numerical solution of odes-ivp

7/30/2019 Numerical Solution of ODEs-IVP

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Prof. A.A. Adesina, ChSE, UNSW 1

NUMERICAL SOLUTION OFORDINARY DIFFERENTIAL

EQUATIONSINITIAL VALUE PROBLEMS


2/32


GENERAL CONSIDERATIONS

The modelling

ofunsteady-state processes

often lead to ODEs

with

time as the independent variable. In other cases, the independent

variable may be a spatial co-ordinate as in the steady-statetreatment of tubular or columnal systems

such as plug flow and

trickle-bed reactors, packed adsorption and absorption towers oreven tubular heat exchangers.

In general, for an nth

order ODE, n conditions are needed tocompletely solve the problem. If ALL these n specifications are

provided at the same value of the independent variable, say, t orx, for all dependent variables, then the ODE is

said to be an INITIAL

VALUE PROBLEM. While this may imply that the value of t or x is

supposed to be the starting point in the system (e.g. t=0 or x=0),there is no mathematical incongruity if

the value oft or x is the end-

point or some other physically admissible point

in the independent

variable range.


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METHODS FOR 1ST ORDER NONLINEARODE - Euler methods

The general 1st

order nonlinear ODE may be written;

The Euler method, one of the earliest techniques for solving ODE

stipulates

that, if we can represent the LHS of Eqn

(7.1) by its first forward finite

difference, then at any position, i, we have,

0 0 0 0

( 7 . 1 )

w i t h t h e i n i t i a l c o n d i t i o n t h a t

a t ( )O

( ,

R

)

x x y y y

d yf x y

d

x y

x

= = =

=

1

2 2 3 3

(7.2)

but from our previous discussion on finite difference operators, we can write,

..... ...2 6

i i i

i ii i

y y y

h D y h D yy hDy

+ =

= + + + (7.3)


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Euler methods

Euler assumed a truncation of the series in Eqn

(7.3) after the 1st

term, thus,

In fact, replacing the term, Dyi

with ffrom Eqn

(7.1), we get the so-called

explicit (forward) Euler method, written,

It is evident that this method is only marginally accurate since

the error is of

order h

2

.

2

2

1

(7.4)so that upon substitution into Eqn (7.2), we receive,

(7

)

( ) )

(

.5

i i

i i i

y hD y O h

y y hD y O h+

= +

= + +

21 ( , ) ( ) (7.6)i i i iy y hf x y O h+ = + +


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Euler methods

If we use the backward finite difference approximation, we can write,

But from earlier notes, we know that

1 1

1 1

(7.7)so that, in terms of backward differences, we have,

(7.8)

i i i i

i i i

y y y y

y y y

+ +

+ +

= =

= +

2 2 3 3

1 11 1

21 1 1

(7.9)

so that combining with Eqns (7.1) & (7.2) we secure,

.............2 6

( (7.10, ) ( ))

i ii i

i i i i

h D y h D yy hDy

y y hf x y O h

+ +

+ +

+ + +

= +

= + +


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Euler methods

Eqn

(7.10)

is commonly called the implicit Euler method

because it

involves the calculation of the function, f, at the unknown value of

yi+1

. We observe that the error in Eqn (7.10) is also of the order of h2.

In principle, implicit equations cannot be solved individually but must

be set up as sets simultaneous algebraic equations. If these are

linear, then we can use any of the techniques we have encounteredbefore (say, generalised inverse matrix approach) to deliver asolution. However, if they are nonlinear, we may use the Jacobian method for solving the set of nonlinear algebraic equations.Regardless, the computation may be time-consuming depending on

how many integration steps we need to do.

Euler simplified this problem by using Eqn

(7.6) to predict yi+1

and

then employing this predicted value

to get a corrected estimate from

Eqn (7.10).


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Euler methods

Eqns

(7.6) and (7.10) constitute the Euler predictor-corrector pair and is more

accurate because it has an error oforder h3

(which can be readily shown by

adding Eqns (7.3) & (7.9)). Thus, the

Euler Predictor-Corrector method

is:

As a result, the Euler Predictor-Corrector method is preferred over the

explicit or implicit method

for the purpose of computation.

( )

( ) [ ]

2

1 Pr

2

1 1 1

(7.11a( , ) ( )

( , ) ( , ) ( )

)

(7.11b2

)

i i i i

i i i i i iCor

y y hf x y O h

hy y f x y f x y O h

+

+ + +

= + +

= + + +


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Runge-Kutta

Methods

The Runge-Kutta

methods are arguably the most widely used

integration techniques for solving ODEs

with initial conditions

(IVP-ODEs). Although the formal derivation for each method issimilar to the Euler method in the sense that we take advantage of

particular approximations of the Taylor series expansion of the

differential operator (cf. Lecture notes on Finite Difference operators

as recalled in Eqns (7.3) & (7.9)), we will NOT provide formalderivations here but simply summarise the necessary equations to

be used for each RK method as displayed in the Tables below.


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Table 1: Summary of Runge-Kutta integration formulas (2nd-4th order)

Orderof theRKMethod

Formula foryi+1 Formula for RKconstants

Order oftheerrorinvloved

Second

1 1 2

1( )

2i iy y k k

+= + + 1

2 1

( , )

( , )

i i

i i

k hf x y

k hf x h y k

== + +

O(h3)

Third1 1 2 3

1( 4 )

6i i

y y k k k+ = + + + 11

2

3 2 1

( , )

,2 2

( , 2 )

i i

i i

i i

k hf x y

h kk hf x y

k hf x h y k k

= = + +

= + +

O(h4)

Fourth1 1 2 3 4

1( 2 2 )

6i i

y y k k k k+ = + + + + 11

2

23

4 3

( , )

,2 2

,2 2

( , )

i i

i i

i i

i i

k hf x y

h kk hf x y

h kk hf x y

k hf x h y k

= = + + = + +

= + +

O(h5)


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Table 2: Runge-Kutta-Gill integration formulas

Orderof theRKMethod

Formula foryi+1 Formula for RK constants Order oftheerrorinvloved

Runge-Kutta-Gill

1 1 2 3 41 1 12 1 2 16 2 2

i iy y k k k k+ = + + + + +

( ) ( )

( )

1

12

31 2

4 32

( , )

,2 2

,2

2 1 2 22 2

,

2 2

22

i i

i i

i

i

i

i

k hf x y

h kk hf x y

hx

k hf k ky

x h

k hf kky

= = + + +

= + + + = + +

O(h5)

The Runge-Kutta-Gill method is the most widely used 4th

order method and the constants are

selected to reduce the amount of storage required in the solution of a large number of

simultaneous 1st

order ODEs. We will discuss application of RK methods to set of ODEs later.


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Table 3: Summary of Runge-Kutta integration formulas (5th order)

Orderof theRKMethod

Formula foryi+1 Formula for RK constants Order oftheerrorinvloved

Fifth1 1 3 4 5 6

1(7 32 12 32 7 )

90i i

y k k k k k+ = + + + + + 11

2

1 23

34

32 45

3 51 2 46

( , )

,2 2

3

,2 16 16

,2 2

63 3 9,4 16 16 16

6 84 12,

7 7 7 7 7

i i

i i

i i

i i

i i

i i

k hf x y

h kk hf x y

h k k

k hf x y

khk hf x y

kh k k

k hf x y

k kk k kk hf x h y

= = + + = + + + = + + = + + + = + + + + +

O(h6)


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Comments on Runge-Kutta methodselection

In general, the 4th

order RK method is more commonly used

because

it has a good accuracy for most engineering situations. However, forstiff ODEs, it may be necessary to reduce the step-size, h, betweensuccessive integration steps. What is the criterion for implementing a

step-size reduction to avoid error propagation through the integration

procedure? Collatz (1960) has recommended that if,

After each integration step, then the step-size, h, should be

decreased

(to probably half its current value).

3 2

2 1

0 . 1 (7 .12)k k

k k


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Illustrative example

Let us find the solution to the IVP

In this case, clearly,

We divide the interval into 10 steps so that x0

=0.0, x1

=0.1..x10

=1.0

with h=0.1. Thus, y(x0

0=1.0. Using the iterative formulas given on

Table 1, we get,

th

(E1.1)

with over the interval 0 x 1 usingy(0 4 order RK method)=1.0

2

1dy y xdx =

+

( , ) 2 1x y y x+


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Results of the 4th

order Runge-Kutta methods for the Illustrative Example

xi R-K R-K-G Exact value of yi

0.0 1.0 1.0 1.00.1 1.01034164 1.01034164 1.01034260

0.2 1.04280472 1.04280472 1.04280663

0.3 1.09971619 1.09971619 1.09971809

0.4 1.18364811 1.18364811 1.18365002

0.5 1.29744053 1.29744148 1.29744339

0.6 1.44423485 1.44423580 1.44423676

0.7 1.62750244 1.62750340 1.62750530

0.8 1.85107803 1.85107994 1.85108185

0.9 2.11920166 2.11920357 2.11920643

1.0 2.43655777 2.43656063 2.43656349


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Predictor-Corrector Methods

The Runge-Kutta

methods are single-step

methods and are good for

solving IVP because they self-starting. However, they may be

plagued with instabilities for highly stiff ODEs which require smallstep-size to overcome numerical instabilities. Predictor-Correctormethods are preferred

for such situations because they deliver more

stable results. They are, however, nonself-starting

and are regarded

as multi-step techniques.

The most common PC methods are summarised

below.


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Table 4: Summary of commonly used Predictor-Corrector Methods

Type of

PC

method

Predictor Equation Corrector Equation Error

involved

Euler ( )1 Pri i iy y hf+ = + ( ) ( )1 12

i i i iCor

hy y f f+ ++ + O(h

2)

Milnes

4thorder

( ) ( )1 3 1 2Pr

42 2

3i i i i i

hy y f f f+ + + ( ) ( )1 1 1 14

3i i i i i

Cor

hy y f f f+ + + + + O(h

5)

Adams-

Moulton

method

( ) ( )1 1 2 3Pr 55 59 37 924i i i i i ihy y f f f f+ + + ( ) ( )1 1 1 29 19 524i i i i i iCor hy y f f f f+ + + + + O(h5)

Milnes

6th order ( ) 1 21 5Pr3 4

11 14 26310 14 11

i i ii i

i i

f f fhy y

f f +

+ = + + ( ) 1 11 2 37 32 12

245 32 7

i i ii iCor

i i

f f fhy y

f f+ + + + = + + +

O(h7)

Note that: fi = f(xi, yi); fi+1 = f(xi+1, yi+1) etc, in the Table above


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Predictor-Corrector Methods

How many times do we iterate on the Corrector equation? Typically, we

employ a termination criterion to determine when to stop. For example,

we can use:

1, 1, 1

1, 1

i+1m number of iterat

ions

perform

ed so far on

is

(7.12)

where

the toleranc

indicates the

usin

1,2,....

Corrector Equat e faciong the and

y

i m i m

i m

y y

ym

+ + + =

-4

,

a small positive number,

tor

10typi or smcal ally, ler.


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Worked Example

Use the Milnes 4th

order PC method to solve

Solution

Assume h=0.1, then let us calculate the 1st three points (y1 ,y2 ,y3 ) bythe single-step method of 4th order R-K technique since the PC isNOT self-starting. Recall that we need, yi-3

to use the Milnes P-Cmethod.

0.8

y(1.0)=3.

(E1)

with over th ra

0.1( )

nge x0 =1.0 to 2

.0e

dy x ydx = +

0.8( , ) 0.1( )f x y x y+


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Starting with the 4th order RK method and using y(1) =3.0; we have, x0=1.0and y0=3.0

1st integration step

0.8

1 0 0

0.8

2 0 0 1

3 0 0 2

4 0 0 3

1 2 3 4

1 0

( , ) 0.1[0.1(1 3) ] 0.0303143( / 2, / 2 ) 0 .1[0.1(1.05 3.01516) ] 0.0307087

( / 2, / 2 ) 0.0307099

( , ) 0.0311043

12 2 0.0307093

63.0

k h f x yk h f x h y k

k h f x h y k

k h f x h y k

y k k k k

y y y

= = + == + + = + == + + == + + =

= + + + = = + =

11 0.1 1

307093

d

1

an

.x = + =

2

nd

integration step0.8

1 1 1

2 1 1 1

3 1 1 2

4 1 1 3

1 2 3

2

2

4

1

( , ) 0.1[0.1(4.13071) ] 0.0311042

( / 2, / 2) 0.0314985

( / 2, / 2 ) 0.0314997

( , ) 0.0311043

12 2 0.0314991

6

3.03

and

1.1

149

0.

9

1

1

1 .

k h f x y

k h f x h y k

k h f x h y k

k h f x h y k

y k k k

x

k

y y y

= = == + + == + + == + + =

= + + + =

= += =

=+

2

Similar repetitive steps give,

3 31.3, 3.09449x y =

At the 4th integration step, we now turn to the 4th order Milnes


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At the 4 integration step, we now turn to the 4 order Milne s

PC. Hence,

( )4 0 3 2 11

2

3

4 3 4 3 2

0.8

4 4

,

4

4

4 0

4Pr : 2 2

3

0.311042

0.318939

0.326834

0.43.0 (0.622084 0.653668 0.318939) 3.127

: ( 4 )

3( , ) 0.1(1.4 3.12758) 0.334728

0.13.09449 (1.961

583

00) 3.15983

hC or y y f f f

f f x y

y

hy y f f f

f

f

f

y

= + + += = +

= + +=

= = + =

==

= + + =

4 ,1 4 ,0

4

0.8

4 4 4 ,0

4 ,1 3 4 3 2

,0

-4

1st iteration step of the Co rrector Equa tion

f ( , ) 0.1(1.4 3.15986) 0.336636

( 4 )3

Check for converge

0.

nc

13.09449 (1.96291) 3.15992

3

e

L et us assu m e th 1

6

at = 0

f x y

hy

y y

f

y

y f f

= = + == + +

+= + =

5 4st

th

3.15992 3.159861.8988 10 10

3.15986

hence convegrence was met after the 1 iteration so we cannow proceed to the 5 integration step.

=


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Solution of higher order ODEs (Solution toSimultaneous 1

st

order ODEs)

In principle, since it is possible to convert an nth

order ODE

to a set ofn 1st

order ODEs.

The approach to solving an nth

order ODE

is therefore exactly the same as solving n 1st

order ODEs. We would 1st

show how this conversion may be carried out and then

proceed with a discussion of the simlutaneous

solution of n 1st

order ODEs.

1 2 1

1 2 1 01 2 1

1

( , ) ( , ) .... ( , ) ( , ) ( , , ,... ) (7.13)

Let us defi

ne new variables as:

(7.1z 4

n n n n

n nn n n n

d y d y d y dy dy d ya x y a x y a x y a x y g x y

dx dx dx dx dx dx

y

+ + + =

=1

2 1 1 2

2

23 2 1 22

2

21 2 1 22

( , , , ..... )

( , , , .....

)

(7.15)

(7.16)

)

( , , , ..... )

n

n

n

nn n nn

dz dyz f x z z z

dx dx

dz d yz f x z z z

dx dx

dz d yz f x z z zdx dx

= = == = =

= = =M

1

11 1 21

1

(7.17)

(7.18)

Introducing these equ ations into the original equation, w e receive

( , , , ..... )

( , , ,...

n

nn n nn

n n

n

n n

dz d yz f x z z z

dx dx

dz d y dy d yg x ydx dx dx dx

= = =

= = 1 21 )= ( , , , ..... ) (7.19)n nf x z z z


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Solution of higher order ODEs (Solution toSimultaneous 1

st

order ODEs)

Eqns

(7.15) to (7.19) constitute the new n 1st

order ODEs

we now

need to solve. The applicable 4th

order Runge-Kutta

constants are:

( )1, 1 2 3 41 1 2

111 122 1 2

221 223 1 2

12 2 1,2,...

6

( , , ,.... ) 1,2,...

, , ,.... 1,2,...2

(7.20

2 2 2

, , ,.

)

(7.21)

(7.22)

...2 2 2 2

i j ij j j j j

j j i i i in

nj j i i i in

nj j i i i in

z z k k k k j n

k hf x z z z j n

kh k kk hf x z z z j n

kh k kk hf x z z z

+ = + + + + == =

= + + + + = = + + + +( )4 1 31 2 32 3

1,2,...

,

(7.23)

(7, ,..... 1,2 .24. ), ..j j i i i in n

j n

k hf x h z k z k z k j n

== + + + + =


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A system ODE is said to be stiff if the so-called stiffness ratio, SR, exceeds

a particular value as indicated below, where SR is defined,

Where i

s

are the eigenvalues

of the ODE system and max |Re(i

)| is the

real part of the maximum eigenvalue of the system. In general, if

a). SR is about an order of magnitude

(10,20, etc), the system is NOT stiff.

b). SR is about three orders of magnitude

(1000, 2000, etc) the system is

STIFF

c). SR is about six orders of magnitude

(106, etc), the system is VERY

STIFF.

The eigenvalues

are obtained from the matrix of the coefficients of the set

of ODEs.

max Re( )

m

in R

(7.e( )

25)i

i

SR=


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In principle, a stiff ODE

is one whose general solution contains an

exponential term, e.g. e

x

forsome constant . When is large andnegative, the ODE is especially troublesome because it permits the solution

to decay quickly to zero.

For instance, the solution to:

i

0 0

x

(7.26)

with , the may be obtained as

(7.27)

where

( , )

eigenvalue, ,

can vary in magnitude at each step of the in .tegration

Eqn (7.26)

( )

y x y

f

y

dy

f x ydx

= ==

will be at some point in the range of integration.

Thus, the s

stiff if i

olution wil

s NE

l be

GATIVE

unstable.

f

y


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Real stability regions for selected solution methods

Method Stability boundary

Explicit Euler -2 h 0

Implicit Euler 0 < h < , for < 0;-2 h 0, for > 0

Euler Predictor-Corrector -1.077 h 02

ndorder Runge-Kutta -2 h 0

3rd order Runge-Kutta -2.5 h 04

thorder Runge-Kutta -2.785 h 0

5th

order Runge-Kutta -5.7 h 0

Adams-Moulton -1.285 h 0


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For a set of n simultaneous ODEs, we can obtain theeigenvalues from the Jacobian matrix of the functions, fiwritten;

1 1 1

1 2

2 2 2

1 2

1 2

.......

........

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .

........

n

n

n n n

n

f f f

y y y

f f f

y y y

J

f f f

y y y


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It is apparent from the foregoing considerations that implicit methods

are recommended for handling the solution to stiff ODEs. This is

why Predictor-Corrector methods are somewhat popular amongchemical engineers

since we often deal with nonlinearity due to the

Arrhenius

or vant

Hoff terms

arising from modeling of non-

isothermal tubular reactors/columns or dynamics of reactive lumped

systems with nonlinear kinetics.


28/32


Example problem

Solve

22 2

2

d yx x y y

d x= + +

for y at x = 2.0 ify = 4.0 at x = 1.0 and

0.5 1.0dy at xdx

= =


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Solution

Let

1

22 2

22

( , , )

( , , )

d yz f x y z

d xd y d z

x x y y f x y zd x d x

= =

= = + + =

Assume h = 0.5. From the initial conditions0 0 01 .0 , 4 .0 , 0 .5x y z= = =

1st

R-K constants

0 0 0

1 1 1

0 0 0

1 1 2

( , , ) 0 .5 0 .5 0 .2 5

( , , ) 0 .5 (1 4 1 6 ) 1 0 .5 0

k h f x y z x

k h f x y z

= = =

= = + + =

2nd

R-K constants

0 0 01 1 1 22 1 1

0

0 1 1

0 1 2

2 1

0 0 01 1 1 22 2 2

2 2

( , , )2 2 2

0 .51 1 .2 5

2 2

0 . 2 54 4 .1 2 5

2 21 0 . 5

0 .5 5 .7 52 2

0 .5 5 .7 5 2 .8 7 5

( , , )2 2 2

0 .5 [ (1 .2 5 ) 4 .1 2 5 1 .2 5 ( 4 .1 2 5 ) ]

1 1 . 8 6 7 2

h k kk h f x y z

hx

ky

kz

k x

h k kk h f x y z

x

= + + +

+ = + =

+ = + =

+ = + =

= =

= + + +

= + +

=

3r R-K constantsh k k


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0 0 02 1 2 23 1 1

0

0 2 1

0 2 2

3 1

0 0 02 1 2 23 2 2

2 2

( , , )2 2 2

0 . 51 1 . 2 5

2 2

2 . 8 7 54 5 . 4 3 7 5

2 2

1 1 . 8 6 7 20 . 5 6 . 4 3 3 62 2

0 . 5 6 . 4 3 3 6 3 . 2 1 6 8

( , , )2 2 2

0 . 5 [ (1 . 2 5 ) 5 . 4 3 7 5 1 . 2 5 ( 5 . 4 3 7 5 ) ]

1 8 . 9 6 2 9

h k kk h f x y z

hx

ky

kz

k

h k kk h f x y z

x

= + + +

+ = + =

+ = + =

+ = + =

= =

= + + +

= + +

=

4th

R-K constants0 0 0

4 1 1 3 1 3 2

0

0

3 1

0

3 2

4 1

0 0 0

4 2 2 3 1 3 2

2 2

1 1 2 1 3 1 4 1

1 2 2 2

( , , )

1 . 5

7 . 2 1 6 8

1 9 . 4 6 2 9

0 . 5 1 9 . 4 6 2 9 9 . 7 3 1 4 5

( , , )

0 . 5 [ ( 7 . 2 1 6 8 ) 1 . 5 7 . 2 1 6 8 (1 . 5 ) ]

3 2 . 5 7 8 7

1[ 2 2 ] 3 . 6 9 4 1 8

6

1 [ 26

k h f x h y k z k

x h

y k

z k

k x

k h f x h y k z k

x

y k k k k

z k k

= + + +

+ =

+ =

+ =

= =

= + + +

= + +

=

= + + + =

= + 3 2 4 2

( 1 ) 0

( 1 ) 0

( 1 ) 0

2 ] 1 7 . 4 5 6 5

7 . 6 9 4 1 8

1 7 . 9 5 6 5

1 . 5

k k

y y y

z z z

x x h

+ + =

= + =

= + =

= + =

We will now usex(1), y

(1), andz

(1)as the starting point for the computation

of ( 2 ) ( 2 ) ( 2 ), ,x y a n d z .


31/32


2nd

integration step

1st

R-K constants

(1) (1) (1)

11 1

(1) (1) (1)

11 2

2 2

( , , ) 0.5 17.9565 8.97824

( , , )

0.5[(7.69418) (7.69418 1.5) (1.5) ] 36.4958

k hf x y z x

k hf x y z

x

= = =

=

= + + =

2

nd

R-K constants

(1) (1) (1)11 1221 1

(1)

(1) 11

(1) 12

21

(1) (1) (1)11 1222 2

2 2

( , , )2 2 2

1.752

12.18332

36.20442

0.5 36.2044 18.1022

( , , )2 2 2

0.5[(1.75) 1.75 12.1833 (12.1833) ]

92.4997

k khk hf x y z

hx

ky

kz

k x

k khk hf x y z

x

= + + +

+ =

+ =

+ =

= =

= + + +

= + +

=

3rd

R-K constants


32/32


3 R K constants

(1 ) (1 ) ( 1 )2 1 2 23 1 1

(1 )

(1 ) 2 1

(1 ) 2 2

3 1

( 1 ) (1 ) ( 1 )2 1 2 23 2 2

2 2

( , , )2 2 2

1 . 7 52

1 6 . 7 4 5 3

2

6 4 . 2 0 6 32

0 . 5 6 4 .2 0 6 3 3 2 . 1 0 3 1 5

( , , )2 2 2

0 .5 [(1 . 7 5 ) 1 . 7 5 1 6 .7 4 5 3 (1 6 . 7 4 5 3 ) ]

1 5 6 . 3 8 6

k khk h f x y z

hx

ky

kz

k x

k khk h f x y z

x

= + + +

+ =

+ =

+ =

= =

= + + +

= + +

=

4th

R-K constants(1 ) (1 ) ( 1 )

4 1 1 3 1 3 2

(1 )

(1 )

3 1

(1 )

3 2

4 1

( 1 ) ( 1 ) ( 1 )

4 2 2 3 1 3 2

2 2

1 1 2 1 3 1 4 1

( , , )

2 . 0

3 9 . 7 9 7 3

1 7 4 . 3 4 20 . 5 1 7 4 . 3 4 2 8 7 . 1 7 1

( , , )

0 . 5 [ ( 2 ) 2 3 9 . 7 9 7 3 ( 3 9 . 7 9 7 3 ) ]

1 6 6 7 . 4 2

1[ 2 2 ] 3 2 .7 6

6

k h f x h y k z k

x h

y k

z kk x

k h f x h y k z k

x

y k k k k

= + + +

+ =

+ =

+ =

= =

= + + +

= + +

=

= + + + =

1 2 2 2 3 2 4 2

( 2 ) (1 )

( 2 ) (1 )

( 2 ) (1 )

1[ 2 2 ] 3 6 6 . 9 4 8

6

4 0 . 4 5 4 2

3 8 4 . 9 0 4

2 . 0

z k k k k

y y y

z z z

x x h

= + + + =

= + =

= + =

= + =

at x = 2.0, y = 40.4542

numerical solution of odes-ivp

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