.02(1)

1!

0.0221y +++++=

02000.2y(1.02)1y ==



Problem (7): Employ Taylor’s method to obtain the approximate value of y at x=0.2 for the differential equation y(0)=0, correct to three places of decimal. Soln: Given data: , h=0.2 From the Taylor’s series, we have ----------(1) Where n=0, 1, 2,……. Put n=0 in Eqn(1) -----------(2)

x0 =0 x1 =0.2

y0 =0 y1 =?

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

x3e2ydx

dy+=

x/ 3e2yy +=

x/ 3e2yy +=x/// 3e2yy +=

x///// 3e2yy +=x///IV 3e2yy +=

33e(2)(0)3e2yy 0x

0/0

0 =+=+=

93e(2)(3)3e2yy 0x/

0

//

00 =+=+=

213e(2)(9)3e2yy 0x//

0

///

00 =+=+=

453e(2)(21)3e2yy 0x///

0

IV

00 =+=+=



Substituting all these values in Eqn(2) we get

∴

_________________________________________________________ Problem (8): Solve for x=1.1 and x=1.2, given y(1)=1 correct to four decimal places by using Taylor’s series method. Soln: Given data: , h=0.1 From the Taylor’s series, we have ----------(1) Where n=0, 1, 2,……. Put n=0 in Eqn(1) -----------(2)

x0 =1 x1 =1.1 x2 =1.2

y0 =1 y1 =? y2 =?

( ) ( ) ( )....(45)

4!

40.2(21)

3!

30.2(9)

2!

20.2(3)

1!

0.201y +++++=

0.811y(0.2)y1 ==

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

3xydx

dy+=

3/ xyy +=

3/ xyy +=2/// 3xyy += 6xyy /////

+= 6yy ///IV+=

53(1)(2)3xyy 22

0

/

0

//

0 =+=+=

116(1)(5)6xyy 0

//

0

///

0 =+=+=

2(1)(1)xyy 33

00

/

0 =+=+=

176116yy ///

0

IV

0 =+=+=



Substituting all these values in Eqn(2) we get Put n=1 in Eqn(1) ---------(3) Substituting all these values in Eqn(3) we get

_________________________________

Problem(9): Solve , y(0)=0 by Taylor’s series method for x=0.2 correct to four decimal places. {Ans: y(0.2)=0.1947} Problem(10): Solve , y(0)=1 by Taylor’s series method for x=0.1 in steps of 0.05 correct to four decimal places. {Ans: y(0.1)=0.9950}

.....///1y

3!

3h//1y

2!

2h/1y

1!

h1y)2f(x2y ++++==

( ) ( ) ( )1.2269....(17)

4!

40.1(11)

3!

30.1(5)

2!

20.1(2)

1!

0.111y =+++++=

2.5579(1.1)(1.2269)xyy 3311

/1 =+=+=

6.18793(1.1)(2.5579)3xyy 221

/1

//1 =+=+=

12.78796(1.1)(6.1879)6xyy 1//1

///1 =+=+=

18.7879612.78796yy ///1

IV1 =+=+=

( ) ( ) ( )

5158.1

...(18.7879)4!

40.1(12.7879)

3!

30.1(6.1879)

2!

20.1(2.5579)

1!

0.12269.12y

=

++++=

2xy-1dx

dy=

0xydx

dy=+



Picard’s method Consider the initial value problem ---------- (1) Integrating Eqn(1) from , we get ---------- (2) Equation (2) is called Integral equation. Such an equation can be solved by successive approximation. The first approximation y1 of y is given by The second approximation is given by Similarly . . . . . . The process of iteration is stopped when the values of and are the same to the desired accuracy. _________________________________________________________ Problem (1): Solve , y (0) =0 by Picard’s method up to third approximation. Soln: Given data: Picard’s iterative formula is given by ------- (1)

00 y)y(x,y)f(x,dx

dy==

xtox0

∫+=x

x0

0

y)dxf(x,yy

∫+=x

x001

0

)dxyf(x,yy

∫+=x

x102

0

)dxyf(x,yy

∫+=x

x203

0

)dxyf(x,yy

∫+=x

x1-n0n

0

)dxyf(x,yy

1-ny ny

2y1dx

dy+=

0y;0x;y1y)f(x, 002 ==+=

∫+=x

x1-n0n

0

)dxyf(x,yy

( ) ( )∫∫∫ +=++=x

0

2

1-n

x

0

x

0

2

1-nn dxy1.dxdxy10y

( )∫+=x

0

21-nn dxyxy



Put n=1 in Eqn(1) Put n=2 in Eqn(1) Put n=3 in Eqn(1) Problem (2): Use Picard’s method to approximate y when x=0.1 & x=0.2 for y(0)=0 by considering third approximation correct to 4 decimal places. Soln: Given data: Picard’s iterative formula is given by --------(1) Step (1): To find y (0.1) Put n=1 in Eqn(1) Put n=2 in Eqn(1)

x0 =0 x1 =0.1 x2 =0.2

y0 =0 y(0.1)=? y(0.2)=?

( ) ( ) xdx0xdxyxyx

0

x

0

201 =+=+= ∫∫

( ) ( )3

xxdxxxdxyxy

3x

0

2x

0

2

12 +=+=+= ∫∫

( )63

x

15

2x

3

xdx

3

xxxdxyxy

753x

0

23x

0

2

23 ++=

++=+= ∫∫

2yxdx

dy+=

2yxy)f(x, +=

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ ++=x

x

2

1-n0n0

dxyxyy

( ) ( ) 0.00502

xdx0x0dxyxyy

0.1

0

20.1

0

x

x

2

0010

=

=++=++= ∫∫



Put n=3 in Eqn(1) Thus 0050.0)1.0( =y .

Step (2): To find y(0.2) Let ---------- (2) Put n=1 in Eqn(2) Put n=2 in Eqn(2) Similarly by putting n=3 in Eqn(2), we obtain Thus

0200.0)2.0( =y _____________________________________________

___________ Problem (3): Use Picard’s method to solve , y(0)=1 for x=0.2 .

x0 =0 x1 =0.2

y0 =1 y(0.2)=?

( ) ( )( ) 0.00502

xdx0.0050x0dxyxyy

0.1

0

20.1

0

2x

x

2

1020

=

=++=++= ∫∫

( ) ( )( ) 0.00502

xdx0050.0x0dxyxyy

0.1

0

20.1

0

2x

x

2203

0

=

=++=++= ∫∫

( ) ( )( ) 02.0dx0.0050x0050.0dxyx0050.0y0.2

0.1

20.2

0.1

2

01 =++=++= ∫∫

1.0x0 = 0.0050y0 =

( )∫ ++=0.2

0.1

2

1-nn dxyx0.0050y

( )( ) ( )( )

( )

0.0200

x0.022

x0.0050

dx0.02x0.0050dxyx0.0050y

0.2

0.1

22

0.2

0.1

20.2

0.1

2

12

=

++=

++=++= ∫∫

( )( ) 0.0200 dx0.02x0.0050y0.2

0.1

2

3 =++= ∫

yxdx

dy 2 −=

yxy)f(x, 2−=



Soln: Given data: Picard’s iterative formula is given by -------(1) Put n=1 in Eqn(1) Put n=2 in Eqn(1) Put n=3 in Eqn(1) Put n=4 in Eqn(1) Similarly Since y4 & y5 are the same up to four places of decimals y(0.2)=0.8355 _________________________________________________________ Problem (4): Solve , y (0) =0 by Picard’s method up to the third approximation. Soln: Given data: Picard’s iterative formula is given by

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ −+=0.2

01-n

2

n dxyx1y

( ) ( ) 0.8026dx1x1dxyx1y0.2

0

20.2

00

2

1 =−+=−+= ∫∫

( ) ( ) 0.8421dx0.8026x1dxyx1y0.2

0

20.2

01

2

2 =−+=−+= ∫∫

( ) ( ) 0.8342dx0.8421x1dxyx1y0.2

0

20.2

02

2

3 =−+=−+= ∫∫

( ) ( ) 0.8358dx0.8342x1dxyx1y0.2

0

20.2

03

2

4 =−+=−+= ∫∫

8355.0y5 = 8355.0y6 =

2xyxdx

dy 2 +=

0y;0x;2xyxy)f(x, 00

2 ==+=

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ −++=x

01n

2

n dx2xyx0y



--------(1) Put n=1 in Eqn(1) Put n=2 in Eqn(1) Put n=3 in Eqn(1) _________________________________________________________ Problem(5): Solve by Picard’s method , y(0)=1 for x=0.1 Correct to four decimal places. Soln: Given data: Picard’s iterative formula is given by ---------(1) Put n=1 in Eqn(1)

x0 =0 x1 =0.1

y0 =1 y(0.1)=?

( )15

x2

3

xdx

3

x2xxdx2xyxy

53x

0

32

x

01

2

2 +=

+=+= ∫∫

( ) ( )3

xdxxdx2xyxy

3x

0

2x

00

21 ∫∫ ==+=

∫ ++=∫ +=

x

0

dx15

5x

23

3x

2x2

xx

0

dx22xy2

x3y

7x105

45x152

3

3xx

0dx6x

1544x

322x3y ++=∫ ++=

xy1dx

dy+=

xy1y)f(x, +=

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ −++=0.1

01nn dxxy11y

( ) ( ) 1.1052

xx1x11dxxy11y

0.1

0

0.1

0

20.1

001 =

++=++=++= ∫∫



Put n=2 in Eqn(1) Put n=3 in Eqn(1) Since y2 & y3 are the same up to four places of decimals y(0.1)=1.1055 Problem (6): Given the differential equation , with the condition y=1 when x=0, use Picard’s method to obtain y for x=0.2 correct to four decimal places. Soln: Given data: Picard’s iterative formula is given by ---------(1) Put n=1 in Eqn(1) Put n=2 in Eqn(1) Similarly,

x0 =0 x1 =0.2

y0 =1 y(0.2)=?

yxdx

dy−=

y-xy)f(x, =

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ −+=0.2

01nn dxy-x1y

( ) ( ) 0.82x2

x11-x1dxy-x1y

0.2

0

0.2

0

20.2

001 =

−+=+=+= ∫∫

( ) ( )

0.8560.82x-2

x1

dx0.82-x1dxy-x1y

0.2

0

2

0.2

0

0.2

012

=

+=

+=+= ∫∫

8500.0y7,n

8500.0 y6,n

8499.0y 5,n

8502.0 y4,n

8488.0 y3,n for

7

6

5

4

3

==

==

==

==

==



Since y6 & y7 are the same up to four places of decimals y(0.2)=0.8500 Problem (7): Given the differential equation , with the condition y=1 when x=0. Use Picard’s method to obtain y for x=0.1 correct to three decimal places. Soln: Given data: Picard’s iterative formula is given by ---------(1) Put n=1 in Eqn(1) Put n=2 in Eqn(1)

x0 =0 x1 =0.1

y0 =1 y(0.1)=?

∫+=x

x1-n0n

0

)dxyf(x,yy

xy

xy

dx

dy

+

−=

xy

x-yy)f(x,

+=

∫

+

−+=

−

−0.1

01n

1nn dx

xy

xy1y

dx1x

211dx

1x

2

1x

1x1dx

1x

21x1

dxx1

1x1dx

x1

x11dx

xy

xy1y

0.1

0

0.1

0

0.1

0

0.1

0

0.1

0

0.1

00

01

∫∫∫

∫∫∫

+−−=

+−

+

+−=

+

−+−=

+

−−+=

+

−+=

+

−+=

( ) ( )[ ] 090.11xlog2x1y0.1

0

0.1

01 =++−=

( )[ ] ( )

1.0917561)2.18(0.0870.11

x1.090xlog2.181dx11.090x

2.181

dx1.090x

1.090)(x-2(1.090)1dx

x1.090

1.0901.090x-1.0901

dxx1.090

x1.0901dx

xy

xy1y

0.1

0

0.1

0

0.1

0

0.1

0

0.1

0

0.1

0

0.1

01

12

=+−=

−++=

−

++=

+

++=

+

+−+=

+

−+=

+

−+=

∫

∫∫

∫∫



Put n=3 in Eqn(1) Since y2 & y3 are the same up to three places of decimals y(0.1)=1.091 _________________________________________________________ Problem(8): Solve , y(0)=1 by Picard’s method up to third approximation and hence find the value of y at x=0.1. Soln: Given data: Picard’s iterative formula is given by ---------(1) Put n=1 in Eqn(1)

( )[ ] ( )

1.091

0.10

x0.10

1.091xlog2.1821

dx0.1

0

11.091x

2.1821

dx0.1

01.091x

1.091)(x-2(1.091)1

dx0.1

0x1.091

1.0911.091x-1.0911

dx0.1

0x1.091

x1.0911

0.1

0

dxx2y

x2y13y

=

−++=

−

++=

+

++=

+

+−+=

+

−+=

+

−+=

∫

∫

∫

∫∫

2xydx

dy−=

2xyy)f(x, −=

∫+=x

x1-n0n

0

)dxyf(x,yy

( )∫ −+= −

x

0

21nn dxxy1y

( ) ( )3

xx1

3

xx1dxx11dxxy1y

3x

0

3x

0

2x

0

2

01 −+=

−+=−+=−+= ∫∫



Put n=2 in Eqn(1) Put n=3 in Eqn(1) ∴ y(0.1)=1.1051 _________________________________________________________ Problem(9): Solve , y(0)=1 for x=0.1 by Picard’s method correct to four decimal places. {Ans: y(0.1)=1.1270} Problem(10): Use Picard’s method to solve , y(0)=0 for x=0.4 {Ans: y(0.4)=0.0214}

( )

60

x

12

x

6

x

2

xx1

dx12

x

3

x

2

xx11

dxx3

x

12

x

2

xx11dxxy1y

5432

x

0

432

x

0

2342

x

0

2

23

−−−++=

−−−++=

−−−+++=−+=

∫

∫∫

2y3xdx

dy+=

22 yxdx

dy+=



Modified Euler’s method Consider first order differential equation

00 y)y(x,y)f(x,dx

dy==

Modified Euler’s formula is given by ___________________________________________________ _________________________________________________________ Problem(1): Determine the value of y for x=0(0.05)0.1 given that

1y(0) y,2xdx

dy=+= , using Modified Euler’s method up to four places of

decimal.

Soln: Given data: ( ) y2xyx,f += , h=0.05

00x =

Modified Euler’s Formula is given by

( ) ( )

( ) ( )

0,1,2,...n

0,1,2,...r______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

=

=+=+

++++=++

Step-(1): (To find y(0.05))1y(x1y == )

Put n=0 in Equations (1) and (2)

( ) ( )

( ) ( ) 0,1,2,...nformula)s(Euler'ny,nxhfny0

1nywhere

,..2,1,0(r)

1ny,1nxfny,nxf

2

hny

1r1n

y

=→+=+

=

++++=++

r

0.051x = 0.12x =

10y = ?1y = ?2y =



( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+

Initial approximation: From Eqn(4)

( ) ( )

[ ] 1.05100.0510y20

x0.051

0y,0xhf0y0

1y

=++=

++=

+=

First approximation: Put r=0 in Eqn(3)

( ) ( )

( ) ( )

( ) ( ) 0513.11.0520.051202

0.051

(0)1

y21x0y2

0x2

0.051

(0)1

y,1xf0y,0xf2

h0y

11

y

=

++++=

++++=

++=

Second approximation: Put r=1 in Eqn(3)

( ) ( )

( ) ( )

( ) ( ) 0513.11.051320.051202

0.051

(1)1

y21x0y2

0x2

0.051

(1)1

y,1xf0y,0xf2

h0y

21

y

=

++++=

++++=

++=

Since( )11

y and ( )21

y are the same correct to four decimal places

y(0.05)=1.0513 Step-(2): (To find y(0.1))2y(x2y == )




( ) ( )

( ) ( ) ______(6)1y,1xhf1y02

ywhere

_______(5)(r)2

y,2xf1y,1xf2

hy1

1r2

y

+=

++=

+


( ) ( )

( ) 1.10391.051320.050.051.0513

1y21

x0.051.05131y,1xhf1y02

y

=

++=

++=+=


( ) ( )

( ) ( )

( ) ( )

1.1054

1.103920.11.051320.052

0.051.0513

(0)2

y21x1y2

1x2

0.051.0513

(0)2

y,2xf1y,1xf2

h1y

12

y

=

++++=

++++=

++=


( ) ( )

( ) ( )

( ) ( ) 1055.11.105420.11.051320.052

0.051.0513

(1)2

y21x1y2

1x2

0.051.0513

(1)2

y,2xf1y,1xf2

h1y

22

y

=

++++=

++++=

++=

Similarly ( )

1055.132

y =

Since ( ) ( )3

2y&

22


y(0.1)=1.1055 Problem(2):

Obtain the solution of the equation yxdx

dy+= with y=1 when x=0 for y

at x=0.6 in steps of 0.3 using Modified Euler’s method correct to four



decimal places.

Soln: Given data: ( ) yxyx,f += , h=0.3

00x =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++

Step-(1): (To find y(0.3))1y(x1y == )


( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+


( ) ( )

[ ] [ ] 1.3100.310y0x0.31

0y,0xhf0y0

1y

=++=++=

+=


( ) ( )

( ) ( )

++++=

++=

(0)1

y1x0y0x2

0.31

(0)1

y,1xf0y,0xf2

h0y

11

y

( ) ( )[ ] 3660.11.30.3102

0.31 =++++=


0.31x = 0.62x =

10y = ?1y = ? 2y =



( ) ( )

( ) ( )

( ) ( ) ( )[ ] 3703.11.36600.3102

0.31

21

y

(1)1

y1x0y0x2

0.31

(1)1

y,1xf0y,0xf2

h0y

21

y

=++++=

++++=

++=

Similarly ( )

3703.13

1y =

Since( ) ( )3

1y&

21


y(0.3)=1.3703 Step-(2): (To find y(0.6))2y(x2y == )


( ) ( )

( ) ( ) ______(6)1y,1xhf1y02

ywhere

_______(5)(r)2

y,2xf1y,1xf2

h0y

1r2

y

+=

++=

+


( ) ( ) ( )

( )[ ] 1.81141.37030.30.31.3703

1y1x0.31.37031y,1xhf1y02

y

=++=

++=+=


( ) ( )

( ) ( )

++++=

++=

(0)2

y2x1y1x2

0.33703.1

(0)2

y,2xf1y,1xf2

h1y

12

y

( ) ( )[ ] 8827.11.81140.61.37030.32

0.33703.1 =++++=

Similarly ( ) ( ) ( )

8869.14

2y,8869.1

32

y,8667.12

2y ===

∴∴∴∴ y(0.6)=1.8869



Problem(3):

Using Modified Euler’s method find y(0.2) given that yxdx

dy+= ; y(0)=1

correct to four decimal places. Soln: Given data: ( ) yxyx,f += , h=0.2

00x = 0.21x =

10y = ? 1y =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++

To find y(0.2))1y(x1y ==


( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+


( ) ( )

[ ] [ ] 1.2100.210y0x0.21

0y,0xhf0y0

1y

=++=++=

+=


( ) ( )

[ ] 24.12.12.0102

0.21

(0)1

y1x0y0x2

0.21

(0)1

y,1xf0y,0xf2

h0y

11

y

=++++=

++++=

++=




( ) ( )

( ) [ ] 244.124.12.0102

0.21

21

y

(1)1

y1x0y0x2

0.21

(1)1

y,1xf0y,0xf2

h0y

21

y

=++++=

++++=

++=

Similarly ( )

2444.13

1y = &

( )2444.1

41

y =

Since ( ) ( )4

1y&

31


y(0.2)=1.2444 Problem(4): Use Modified Euler’s method to find the approximate value of y(1.1) for

the solution of the initial value problem 2xydx

dy= , y(1)=1 correct to three

decimal places. Perform two iterations. Soln: Given data: ( ) 2xyyx,f = , h=0.1

10x = 1.11x =

10y = ?1y =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++



( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+




( ) ( )

[ ] [ ] 1.2)1)(1)(2(0.110y02x0.11

0y,0xhf0y0

1y

=+=+=

+=


( ) ( )

[ ] 1.232)2(1.1)(1.22(1)(1)2

0.11

(0)1

y12x0y02x2

0.11

(0)1

y,1xf0y,0xf2

h0y

11

y

=++=

++=

++=


( ) ( )

( ) [ ] 1.235532)2(1.1)(1.22(1)(1)2

0.11

21

y

(1)1

y12x0y02x2

0.11

(1)1

y,1xf0y,0xf2

h0y

21

y

=++=

++=

++=

∴ The value of y(1.1) after two iteration is y(0.2)=1.2355 Problem(5): Find y(1.2) and y(1.4) by Modified Euler’s method given that

3xx

2y

dx

dy+= , y(1)=0.5 correct to three decimal places.

Soln: Given data: ( ) 3xx

2yyx,f += , h=0.2

10x = 1.21x = 1.41x =

0.50y = ?1y = ?1y =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++



Step(1): To find y(1.2))1y(x1y ==


( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+


( ) ( )

( )

( )

0.9

311

2(0.5)0.20.5

30x

0x02y

0.20.5

0y,0xhf0y0

1y

=

++=

++=

+=


( ) ( )

( ) ( )

( )( )

( )( )

0227.1

31.21.2

0.92311

0.52

2

0.20.5

31x

1x

(0)1

2y30x

0x02y

2

0.20.5

(0)1

y,1xf0y,0xf2

h0y

11

y

=

++++=

++++=

++=




( ) ( )

( ) ( )

( )( )

( )( )

043.1

31.21.2

1.02272311

0.52

2

0.20.5

31x

1x

(1)1

2y30x

0x02y

2

0.20.5

(1)1

y,1xf0y,0xf2

h0y

21

y

=

++++=

++++=

++=

Similarly ( )

046.13

1y = and

( )046.1

41

y =

Since ( ) ( )4

1y&

31


y(1.2)=1.2444 Step(2): To find y(1.4))2y(x2y ==


( ) ( )

( ) ( ) ______(6)1y,1xhf1y02

ywhere

5)_______((r)2

y,2xf1y,1xf2

h1y

1r2

y

+=

++=

+


( ) ( )

( )

( )

1.740

31.21.2

2(1.046)0.2046.1

31x

1x12y

0.2046.1

1y,1xhf1y02

y

=

++=

++=

+=




( ) ( )

( ) ( )

( )( )

( )( )

916.1

31.41.4

1.74231.21.2

1.0462

2

0.2046.1

32x

2x

(0)2

2y31x

1x12y

2

0.2046.1

(0)2

y,2xf1y,1xf2

h1y

12

y

=

++++=

++++=

++=

Similarly ( )

941.12

2y =

( )

944.132

y =,

( )945.1

42

y =,

( )945.1

52

y =

Since ( ) ( )5

2y&

42

y are the same correct to three decimal places

y(1.4)=1.945

Problem(6):

solve y1dx

dy−= , y(0)=0 by Modified Euler’s method for x=0.1 correct to

four decimal places. Soln: Given data: ( ) y1yx,f −= , h=0.1

00x = 0.11x =

00y = ?1y =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++





( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+


( ) ( )

[ ] 1.0010.10

]01[00y,0xhf0y0

1y

=−+=

−+=+= yhy


( ) ( )

[ ] 095.01.01012

0.10

(0)1

y1012

h0

(0)1

y,1xf0y,0xf2

h0y

11

y

=−+−+=

++−+=

++= yy


( ) ( )

0952.0]095.0101[2

0.10

(1)1

y10y-12

h0y

(1)1

y,1xf0y,0xf2

h0y

21

y

=−+−+=

−++=

++=

Similarly ( )

0952.03

1y = ,

Since ( ) ( )4

1y&

31


y(0.1)=0.0952 Problem(7): Use Modified Euler’s method to solve the differential equation

2yxdx

dy+= ,y(0)=1 for x=0.2 in steps of 0.1 correct to three

decimal places.

Soln: Given data: ( ) 2yxyx,f += , h=0.1

00x = 0.11x = 0.22x =



10y = 1.11741y = 1.27622y =


( ) ( )

( ) ( ) ______(2)ny,nxhfny0

1nywhere

1)_______((r)

1ny,1nxfny,nxf

2

hny

1r1n

y

+=+

++++=++

Step(1): To find y(0.1))1y(x1y ==


( ) ( )

( ) ( ) ______(4)0y,0xhf0y0

1ywhere

3)_______((r)1

y,1xf0y,0xf2

h0y

1r1

y

+=

++=

+


( ) ( )

( ) ( ) 1.12100.1120y0x0.11

0y,0xhf0y0

1y

=

++=

++=

+=


( ) ( ) ( )

( ) ( ) 1155.121.11.02102

0.11

2(0)1

y1x20y0x

2

h0y

(0)1

y,1xf0y,0xf2

h0y

11

y

=

++++=

++++=

++=


( ) ( )

( )

1172.1]2)1155.1(1.02)1(0[2

0.11

2(1)1

y1x20y0x

2

h0y

(1)1

y,1xf0y,0xf2

h0y

21

y

=++++=

++++=

++=



Similarly ( )

1174.13

1y = ,

( )1174.1

41

y =

Clearly ( ) ( )4

1y&

31

y are same correct to four decimal places.

∴∴∴∴ y(0.1)=1.1174 Step-(2): (To find y(0.2))2y(x2y == )


( ) ( )

( ) ( ) ______(6)1y,1xhf1y02

ywhere

_______(5)(r)2

y,2xf1y,1xf2

h1y

1r2

y

+=

++=

+


( ) ( ) ( )

( ) 2522.121.11741.00.11.1174

21y1xh1y1y,1xhf1y

02

y

=

++=

++=+=


( ) ( )

( )

( ) ( ) 1.273221.25220.221.11740.12

0.11.1174

2(0)2

y2x21y1x

2

h1y

(0)2

y,2xf1y,1xf2

h1y

12

y

=

++++=

++++=

++=

Similarly ( ) ( ) ( )

2762.14

2y,2762.1

32

y,2758.12

2y ===

Since( )32

y and( )42


y(0.2)=1.2762 Problem(8): Find y(4.4) by Modified Euler’s method taking h=0.2 from the differential

equation 5x

2y-2

dx

dy= , y(4)=1 correct to Four decimal places.

{Ans:y(4.4)=1.0187}



Problem(9):

Solve y2xdx

dy+= , y(0)=1 for x=0.02 taking h=0.01 by Modified Euler’s

method correct to Four decimal places. Carry out two iterations after each step. {Ans:y(0.02)=1.020} ___________________________________________________________

Runge-Kutta Method of 4th order

Consider 0y)0y(x,y)f(x,dx

dy==

The Runge-Kutta method of 4th

order is given by

[ ]

( )

( )3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

++=

++=

++=

=

++++=+

Problem(1): By employing Runge-Kutta method of fourth order solve the differential

equation y6x/2y =− ,y(0)=1 for x=0.2 in steps of 0.1 correct to four

decimal places.

Soln: Given data: ( )2

y3xyx,f += , h=0.1

00x = 0.11x = 0.21x =

10y = 1.06641y = 1670.12y =

The Runge-Kutta method of 4th order is given by



[ ]

( )

( )

++=

++=

++=

=

++++=+

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

-------------------- (1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= --------------------------- (2)

( )

( )

++=

++=

++=

=

3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

( )

0.052

13(0)0.1

20y

03xh

0y,0xhf1k

=

+=

+=

=

0.06622

0.051

2

1

2

0.1030.1

21k

0y2

1

2

h0x3h

21k

0y,2

h0xhf2k

=

++

+=

++

+=

++=

0.06662

0.06621

2

1

2

0.1030.1

22k

0y2

1

2

h0x3h

22k

0y,2

h0xhf3k

=

++

+=

++

+=

++=



( )

( ) ( )

( ) ( ) 0.08330.066612

10.1030.1

3k0y2

1h0x3h

3k0yh,0xhf4k

=

+++=

+++=

++=

Substituting all these values in Eqn(2), we get

[ ] 1.06640.08332(0.0666)2(0.0662)0.056

111y =++++=

Put n=1 in Eqn(1)

[ ]4k32k22k1k6

11y2y ++++= ---------------------- (3)

Where

( )

( )3k1yh,1xhf4k

22k

1y,2

h1xhf3k

21k

1y,2

h1xhf2k

1y,1xhf1k

++=

++=

++=

=

( )

0.08332

1.06643(0.1)0.1

21y

13xh

1y,1xhf1k

=

+=

+=

=

0.10042

0.08331.0664

2

1

2

0.10.130.1

21k

1y2

1

2

h1x3h

21k

1y,2

h1xhf2k

=

++

+=

++

+=

++=

0.10082

0.10041.0644

2

1

2

0.10.130.1

22k

1y2

1

2

h1x3h

22k

1y,2

h1xhf3k

=

++

+=

++

+=

++=



( )

( ) ( )

( ) ( ) 0.11830.10081.06642

10.10.130.1

3k1y2

1h1x3h

3k1yh,1xhf4k

=

+++=

+++=

++=


[ ]

1.1670

0.11832(0.1008)2(0.1004)0.08336

11.06642y

=

++++=

Problem(2): Apply Runge-Kutta method of fourth order to find an approximate value

of y(0.1) and y(0.2) of 2yxdx

dy+= , y(0)=1 correct to three decimal

places.

Soln: Given data: ( ) 2yxyx,f += , h=0.1

00x = 0.11x = 0.22x =

10y = ?1y = ?2y =


[ ]

( )

( )

++=

++=

++=

=

++++=+

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

-------------------------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= --------------------------(2)



( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

( ) ( ) 0.12100.120y0xh

0y,0xhf1k

=

+=

+=

=

0.11522

2

0.11

2

0.100.1

2

21k

0y2

h0xh

21k

0y,2

h0xhf2k

=

++

+=

++

+=

++=

0.11682

2

0.11521

2

0.100.1

2

22k

0y2

h0xh

22k

0y,2

h0xhf3k

=

++

+=

++

+=

++=

( )

( ) ( )

( ) ( ) 0.134720.116810.100.1

23k0yh0xh

3k0yh,0xhf4k

=

+++=

+++=

++=


[ ] 1.11640.13472(0.1168)2(0.1152)0.16

111y =++++=

Put n=1 in Eqn(1) [ ]4k32k22k1k6

11y2y ++++= -----------(3)

Where



( )

( )3k1yh,1xhf4k

22k

1y,2

h1xhf3k

21k

1y,2

h1xhf2k

1y,1xhf1k

++=

++=

++=

=

( )

( ) ( ) 0.134621164.1(0.1)0.121y1xh

1y,1xhf1k

=

+=

+=

=

++

+=

++=

2

21k

1y2

h1xh

21k

1y,2

h1xhf2k

0.15512

2

0.13461.1164

2

0.10.10.1 =

++

+=

0.15752

2

0.15511.1164

2

0.10.10.1

2

22k

1y2

h1xh

22k

1y,2

h1xhf3k

=

++

+=

++

+=

++=

( )

( ) ( )

( ) ( ) 0.182220.15751.11640.10.10.1

23k1yh1xh

3k1yh,1xhf4k

=

+++=

+++=

++=


[ ]

27341.

0.18222(0.1575)2(0.1551)0.13466

11.11642y

=

++++=



Problem(3): Use Runge-Kutta method of fourth order to approximate y when x=0.1,

given that y=1 when x=0 and yxdx

dy+=

Soln: Given data: ( ) yxyx,f += , h=0.1

00x = 0.11x =

10y = ?1y =


[ ]

( )

( )3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

++=

++=

++=

=

++++=+

------------------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= -------------------(2)

( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

( )[ ] [ ] 0.1100.10y0xh

0y,0xhf1k

=+=+=

=



0.112

0.11

2

0.100.1

21k

0y2

h0xh

21k

0y,2

h0xhf2k

=

++

+=

++

+=

++=

1105.02

0.111

2

0.100.1

22k

0y2

h0xh

22k

0y,2

h0xhf3k

=

++

+=

++

+=

++=

( )

( ) ( )[ ]( ) ( )[ ] 0.12100.110510.100.1

3k0yh0xh

3k0yh,0xhf4k

=+++=

+++=

++=


[ ]

11031.

0.12102(0.1105)2(0.11)0.16

111y

=

++++=

Problem(4): Use Runge-Kutta method of fourth order to obtain an approximation to

y(1.5) for the solution of 2xydx

dy= ;y(1)=1 correct to four decimal places.

Soln: Given data: ( ) 2xyyx,f = , h=0.5

10x = 1.51x =

10y = ?1y =




[ ]

( )

( )

++=

++=

++=

=

++++=+

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

------------------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= --------------------------(2)

( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

[ ] [ ] 12(1)(1)0.50y02xh

0y,0xhf1k

===

=

1.8752

11

2

0.5120.5

21k

0y2

h0x2h

21k

0y,2

h0xhf2k

=

+

+=

+

+=

++=

4218.22

1.8751

2

0.5120.5

22k

0y2

h0x2h

22k

0y,2

h0xhf3k

=

+

+=

+

+=

++=



( )

( )( )[ ]( )( )[ ] 5.13272.421810.5120.5

3k0yh0x2

3k0yh,0xhf4k

=++=

++=

++=

h


[ ]

4543.3

5.13272(2.4218)2(1.875)16

111y

=

++++=

Problem(5): Obtain the values of y at x=0.1, 0.2 using Runge-Kutta method of 4th

order for the differential equation y/y −= ; y(0)=1 correct to four decimal

places. Soln: Given data: ( ) -yyx,f = , h=0.1

00x = 0.11x = 0.22x =

10y = ?1y = ?2y =


[ ]

( )

( )

++=

++=

++=

=

++++=+

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

-----------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= -------------(2)

( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=



( )

[ ] [ ] -0.11-0.10y-h

0y,0xhf1k

===

=

095.02

0.1-10.1

21k

0yh

21k

0y,2

h0xhf2k

−=

−=

+−=

++=

0952.02

0.09510.1

22k

0yh

22k

0y,2

h0xhf3k

−=

−−=

+−=

++=

( )

( )[ ]( )[ ] -0.09040.0952-10.1

3k0yh

3k0yh,0xhf4k

=−=

+−=

++=


[ ] 9048.00.0904-2(-0.0952)2(-0.095)0.1-6

111y =+++=


11y2y ++++= -------------(3)

( )

( )3k1yh,1xhf4k

22k

1y,2

h1xhf3k

21k

1y,2

h1xhf2k

1y,1xhf1k

++=

++=

++=

=

( )

( )[ ] [ ] -0.09049048.00.11y-h

1y,1xhf1k

=−==

=



-0.08592

0.09040.90480.1

21k

1yh

21k

1y,2

h1xhf2k

=

−−=

+−=

++=

-0.08612

0.08590.90480.1

22k

1yh

22k

1y,2

h1xhf3k

=

−−=

+−=

++=

( )

( )[ ]( )[ ] 0.08180.0861-0.90480.1

3k1yh

3k1yh,1xhf4k

−=−=

+−=

++=


[ ]

8187.0

0.0818-2(-0.0861)2(-0.0859)0.0904-6

10.90482y

=

+++=

Problem(6): By using the Runge-Kutta method of fourth order find the approximate

values of y(0.5) and y(1), given that yx

1

dx

dy

+= y(0)=1 correct to four

decimal places.

Soln: Given data: yx

1

dx

dy

+= , h=0.5

00x = 0.51x = 12x =

10y = ?1 =y ?2y =




[ ]

( )

( )

++=

++=

++=

=

++++=+

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

---------------------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= -----------------------(2)

( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

0.510

10.5

0y0x

1h

0y,0xhf1k

=

+=

+=

=

3333.0

2

0.51

2

5.00

10.5

21k

0y2

h0x

1h

21k

0y,2

h0xhf2k

=

+++

=

+++

=

++=

3529.0

2

0.33331

2

0.50

10.5

22k

0y2

h0x

1h

22k

0y,2

h0xhf3k

=

+++

=

+++

=

++=



( )

0.2698

0.352910.50

10.5

3k0yh0x

1h

3k0yh,0xhf4k

=

+++=

+++=

++=


[ ] 3570.10.26982(0.3529)2(0.3333)0.56

111y =++++=


11y2y ++++= --------------(3)

( )

( )3k1yh,1xhf4k

22k

1y,2

h1xhf3k

21k

1y,2

h1xhf2k

1y,1xhf1k

++=

++=

++=

=

( )

0.26921.3570.5

10.5

1y1x

1h

1y,1xhf1k

=

+=

+=

=

2230.0

2

0.26921.3570

2

5.00.5

10.5

21k

1y2

h1x

1h

21k

1y,2

h1xhf2k

=

+++

=

+++

=

++=

2253.0

2

0.22301.357

2

0.50.5

10.5

22k

1y2

h1x

1h

22k

1y,2

h1xhf3k

=

+++

=

+++

=

++=



( )

0.1936

0.22531.3570.50.5

10.5

3k1yh1x

1h

3k1yh,1xhf4k

=

+++=

+++=

++=


[ ]

5835.1

0.19362(0.2253)2(0.2230)0.26926

11.35702y

=

++++=

Problem(7): By using the Runge-Kutta method of fourth order solve the initial value

problem 2yx3e/y += ; y(0)=0 at x=0.1 correct to three decimal places.

Soln: Given data: ( ) 2yx3eyx,f += , h=0.1

00x = 0.11x =

00y = ?1y =

The Runge-Kutta method of 4th

order is given by

[ ]

( )

( )3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

++=

++=

++=

=

++++=+

---------------(1)

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= -------------------(2)



( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

0.32(0)03e0.1

02yx

3eh

0y,0xhf1k

0

=

+=

+=

=

0.345

2

0.302

)2

0.1(0

3e0.521k

0y2)

2

h(x

3eh

21k

0y,2

h0xhf2k

0

=

++

+=

++

+=

++=

0.349

2

0.34502

)2

0.1(0

3e0.122k

0y2)

2

h(x

3eh

22k

0y,2

h0xhf3k

0

=

++

+=

++

+=

++=

( )

( ) ( )

401.0

349.002)1.0(0

3e1.03k0y2h)(x

3eh

3k0yh,0xhf4k

00

=

++

+=

++

+=

++=


[ ]

348.0

0.4012(0.349)2(0.345)0.36

101y

=

++++=

Problem(8): Obtain the value of y at x=0.2 using Runge-Kutta method of 4th order for



the differential equation xy

x-y/y+

= ; y(0)=1 correct to four decimal

places.

Soln: Given data: ( )xy

x-yyx,f

+= , h=0.2

00x = 0.21x =

10y = ?1y =


[ ]

( )

( )

)1.(..............................

3knyh,nxhf4k

22k

ny,2

hnxhf3k

21k

ny,2

hnxhf2k

ny,nxhf1kwhere

4k32k22k1k6

1ny1ny

++=

++=

++=

=

++++=+

Put n=0 in Eqn(1)

[ ]4k32k22k1k6

10y1y ++++= -----------------(2)

( )

( )3k0yh,0xhf4k

22k

0y,2

h0xhf3k

21k

0y,2

h0xhf2k

0y,0xhf1kwhere

++=

++=

++=

=

( )

0.201

0-10.2

0x0y0x0y

h

0y,0xhf1k

=

+=

+

−=

=



0.1666

2

0.20

2

0.21

2

0.20

2

0.21

0.2

2

h0x

21k

0y

2

h0x

21k

0y

h

21k

0y,2

h0xhf2k

=

++

+

+−

+

=

++

+

+−

+

=

++=

0.1661

2

0.20

2

0.16661

2

0.20

2

0.16661

0.2

2

h0x

22k

0y

2

h0x

22k

0y

h

22k

0y,2

h0xhf3k

=

++

+

+−

+

=

++

+

+−

+

=

++=

( )

( ) ( )( ) ( )

( ) ( )( ) ( )

1414.0

2.001661.01

2.001661.010.2

h0x3k0y

h0x3k0y0.2

3k0yh,0xhf4k

=

+++

+−+=

+++

+−+=

++=


[ ]

1678.1

0.14142(0.1661)2(0.1666)0.26

111y

=

++++=

Problem(9):

Evaluate y(1.1) by fourth order Runge-Kutta method given that

2x

1

x

y/y =+ , y(1)=1 correct to four decimal places.

{Ans: y(1.1)=0.9958}

Problem(10):

Using Runge-Kutta method of fourth order solve 2x2y

2x2y

dx

dy

+

−=

y(0)=1 at x=0.2 and 0.4.

{Ans: y(0.2)=1.1959,

y(0.4)=1.3751}



Numerical Methods Predictor-Corrector methods

(Multi Step Methods)

The methods in which the construction of involves the use of not

only

the solution but also some of its predecessors

are called Multi step methods.

Milne’s Predictor-Corrector Method Consider the differential equation

Milne’s predictor and corrector formula is given by Problem(1)

Find y(2) if y(x) is the solution of , given that y(0)=2,

y(0.5)=2.636, y(1)=3.595, y(1.5)=4.968 using Milne’s Predictor-Corrector

method correct to four decimal places.

Soln: Given data , h=0.5

x0 = 0 x1 = 0.5 x2 = 1.0 x3 = 1.5 x4 = 2.0

y0 = 0 y1 = 2.636 y2 = 3.595 y3 = 4.968 y4 = ?

Milne’s Predictor formula is given by

1ny +

ny ,....etc)2-ny,1-n y(i.e

00 y)y(x;y)f(x,dx

dy==

( )

( ) ( ) ( )

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f:Note

(r)4

y,4xf(r)4

f ,3y,3xf3f ,2y,2xf2f,1y,1xf1f where

formula Corrector(2)(r)4

f34f2f3

h2y

1)(rc4,

y

formula Predictor(1)32f2f12f3

4h0yp4,y

≠==

=

====

→−−−−−

+++=

+

→−−−−−+−+=

2

yx

dx

dy +=

2

yxy)f(x,

+=

( ) (1)2ff2f3

4hyy 3210p4, −−−−−+−+=



Substituting all the values in eqn(1) we get,

Milne’s Corrector formula is given by

First improvement: Put r=0 in eqn(2)

Second improvement: Put r=1 in eqn(2)

2

yx)y,f(xf ii

iii

+==

ix iy

0.5x1 =

1x2 =

1.5x3 =

636.2y1 =

595.3y2 =

968.4y3 =

568.12

2.6360.5

2

yxf 111 =

+=

+=

2975.22

3.5951

2

yxf 222 =

+=

+=

234.32

4.9681.5

2

yxf 333 =

+=

+=

{ } 871.6)234.3(22975.2)568.1(23

4(0.5)2y p4, =+−+=

( ) ( )(r)44

(r)4

(r)4322

1)(rc4, y,xff where)2(f4ff

3

hyy =−−−−−+++=+

( )

( ) ( )

( ) 6.87314.43554(3.234)2.29753

0.53.595y

4.43552

6.8712

2

yxy,xfy,xff

Where,f4ff3

hy y

(1)

c4,

p4,4

p4,4

(0)

44

(0)

4

(0)4322

(1)c4,

=+++=∴

=+

=+

===

+++=

( )

( ) ( )

( )

6.8733

4.43654(3.234)2.29753

0.53.595y

4.43652

6.87312

2

yxy,xfy,xff

where

f4ff3

hy y

(2)

c4,

(1)c4,4(1)

c4,4(1)44

(1)4

(1)

4322

(2)

c4,

=

+++=∴

=+

=

+===

+++=



Third improvement: Put r=2 in eqn(2)

Since are the same up to four decimal places

y(2)=6.8733

Problem(2):

Use Milne’s method to find y(0.3) from , y(0)=1 after

computing y(-0.1),y(0.1) and y(0.2) by Taylor’s series method correct to

four decimal places..

Soln: Given data , h=0.1, ,

We shall first find y(-0.1),y(0.1) and y(0.2) by Taylor’s series method.

By Taylor’s series method, we have

_____(1)

Put n=0 in eqn(1) ____(2)

22 yxdx

dy+=

( )

( ) ( )

( ) 6.87334.43664(3.234)2.29753

0.53.595y

4.43662

6.87332

2

yxy,xfy,xff

where

f4ff3

hy y

(3)c4,

(2)

c4,4(2)

c4,4

(2)

44

(2)

4

(2)

4322

(3)

c4,

=+++=∴

=+

=+

===

+++=

(3)c4,

(2)c4, y& y

22 yxy)f(x, += 0x0 = 2y0 =

.....y3!

hy

2!

hy

1!

hy)f(xy

///

n

3//

n

2/

nn1n1n ++++== ++

( )( )( ) ( )///////////////IV

2//////

///

22/

y3yyy2y2yyyyy2y

yyy22y

2yy2xy

yxyGiven

+=++=

++=

+=∴

+=

.....y3!

hy

2!

hy

1!

hy)f(xy

///

0

3//

0

2/

0011 ++++==

( )( )( ) 28yy3yy2y

8yyy22y

2y2y2xy

1yxy

//0

/0

///00

IV0

2/0

//00

///0

/000

//0

20

20

/0

=+=

=++=

=+=

=+=




∴ y(0.1)=1.1114

similarly y(-0.1)=0.9087, y(0.2)=1.2529

Thus


Substituting all the values in eqn(3) we get, Milne’s Corrector formula is given by First improvement: Put r=0 in eqn(4)

( ) ( ) ( )

1.1114

.......(28)4!

0.1(8)

3!

0.1(2)

2!

0.1(1)

1!

0.11y

432

1

=

+++++=

0.1x0 −=

1y1 =

0x1 = 0.1x2 = 0.2x3 = 0.3x4 =

9087.0y0 = 1114.1y2 = 2529.1y3 = ?y4 =

( ) (3)32f2f12f3

4h0yp4,y −−−−−+−+=

( ) ( )2

i

2

iiii yx)y,f(xf +==ixiy

0x1 = 1y1 = ( ) ( ) ( ) ( ) 110yxf222

1

2

11 =+=+=

0.1x2 = 1114.1y2 = ( ) ( ) 2452.1yxf2

2

2

22 =+=

0.2x3 = 2529.1y3 = ( ) ( ) 6097.1yxf2

3

2

33 =+=

{ } 4385.1)6097.1(22452.1)1(23

4(0.1)0.9087y p4, =+−+=

0,(r)

c4,y

(r)4

fp4,y(0)4

f

(r)4

y,4xf(r)4

f where)4((r)4

f34f2f3

h2y

1)(rc4,

y

≠==

=−−−−−

+++=

+

rand



( ) ( ) ( ) ( )

( ) 4395.11592.24(1.6097)1.24523

0.11.1114

(1)c4,

y

1592.221.438520.32p4,y2

4x(0)4

f

(0)4

y,4xf(0)4

f Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=+=+=∴

=

+++=


( )

( ) ( )

( )

4396.1(3)

c4,ysimilarly

1.43961621.24(1.6097)1.24523

0.11.1114

(2)c4,

y

2.162121.439520.3

2(1)

c4,y2

4x(1)

c4,y,4xf

(1)4

y,4xf(1)4

f

where(1)4

f34f2f3

h2y

(2)c4,

y

=

=+++=∴

=+=

+=

=

=

+++=

Since (3)

c4, y&

(2)c4,

y are the same up to four decimal places

y(0.3)=1.4396 Problem(3):

Using Milne’s method find y(4.4) given that 022y/5xy =−+

y(4)=1, y(4.1)=1.0049, y(4.2)=1.0097, y(4.3)=1.0143 correct to four decimal places.

Soln: 0.1h,5x

2y2y)f(x, :data Given =

−=


( ) (1)32f2f12f3

4h0yp4,y −−−−−+−+=

40x =

0049.11y =

4.11x = 4.22x = 4.33x = 4.44x =

10y = 0097.12y = 0143.13y = ?4y =




{ } 0816.1)0451.0(20466.0)0483.0(23

4(0.1)1p4,y =+−+=


=−−−−−

+++=

+ (r)4

y,4xf(r)4

f where)2((r)4

f34f2f3

h2y

1)(rc4,

y

0,(r)

c4,y

(r)4

fp4,y(0)4

f ≠== rand


( )

i5x

2iy2

)iy,if(xif−

==ix iy

4.11x =

4.22x =

4.33x =

0049.11y =

0097.12y =

0143.13y =

( ) ( )0.0483

5(4.1)

21.00492

15x

21y2

1f =−

=−

=

( ) ( )0.0466

5(4.2)

21.00972

25x

22y2

2f =−

=−

=

( ) ( )0.0451

5(4.3)

21.01432

35x

23y2

3f =−

=−

=



( )( ) ( )

( ) 0187.10.04374(0.0451)0.04663

0.11.0097

(1)c4,

y

0437.05(4.4)

21.01862

45x

2p4,y2

p4,y,4xf(0)4

y,4xf(0)4

f

Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=−

=−

==

=

+++=


( )

( ) 0187.10437.04(0.0451)0.04663

0.11.0097

(2)c4,

y

0437.05(4.4)

21.01872

45x

2(1)

c4,y2

(1)c4,

y,4xf(1)4

y,4xf(1)4

f

where(1)4

f34f2f3

h2y

(2)c4,

y

=+++=∴

=−

=

−

=

=

=

+++=

Since (2)

c4, y&

(1)c4,


y(4.4)=1.0187 Problem(4):

Given 2y2x12

1

dx

dy

+= and y(0)=1, y(0.1)=1.06, y(0.2)=1.12,

y(0.3)=1.21. Evaluate y(0.4) by Milne’s Predictor-Corrector method.

Soln : 0.1h,2y2x12

1y)f(x, :data Given =

+=


00x =

06.11y =

0.11x = 0.22x = 0.33x =

10y = 12.12y = 21.13y = ?4y =

0.44x =



( ) (1)32f2f12f3

4h0yp4,y −−−−−+−+=


{ } 2771.1)7979.0(26522.0)5674.0(23

4(0.1)1p4,y =+−+=


)2((r)4

f34f2f3

h2y

1)(rc4,

y −−−−−

+++=

+

0,(r)

c4,y

(r)4

fp4,y(0)4

f ,(r)4

y,4xf(r)4

f e wher ≠==

= rand


2i

y2i

x12

1)iy,if(xif

+==

ix iy

0.11x =

0.22x =

0.33x =

06.11y =

12.12y =

21.13y =

5674.021

y21

x12

11f =

+=

6522.022

y22

x12

12f =

+=

7979.023

y23

x12

13f =

+=



( ) ( )

( ) 2796.10.94594(0.7979)0.65223

0.11.12

(1)c4,

y

9459.022771.124.012

12p4,

y24

x12

1(0)4

y,4xf(0)4

f

Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=

+=

+=

=

+++=


( ) ( )

( )

1.2797(3)

c4, ySimilarly

1.27970.94964(0.7979)0.65223

0.11.12

(2)c4,

y

0.949621.279620.412

1(1)c4,

y24

x12

1(1)4

y,4xf(1)4

f

Where,(1)4

f34f2f3

h2y

(2)c4,

y

=

=+++=∴

=

+=

+=

=

+++=

Since (3)

c4, y&

(2)c4,


y(0.4)=1.2797 Problem(5):

Using Milne’s predictor-corrector method solve 2y2ydx

dy−=

y(0)=1 for x=0.2 if y(0.05)=1.0499, y(0.1)=1.0996, y(0.15)=1.1488 correct to four decimal places.

Soln: 0.05h,2y2yy)f(x, :data Given =−=


( ) (1)32f2f12f3

4h0yp4,y −−−−−+−+=

00x =

0499.11y =

0.051x = 0.12x = 0.153x = 0.44x =

10y = 0996.12y = 1488.13y = ?4y =

2i

y-i2y)iy,if(xif ==ix iy

0.051x = 0499.11y = 9975.021

y-12y1f ==



9960.02

2y-22y2f ==


{ } 1969.1)9778.0(29960.0)9975.0(23

4(0.05)1p4,y =+−+=


0,(r)

c4,y

(r)4

fp4,y(0)4

f ,(r)4

y,4xf(r)4

f where

)2((r)4

f34f2f3

h2y

1)(rc4,

y

≠==

=

−−−−−

+++=

+

rand


( ) ( ) ( )

( ) 1.19740.96124(0.9778)0.99603

0.051.0996

(1)c4,

y

0.961221.19691.196922p4,yp4,2y

(0)4

y,4xf(0)4

f

Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=−=−=

=

+++=


( ) ( )

( )

1.1974

0.96104(0.9778)0.99603

0.051.0996

(2)c4,

y

0.961021.19741.197422

(1)c4,

y(1)

c4,2y

(1)4

y,4xf(1)4

f

Where,(1)4

f34f2f3

h2y

(2)c4,

y

=

+++=∴

=−=

−=

=

+++=

Since (2)

c4, y&

(1)c4,


y(0.2)=1.1974

0.12x =

0.153x =

0996.12y =

1488.13y = 9778.023

y-32y3f ==



Problem(6):

Solve the initial value problem 1y(0);2xy1dx

dy=+=

for x=0.4 by Milne’s predictor and corrector method correct to three decimal places, given that

Soln: 0.1h,2y1y)f(x, :data Given =+= x


( ) (1)32f2f12f3

4h0yp4,y −−−−−+−+=

( )( ) 122.12105.11.0121

y1x11f =+=+=

( )( ) 299.12223.12.0122

y2x12f =+=+=

( )( ) 1.55021.3550.3123

y3x13f =+=+=


{ } 526.1)550.1(2299.1)122.1(23

4(0.1)1p4,y =+−+=

x 0.1 0.2 0.3

y 1.105 1.223 1.355

00x =

105.11y =

0.11x = 0.22x = 0.33x = 0.44x =

10y = 223.12y = 355.13y = ?4y =

2i

yix1)iy,if(xif +==ix iy

0.11x =

0.22x =

0.33x =

105.11y =

223.12y =

355.13y =




0,(r)

c4,y

(r)4

fp4,y(0)4

f ,(r)4

y,4xf(r)4

f where

)2((r)4

f34f2f3

h2y

1)(rc4,

y

≠==

=

−−−−−

+++=

+

rand


( ) ( )( )

( ) 1.5371.9314(1.550)1.2993

0.11.223

(1)c4,

y

1.93121.5260.412p4,y4x1

(0)4

y,4xf(0)4

f

Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=+=+=

=

+++=


( )( )

( )

1.537

1.9444(1.550)1.2993

0.11.223

(2)c4,

y

944.121.5370.412

(1)c4,

y4x1(1)4

y,4xf(1)4

f

Where,(1)4

f34f2f3

h2y

(2)c4,

y

=

+++=∴

=+=

+=

=

+++=

Since (2)

c4, y&

(1)c4,


y(0.4)=1.537 Problem(7):

Part of a Numerical solution of ( ) ( )y0.1x0.2dx

dy+= is shown in the

following table.

x 0.00 0.05 0.10 0.15

y 2.0000 2.0103 2.0211 2.0323



Use Milne’s Predictor and corrector method to find the next entry in the table, correct to four decimal places. Soln: ( ) ( ) 0.05h,y0.1x0.2y)f(x, :data Given =+=


( ) (1)32f2f12f3

4h0yp4,y −−−−−+−+=

( ) ( ) iy0.1ix2.0)iy,if(xif +==

( ) ( ) 2110.00103.21.005.02.01f =+=

( ) ( ) 2221.00211.21.01.02.02f =+=

( ) ( ) 2332.00323.21.015.02.03f =+=


{ } 0444.2)2332.0(22221.0)2110.0(23

4(0.05)1p4,y =+−+=


0,(r)

c4,y

(r)4

fp4,y(0)4

f ,(r)4

y,4xf(r)4

f where

)2((r)4

f34f2f3

h2y

1)(rc4,

y

≠==

=

−−−−−

+++=

+

rand


00x =

0103.21y =

0.051x = 0.12x = 0.153x = 0.24x =

20y = 0211.22y = 0323.23y = ?4y =

ix iy

0.11x =

0.22x =

0.33x =

105.11y =

223.12y =

355.13y =



( )

( )

( ) 0444.20.24444(0.2332)0.22213

0.052.0211

(1)c4,

y

0.24444(0.1)2.0440.22.0

p4,(0.1)y4x2.0(0)4

y,4xf(0)4

f

Where,(0)4

f34f2f3

h2y

(1)c4,

y

=+++=∴

=+=

+=

=

+++=

Since (1)

c4, y&p4, y are the same up to four decimal places

y(0.2)=2.0444 Problem(8): Determine the value of y(0.4) using Milne’s predictor and corrector method correct to four decimal places. Given that

1y(0);2yxy/y =+= Use Taylor’s series method to get the values of

y(0.1),y(0.2) and y(0.3). {Ans: y(0.1)=1.1167, y(0.2)=1.2767, y(0.3)=1.5023

1.8376}y(0.4)

(4)c4,

y1.8376(3)

c4,y1.8375,

(2)c4,

y1.8369,(1)

c4, y1.8397,p4, y

=∴

=====

Problem(9):

By using the Milne’s predictor-corrector method find an approximate

solution of the equation 0x,x

2y/y ≠= at the point x=2 given that y(1)=2,

y(1.25)=3.13, y(1.5)=4.5 , y(1.75)=6.13.

{Ans:

8.00}y(2)

8.00(2)

c4,y8.00,

(1)c4,

y8.01,p4, y

=∴

===



Adam-Bashforth Predictor-Corrector Method

Consider the differential equation 0y)0y(x;y)f(x,dx

dy==

Adam-Bashforth Predictor-Corrector formula is given by

( )

( ) ( ) ( )

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f:Note

(r)4

y,4xf(r)4

f ,3y,3xf3f ,2y,2xf2f,1y,1xf1f where

formulaCorrector1f25f319f(r)4

9f24

h3y

1)(rc4,

y

formulaPredictor09f137f259f355f24

h3yp4,y

≠==

=

====

→

+−++=

+

→−+−+=

Problem(1):

Solve for y(2) given that ;2

yx

dx

dy += y(0)=2, y(0.5)=2.636, y(1.0)=3.595,

y(1.5)=4.968 by Adam-Bashforth Predictor Corrector method correct to

four decimal places.

Soln: 0.5h,2

yxy)f(x, :data Given =

+=

00x = 0.51x = 1.02x = 1.53x = 2.04x =

636.21y = 595.32y = 968.43y = ?4y =

Adam’s Predictor formula is given by

( ) -(1)-------09f137f259f355f24

h3yp4,y −+−+=

20y =



ix

iy 2iyix

)iy,if(xif+

==

00x =

20y = 1

2

20

20y0x

)0y,0f(x0f =+

=+

==

0.51x =

636.21y =

568.1

2

636.25.0

21y1x

)1y,1f(x1f

=

+=

+==

12x =

595.32y = 2975.2

2

595.30.1

22y2x

)2y,2f(x2f

=

+=

+==

1.53x =

968.43y =

234.3

2

968.45.1

23y3x

)3y,3f(x3f

=

+=

+==

( ) ( ) ( ) ( )( )

8707.6

191.568372.2975593.2345524

0.54.968p4,y

becomes Eqn(1)

=

−+−+=

∴

Adam-Bashforth Corrector formula is given by

(2)1f25f319f(r)4

9f24

h3y

1)(rc4,

y −−−−−−−

+−++=

+

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=




( )

( ) ( ) ( )( )

6.8730

1.5682.297553.234194.4353924

0.54.968

(1)c4,

y

4.43532

6.87072

2

p4,y4xp4,y,4xf

(0)4

y,4xf(0)4

f where

1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

+−++=∴

=+

=+

==

=

+−++=


( ) ( ) ( )( )

6.8733

1.5682.297553.234194.4365924

0.54.968

(2)c4,

y

4.4365

2

6.87302

2

(1)c4,

y4x(1)

c4,y,4xf

(1)4

y,4xf(1)4

f where

1f25f319f(1)4

9f24

h3y

(2)c4,

y

=

+−++=∴

=

+=

+=

=

=

+−++=

Third improvement: Put r=2 in eqn(2)

( ) ( ) ( )( )

6.8733

1.5682.297553.234194.4366924

0.54.968

(3)c4,

y

4.43662

6.87332

2

(2)c4,

y4x(2)

c4,y,4xf

(2)4

y,4xf(2)4

f

1f25f319f(2)4

9f24

h3y

(3)c4,

y

=

+−++=∴

=+

=+

=

=

=

+−++=

Since (3)

c4, y&

(2)c4,


y(2)=6.8733



Problem(2):

Obtain the solution of the initial value problem 2xy2xdx

dy=− , y(1)=1 at

x=1(0.1)1.3 by Taylor’s series method and at x=1.4 by Adam’s-Bashforth

method correct to four decimal places.

Soln: 0.1h1,0y1,0xy),(12xy2x2xy)f(x,Given ===+=+=

By Taylor’s series method, we have

.....///ny

3!

3h//ny

2!

2h/ny

1!

hny)1nf(x1ny ++++=+=+ _____(1)

/y2xy)2x(1//y

y)(12x/yGiven

++=∴

+=

///y2x//6xy/6yIVySimilarly,

//y2xy)2(1/4xy///y

++=

+++=

Put n=0 in eqn(1) .....///0y

3!

3h//0y

2!

2h/0y

1!

h0y)1f(x1y ++++== ------

(2)

661x186x1x66x2///0

y20

x//0

y06x/0

6yIV0

y

181x61)2(14x1x2//0

y20

x)0y2(1/0

y04x///0

y

61x21)2(1)(1/0

y20

x)0y(102x//0

y

21)1(1)0y(120

x/0

y

=++=++=

=+++=+++=

=++=++=

=+=+=


( ) ( ) ( )

1.2332

.......(66)24

40.1(18)

3!

30.1(6)

2!

20.1(2)

1!

0.111y

=

+++++=

Put n=1 in eqn(1) .....///1y

3!

3h//1y

2!

2h/1y

1!

h1y2y ++++= ----------(3)



( )

( )

99.5907///1

y21

x//1

y16x/1

6yIV1

y

25.8968//1

y21

x)1y2(1/1

y14x///1

y

7.8853

2.702121.11.2332)2(1.1)(1/1

y21

x)1y(112x//1

y

2.70211.2332)(121.1)1y(121

x/1

y

=++=

=+++=

=

++=++=

=+=+=


( ) ( )

( )

1.5475

.......(99.5907)4!

40.1

(25.8968)3!

30.1(7.8853)

2!

20.1(2.7021)

1!

0.11.23322y

=

++

+++=

similarly 9785.13y =

Thus

5475.12y = 9785.13y =

Adam Predictor formula is given by

( )09f137f259f355f24

h3yp4,y −+−+= ---------(4)

( ) ( )iy12ix)iy,if(xif +==

( ) ( ) ( ) ( ) 211210y120x0f =+=+=

( ) ( ) ( ) ( )

7021.2

1.2332121.11y121x1f

=

+=+=

( ) ( ) ( ) ( )

6684.3

1.5475121.22y122x2f

=

+=+=

( ) ( ) ( ) ( )

0336.5

1.9785121.33y123x3f

=

+=+=

10x =

2332.11y =

1.11x = 1.22x = 1.33x = 1.44x =

10y = ?4y =

ix iy

10x = 10y =

1.11x = 2332.11y =

1.22x = 5475.12y =

1.33x = 9785.13y =



( ) ( ) ( ) ( )( )

8707.6

191.568372.2975593.2345524

0.54.968p4,y

becomes Eqn(4)

=

−+−+=

∴


+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y ---------(5)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( ) ( ) ( ) ( ) ( )

7.0005

2.5717121.4p4,y124xp4,y,4xf0

4y,4xf

(0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

+=+==

=

+−++=

( ) ( ) ( )( )

2.5743

2.70213.668455.0336197.0005924

0.11.9785

(1)c4,

y

=

+−++=∴


( ) ( ) ( )

2.5745(3)

c4,ySimilarly 2.5745

(2)c4,

y

7.0056

2.5743121.4(1)

c4,y12

4x(1)

c4,y,4xf

(1)4

y,4xf(1)4

f

where1f25f319f(1)4

9f24

h3y

(2)c4,

y

==∴

=

+=

+=

=

=

+−++=

Since (3)

c4, y&

(2)c4,


y(1.4)=2.5745



Problem(3):

Solve for y(0.4) given that ;2yxdx

dy−= y(0)=1, y(0.1)=0.9117,

y(0.2)=0.8494, y(0.3)=0.8061 by Adam-Bashforth Predictor Corrector method correct to four decimal places.

Soln 0.1h,2yxy)f(x, :dataGiven : =−=


( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

( )2iy-ix)iy,if(xif ==

( ) ( ) 121-020y-0x0f −===

( ) ( )

7311.0

20.9117-0.121y-1x1f

−=

==

( ) ( )

0.5214

20.8494-0.222y-2x2f

−=

==

( ) ( )

0.3497

20.8061-0.323y-3x3f

−=

==

( ) ( ) ( ) ( )( )

7789.0

1-90.7311-370.5214-590.3497-5524

0.10.8061p4,y

becomes Eqn(1)

=

−+−+=

∴


00x =

10y = 9117.01y =

0.11x = 0.22x =

8494.02y =

0.33x =

8061.03y =

0.44x =

?4y =

0.33x = 8061.03y =

0.22x = 8494.02y =

0.11x = 9117.01y =

00x = 10y =

ix iy



+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y ---------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( ) ( ) ( )

( ) ( ) ( )

0.7784

0.7311

0.5214-50.3497-190.2066-9

24

0.10.8061

(1)c4,

y

0.206620.77890.42p4,y4xp4,y,4xf

(0)4

y,4xf(0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

−

−++=∴

−=−=−==

=

+−++=


( )

( ) ( ) ( )( )

0.7785(3)

c4,ySimilarly

0.7785

0.73110.5214-50.3497-190.2059-924

0.10.8061

(2)c4,

y

0.205920.77840.42

(1)c4,

y4x(1)

c4,y,4xf

(1)4

f

where1f25f319f(1)4

9f24

h3y

(2)c4,

y

=

=

−−++=∴

−=−=

−=

=

+−++=

Since (3)

c4, y&

(2)c4,


y(0.4)=0.7785 Problem(4):

Solve ;2

xy

dx

dy= for x=0.4 using Adam-Bashforth Predictor Corrector

method correct to four decimal places. Given that y(0)=1, y(0.1)=1.01, y(0.2)=1.022, y(0.3)=0.1023.



Soln: 0.1h,2

xyy)f(x, :dataGiven ==


( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

ix iy

2iyix

)iy,if(xif ==

00x = 10y = 0

2

0(1)

20y0x

0f ===

0.11x = 01.11y = 0505.0

2

0.1(1.01)

21y1x

1f ===

0.22x = 022.12y = 1022.0

2

0.2(1.022)

22y2x

2f ===

0.33x = 023.13y = 1534.0

2

0.3(1.023)

23y3x

3f ===

( ) ( ) ( ) ( )( )

0408.1

090.0505370.1022590.15345524

0.11.023p4,y

becomes Eqn(1)

=

−+−+=

∴


+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y ---------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=

00x =

10y = 01.11y =

0.11x = 0.22x =

022.12y =

0.33x = 0.44x =

023.13y = ?4y =




( )

( ) ( ) ( )

1.0410

0505.0

0.102250.1534190.20819

24

0.1023.1

(1)c4,

y

2081.02

)0.4(1.0408

2

p4,y4xp4,y,4xf0

4y,4xf

(0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

+

−++=∴

====

=

+−++=


( ) ( ) ( )

1.0410

0505.0

0.102250.1534190.20829

24

0.1023.1

(2)c4,

y

2082.02

)0.4(1.0410

2

(1)c4,

y4x(1)

c4,y,4xf

(1)4

f

where1f25f319f(1)4

9f24

h3y

(2)c4,

y

=

+

−++=∴

===

=

+−++=

Since (2)

c4, y&

(1)c4,


y(0.4)=1.0410 Problem(5):

Solve ;yx

1

dx

dy

+= for x=0.8 using Adam-Bashforth Predictor Corrector

method correct to four decimal places. Given that y(0)=2, y(0.2)=2.0932, y(0.4)=2.1754, y(0.6)=2.2492.

Soln:

0.2h,yx

1y)f(x, :data Given =

+=

2492.23y =

Adam- Predictor formula is given by

( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

00x =

0932.21y =

0.21x = 0.42x =

20y = 1754.22y =

0.33x = 0.44x =

?4y =



ix

iy iyix

1)iy,if(xif

+==

00x =

20y =

5.020

1

0y0x

10f =

+=

+=

0.21x =

0932.21y =

4360.00932.22.0

1

1y1x

11f =

+=

+=

0.42x =

1754.22y =

3882.01754.24.0

1

2y2x

12f =

+=

+=

0.63x =

2492.23y = 3509.02492.26.0

1

3y3x

13f =

+=

+=

( ) ( ) ( ) ( )( )

2.3160

0.590.4360370.3882590.35095524

0.22.2492p4,y

becomes Eqn(1)

=

−+−+=

∴


+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y ---------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( )

( ) ( ) ( )( )

2.3162

0.43600.388250.3509190.3209924

0.22.2492

(1)c4,

y

0.32092.31600.8

1

p4,y4x

1p4,y,4xf0

4y,4xf

(0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

+−++=∴

=+

=+

==

=

+−++=




( ) ( ) ( )( )

2.3162

0.43600.388250.3509190.3209924

0.22.2492

(2)c4,

y

0.32092.31620.8

1

(1)c4,

y4x

1(1)c4,

y,4xf(1)4

f

where1f25f319f(1)4

9f24

h3y

(2)c4,

y

=

+−++=∴

=+

=

+

=

=

+−++=

Since (2)

c4, y&

(1)c4,


y(0.8)=2.3162 Problem(6): Using Adam-Bashforth Predictor Corrector method evaluate y(1.4) if y

satisfies 2x

1

x

y

dx

dy=+ and y(1)=1, y(1.1)=0.996, y(1.2)=0.986,

y(1.3)=0.972 correct to three decimal places.

Soln: 0.1h,2x

xy1xy

2x

1y)f(x, :dataGiven =

−=−=

Adam-Bashforth Predictor formula is given by

( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

10x =

996.01y =

1.11x = 1.22x =

10y = 986.02y =

1.33x = 1.44x =

972.03y = ?4y =



ix

iy ( )2ix

iyix-1)iy,if(xif ==

00x =

10y = ( ) ( )0

21

1x1-1

20x

0y0x-10f ===

.111x =

996.01y = ( ) ( )

0.079

21.1

1.1x0.996-1

21x

1y1x-11f

−=

==

1.22x =

986.02y = ( ) ( )

0.127

21.2

1.2x0.986-1

22x

2y2x-12f

−=

==

1.33x =

972.03y = ( ) ( )

155.0

21.3

1.3x0.972-1

23x

3y3x-13f

−=

==

( ) ( ) ( ) ( )( )

955.0

090.079-370.127-590.155-5524

0.10.972p4,y

becomes Eqn(1)

=

−+−+=

∴

Adam’s-Bashforth Corrector formula is given by

+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y --------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( )( ) ( )

( ) ( ) ( )( ) 955.00.0790.171-50.155-190.171-924

0.10.972

(1)c4,

y

-0.17121.4

)1.4)(0.955-1

24x

p4,y4x-1p4,y,4xf)

(0)4

y,4((0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=−−++=∴

=====

+−++=

xf

Since (1)

c4, y&p4,y are the same up to four decimal places

y(1.4)=0.955



Problem(7): Using Adam-Bashforth Predictor Corrector method obtain the solution of

y2xdx

dy−= at x=0.4 correct to four places of decimals given that

x: 0 0.1 0.2 0.3

y: 1 0.9051 0.8212 0.7491

Soln: 0.1hy,-2xy)f(x, :dataGiven ==

Adam-Bashforth Predictor formula is given by

( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

ix

iy

( ) iy-2ix)iy,if(xif ==

00x =

10y =

( ) ( ) 11-200y-20x0f −===

.101x =

9051.01y =

( ) ( ) 8951.09051.0-20.11y-21x1f −===

0.22x =

8212.02y =

( ) ( ) 7812.08212.0-20.22y-22x2f −===

0.33x =

7491.03y =

( ) ( ) 6591.07491.0-20.33y-23x3f −===

00x =

9051.01y =

0.11x = 0.22x =

10y = 8212.02y =

0.33x = 0.44x =

7491.03y = ?4y =



( ) ( ) ( ) ( )( )

6896.0

1-90.8951-370.7812-590.6591-5524

0.10.7491p4,y

becomes Eqn(1)

=

−+−+=

∴


+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y --------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( ) ( ) ( )

( ) ( ) ( )( )

.68960

0.89510.7812-50.6591-190.5296-924

0.10.7491

(1)c4,

y

-0.52960.6896-20.4p4,y-24xp4,y,4xf

(0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

=

−−++=∴

====

+−++=

Since (1)

c4, y&p4,y are the same up to four decimal places

y(0.4)=0.6896 Problem(8): Using Adam-Bashforth Predictor Corrector method obtain the solution of

yx2edx

dy−= for x=0.4 under the conditions y(0)=2, y(0.1)=2.010,

y(0.2)=2.040 and y(0.3)=2.090 correct to four decimal places.

Soln: 0.1h-y,x2ey)f(x, :dataGiven ==

Adam’s Predictor formula is given by

( )09f137f259f355f24

h3yp4,y −+−+= --------(1)

00x =

010.21y =

0.11x = 0.22x =

20y = 040.22y =

0.33x = 0.44x =

090.23y = ?4y =



ix

iy

iyx

2e)iy,if(xifi −==

00x =

20y =

0202e0yx

2e0f0i =−=−=

.101x =

010.21y =

0.20032.0100.12e1yx

2e1fi =−=−=

0.22x =

040.22y =

0.40282.0400.22e2yx

2e2f2 =−=−=

0.33x =

090.23y =

0.60972.0900.32e3yx

2e3f3 =−=−=

( ) ( ) ( ) ( )( )

2.1615

090.2003370.4028590.60975524

0.12.090p4,y

becomes Eqn(1)

=

−+−+=

∴


+−++=

+1f25f319f

(r)4

9f24

h3y

1)(rc4,

y --------(2)

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

ywhere(0)4

y,4xf(0)4

f ≠==

=


( )

( ) ( ) ( )( )

2.1615

0.20030.402850.6097190.8221924

0.12.090

(1)c4,

y

2.16150.42ep4,yx

2ep4,y,4xf)(0)4

y,4((0)4

f

where1f25f319f(0)4

9f24

h3y

(1)c4,

y

4

=

+−++=∴

−=−===

+−++=

xf

Since (1)

c4, y&p4, y are the same up to four decimal places

y(0.4)=2.1615



Problem(8): Using Adam-Bashforth Predictor Corrector method obtain the solution of

2yxdx

dy−= at x=0.8 correct to four places of decimals given that

x: 0 0.2 0.4 0.6

y: 0 0.0200 0.0795 0.1762

{Ans: y(0.8)=0.2416}

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