7/25/2019 Piccards'

1/7

JUST THE MATHS

UNIT NUMBER

17.7

NUMERICAL MATHEMATICS 7

(Numerical solution)of

(ordinary differential equations (B))

by

A.J.Hobson

17.7.1 Picards method17.7.2 Exercises17.7.3 Answers to exercises

7/25/2019 Piccards'

2/7

UNIT 17.7 - NUMERICAL MATHEMATICS 7

NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL

EQUATIONS (B)

17.7.1 PICARDS METHOD

This method of solving a differential equation approximately is one of successive approxi-mation; that is, it is an iterative method in which the numerical results become more andmore accurate, the more times it is used.

An approximate value of y (taken, at first, to be a constant) is substituted into the righthand side of the differential equation

dy

dx =f(x, y).

The equation is then integrated with respect to x giving y in terms ofx as a second approx-imation, into which given numerical values are substituted and the result rounded off to anassigned number of decimal places or significant figures.

The iterative process is continued until two consecutive numerical solutions are the samewhen rounded off to the required number of decimal places.

A hint on notationImagine, for example, that we wished to solve the differential equation

dy

dx= 3x2,

given that y= y0= 7 when x= x0= 2.

This ofcourse can be solved exactly to give

y = x3 +C,

which requires that7 = 23 +C.

Hence,y 7 =x3 23;

or, in more general termsy y0=x

3 x3

0.

1

7/25/2019 Piccards'

3/7

Thus, y

y0

dy=

x

x0

3x2dx.

In other words,

x

x0

dy

dxdx=

x

x0

3x2dx.

The rule, in future, therefore, will be to integrate both sides of the given differential equationwith respect to x, from x0 to x.

EXAMPLES

1. Given thatdy

dx=x+y2,

and that y = 0 when x = 0, determine the value of y when x = 0.3, correct to fourplaces of decimals.

Solution

To begin the solution, we proceed as follows:

x

x0

dy

dxdx=

x

x0

(x+y2)dx,

where x0= 0.

Hence,

y y0=

x

x0

(x+y2)dx,

where y0= 0.

That is,

y=

x

0

(x+y2)dx.

(a) First IterationWe do not know y in terms of x yet, so we replace y by the constant value y0 in thefunction to be integrated.

The result of the first iteration is thus given, at x= 0.3, by

y1=

x

0

xdx= x2

2 0.0450

2

7/25/2019 Piccards'

4/7

(b) Second IterationNow we use

dy

dx=x+y2

1=x+

x4

4

.

Therefore, x

0

dy

dxdx=

x

0

x+

x4

4

dx,

which gives

y 0 = x2

2 +

x5

20.

The result of the second iteration is thus given by

y2= x2

2 +

x5

20 0.0451

at x= 0.3.

(c) Third IterationNow we use

dy

dx=x+y2

2

=x+x4

4 +

x7

20+

x10

400.

Therefore,

x0

dy

dx dx= x0

x+

x4

4 +

x7

20+

x10

400

dx,

which gives

y 0 = x2

2 +

x5

20+

x8

160+

x11

4400.

The result of the third iteration is thus given by

y3= x2

2 +

x5

20+

x8

160+

x11

4400 0.0451 at x= 0.3

Hence, y= 0.0451, correct to four decimal places, at x= 0.3.

2. Ifdy

dx= 2

y

x

and y = 2 when x = 1, perform three iterations of Picards method to estimate a valuefory whenx = 1.2. Work to four places of decimals throughout and state how accurateis the result of the third iteration.

3

7/25/2019 Piccards'

5/7

Solution

(a) First Iteration

x

x0

dydx

dx=

x

x0

2 y

x

dx,

where x0= 1.

That is,

y y0=

x

x0

2

y

x

dx,

where y0= 2.

Hence,

y 2 =

x

1

2

y

x

dx.

Replacing y byy0= 2 in the function being integrated, we have

y 2 =

x

1

2

2

x

dx.

Therefore,y= 2 + [2x 2 ln x]x

1

= 2 + 2x 2 ln x 2 + 2 ln 1 = 2(x ln x).

The result of the first iteration is thus given by

y1= 2(x ln x) 2.0354,when x= 1.2.

(b) Second IterationIn this case we use

dy

dx= 2

y1

x = 2

2(x ln x)

x =

2 ln x

x .

Hence, x

1

dy

dxdx=

x

1

2lnx

x dx.

That is,

y 2 =(ln x)2

x

1= (ln x)2.

The result of the second iteration is thus given by

y2= 2 + (ln x)2 2.0332,

4

7/25/2019 Piccards'

6/7

when x= 1.2.

(c) Third IterationFinally, we use

dydx

= 2 y2

x = 2 2

x

(ln x)2

x .

Hence, x

1

dy

dxdx=

x

1

2

2

x

(lnx)2

x

dx.

That is,

y 2 =

2x 2 ln x

(ln x)3

3

x

1

= 2x 2 ln x (ln x)3

3 2.

The result of the third iteration is thus given by

y3= 2x 2 ln x (ln x)3

3 2.0293,

when x= 1.2.

The results of the last two iterations are identical when rounded off to two places ofdecimals, namely 2.03. Hence, the accuracy of the third iteration is two decimal place

accuracy.

17.7.2 EXERCISES

1. Use Picards method to solve the differential equation

dy

dx=y +ex

at x= 1, correct to two significant figures, given that y = 0 when x= 0.

2. Use Picards method to solve the differential equationdy

dx=x2 +

y

2

at x= 0.5, correct to two decimal places, given that y= 1 when x= 0.

5

7/25/2019 Piccards'

7/7

3. Given the differential equationdy

dx= 1 xy,

where y(0) = 0, use Picards method to obtain y as a series of powers ofx which willgive two decimal place accuracy in the interval 0 x 1.

What is the solution when x= 1 ?

17.7.3 ANSWERS TO EXERCISES

1.

y(1) 2.7

2.

y(0.5) 1.33

3.

y = x x3

3 +

x5

15

x7

105+

x9

945 ......

y(1) 0.72

6