A textbook of Computer Based Numerical and Statiscal Techniques part 36 docx

Using Picard’s method of successive approximation to with y0 = 0, obtain y.25, y.3 and y1 correct to 3 decimal places... Now we shall obtain exact value.. When x=0.1,y=1.11034 These resu

( ).4 21 2 1.084( ) ( ) (.2 2 8536) ( ) (.2 3 4.67744) ( ) (.2 4 6.992576) ( ) (.2 5 40.21655)

= 45068 = 451 Ans

7.3 PICARD’S METHOD OF SUCCESSIVE APPROXIMATIONS

Consider first order differential equation

( ),

dy

f x y

with the initial condition

y y= 0 at x x= 0 Integrating (1) with respect to x between x0 and x, we have

( )

,

y dy= x f x y dx

or ( )

0

0 x ,

x

Now, we solve (2) by the method of successive approximation to find out the solution of (1) The first approximate solution (approximation) y1 of y is given by

0

x

y = +y ∫ f x y dx Similarly, the second approximation y2 is given by

y2 = y0+z f x y dx

x

x

, 1

0 b g

for the nth aproximation y n is given by

y n = y0 + f x y n dx

x

x

, –1

0 b g

with y x( )0 =y0

Hence, this method gives a sequence of approximation y y1, 2 y n and it can be proved f x y( ), is bounded in some regions containing the point ( ,x y0 0) and if f x y( ), satisfies the Lipchitz condition, namely

f x y −f x y ≤k y y− where k is a constant Then the sequence

1, 2

y y converges to the sol (2).

Example 6 Use Picard’s method to obtain y for x = 0.2 Given

dy

x y

dx= − with initial condition y = 1 when x = 0.

Sol Here, f x y( ), = −x y, x0=0, y0=1

We have first approximation,

0x , 1 0x 1

y = +y ∫ f x y dx= +∫ x− dx = 1 – x + x

2 2

Second approximation,

y2= +y0 ∫0x f x y dx( , 1) = +1 ∫0x(x y dx− 1)

2

0

2

∫

= − +1 2− 3

6

x

x x

Third approximation,

y3 = y0 x f x y dx2 x x y dx

6

2 3 0

zx x– x–x x dx

= 1 – x + x2 – x x

3 4

6 + 24

Fourth approximation,

y4 = 0 ( 3) ( 3)

0x , 1 0x

y +∫ f x y dx = +∫ x y dx−

=

3 4 2

0

3 24

∫

=

2

1

x x

Fifth Approximation,

y5 = 0 ( 4) ( 4)

0x , 1 0x

y +∫ f x y dx = +∫ x y− dx

=

2 0

3 12 120

∫

=

2

1

x x

1 82,

y = y2=.83867, y3=.83740

y4=.83746, y5=.83746

Thus, y=.837 when x=.2 Ans

Example 7 Find the solution of dy =1+ xy, y(0) = 1

dx which passess through (0, 1) in the interval

(0, 0.5) such that the value of y is correct to three decimal places (use the whole interval as one interval only) Take h = 0.1.

Sol The given initial value problem is

dy

f x y xy y

i.e., y y= =0 1 at x = x0 = 0

Here,

2

1 1

2

x

y = + +x

= + + 2 + 3+ 4

2 1

x x x

3 1

4 3

105 384

y =y + + When x=0,y=1.000

x=0.1,y1=1.105,y2=1.1053

x=0.2,y1+1.220,y2=1.223=y3 (correct up to 3 decimals)

x=0.3, y=1.355 as y2=1.355=y3 (correct up to 3 decimals)

x=0.4,y=1.505

x=0.5,y=1.677 as y4= =y3 1.677 Thus,

We have numerically solved the given differential equation for x=0.1, 0.2, 0.3, 0.4 and 0.5

Example 8 Using Picard’s method of successive approximation to with y(0) = 0, obtain y(.25), y(.3) and y(1) correct to 3 decimal places.

Sol Here, ( ) =

+

2 2

1

x

f x y

y x0=0 and y0=0 The first approximation y1 of y is given by

1 0 0

0 1 3

y = + dx=

+

∫

Similarly, the second approximation y2 of y is given by

2

3 0

0

x

= +

+

3

x

−

=

3

3 1 3

x x 

3

3

x

≈

We see that y1 and y2 agree to first term, namely

3

3

x

Neglecting

9

81

x

, we obtain the range

in which the result is correct to 3 decimal places, i.e., we put

1 9 .0005

81x ≤

which yield x ≤ 7

Hence, we obtain ( ) ( )1 3

3

( ) ( )1 3

3

( ) ( )1 3 1 1 3

Example 9 Use Picard’s method to obtain y for x = 0.1 Given that dy =3x + y ; y =1 2

dx at x = 0.

f x y x y x y

First approximation, y1= +y0 ∫0x f x y( , 0)dx

0

1 x 3x 1 dx

= 1 + x + 3

2

2

x

1

y = + +x x + x + x + x

y = + +x x + x + x + x + x + x

32x 240x 400x 4400x

When, x=0.1, we have

1 1.115,

y = y2=1.1264, y3=1.12721

Thus, y=1.127 when x=0.1 Ans

Example 10 Obtain y when x = 0.1, x = 0.2 Given that dy x y;

dx= + y(0) = 1, Check the result with exact value.

Sol We have, dy f x y( ), x y x; 0 0

Now First approximation,

1 0

2

y = +∫ +x dx= + +x

Second approximation,

2 2

0

∫

Third approximation,

3 4 2

3 1

x x

y = + + +x x + When x=.1, y1=1.105

2 1.11016

y =

3 1.11033

3 1.2427

y =

We can continue further to get the better approximations Now we shall obtain exact value

dy y x

dx− = is the given differential equation General sol is

ye−x= −e−x(1+ +x) c (IF e= −x) Putting y=1,x=0 we obtain, c=2

When x=0.1,y=1.11034

These results reveal that the approximations obtained for x=0.1 is correct to four decimal places while that for x=0.2is correct to 3 decimal places

Example 11 If dy dx y x=y x−+ Find the value of y at x = 0.1 using Picard’s method Given that y(0) = 1.

Sol First approximation,

0

1 0

1 1

1

0

2

1

x

dx x

+

∫

1 x 2log 1 x

Second approximation,

2

0

1 2log 1

x

= + −

∫

which is difficult to integrate

Thus, when, x=0.1,y1= −1 0.1 2 log 1.1+ ( )=0.9828 Ans

Here in this example, only I approximation can be obtained and so it gives that approximate

value of y for x=0.1

Example 12 Find the series expansion that gives y as a function of x in the neighbourhood of

x = 0, when dy x2 y2,

dx= + with y(0) = 0.

0

f x y = +x y x = and y0=0 The nth approximation y n of y is given by

0

x

y = +y ∫ f x y − dx

As an initial approximation, it is given that y0=0 Then, the first approximation y1 is given by

2 1

0

3

y = +∫ x + dx = ⋅ Similarly, the second approximation y2 is given by

2

2 2

0

0

y x   dx

∫

Likewise, the higher order approximations are given as

2

2 3

0

2

∫

y x

dx

x

0

0

2

2079 59535

F

z

=1 + + + +

3

1 63

2 2079

13 218295

If the series is truncated after the third term and used to approximate y to 4 decimal places,

then using the first neglected term, namely 13

218295

15

x as an approximation of the error, we have

13

218295

15

x ≤ 00005

Taking logarithm, we obtain

15 log x ≤ log .00005 218295

13

or x ≤ 988

Thus, x x x

3 7

11

3 63

2 2079 + + represents y correct to 4 decimal places In the range |x| ≤ 988

i.e –0.988 ≤ x ≤ 0.988

Example 13 Integrate the differential equation dy

dx = x sin πy with y = 1

2 at x = 0, by Picard’s method of successive approximations.

y = y 0 + f x y dx

x

x

,

b g

0

Sol The first approximation y1 of y is obtained by substituting y = 1

2 In the right hand

member of (1) i.e., we have

y1 = 1 2

1 2

0

+zx xsinFHG IKJπ dx = 1

2 2

2 +x

Similarly, the second and third approximation y2 and y3 are given as

y2 = 1 2

1 2

2

0

T|

U V|

W|

zx xsin π.e jx dx = 1

2 +zx xcosπx dx

2 4 0

2 2 48

2 2 6 + x – π x +

2

1

1

0 0

RS|

1

2 2 6 2 0

RS|

2 2 48

2 2 6 + x – π x +

We observe that y2 agree with y3 upto and including term in x6

We can use the relation y = 1

2 2

2 +x with the knowledge that the error is approximately –x

6

5 .

Thus, we can find y1 and y2 correct to 4 decimal places with h = 0.1.

7.4 EULER’S METHOD

The oldest and simplest method was derived by Euler In this method, we determine the change

∆y is y corresponding to small increment in the argument x Consider the differential equation

dy f x y( ),

with the initial condition y x( )0 =y0

Integrating (1) w.r.t x between x0 and x1, we get

( )

,

y dy= x f x y dx

0

x

Now, replacing f x y( ), by the approximation f x y( 0, 0), we get

0

1 0 x 0, 0

x

y = +y ∫ f x y dx

= +y0 f x y( 0, 0)(x1−x0)

y1= +y0 hf x y( 0, 0) (3 x1 – x0 = ∆x = h)

This is the formula for first approximation y1 of y.

Similarly, second approximation y2 is given by

y2= +y1 hf x y( 1, 1)

In general,

y n+1= +y n hf x y( n, n)

7.5 EULER’S MODIFIED METHOD

Instead of approximating f x y( ), by f x y( 0, 0) in equation (2) Let the integral is approximated by Trapezoidal rule to botain

1 0 [ ( 0, 0) ( 1 1)]

2

h

f x y f x y

We obtain the iteration formula,

2

n

y + = +y f x y +f x y 

where, y1(n) is the nth approximation to y1

The above iteration formula can be started by y1(1) from Euler’s method

y1(0) = y0 + h(x0,y0)

Example 14 Using Euler’s method, compute y(0.5) for differential equation

dy y2 x2,

dx= − with y = 1 when x = 0

Sol Let 0.5 0.1

5

h= =

0 0, 0 1, ,

x = y = f x y =y − x

Using Euler’s method we have

y n+1= +y n hf x y( n, n)

But considering n=0,1, 2, in succession, we get

1 0 0, 0

y = +y hf x y

= +1 0.1 1( 2− =0) 1.10000

2 1 1, 1

y = +y hf x y

1.10000 0.1 1.10000 0.1 =1.22000

y3= +y2 hf x y( 2, 2)

1.22000 0.1 1.22 0.2 

4 3 3, 3

y = +y hf x y

=1.36484 0.1[(1.36484) _ 0.3 ]+ 2 ( )2 =1.54212

5 4 4, 4

y =y +hf x y

1.54212 0.1 1.54212 0.4 =1.76393 Hence, the value of y at x=0.5 is 1.76393. Ans

Example 15 Using Euler’s method, compute y(0.04) for the differential equation.

y 1 = –y with y(0) = 1 (Take h = 0.01)

Sol Using Euler’s method

y n+1= +y n hf x y( n, n)

By considering n=0,1, 2, in succession, we obtain

1 0 0, 0

y = +y hf x y

= +1 0.01 1( )− =0.99

2 1 1, 1

y = +y hf x y

=0.99 0.01 0.99+ (− )=0.9801

3 0.9801 0.01 0.9801 0.970299

4 0.970299 0.01 0970299 0960596

Hence, the value of y(0.04) is 0.960596 Ans

Example 16 Find the solution of differential equation

dy xy

dx= with y(1) = 5

in the interval 1,1.5] using h = 0.1.

Sol As per given we have

1 1,

x = y0=5, f xy( )=xy

Using Euler’s method

y n+1= +y n hf x y( n, n)

Now, by considering n=0,1, 2 in succession, we get

For n=0 y1= +y0 0.1f x y( 0, 0)

= +5 0.1 1 5( )× =5.5 For n=1 y2= +y1 0.1f x y( 1, 1)

=5.5 0.1 1.1 5.5+ ( × )=6.105 For n=2 y3= +y2 0.1f x y( 2, 2)

=6.105 0.1 1.2 6.105+ ( × )=6.838 For n=3 y4= +y3 0.1f x y( 3, 3)

=6.838 0.1 1.3 6.838+ ( × )=7.727 For n=4 y5 = y4 + 0.1 f(x4, y4)

=7.727 0.1 1.4 7.727+ ( × )=8.809 Hence, the value of y( )1.5 is 8.809 Ans

Example 17 Given y y x

y x

−

′ = + with y 0 = 1 find y for x = 0.1 in four steps by Euler’s method.

Sol Let 0.1 0.025

4

h= = , given y0=1, where x=0

We know that

y n+1= +y n hf x y( n, n)

By putting n=0,1, 2, 3,we obtain

y1= +y0 hf x y( 0, 0)

(1 0)

1 0.025

1 0

−

= +

+ =1.025

Again, y2=y1 +hf x y( 1, 1)

(1.025 0.025)

1.025 0.025

1.025 0.025

−

+ (where x1= + = +x0 h 0 0.025 ⇒ =x1 0.025) = 1.0488

⇒ y2 = 1.0488

Now again y3 = y2 + hf(x2, y2) (where x2 = x0 + 2h = 0 + 2 × 0.025 = 0.05)

=1.0488 + 0.025 1 0488 0 05

1 0488 0 05

4 3 3, 3

y = +y hf x y (where x3=x0+3h= + ×0 3 0.025 0.075= )