Unit2 vrs

Picard’s Method for Simultaneous and second order ODE 1

Dr. V. Ramachandra Murthy

Unit-II: Numerical Methods-II

Numerical solution of simultaneous first order differential equations

Picard’s method

__________________________________________________________

Problem (1):

Soln:

x0=0 x1=0.1

y0=2 y(x1)=?

z0=1 z(x1)=?

0000 z)z(xandy)y(xconditionsinitialwith

(2)z)y,φ(x,dx

dz

(1)z)y,f(x,dx

dy

==

−−−−−−=

−−−−−−=

1,2,3.....n

(4))dxz,yφ(x,zz

(3))dxz,yf(x,yy

by given is formula iterative sPicard' The

x

x1-n1-n0n

x

x1-n1-n0n

0

0

=

−−−−+=

−−−−+=

∫

∫


(2)z)y,φ(x,dx

dz (1)z)y,f(x,

dx

dy

equationaldifferentiussimultaneoorderfirstaConsider

==

−−−−−−=−−−−−−=

z(0.1)andy(0.1)ofvaluesthetheobtainHence1.z(0)2,y(0);yxdx

dz

;zxdx

dyofionapproximatsecondthefindMethodsPicard'Using

2==−=

+=

( ) ( ) 2yxzy,x,φ;zxzy,x,fGiven −=+=



( )

( )( )

( ) ( )

2

xx2x

2

x2y

dx1x2dxzxy y

Eqn(1) in 1n Put

)1(

dxy-xz)dxz,yφ(x,zz

dxzxy)dxz,yf(x,yy


2x

0

2

1

x

0

x

x001

x

x

2

1-n0

x

x1-n1-n0n

x

x1-n0

x

x1-n1-n0n

0

00

00

++=

++=∴

++=++=

=

−−−−

+=+=

++=+=

∫∫

∫∫

∫∫

( ) ( )

( )

(2)6

3x

2

2x3x2

x

0

dx2

2x3x122 y

x

0

dx2

2x4x1x2

x

x

dx1zx0y2 y

Eqn(1) in 2n Put

2

2x4x1

x

0

4x2

2x1

x

0

dx4-x1x

x

dx20y-x0z1z

0

0

−−−+−+=

+−+=

+−++=++=

=

+−=

−+=

+=

+=

∫

∫∫

∫∫

( )

(3)20

5x

4

4x3x2

2x34x1

x

0

dx4

4x3x23x3x41

x

0

dx22x3x4x4

4x2x4-x1

x

0

dx

2

2

2xx2-x1

x

0x

dx21y-x0z2z

−−−−−−

++++−=

∫

++++−=

∫

++++++=

∫

+++=∫

+=



( ) ( )1151.2

6

30.1

2

20.13)1.0(2 y(0.1)

Eqn(2) From

=+−+=

( )( )

( ) ( )0.5839

20

50.1

4

40.130.12

20.134(0.1)1z(0.1)

Eqn(3) From

=

++++−=

Problem (2):

Soln:

( ) 0.xwhen2

1z & 1y;zy3x

dx

dz;z

dx

dythatgiven

zandytoionapproximatsecondthefindtoMethodsPicard'Apply

===+==

( ) ( )

1/2z&1y,0x

z)(yxzy,x,φ;zzy,x,fGiven

000

3

===

+==

( )

( ) 4x

0

3x

x00

3

01

x

0

x

x001

x

x

x

x1-n1-n

3

01-n1-n0n

x

x1-n0

x

x1-n1-n0n

x8

3

2

1dx

2

11x

2

1dxzyxzz

2

x1dx

2

11dxzy y

Eqn(1) in 1n Put

)1(

dxzyxz)dxz,yφ(x,zz

dxzy)dxz,yf(x,yy


0

0

0 0

00

+=

++=++=

+=+=+=

=

−−−

++=+=

+=+=

∫∫

∫∫

∫ ∫

∫∫

( )

85

4

x

0

43

x

x11

3

02

5x

0

4x

x102

x64

3

10

xx

8

3

2

1

dxx8

3

2

1

2

x1x

2

1

dxzyxzz

x40

3

2

x1dxx

8

3

2

11dxzy y

Eqn(1) in 2n Put

0

0

+++=

++

++=

++=

++=

++=+=

=

∫

∫

∫∫



__________________________________________________________

Problem (3):

Soln:

x0=0 x1=0.1 x2=0.2

y0=0 y(x1)=? y(x2)=?

z0=1 z(x1)=? z(x2)=?

576

3x

60

x

40

3x

2

x1

dx64

3x

10

x

8

3x

2

11dxzy y

Eqn(1) in 3n Put

965

x

0

854x

x203

0

++++=

++++=+=

=

∫∫

( )

256

x

360

x7x

64

3

10

xx

8

3

2

1

dx64

x3

10

xx

8

3

2

1

40

x3

2

x1x

2

1

dxzyxzz

1298

54

x

0

854

53

x

x22

3

030

+++++=

++++

+++=

++=

∫

∫

( ) ( ) yxzy,x,φ;zxzy,x,fGiven −=+=

1.z(0)and

0y(0)y,xdx

dz;zx

dx

dythatgivenplaces decimal three to correct

2.0&0.1xforionapproximatsuccessiveMethodsPicard'usingbySolve

=

=−=+=

=

( )

( )

( ) ( )

( ) ( ) 1.0052

x1dx0-x1 dxy-xzz

0.105x2

xdx1x0dxzxyy

Eqn(1) in1nPut

)1(

dxy-xz)dxz,yφ(x,zz

dxzxy)dxz,yf(x,yy


0.1

0

20.1

0

x

x001

0.1

0

0.1

0

2x

x001

x

x1-n0

x

x1-n1-n0n

x

x1-n0

x

x1-n1-n0n

0

0

00

00

=

+=+=+=

=

+=++=++=

=

−−−−

+=+=

++=+=

∫∫

∫∫

∫∫

∫∫



Now to find y(0.2) and z(0.3)

( ) ( )

( ) ( ) 0.994x0.1052

x1dx0.105-x1 dxy-xzz

0.105 x1.0052

xy

dx1.005x0dxzxyy

Eqn(1) in2nPut

0.1

0

20.1

0

x

x102

0.1

0

2

2

0.1

0

x

x102

0

0

=

−+=+=+=

=

+=

++=++=

=

∫∫

∫∫

( ) ( )

( ) ( )

0.994z(0.1)0.105,y(0.1)Thus

0.994dx0.105x1dxyxzz

0.105dx0.994x0dxzxyy

getweEqn(1), in3nPut

x

x

0.1

0203

0.1

0

x

x203

0

0

==

=−+=−+=

=++=++=

=

∫ ∫

∫∫

( ) ( )

( ) ( )

( ) ( )

( ) ( ) 0.219dx0.987x0.105dxzxyy

Eqn(1)in3nPut

0.987dx0.219-x0.994 dxy-xzz

0.219dx0.998x0.105dxzxyy

Eqn(1)in2nPut

0.998dx0.105-x0.994 dxy-xzz

0.2

0.1

x

x203

0.2

0.1

x

x102

0.2

0.1

x

x102

0.2

0.1

x

x001

0

0

0

0

=++=++=

=

=+=+=

=++=++=

=

=+=+=

∫∫

∫∫

∫∫

∫∫

( ) ( ) 219.0dx0.994x0.105dxzxyy

Eqn(1)in1nPut

0.994z&0.105y,0.1xLet

0.2

0.1

x

x001

000

0

=++=++=

=

===

∫∫



0.987z(0.2)places

decimalthreetocorrectsametheare3zand2zSimilarly

0.219y(0.2)

placesdecimalthreetocorrectsametheare3yand2y Since

=

=

Problem (4):

Soln:

x0=0 x1=0.1

y0=1 y(x1)=?

z0=1 z(x1)=?

( ) ( )

0.987

0.2

0.1

0.219x-2

2x0.9943z

0.2

0.1

dx0.219-x0.994 x

x

dx2y-x0z3z

0

=

+=∴

+=+= ∫∫

1.z(0)1,y(0)

withxydx

dzandxz

dx

dythatgivenplacesdecimalfourtocorrect

0.1xfor ionapproximatsuccessiveofMethodsPicard'usingbySolve

==

+=−=

=

( ) ( ) xyzy,x,φ;xzzy,x,fGiven +=−=

( )

( )

( ) ( )

( ) ( ) 1.1052

xx1dxx11 dxxyzz

1.0952

xx1dxx-11dxxzyy

Eqn(1) in1nPut

)1(

dxxyz)dxz,yφ(x,zz

dxxzy)dxz,yf(x,yy


0.1

0

20.1

0

x

x001

0.1

0

0.1

0

2x

x001

x

x1-n0

x

x1-n1-n0n

x

x1-n0

x

x1-n1-n0n

0

0

00

00

=

++=++=++=

=

−+=+=−+=

=

−−−−

++=+=

−+=+=

∫∫

∫∫

∫∫

∫∫



( ) ( )

1.1155

0.1

02

2x1.1055x13z

0.1

0

dxx1.10551 x

x

dxx2y0z3z

0

=

++=∴

++=++= ∫∫

__________________________________________________________

( ) ( )

( ) ( )

1.11452

x1.095x1z

dxx1.0951 dxxyzz

1.10552

x1.105x1y

dxx-1.1051dxxzyy

Eqn(1) in2nPut

0.1

0

2

2

0.1

0

x

x102

0.1

0

2

2

0.1

0

x

x102

0

0

=

++=∴

++=++=

=

−+=∴

+=−+=

=

∫∫

∫∫

( ) ( )

1.10642

x1.1145x1y

dxx-1.11451dxxzyy

Eqn(1) in3nPut

0.1

0

2

3

0.1

0

x

x203

0

=

−+=∴

+=−+=

=

∫∫

1.1156z(0.1)

places,decimalfourtoupsamethearezandzsinceAlso

1.1065y(0.1)

placesdecimalfourtoupsametheareyandySince

1.1156z ,1.1065y:5n when

1.1156z ,1.1065y:4n when

Similarly

54

54

55

44

=

=

===

===



Problem (5):

Soln:

t0=0 t1=0.1

x0=1 x( t0)=?

y0=-1 z( t0)=?

0.1taty&xofvalues

theDeduce1.y(0);xtydt

dyand1x(0);txy

dt

dx equations

thesatisfy y and x that Giveny.andxofvaluesthetoionapproximat

ordersecondthefindionapproximatsuccessiveofMethodsPicard'Using

=

−=+==+=

( ) ( ) xtyyx,t,φ;txyyx,t,fGiven +=+=

( )

( )

( ) ( )

( ) ( )2

tt1dx1t-1-dtxtyyy

2

tt-1

2

tt-1dtt1-1dttyxxx

Eqn(1) in1nPut

dtxtyy)dty,xφ(t,yy

dttyxx)dty,xf(t,xx


2t

0

t

t0001

2t

0

t

0

2t

t0001

t

t1-n1-n0

t

t1-n1-n0n

t

t1-n1-n0

t

t1-n1-n0n

0

0

00

00

−+−=++=++=

+=

++=++=++=

=

++=+=

++=+=

∫∫

∫∫

∫∫

∫∫

( )

( )

(3)8

t

2

ttt1-dt

2

t-

2

3t2t-11

dt2

tt-1

2

tt1t1-dtxtyyy

(2)20

t

4

t

3

2t

2

3tt-1dt

4

t-t2t-3t1-1

dtt2

tt1

2

tt-11dttyxxx

Eqn(1) in2nPut

432

t

0

32

t

0

22t

t1102

5432t

0

432

t

0

22t

t1102

0

0

−−−

−+−+=

++−=

++

−+−+=++=

−−

−+−+=

+++=

+

−+−

++=++=

=

∫

∫∫

∫

∫∫



__________________________________________________________

Numerical solution of Second order Ordinary differential equations

__________________________________________________________

Problem (1):

-0.9095.1) y(0

0.9143.1) x(0

(3)&(2) EquationsFrom

=

=

( )

( )/0000

/

/

/

00

/

00

///

y)z(x&y)y(xwith

(3)zy,x,FzbecomesEqn(1)then

(2),zySet

y)(xyandy)y(xconditionsinitial with

(1)-----yy,x,F y

formthe of O.D.E order second a solve to wishweSuppose

==

−−−−−=

−−−−−=

==

=

/

0000 y)z(x&y)y(xwithz)y,F(x,dx

dz&z

dx

dyi.e,

Eqn(3).&Eqn(2)bygivensO.D.E' ofsystemasolvingofproblema

toreducedis(1)formofO.D.EordersecondsolvingofproblemtheThus

====

places.decimalthreetocorrect(1)y1y(1);0xdx

dyy

dx

yd

equationaldifferenti the of 1.2 xand 1.1 xat solution

the toionapproximatordersecondtheobtainMethodsPicard'Using

/32

2

2

===−+

==



( ) ( ) z2y3xzy,x,φ&zzy,x,f

1z(1)1,y(1)conditionsinitialwith03xz2ydx

dz

becomesequationaldifferentigiventhethen,zdx

dySet

:ln

−==∴

===−+

=

So

x0=1 x1=1.1 x2=1.2

y0=1 y(x1)=? y(x2)=?

z0=1 z(x1)=? z(x2)=?

( ))1(

dxzy-xz)dxz,yφ(x,zz

dxzy)dxz,yf(x,yy


x

x

x

x1-n

21-n

301-n1-n0n

x

x1-n0

x

x1-n1-n0n

0 0

00

−−−

+=+=

+=+=

∫ ∫

∫∫

( )

( ) ( )

( ) ( )

1.1061y

1.0160x1dx1.01601dxzyy

Eqn(1) in2nPut

1.0160z

x4

x1dx1-x1dxzy-xzz

1.1x11dx1dxzyy

Eqn(1) in1nPut

2

1.1

1

1.1

1

x

x102

1

1.1

1

1.1

1

43

x

x0

20

301

1.1

1

1.1

1

x

x001

0

0

0

=∴

+=+=+=

=

=∴

−+=+=+=

=+=+=+=

=

∫∫

∫∫

∫∫

( ) ( )

0.99301.2293x4

x1

dx1.2293-x1dxzy-xzz

1.1

1

4

1.1

1

3x

x1

2

1

3

020

=

−+=

+=+= ∫∫



( ) 1.20091.21.1

0.9930x1.10161y

1.2

1.1

0.9930dx1.1016x

x

dx0z0y1y

Eqn(1) in1nPut

0.99300z;1.10160y;1.10xLet

y(1.2)find To

0

=+=∴

+=+=

=

===

∫∫

Problem(2):

Soln:

x0=0 x1=0.1

y0=0.5 y(x1)=?

z0=0.1 z(x1)=?

1.20401.21.1)0248.1(1.10162y

1.2

1.1

1.0248dx1.1016x

x

dx1z0y2y

Eqn(1) in2nPut

1.02481.2

1.1

dx1.2050-3x0.9930x

x

dx0z20

y-3x0z1z

0

0

=+=∴

+=+=

=

=

+=

+=

∫∫

∫∫

decimal. of places four tocorrect

0.1(0) y0.5,y(0)with0ydx

dy2x

dx

yd equation aldifferenti the of

0.1 xat solution eapproximatorderthirdtheobtainMethodsPicard'Using

/

2

2

===++

=

( ) ( ) ( )2xzy-zy,x,φ&zzy,x,f

0.1z(0)0.5,y(0)conditionsinitialwith0y2xzdx

dz


dySet

+==∴

===++

=



)1(x

x

x

x

dx1-n2xz1-n

y0z)dx1-nz,1-nyφ(x,0znz

x

x

dx1-nz0yx

x

)dx1-nz,1-nyf(x,0yny


0 0

00

−−−

+−=+=

+=+=

∫ ∫

∫∫

__________________________________________________________

Problem (3):

( ) ( )

( ) ( )( )

( ) ( ) 0.50490.049x0.5dx0.0490.5dxzyy

Eqn(1) in2nPut

0.0492

x0.20.5x-0.1z

dx(0.1)x20.5-0.1dx2xzyzz

0.510.1x0.5dx0.10.5dxzyy

Eqn(1) in1nPut

0.1

0

0.1

0

x

x102

0.1

0

2

1

0.1

0

x

x0001

0.1

0

0.1

0

x

x001

0

0

0

=+=+=+=

=

=

+=∴

+=+−+=

=+=+=+=

=

∫∫

∫∫

∫∫

( ) ( )

0.5048y(0.1)

0.50480.0485x0.5dx0.04850.5dxzyy

Eqn(1) in3nPut

0.1

0

0.1

0

x

x203

0

=∴

=+=+=+=

=

∫∫

( ) ( )

0.04852

x0.0980.51x-0.1z

dx0.098x0.51-0.1dx2xzyzz

0.1

0

2

2

0.1

0

x

x1102

0

=

+=∴

+=+−+= ∫∫

0.1(0) yand

1y(0)with06ydx

dy3x

dx

yd equation aldifferenti the of 0.2x

at solutioneapproximatorderthirdthefindMethodsPicard'Employing

/

2

2

=

==−+=



Soln:

( ) ( ) 1.25881.294x1dx1.2941dxzyy

Eqn(1) in2nPut

0.2

0

0.2

0

x

x102

0

=+=+=+=

=

∫∫

x0=0 x1=0.2

y0=1 y(x1)=?

z0=0.1 z(x1)=?

( ) ( ) 3xz-6yzy,x,φ&zzy,x,f

0.1z(0)1,y(0)conditionsinitialwith06y3xzdx

dz


dySet

==∴

===−+

=

)1(x

0x

x

0x

dx1-n3xz-1-n

6y0z)dx1-nz,1-nyφ(x,0znz

x

0x

dx1-nz0yx

0x



−−−

∫ ∫+=+=

∫+∫ =+=

( ) ( )

( ) ( )( )

1.2942

x0.36x0.1z

dxx0.360.1dx3xz6yzz

1.020.1x1dx0.11dxzyy

Eqn(1) in1nPut

0.2

0

2

1

0.2

0

x

x0001

0.2

0

0.2

0

x

x001

0

0

=

−+=∴

−+=−+=

=+=+=+=

=

∫∫

∫∫

( ) ( )

1.2492y(0.2)

1.24921.2463x1dx1.24631dxzyy

Eqn(1) in3nPut

0.2

0

0.2

0

x

x203

0

=∴

=+=+=+=

=

∫∫

( ) ( )

1.24632

x3.8826.12x0.1z

dx3.882x-6.120.1dx3xz6yzz

0.2

0

2

2

0.2

0

x

x1102

0

=

−+=∴

+=−+= ∫∫



__________________________________________________________

Problem (4):

Soln:

x0=0 x1=0.2

y0=1 y(x1)=?

z0=0 z(x1)=?

0.xwhen0dx

dy ,1ywith0y

dx

dyx

dx


0.2 xat solutioneapproximatorderthirdthefindMethodsPicard'Using

2

2

====−+

=

( ) ( ) xz-yzy,x,φ&zzy,x,f

0z(0)1,y(0)conditionsinitialwith0yxzdx

dz


dySet

==∴

===−+

=

)1(x

0x

x

0x

dx1-nz-1-n

y0z)dx1-nz,1-nyφ(x,0znz

x

0x

dx1-nz0yx

0x



−−−

∫ ∫

+=+=

∫+∫ =+=

x

( )

( ) ( )

( ) ( ) 1.040.20

0.2x10.2

0

dx0.21x

0x

dx1z0y2y

Eqn(1) in2nPut

0.2

0

0.2dx010x

0x

dx0xz0y0z1z

10.2

0

dx01x

0x

dx0z0y1y

Eqn(1) in1nPut

=+∫ =+∫ =+=

=

∫ =−+=∫ −+=

∫ =+∫ =+=

=



( ) ( )

( ) ( )

1.0392y

1.03920.20

0.196x10.2

0

dx0.1961x

x

dx2z0y3y

Eqn(1) in3nPut

0.196

0.2

02

2x0.2x2z

0.2

0

dx0.2x-10x

x

dx1xz1y0z2z

0

0

=∴

=+=+=+=

=

=

−=∴

+=−+=

∫∫

∫∫

_________________________________________________________

Problem (5):

Soln:

x0=0 x1=0.1 x2=0.2

y0=1 y(x1)=? y(x2)=?

z0=0.5 z(x1)=? z(x2)=?

0.5(0)y1,y(0)conditions

initialwith0xydx

yd equationaldifferenti the of 0.2 xand 0.1x

whensolutioneapproximatorderthirdthefindMethodsPicard'usingBy

/

2

2

==

=+==

( ) ( ) -xyzy,x,φ&zzy,x,f

0.5z(0)1,y(0)conditionsinitialwith0xydx

dz


dySet

==∴

===+

=

)1(x

0x

x

0x

dx1-nxy0z)dx1-nz,1-nyφ(x,0znz

x

0x

dx1-nz0yx

0x



−−−

∫ ∫ −+=+=

∫+∫ =+=



( ) ( )

( ) ( )

( ) ( ) 1.04950.10

0.495x10.1

0

dx0.4951x

x

dx1z0y2y

Eqn(1) in2nPut

0.495

0.1

02

2x0.1

0

0.5dxx0.5x

x

dx0xy0z1z

1.050.10

0.5x10.1

0

dx0.51x

x

dx0z0y1y

Eqn(1) in1nPut

0

0

0

=+=+=+=

=

=

−=−=−=

=+=+=+=

=

∫∫

∫∫

∫∫

To find y (0.2):

( ) ( )

( ) ( )

( ) ( )

0.4947z(0.1)

0.4947

2

x1.04950.5dx1.0495x0.5dxxyzz

1.0494y(0.1)

1.04940.4947x1dx0.49471dxzyy

Eqn(1) in3nPut

0.4947

2

x1.050.5dx1.05x0.5dxxyzz

0.1

0

20.1

0

x

x203

0.1

0

0.1

0

x

x203

0.1

0

20.1

0

x

x102

0

0

0

=∴

=

−=−=−=

=∴

=+=+=+=

=

=

−=−=−=

∫∫

∫∫

∫∫

( )

( ) 1.09880.4947x1.0494y

dx0.49471.0494dxzyy

Eqn(1) in1nPut

0.4947z&1.0494y,0.1 xLet

0.2

0.11

0.2

0.1

x

x001

000

0

=+=∴

+=+=

=

===

∫∫



( ) ( )

0.4789

0.2

0.12

2x1.04940.49471z

0.2

0.1

dx1.0494x0.4947x

x

dx0xy0z1z

0

=

−=∴

−=−= ∫∫

__________________________________________________________

Problem (6):

Soln:

x0=0 x1=0.2

y0=1 y(x1)=?

z0=0 z(x1)=?

( )

( )

( )

1.0972y(0.2)

1.0972dx0.47821.0494dxzyy

Eqn(1) in3nPut

0.4782dx1.0988x-0.4947dxxyzz

1.0972dx0.47891.0494dxzyy

Eqn(1) in2nPut

0.2

0.1

x

x203

0.2

0.1

x

x102

0.2

0.1

x

x102

0

0

0

=∴

=+=+=

=

==−=

=+=+=

=

∫∫

∫∫

∫∫

0.(0)y1,y(0)withydx

dyx

dx


0.2 xat solutioneapproximatorderthirdthefindMethodsPicard'Using

/2

2

2

2

==−

=

=

( ) ( ) 22

22

yxzzy,x,φ&zzy,x,f

0z(0)1,y(0)conditionsinitialwithyxzdx

dz


dySet

−==∴

==−=

=



)1(x

x

x

x

dx21-n

y-21-n

xz0z)dx1-nz,1-nyφ(x,0znz

x

x

dx1-nz0yx

x



0 0

00

−−−

+=+=

+=+=

∫ ∫

∫∫

_________________________________________________________

( )

( ) ( )( ) ( )

( ) ( )

( ) ( )( ) ( )( )∫∫

∫∫

∫∫

∫∫

=−+=−+=

=−=+=+=

=

=+=−+=

=+=+=

=

0.2

0

2x

x

2

1

2

102

0.2

0

0.2

0

x

x102

0.2

0

x

x

2

0

2

001

0.2

0

x

x001

-0.1992dx10.2-x0dxyzxzz

0.960.2x1dx0.2-1dxzyy

Eqn(1) in2nPut

-0.2dx1-00dxyzxzz

1dx01dxzyy

Eqn(1) in1nPut

0

0

0

0

( )

( )

0.9601y(0.2)

0.96010.1992x1

dx0.1992-1dxzyy

Eqn(1) in3nPut

0.2

0

0.2

0

x

x203

0

=∴

=−=

+=+=

=

∫∫



Numerical solution of simultaneous first order differential equations

Consider the first order simultaneous differential equation

Runge-Kutta Method of 4th Order

__________________________________________________________

Problem (1):

Soln:

x0=0 x1=0.1


(2)z)y,φ(x,dx

dz

(1)z)y,f(x,dx

dy

==

−−−−−−=

−−−−−−=

[ ]

[ ]

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001

432101

432101

lz,kyh,xhφl;lz,kyh,xhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where;l2l2ll6

1zz

k2k2kk6

1yy

+++=+++=

+++=

+++=

+++=

+++=

==

++++=

++++=

places.decimalfour

to correct 0.1xpointthe at1z(0)0,y(0);y-zdx

dz;zy

dx

dy

equationsofsystemthesolveorder4thofMethodKutta-RungeUsing

====+=

( ) ( ) 0.1h;yzzy,x,φ;zyzy,x,fGiven =−=+=



y0=0 y1=?

z0=1 z1=?


[ ]

[ ]

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001

432101

432101


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

+++=+++=

+++=

+++=

+++=

+++=

==

−−−−−−

++++=

++++=

( ) [ ] [ ]

( ) [ ] [ ]

0.112

0.11

2

0.100.1

21l

0z21k

0yh21l

0z,21k

0y,2

h0xhf2k

0.1010.10y0zh0z,0y,0xhφ1l

0.1100.10z0yh0z,0y,0xhf1k

=

+++=

+++=

+++=

=−=−==

=+=+==

0.11052

0.11

2

0.1100.1

2

lz

2

kyh

2

lz,

2

ky,

2

hxhfk

0.12

0.10

2

0.110.1

2

ky

2

lzh

2

lz,

2

ky,

2

hxhφl

20

20

20

2003

10

10

10

1002

=

+++=

+++=

+++=

=

+−+=

+−+=

+++=

0.09952

0.110

2

0.110.1

2

ky

2

lzh

2

lz,

2

ky,

2

hxhφl 2

02

02

02

003

=

+−+=

+−+=

+++=



Substituting all these values in Eqn(1), we get

__________________________________________________________

Problem (2):

Soln:

x0=0 x1=0.1

y0=2 y1=?

z0=1 z1=?

( ) [ ]

[ ]

( ) ( )[ ]

( )[ ] 0.09890.110500.099510.1

kylzhlz,kyh,xhφl

0.1210.099510.110500.1

lzkyhlz,kyh,xhfk

3030303004

3030303004

=+−+=

+−+=+++=

=+++=

+++=+++=

[ ]

( ) ( )[ ]

[ ]

( ) ( )[ ]

1.0996z(0.1)

1.09960.09890.099520.120.16

11

l2l2ll6

1zz

0.1103y(0.1)

0.11030.1210.110520.1120.16

10

k2k2kk6

1yy

432101

432101

=∴

=++++=

++++=

=∴

=++++=

++++=

places.decimalfourtocorrect

0.1xfor1z(0)2,y(0);y-xdx

dz;zx

dx

dyequations

ofsystemthesolveorder4ofMethodKutta-RungeUsing

2

th

====+=

( ) ( ) 0.1h ;yxzy,x,φ;zxzy,x,fGiven 2=−=+=




( ) [ ] [ ]

( ) ( )[ ] [ ]

0.0852

0.4-1

2

0.100.1

2

lz

2

hxh

2

lz,

2

ky,

2

hxhfk

0.4200.1yxhz,y,xhφl

0.1100.1zxhz,y,xhfk

100

10

1002

22

000001

000001

=

++=

+++=

+++=

−=−=−==

=+=+==

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

+++=+++=

+++=

+++=

+++=

+++=

==

[ ]

[ ]

)1(

l2l2ll6

1zz

k2k2kk6

1yy

432101

432101

−−−−−−

++++=

++++=

0.08422

0.4152-1

2

0.100.1

2

lz

2

hxh

2

lz,

2

ky,

2

hxhfk

0.41522

0.12

2

0.110.1

2

ky

2

hxh

2

lz,

2

ky,

2

hxhφl

200

20

2003

2

2

100

10

1002

=

+++=

+++=

+++=

−=

+−+=

+−+=

+++=

( ) [ ]lzhxhlz,kyh,xhfk

0.41212

0.0852

2

0.100.1

2

ky

2

hxh

2

lz,

2

ky,

2

hxhφl

2

2

200

20

2003

+++=+++=

−=

+−+=

+−+=

+++=



Substituting all these values in Eqn (1)

__________________________________________________________

Problem (3):

Soln:

x0=0 x1=0.3

y0=0 y1=?

z0=1 z1=?

( ) ( )[ ]( )[ ] 0.42430.084220.100.1

kyhxhlz,kyh,xhφl

2

2

300303004

−=+−+=

+−+=+++=

[ ]

( ) ( )[ ]

[ ]

( ) ( )[ ]

0.5868z(0.1)

0.58680.42430.4121-20.4152-20.4-6

11

l2l2ll6

1zz

2.0845y(0.1)

2.08450.06870.084220.085520.16

12

k2k2kk6

1yy

432101

432101

=∴

=−+++=

++++=

=∴

=++++=

++++=

0. x when1z and 0 yare values initial The

places.decimalfourtoorrectMethodKutta-Rungeorderfourth

using0.3xforxydx

dz;xz1

dx

dyequationsaldifferentitheSolve

===

=−=+=

c

( ) ( ) 0.3h;xyzy,x,φ;xz1zy,x,fGiven =−=+=




[ ]

[ ]

( ) ( )

+++=

+++=

==

−−−−−−

++++=

++++=

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

10

1002

10

1002

00010001

432101

432101

( ) ( )303004303004

20

2003

20

2003


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

+++=+++=

+++=

+++=

( ) [ ] [ ]

( ) [ ] [ ]

0.3452

01

2

0.3010.3

2

lz

2

hx1h

2

lz,

2

ky,

2

hxhfk

00x0-0.3yx-hz,y,xhφl

0.30x110.3zx1hz,y,xhfk

100

10

1002

000001

000001

=

+

++=

+

++=

+++=

====

=+=+==

0.34482

0.0067-1

2

0.3010.3

2

lz

2

hx1h

2

lz,

2

ky,

2

hxhfk

0.00672

0.30

2

0.300.3

2

ky

2

hxh

2

lz,

2

ky,

2

hxhφl

200

20

2003

100

10

1002

=

+

++=

+

++=

+++=

−=

+

+−=

+

+−=

+++=

( ) ( )( )[ ]

( )( )[ ] 0.38930.0077-10.3010.3

3l0zh0x1h3l0z,3k0yh,0xhf4k

0.00772

0.3450

2

0.303.0

22k

0y2

h0xh

22l

0z,22k

0y,2

h0xhφ3l

=++=

+++=+++=

−=

+

+−=

+

+−=

+++=



Substituting all these values in Eqn (1), we get

__________________________________________________________

Problem (4):

Soln:


t0=0 t1=0.1

x0=1 x1=?

y0=1 y1=?

[ ]

( ) ( )[ ]

[ ]

( ) ( )[ ]

0.9900z(0.3)

0.99000.03100.0077-20.0067-206

11

l2l2ll6

1zz

0.3448y(0.3)

0.34480.38930.344820.34520.36

10

k2k2kk6

1yy

432101

432101

=∴

=−+++=

++++=

=∴

=++++=

++++=

0.t when1 y and 1xthatgivenMethodKutta-Rungeorderfourth

using0.1tattxdt

dy;ty

dt

dxequationsofsystemtheSolve

===

=+=−=

( ) ( ) 0.1h;txyx,t,φ;t-yyx,t,fGiven =+==

[ ]

[ ]

where

)1(

4l32l22l1l6

10z1y

4k32k22k1k6

10y1x

−−−−−−

++++=

++++=



( ) [ ] [ ]

( ) [ ] [ ]

0.12

0.10

2

0.110.1

2

ht

2

lyh

2

ly,

2

kx,

2

hthfk

0.1010.1txhy,x,thφl

0.10-10.1t-yhy,x,thfk

01

01

01

002

000001

000001

=

+−

+=

+−

+=

+++=

=+=+==

====

0.10052

0.10

2

0.1110.1

2

ht

2

lyh

2

ly,

2

kx,

2

hthfk

0.112

0.10

2

0.110.1

2

ht

2

kxh

2

ly,

2

kx,

2

hthφl

02

02

02

003

01

01

01

002

=

+−+=

+−+=

+++=

=

++

+=

++

+=

+++=

( ) ( ) ( )[ ]

( )[ ]

( ) ( ) ( )[ ]htkxhly,kxh,thφl

0.1010.100.1110.1

htlyhly,kxh,thfk

0.112

0.10

2

0.110.1

2

ht

2

kxh

2

ly,

2

kx,

2

hthφl

030303004

030303004

02

02

02

003

+++=+++=

=+−+=

+−+=+++=

=

++

+=

++

+=

+++=




__________________________________________________________

Problem (5):

Soln:


x0=0 x1=0.1

y0=1 y1=?

z0=1 z1=?

[ ]

( ) ( )[ ] 1.10030.1010.100520.120.16

11

k2k2kk6

1xx 432101

=++++=

++++=

[ ]

( ) ( )[ ]

1.1100y(0.1) & 1.1003x(0.1)

1.11000.12000.1120.1120.16

11

l2l2ll6

1yy 432101

==∴

=++++=

++++=

places.decimalfourto

correctMethodKutta-Rungeorderfourthusing1z(0)y(0);0z-dx

dy

;02z4ydx

dzthatgiven0.1xatequationsaldifferentitheSolve

===

=++=

( ) ( ) 0.1h;2z-4yzy,x,φ;zzy,x,fGiven =−==

[ ]

[ ]

( ) ( )00010001

432101

432101

lkhlkh

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

==

−−−−−−

++++=

++++=



( ) [ ] [ ]

( ) [ ] [ ] -0.62x1-4x1-0.12z4y-hz,y,xhφl

0.110.1zhz,y,xhfk

000001

00001

==−==

====

0.072

0.6-10.1

2

lzh

2

lz,

2

ky,

2

hxhfk 1

01

01

002

=

+=

+=

+++=

0.0722

0.56-10.1

2

lzh

2

lz,

2

ky,

2

hxhfk

0.562

0.6-12

2

0.1140.1

2

lz2

2

ky4h

2

lz,

2

ky,

2

hxhφl

20

20

2003

10

10

10

1002

=

+=

+=

+++=

−=

+−

+−=

+−

+−=

+++=

( ) ( )[ ]

[ ]

( ) ( ) ( )[ ]

( ) ( )[ ] 0.51720.558120.072140.1

lz2ky4hlz,kyh,xhφl

0.04420.55810.1

lzhlz,kyh,xhfk

0.5582

0.56-12

2

0.07140.1

2

lz2

2

ky4h

2

lz,

2

ky,

2

hxhφl

3030303004

30303004

20

20

20

2003

−=−−+−=

+−+−=+++=

=−=

+=+++=

−=

+−

+−=

+−

+−=

+++=




Problem (6):

Problem (7):

__________________________________________________________

[ ]

( ) ( )[ ]

[ ]

( ) ( )[ ]

.44110z(0.1)

0.44110.51720.558-20.56-20.6-6

11

l2l2ll6

1zz

1.0713y(0.1)

1.07130.04420.07220.0720.16

11

k2k2kk6

1yy

432101

432101

=∴

=−+++=

++++=

=∴

=++++=

++++=

{ }0.9998z(0.1)0.1050,y(0.1):Ans

places.decimalfourtocorrectMethodKutta-Runge

using0.1xfor1z(0)0,y(0)giveny-xdx

dz;zx

dx

dySolve

==

====+=

( )

{ }0.51z(0.1)1.050,y(0.1):Ans

places.decimalfourtocorrectMethodKutta-Runge

using0.1xfor1/2z(0)1,y(0)givenzy3xdx

dz;z

dx

dySolve

==

===+==




__________________________________________________________

Problem (1):

( )

( ) /0000

/

/

/

00

/

00

///

y)z(x&y)y(x with(3)zy,x,Fz

becomesEqn(1)then(2),zySet

y)(xyandy)y(xconditionsinitialwith

(1)-----yy,x,Fy

formthe of O.D.E order Second a Consider

==−−−−−=

−−−−−=

==

=

/

0000 y)z(x&y)y(xwithz)y,F(x,dx

dz&z

dx

dyi.e,

Eqn(3).&Eqn(2)bygivensO.D.E'ofsystemasolvingofproblem a


====

0(0)/y1,y(0)areconditions

Initialplaces.decimalfourto correct0.2xfor2y

2

dx

ydx

2dx

y2d

equationaldifferentithesolveorderfourthofmethodKuttaRungeUsing

==

=−

=

−



Soln:

h=0.1


x0=0 x1=0.2

y0=1 y1=?

z0=0 -------

( ) ( ) 22

22

yxzzy,x,φ&zzy,x,f

0z(0)1,y(0)conditionswithyxzdx

dz

becomesequationgiventhethen,zdx

dySet

−==∴

==−=

=

[ ]

[ ]

)1(

l2l2ll6

1zz

k2k2kk6

1yy

432101

432101

−−−−−−

++++=

++++=

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

+++=+++=

+++=

+++=

+++=

+++=

==

( ) [ ] [ ]

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

0.022

0.2-00.2

2

lzh

2

lz,

2

ky,

2

hxhfk

0.21000.2yzxhz,y,xhφl

000.2zhz,y,xhfk

10

10

1002

222

0

2

000001

00001

−=

+=

+=

+++=

−=−=−==

====




__________________________________________________________

Problem (2):

0.01992

0.1998-00.2

2

lzh

2

lz,

2

ky,

2

hxhfk

0.19982

01

2

0.2-0

2

0.200.2

2

ky

2

lz

2

hxh

2

lz,

2

ky,

2

hxhφl

20

20

2003

22

2

10

2

100

10

1002

−=

+=

+=

+++=

−=

+−

+

+=

+−

+

+=

+++=

0.19582

0.02-1

2

0.1998-0

2

0.200.2

2

ky

2

lz

2

hxh

2

lz,

2

ky,

2

hxhφl

22

2

20

2

200

20

2003

−=

+−

+

+=

+−

+

+=

+++=

( ) ( )[ ] 0.0391lzhlz,kyh,xhfk 30303004 −=+=+++=

[ ]

( ) ( )[ ]

9801.0y(0.2)

9801.00391.00.0199-20.02-206

11

k2k2kk6

1yy 432101

=

=−+++=

++++=

0.5(0)/y1,y(0)areconditionsInitial

places.decimalfour to correct0.1xfordx

dyy3x

2dx

y2d


==

=

+=

−



Soln:


x0=0 x1=0.1

y0=1 y1=?

z0=0.5 -------

( ) ( ) 0.2h;z)(yxzy,x,φ&zzy,x,f

0.5z(0)1,y(0)conditionswithz)(yxdx

dz


dySet

3

3

=+==∴

==+=

=

[ ]

[ ]

( ) ( )

+++=

+++=

==

−−−−−−

++++=

++++=

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

10

1002

10

1002

00010001

432101

432101

( ) ( )303004303004

20

2003

20

2003


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

+++=+++=

+++=

+++=

( ) [ ] [ ]

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

0.052

00.50.1

2

lzh

2

lz,

2

ky,

2

hxhfk

00.5100.1zyxhz,y,xhφl

0.050.50.1zhz,y,xhfk

10

10

1002

00

3

00001

00001

=

+=

+=

+++=

=+=+==

====



05.02

00.50.1

2

lzh

2

lz,

2

ky,

2

hxhfk

0000.02

00.5

2

0.051

2

0.100.1

2

ky

2

lz

2

hxh

2

lz,

2

ky,

2

hxhφl

20

20

2003

3

10

10

3

0

10

1002

=

+=

+=

+++=

=

+++

+=

++

+

+=

+++=


__________________________________________________________

Problem (3):

( ) ( )[ ] 0.05lzhlz,kyh,xhfk

0.00002

00.5

2

0.051

2

0.100.1

2

lz

2

ky

2

hxh

2

lz,

2

ky,

2

hxhφl

30303004

3

20

20

3

0

20

2003

=+=+++=

=

+++

+=

+++

+=

+++=

[ ]

( ) ( )[ ]

0500.1y(0.1)

0500.105.00.0520.0520.056

11

k2k2kk6

1yy 432101

=∴

=++++=

++++=

places.

decimalfour to correct0.1xfor 1/2(0)/y0.2, y(0)0,4ydx

dy2x

2dx

y2d


====−+

−



Soln:


x0=0 x1=0.1

y0=0.2 y1=?

z0=0.5 -------

( ) ( ) 0.1h;2xz4yzy,x,φ&zzy,x,f

0.5z(0) 0.2,y(0)conditionswith2xz-4ydx

dz


dySet

=−==∴

===

=

[ ]

[ ]

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001

432101

432101


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

+++=+++=

+++=

+++=

+++=

+++=

==

−−−−−−

++++=

++++=

( ) [ ] [ ]

( ) [ ] [ ]

0.0542

0.080.50.1

21l

0zh21l

0z,21k

0y,2

h0xhf2k

0.082x0x0.54x0.20.10z02x04yh0z,0y,0xhφ1l

0.050.50.10zh0z,0y,0xhf1k

=

+=

+=

+++=

=−=−==

====

0.08462

0.080.5

2

0.102

2

0.050.240.1

21l

0z2

h0x2

21k

0y4h

21l

0z,21k

0y,2

h0xhφ2l

=

+

+−

+=

+

+−

+=

+++=




Problem (4):

( ) ( )[ ] 0.05853l0zh3l0z,3k0yh,0xhf4k

0.08532

0.08460.5

2

0.102

2

0.0540.240.1

22l

0z2

h0x2

22k

0y4h

22l

0z,22k

0y,2

h0xhφ3l

=+=+++=

=

+

+−

+=

+

+−

+=

+++=

( ) ( )[ ]

0.2542y(0.1)

0.25420.05850.054220.05420.056

12.01y

=∴

=++++=

order.fourthofmethodKutta-Runge using placesdecimalfour to correct

y(0.1)ofvaluetheFind.0.1xfor5(0)y10,y(0),ydx

yd Given /3

2

2

====



Soln:


x0=0 x1=0.1

y0=10 y1=?

z0=5 -------

( ) ( ) 0.1h;yzy,x,φ&zzy,x,f

5z(0)10,y(0)conditionswithydx

dz


dySet

3

3

===∴

===

=

[ ]

[ ]

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001

432101

432101


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

+++=+++=

+++=

+++=

+++=

+++=

==

−−−−−−

++++=

++++=



( ) [ ] [ ]

( ) ( ) ( ) 1003100.130yh0z,0y,0xhφ1l

0.550.10zh0z,0y,0xhf1k

=

=

==

====

5.52

10050.1

21l

0zh21l

0z,21k

0y,2

h0xhf2k

=+=

+=+++=


5.88442

107.689050.1

2

lzh

2

lz,

2

ky,

2

hxhfk

107.68902

0.5100.1

2

kyh

2

lz,

2

ky,

2

hxhφl

20

20

2003

3

3

10

10

1002

=

+=

+=

+++=

=

+=

+=

+++=

( ) ( )[ ] 2267.213l0zh3l0z,3k0yh,0xhf4k

2671.2073

2

5.5100.1

3

22k

0yh

22l

0z,22k

0y,2

h0xhφ3l

=+=+++=

=

+=

+=

+++=

[ ]

( ) ( )[ ]

17.4159y(0.1)17.4159

21.22675.884425.520.56110

4k32k22k1k61

0y1y

=∴

=

++++=

++++=



Problem (5):

Soln:


x0=0 x1=0.2

y0=1 y1=?

z0=0 -------

places.decimalfourcorrect0(0)y1,y(0);dx

dyxy

dx

ydthat

given ,(0.2)yandy(0.2)findorderfourthofmethodKuttaRungeUsing

/

2

2

/

==+=

−

( ) ( ) 0.2h;xzyzy,x,φ&zzy,x,f

0z(0)1,y(0)conditionswithxzydx

dz


dySet

=+==∴

==+=

=

[ ]

[ ]

( ) ( )

( ) ( )303004303004

20

2003

20

2003

10

1002

10

1002

00010001

432101

432101


2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

2

lz,

2

ky,

2

hxhφl;

2

lz,

2

ky,

2

hxhfk

z,y,xhφl;z,y,xhfk

where

)1(

l2l2ll6

1zz

k2k2kk6

1yy

+++=+++=

+++=

+++=

+++=

+++=

==

−−−−−−

++++=

++++=

( ) [ ] [ ]

( ) [ ] [ ]

0.022

0.200.2

2

lzh

2

lz,

2

ky,

2

hxhfk

0.20x010.2zxyhz,y,xhφl

000.2zhz,y,xhfk

10

10

1002

0000001

00001

=

+=

+=

+++=

=+=+==

====



Substituting all these values in Eqn (1)

0.02022

0.20200.2

2

lzh

2

lz,

2

ky,

2

hxhfk

0.2022

0.20

2

0.20

2

010.2

2

lz

2

hx

2

kyh

2

lz,

2

ky,

2

hxhφl

20

20

2003

100

10

10

1002

=

+=

+=

+++=

=

+

++

+=

+

++

+=

+++=

( ) ( )[ ]

( ) ( ) ( )( )[ ]

( ) ( )( )[ ] 0.21220.204000.200.020210.2

lzhxkyhlz,kyh,xhφl

0.0408lzhlz,kyh,xhfk

0.20402

0.2020

2

0.20

2

0.0210.2

2

lz

2

hx

2

kyh

2

lz,

2

ky,

2

hxhφl

30030303004

30303004

200

20

20

2003

=++++=

++++=+++=

=+=+++=

=

+

++

+=

+

++

+=

+++=

[ ]

( ) ( )[ ]

( ) [ ]

( ) ( )[ ]

( ) 0.20400.2y

0.21220.204020.20220.26

10

l2l2ll6

1zz(0.2)0.2y

1.0202y(0.2)

1.02020.04080.020220.02206

11

k2k2kk6

1yy

/

43210

/

432101

=∴

++++=

++++==

=

=++++=

++++=



Problem (6):

1.1380}y(0.2):{Ans

0.1(0)y1,y(0)are

conditionsInitialplaces.decimalfourcorrect0.2xfor06ydx

dy3x

dx

yd


/

2

2

=

==

==−+

−

Problem (7):

1.0053}y(0.1):{Ans

places.

decimalfourcorrect0(0)y1,y(0);12xydx

dyx

dx

ydthatgiven

,0.1xatsolutionthefindorderfourthofmethodKuttaRungeUsing

/2

2

2

=

===−−

=−

__________________________________________________________




using Milne’s method

__________________________________________________________

Milne’s Predictor-Corrector Method

( )

( ) /0000

/

/

/

00

/

00

///

y)z(x&y)y(x with(3)zy,x,φz

becomesEqn(1)then(2),zySet

y)(xyandy)y(xconditionsinitialwith

(1)-----yy,x,Fy

formthe of O.D.E order Second a Consider

==−−−−−=

−−−−−=

==

=

/

0000 y)z(x&y)y(xwithz)y,φ(x,dx

dz&z

dx

dyi.e,

Eqn(3).&Eqn(2)bygivensO.D.E'ofsystemasolvingofproblema


====

( )

( )

( )

( )( )( )

( )( )

( )(r)

4

(r)

44

r/

4

(r)

c4,

(r)

4p4,

(0)

4

(r)

c4,

(r)

4p4,

(0)

4

r/4

/3

/22

1)(rc4,

(r)4322

1)(rc4,

/3

/2

/10p4,

3210p4,

z,y,xφz;0rforzz& zz

0rforyy& yy where

formula Corrector(2)

z4zz3

hzz

z4zz3

hyy

formula Predictor(1)

2zz2z3

4hzz

2zz2z3

4hyy

=≠==

≠==

→−−

+++=

+++=

→−−

+−+=

+−+=

+

+



j

Problem (1):

x 0 0.2 0.4 0.6

y 0 0.02 0.0795 0.1762

y/ 0 0.1996 0.3937 0.5689

Soln:

x 0 0.2 0.4 0.6 0.8

y 0 0.02 0.0795 0.1762 ?

z 0 0.1996 0.3937 0.5689 -

Milne’s Predictor formula

and

;dx

dy2y1

2dx

y2d thativeny(0.8).computetomethodsMilne'Apply −=G

( ) ( )

Thus

0.2h2yz;1zy,x,φ&zzy,x,f

0z(0)0,y(0)conditionswith2yz1dx

dz


dySet

=−==∴

==−=

=

02.0y1 =1996.0z1 =

0795.0y2 =

1762.0y3 =

3937.0z2 =

5689.0z3 =

2yz-1dx

dzz /

==

9920.0zy21z 11/1 =−=

9374.0zy21z 22/2 =−=

7995.0zy21z 33

/

3 =−=

)y(xy ii =)z(xz ii =



Milne’s Corrector formula

First improvement: {Put r = 0 in Eqn (2)}

( )

( )(1)

2zz2z3

4hz

2zz2z3

4hyy

/

3

/

2

/

10p4,

3210p4,

−−−−−

+−+=

+−+=

z

( ) ( ) ( )( )

( ) ( ) ( )( )

0.7055

0.799520.93740.992023

4(0.2)0z

0.3049

0.568920.39370.199623

4(0.2)0y

p4,

p4,

=

+−+=

=

+−+=∴

( )

( )( )( )

( )( )

( )(r)

4

(r)

44

r/

4

(r)

c4,

(r)

4p4,

(0)

4

(r)

c4,

(r)

4p4,

(0)

4

r/

4

/

3

/

22

1)(r

c4,

(r)

4322

1)(r

c4,

z,y,xφz

0rforzz& zz

0rforyy& yy

where

(2)

z4zz3

hzz

z4zz3

hyy

=

≠==

≠==

−−−−

+++=

+++=

+

+

( ) ( )

( )( )

( )( )( )

( )( )

( ) ( )

( )( )

0.7074

0.56980.799540.93743

0.20.3937z

0.5698z2y1z,y,xφz,y,xφz

where

z4zz3

hzz

0.30450.70550.568940.39373

0.20.0795

z4zz3

hyz4zz

3

hyy

(1)c4,

p4,p4,p4,p4,4

(0)

4

(0)

44

0/

4

0/

4

/

3

/

22

(1)

c4,

p4,322(0)4322

(1)c4,

=

+++=∴

=−===

+++=

=+++=

+++=+++=



Second improvement: {Put r = 1 in Eqn (2)}

Third improvement: {Put r = 2 in Eqn (2)}

__________________________________________________________

Problem (2):

x 1 1.1 1.2 1.3

y 2 2.2156 2.4649 2.7514

y/ 2 2.3178 2.6725 3.0657

Soln:

( )( )( )

( )( )

( ) ( )

( )( )

0.7073

0.56920.799540.93743

0.20.3937z

0.5692z2y1z,y,xφz,y,xφz

where

z4zz3

hzz

(2)c4,

(1)

c4,

(1)

c4,

(1)

c4,

(1)

c4,4

(1)

4

(1)

44

1/

4

1/

4

/

3

/

22

(2)

c4,

=

+++=∴

=−===

+++=

( ) ( )

( )( ) 0.30460.70740.568940.39373

0.20.0795

z4zz3

hyz4zz

3

hyy (1)

c4,322

(1)

4322

(2)

c4,

=+++=

+++=+++=

( ) ( )

0.3046y(0.4)

placesdecimalfourtocorrectsamethearey&ySince

0.3046

z4zz3

hyz4zz

3

hyy

(3)

c4,

(2)

c4,

(2)

c4,322

(2)

4322

(3)

c4,

=

=

+++=+++=

anddx

dyx4

dx

yd2thatgiveny(1.4)computetomethodsMilne'Apply

2

2

+=

( ) ( )

Thus

0.1h;2

z2xzy,x,φ&zzy,x,f

2z(1)2,y(1)conditionswith2

z2xdx

dz


dySet

=+==∴

==+=

=





x 1 1.1 1.2 1.3 1.4

y 2 2.2156 2.4649 2.7514 ?

z 2 2.3178 2.6725 3.0657 -

2156.2y1 = 3178.2z1 =

4649.2y2 =

7514.2y3 =

6725.2z2 =

0657.3z3 =

2

z2x

dx

dzz /

+==

3.35892

z2xz 1

1/1 =+=

7362.32

zx2z 2

2/2 =+=

1328.42

zx2z 3

3/3 =+=


( )

( )(1)

2zz2z3

4hz

2zz2z3

4hyy

/

3

/

2

/

10p4,

3210p4,

−−−−−

+−+=

+−+=

z

( ) ( ) ( )( )

( ) ( ) ( )( )

3.4996

4.132823.73623.358923

4(0.1)2z

3.0793

3.065722.67252.317823

4(0.1)2y

p4,

p4,

=

+−+=

=

+−+=∴

( )

( )( )( )

( )( )

( )(r)

4

(r)

44

r/

4

(r)

c4,

(r)

4p4,

(0)

4

(r)

c4,

(r)

4p4,

(0)

4

r/4

/3

/22

1)(rc4,

(r)

4322

1)(r

c4,

z,y,xφz

0rforzz& zz

0rforyy& yy where

(2)

z4zz3

hzz

z4zz3

hyy

=

≠==

≠==

−−−−

+++=

+++=

+

+





__________________________________________________________

Problem (3):

( )( )( )

( )( ) ( ) ( )

( )( ) 3.49974.54984.132843.73623

0.12.6725z

4.54982

z2xz,y,xφz,y,xφz

where

z4zz3

hzz

(1)

c4,

p4,

4p4,p4,4

(0)

4

(0)

44

0/

4

0/

4

/

3

/

22

(1)

c4,

=+++=∴

=+===

+++=

( ) ( )

( )( ) 3.07943.49963.065742.67253

0.12.4649

z4zz3

hyz4zz

3

hyy p4,322

(0)4322

(1)c4,

=+++=

+++=+++=

( ) ( )

( )( )

0794.3y(1.4)


3.0794

4997.33.065746725.23

0.14649.2

z4zz3

hyz4zz

3

hyy

(2)

c4,

(1)

c4,

(1)

c4,322

(1)

4322

(2)

c4,

=

=

+++=

+++=+++=

ion.approximatthirdwithmethodsPicard'using(0.3)yandy(0.3)(0.2),y

y(0.2),(0.1),yy(0.1),findingafter methodsMilne' by placesdecimal

fourcorrect y(0.4)Find0.(0)y,1y(0),0ydx

dyx

dx

ydGiven

//

/

/

2

2

===++

( ) ( ) ( ) 1.0,xzyzy,x,φ&zzy,x,f

0z(0)1,y(0)conditionswith0yxzdx

dz


dySet

=+−==∴

===++

=

h



Soln:

x0=0 x1=0.1 x2=0.2 x3=0.3 x4=0.4 y0=1 y(x1)=? y(x2)=? y(x3)=? y(x4)=? z0=0 z(x1)=? z(x2)=? z(x3)=? -----

Picard’s method Milne’s method

Step (1): To find y(0.1) & z(0.1)

( )

( ))1(

dxzyz)dxz,yφ(x,zz

dxzy)dxz,yf(x,yy


x

x1-n1-n0

x

x1-n1-n0n

x

x1-n0

x

x1-n1-n0n

00

00

−−−−

+−=+=

+=+=

∫∫

∫∫

x

( ) ( )

( ) ( )( ) 0.1dx0x10 dxxzyzz

1dx01dxzyy

Eqn(1) in1nPut

0.1

0

x

x0001

0.1

0

x

x001

0

0

−=+−=+−+=

=+=+=

=

∫∫

∫∫

( ) ( )

( ) ( )( )

( ) ( )

( ) ( )( )

0.0985z(0.1)&0.9900y(0.1)

0.0985dx0.0995-x0.99-0dxxzyzz

0.9900dx0.0995-1dxzyy

Eqn(1) in3nPut

0.0985dx0.1-x1-0 dxxzyzz

0.99dx0.1-1dxzyy

Eqn(1) in2nPut

0.1

0

x

x2203

0.1

0

x

x203

0.1

0

x

x1102

0.1

0

x

x102

0

0

0

0

−==∴

−=+=+−+=

=+=+=

=

−=+=+−+=

=+=+=

=

∫∫

∫∫

∫∫

∫∫



Step (2): To find y(0.2) & z(0.2)

Step (3): To find y(0.3) & z(0.3)

( ) ( )

( )

( )( ) 0.1960dx0.0985-x0.990.0985-

dxxzyzz

0.9801dx0.0985-0.99dxzyy

Eqn(1) in1nPut

0.0985z,0.99y,0.1xLet

0.2

0.1

x

x0001

0.2

0.1

x

x001

000

0

0

−=+−=

+−+=

=+=+=

=

−===

∫

∫

∫∫

( ) ( )

( ) ( )( )

( ) ( )

( ) ( )( )

0.1926z(0.2) &0.9706y(0.2)

0.1926

dx0.1935-x0.9704-0.0985dxxzyzz

0.9706dx0.1935-0.99dxzyy

Eqn(1) in3nPut

0.1935

dx0.1960-x0.9801-0.0985 dxxzyzz

0.9704dx0.1960-0.99dxzyy

Eqn(1) in2nPut

0.2

0.1

x

x2203

0.2

0.1

x

x203

0.2

0.1

x

x1102

0.2

0.1

x

x102

0

0

0

0

−==∴

−=

+−=+−+=

=+=+=

=

−=

+−=+−+=

==+=

=

∫∫

∫∫

∫∫

∫∫

( ) ( )

( )

( )( )0.3

x

x

dx0xz0y0z1z

0.95130.3

0.2

dx0.1926-0.9706x

x

dx0z0y1y

Eqn(1) in1nPut

0.19260z,0.97060y,0.20xLet

0

0

+−+=

=+=+=

=

−===

∫

∫∫



( ) ( )

( ) ( )( )

( ) ( )

( ) ( )( )

0.2797z(0.3) &0.9425y(0.3)

0.2797

dx0.2806-x0.9421-0.1926dxxzyzz

0.9425dx0.2806-0.9706dxzyy

Eqn(1) in3nPut

0.2806

dx0.2848-x0.9513-0.1926 dxxzyzz

0.9421dx0.2848-0.9706dxzyy

Eqn(1) in2nPut

0.3

0.2

x

x2203

0.3

0.2

x

x203

0.3

0.2

x

x1102

0.3

0.2

x

x102

0

0

0

0

−==∴

−=

+−=+−+=

=+=+=

=

−=

+−=+−+=

==+=

=

∫∫

∫∫

∫∫

∫∫

9900.0y1 = 0985.0z1 −=

xz)(ydx

dzz /

+−==

-0.9801

0.0985)-0.1x-(0.99

)zx(yz 111

/

1

=

+=

+−=







9706.0y2 =

9425.0y3 =

1926.0z2 −=

2797.0z3 −=

-0.9320

0.1926)-0.2x-(0.9706

)zx(yz 222

/

2

=

+=

+−=

-0.8585

0.2797)-0.3x-(0.9425

)zx(yz 333

/

3

=

+=

+−=

( )

( )(2)

2zz2z3

4hz

2zz2z3

4hyy

/

3

/

2

/

10p4,

3210p4,

−−−−−

+−+=

+−+=

z

( ) ( ) ( )( )

( ) ( ) ( )( )

0.3660

0.8585-20.93200.980123

4(0.1)0z

0.9248

0.2797-20.19260.098523

4(0.1)1y

p4,

p4,

−=

+−−−+=

=

+−−−+=∴

( )

( )( )( )

( )( )

( )(r)

4

(r)

44

r/

4

(r)

c4,

(r)

4p4,

(0)

4

(r)

c4,

(r)

4p4,

(0)

4

r/4

/3

/22

1)(rc4,

(r)4322

1)(rc4,

z,y,xφz

0rforzz& zz

0rforyy& yy where

(3)

z4zz3

hzz

z4zz3

hyy

=

≠==

≠==

−−−−

+++=

+++=

+

+




Third improvement: {Put r = 2 in Eqn (3)}

( ) ( )

( )( )

( )( )( )

( )( )

( ) ( )

( )( )

0.3641

0.77840.8585-40.93203

0.1-0.1926z

0.7784)zx(yz,y,xφz,y,xφz

where

z4zz3

hzz

0.91460.36600.2797-40.19263

0.10.9706

z4zz3

hyz4zz

3

hyy

(1)c4,

p4,4p4,p4,p4,4

(0)

4

(0)

44

0/

4

0/4

/3

/22

(1)c4,

p4,322(0)4322

(1)c4,

−=

−+−+=∴

−=+−===

+++=

=−+−+=

+++=+++=

( ) ( )

( )( )

( )( )( )

( )( ) ( ) ( )

( )( ) 0.36380.76900.8585-40.93203

0.1-0.1926z

0.7690)zx(yz,y,xφz,y,xφz

where

z4zz3

hzz

0.91480.36410.2797-40.19263

0.10.9706

z4zz3

hyz4zz

3

hyy

(1)

c4,

(1)c4,4

(1)c4,

(1)c4,

(1)c4,4

(1)4

(1)44

1/4

1/4

/3

/22

(2)c4,

(1)c4,322

(1)4322

(2)c4,

−=−+−+=∴

−=+−===

+++=

=−+−+=

+++=+++=

( ) ( )

( )( )

0.9148y(0.4)


0.9148

0.36380.2797-40.19263

0.10.9706

z4zz3

hyz4zz

3

hyy

(3)c4,

(2)c4,

(2)c4,322

(2)4322

(3)c4,

=

=

−+−+=

+++=+++=



__________________________________________________________

Problem (4):

Soln:

x0=0 x1=0.1 x2=0.2 x3=0.3 x4=0.4 y0=1 y(x1)=? y(x2)=? y(x3)=? y(x4)=? z0=0.1 z(x1)=? z(x2)=? z(x3)=? -----

Picard’s method Milne’s method

Step (1): To find y(0.1) & z(0.1)

ion.approximatordersecondsPicard'of

aidthewith(0.3)yandy(0.3)(0.2),y y(0.2),(0.1),yy(0.1),evaluating

after0.1(0)y,1y(0);06ydx

dy3x

dx

ydproblemtheof0.4x

pointtheatsoloutioneapproximatanobtainmethodsMilne'theUsing

///

/

2

2

===−+=

( ) ( ) 3xz6yzy,x,φ&zzy,x,f

0.1z(0)1,y(0)conditionswith06y-3xzdx

dz


dySet

−==∴

===+

=

( )

)1(x

x

)dx1-n3xz-16(0zx

x

)dx1-nz,1-nyφ(x,0znz

x

x

dx1-nz0yx

x



00

00

−−−−

−−=+=

+=+=

∫∫

∫∫

ny

( ) ( )

( ) ( ) ( )( ) 0.69850.1

dxx0.13-160.1 x

dx0xz30y60z1z

1.010.1

0

dx0.11x

x

dx0z0y1y

Eqn(1) in1nPut

0

=+=++=

=+=+=

=

∫∫

∫∫



Step (2): To find y(0.2) & z(0.2)

Step (2): To find y(0.2) & z(0.2)

( ) ( )

( )

( ) ( )( )

0.6955z(0.1)

1.0698y(0.1)

0.69550.1

0

dxx0.698531.0160.1

x

x

dx13xz16y0z2z

1.06980.1

0

dx0.69851x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

=

=∴

=−+=

−+=

=+=+=

=

∫

∫

∫∫

( ) ( )

( )

( ) ( )( ) 1.3060dxx0.695531.069860.6955

dx3xzy6zz

1.1393dx0.69551.0698dxzyy

Eqn(1) in1nPut

0.6955z,1.0698y,0.1xLet

0.2

0.1

x

x0001

0.2

0.1

x

x001

000

0

0

=−+=

−+=

=+=+=

=

===

∫

∫

∫∫

( ) ( )

( ) ( ) ( )( )

1.2004y(0.2)

1.32030.2

0.1

dxx1.306031.139360.6995x

x

dx13xz16y0z2z

0.2

0.1

1.2004dx1.30601.0698x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

=∴

=−+=−+=

=+=+=

=

∫∫

∫∫



Step (3): To find y(0.3) & z(0.3)

Step (3): To find y(0.3) & z(0.3)

( ) ( )

( )

( ) ( )( ) 1.9415dxx1.320331.200461.3203

dx3xzy6zz

1.3324dx1.32031.2004dxzyy

Eqn(1) in1nPut

1.3203z,1.2004y,0.2xLet

0.3

0.2

x

x0001

0.3

0.2

x

x001

000

0

0

=−+=

−+=

=+=+=

=

===

∫

∫

∫∫

( ) ( )

( )

( ) ( )( )

1.9741z(0.3) &3945.1y(0.3)

1.9741

0.3

0.2

dxx1.941531.332461.3203

x

x

dx13xz16y0z2z

3945.1

0.3

0.2

dx1.94151.2004x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

==∴

=

−+=

−+=

=

+=+=

=

∫

∫

∫∫





0698.1y1 = 6955.0z1 =

2004.1y2 =

3945.1y3 =

3203.1z2 =

9741.1z3 =

xz3y6dx

dzz /

−==

6.2101

955)3(0.1)(0.66(1.0698)

z3x6yz 111

/

1

=

−=

−=

59036

97411303394516

36 3333

.

).)(.().(

zxyz /

=

−=

−=

6.4102

203)3(0.2)(1.36(1.2004)

z3x6yz 222

/

2

=

−=

−=


( )

( )(2)

2zz2z3

4hz

2zz2z3

4hyy

/

3

/

2

/

10p4,

3210p4,

−−−−−

+−+=

+−+=

z

( ) ( ) ( )( )

( ) ( ) ( )( )

2.6587

6.590326.41026.210123

4(0.1)0.1p4,z

1.5358

1.974121.32030.695523

4(0.1)1p4,y

=

+−+=

=

+−+=∴

( )

( )( )( )

( )( )

( )(r)

4

(r)

44

r/

4

(r)

c4,

(r)

4p4,

(0)

4

(r)

c4,

(r)

4p4,

(0)

4

r/

4

/

3

/

22

1)(r

c4,

(r)

4322

1)(r

c4,

z,y,xφz

0rforzz& zz

0rforyy& yy

where

(3)

z4zz3

hzz

z4zz3

hyy

=

≠==

≠==

−−−−

+++=

+++=

+

+



First improvement: {Put r = 0 in eqn(3)}

( )

( )( )

( )

+++=

=+++=

+++=

+++=

0/4

z/3

4z/2

z3

h2z

(1)c4,

z

1.59622.65871.974141.32033

0.11.2004

p4,z34z2z3

h2y

(0)4

z34z2z3

h2y

(1)c4,

y

( )( )

( )( )

2.6135

6.02446.590346.41023

0.11.3203

(1)c4,

z

6.0244)p4,z43xp4,6yp4,z,p4,y,4xφ(0)4

z,(0)4

y,4xφ0

/4

z

where

=

+++=∴

=−==

=

Second improvement: {Put r = 1 in eqn(3)}

( )( )

( )

( )

( )( )

2.6274

6.44106.590346.41023

0.11.3203

(2)c4,

z

6.4410)(1)

c4,z43x

(1)c4,

6y(1)

c4,z,

(1)c4,

y,4xφ(1)4

z,(1)4

y,4xφ1

/4

z

where

1/4

z/3

4z/2

z3

h2z

(2)c4,

z

1.59472.61351.974141.32033

0.11.2004

(1)c4,

z34z2z3

h2y

(1)4

z34z2z3

h2y

(2)c4,

y

=

+++=∴

=−=

=

=

+++=

=+++=

+++=

+++=



Third improvement: {Put r = 2 in eqn(3)}

( )( )

( )

( )

( )( )

2.6265

6.41536.590346.41023

0.11.3203

(2)c4,

z

6.4153)(2)

c4,z43x

(2)c4,

6y(2)

c4,z,

(2)c4,

y,4xφ(2)4

z,(2)4

y,4xφ2

/4

z

where

2/4

z/3

4z/2

z3

h2z

(3)c4,

z

1.59522.62741.974141.32033

0.11.2004

(2)c4,

z34z2z3

h2y

(2)4

z34z2z3

h2y

(3)c4,

y

=

+++=∴

=−=

=

=

+++=

=+++=

+++=

+++=

Fourth improvement: {Put r = 3 in eqn(3)}

1.5952y(0.4)

placesdecimalfourtocorrectsametheare(4)

c4,y&

(3)c4,

ySince

1.5952

(3)c4,

z34z2z3

h2y

(3)4

z34z2z3

h2y

(4)c4,

y

=

=

+++=

+++=

Problem(5):

ion.approximatsecondwithmethodsPicard'

using(0.3)/yandy(0.3)(0.2),/y y(0.2),(0.1),/yy(0.1),

findingafter methodsMilne' by placesdecimalfourcorrect

y(0.4)Find1.(0)/y,2y(0),2yx12dx

y2dGiven ==++=



Soln:

( ) ( ) 2yx1zy,x,φ&zzy,x,f

1z(0)2,y(0)conditionswith2yx1dx

dz


dySet

++==∴

==++=

=

Using Picard’s method

Using Milne’s method

( )

( )

)1(x

x

dx21-nyx10z

x

x

)dx1-nz,1-nyφ(x,0znz

x

x

dx1-nz0yx

x



00

00

−−−−

+++=+=

+=+=

∫∫

∫∫

Step(1): To find y(0.1) & z(0.1)

( ) ( )

( ) ( ) 1.5050.1

0

dx22x11 x

x

dx20yx10z1z

2.10.1

0

dx12x

x

dx0z0y1y

Eqn(1) in1nPut

0

0

=

+++=

+++=

=+=+=

=

∫∫

∫∫

00x =

20y =

10z =

1.01x =

?)1y(x =

?)1z(x =

2.02x =

?)2y(x =

?)2z(x = ?)3z(x =

?)3y(x = ?)4y(x =

3.03x = 4.04x =



( ) ( )

( )

( )

1.546z(0.1)&2.1505y(0.1)

1.5460.1

0

dx22.1x11

x

x

dx21yx10z2z

2.1505

0.1

0

dx1.5052x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

==∴

=

+++=

+++=

=

+=+=

=

∫

∫

∫∫



Step(2): To find y(0.2) & z(0.2)

( ) ( )

( )

( ) 2.12340.2

0.1

dx22.1505x11.546

x

x

dx20yx10z1z

2.30510.2

0.1

dx1.5462.1505x

x

dx0z0y1y

Eqn(1) in1nPut

1.5460z,2.15050y,0.10xLet

0

0

=

+++=

+++=

=+=+=

=

===

∫

∫

∫∫

( ) ( )

( ) ( )

2.1923z(0.2) &2.3628y(0.2)

2.1923

0.2

0.1

dx22.3051x11.546 x

x

dx21yx10z2z

2.3628

0.2

0.1

dx2.12342.1505x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

==∴

=

+++=

+++=

=

+=+=

=

∫∫

∫∫



Step(3): To find y(0.3) & z(0.3)

( ) ( ) 2.58200.3

0.2

dx2.19232.3628x

x

dx0z0y1y

Eqn(1) in1nPut

2.19230z,2.36280y,0.20xLet

0

∫∫ =+=+=

=

===

( )

( ) 2.87550.3

0.2

dx22.3628x12.1923

x

x

dx20yx10z1z

0

=

+++=

+++=

∫

∫

( ) ( )

( ) ( )

2.9839z(0.3) &2.6503y(0.3)

2.9839

0.3

0.2

dx22.5820x12.1923 x

x

dx21yx10z2z

2.6503

0.3

0.2

dx2.87552.3628x

x

dx1z0y2y

Eqn(1) in2nPut

0

0

==∴

=

+++=

+++=

=

+=+=

=

∫∫

∫∫



2yx1dx

dz/z ++==

( )

( )

5.7246

22.15050.11

21y1x1/

1z

=

++=

++=

( )

( )

6.7828

22.36280.21

22y2x1/

2z

=

++=

++=

( )

( )

8.3240

22.65030.31

23y3x1/

3z

=

++=

++=


( )(2)

/3

2z/2

z/1

2z3

4h0p4,z

32z2z12z3

4h0yp4,y

−−−−−

+−+=

+−+=

z

( ) ( ) ( )( )

( ) ( ) ( )( )

3.8419

8.324026.78285.724623

4(0.1)1p4,z

2.9157

2.983922.19231.546023

4(0.1)2p4,y

=

+−+=

=

+−+=∴

1505.21y = 5460.11z =

3628.22y =

)iy(xiy = )iz(xiz =

6503.23y =

1923.22z =

9839.23z =



( )

( )

=

≠==

≠==

−−−−

+++=

+

+++=+

(r)4

z,(r)4

y,4xφr

/4

z

0rfor(r)

c4,z

(r)4

z& p4,z(0)4

z

0rfor(r)

c4,y

(r)4

y& p4,y(0)4

y

where

(3)

r/4

z/3

4z/2

z3

h2z

1)(rc4,

z

(r)4

z34z2z3

h2y

1)(rc4,

y

:formula corrector sMilne'

First improvement: {Put r = 0 in eqn(3)}

( )

( )( )

( )

+++=

=+++=

+++=

+++=

0/4

z/3

4z/2

z3

h2z

(1)c4,

z

2.96183.84192.983942.19233

0.12.3628

p4,z34z2z3

h2y

(0)4

z34z2z3

h2y

(1)c4,

y

( )( ) ( )

( )( )

3.8583

9.90138.324046.78283

0.12.1923

(1)c4,

z

9.90132p4,y4x1p4,z,p4,y,4xφ

(0)4

z,(0)4

y,4xφ0

/4

z

where

=

+++=∴

=++==

=



Second improvement: {Put r = 1 in eqn(3)}

( )( )

( )

( )

8673.3(2)

c4,z

1722.10)2

(1)c4,

y4x(1(1)

c4,z,

(1)c4,

y,4xφ(1)4

z,(1)4

y,4xφ1

/4

z

where

1/4

z/3

4z/2

z3

h2z

(2)c4,

z

9623.28583.32.983941923.23

0.13628.2

(1)c4,

z34z2z3

h2y

(1)4

z34z2z3

h2y

(2)c4,

y

=∴

=

++=

=

=

+++=

=+++=

+++=

+++=

Similarly

2.9626y(0.4)

placesdecimalfourtocorrectsametheare(4)

c4,y&

(3)c4,

ySince

2.9626(4)

c4, y

Eqn(3)in3rPut

3.8674(3)

c4,z

2.9626(3)

c4,y

Eqn(3)in2rPut

=

=

=

=

=

=



Problem(6):

1.5752}y(0.4):{Ans

ion.approximatsecondwithmethodsPicard'

using(0.3)/yandy(0.3)(0.2),/y y(0.2),(0.1),/yy(0.1),

findingafter methodsMilne' by placesdecimalfourcorrect

y(0.4)Find1.(0)/y,1y(0),dx

dy1

2dx

y2dGiven

=

==+=

Problem(7):

andx2edx

dy

2dx

y2d

thatgiveny(0.4)computetomethodsMilne'Apply

=+

x 0 0.1 1.2 1.3

y 2 2.01 2.04 2.09

y/

0 0.2 0.4 0.6

2.16}y(0.4):{Ans =

Unit2 vrs

Engineering

Transcript of Unit2 vrs