The document discusses numerical methods for solving ordinary differential equations (ODEs), including Taylor's series method and Picard's method. It provides examples of applying Taylor's series method to approximate solutions of first order ODEs at different values of x to 4-5 decimal places of accuracy. The examples given include solving ODEs with initial conditions and computing solutions at multiple x values by taking terms from the Taylor series expansion.

1. 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.
0
dx
yd
....,,.........
dx
yd
,
dx
yd
,
dx
dy
y,x,φ n
n
3
3
2
2
=





( ) 0c....,,.........c,c,cy,x,ψ n321 =

2. Taylor’s and Picard’s methods 3
Dr. V. Ramachandra Murthy
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,...,1x,0x
m10 y........,,y,y
00 y)y(x =
[ ]mx,0x
Approximate
solution
Exact solution
ix
m10 y......,,y,y
00 y)y(x,y)f(x,
dx
dy
==

3. Taylor’s and Picard’s methods 4
Dr. V. Ramachandra Murthy
• 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
0
0
//
2
0
0
/0
0 +
−
+
−
+
−
+=
( ) ( ) ( ) .....y
3!
xx
y
2!
xx
y
1!
xx
yy(x)
///
0
3
0//
0
2
0/
0
0
0 +
−
+
−
+
−
+=
hxxx 01 +==
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
1
3
//
1
2
/
1122 ++++==
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
),........(xfy),(xfy),f(xy n
////
nn
//
nnn ===
0y(1)y,x
dx
dy
=+=

4. Taylor’s and Picard’s methods 5
Dr. V. Ramachandra Murthy
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/
+=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
yxy/
+=
///
y1y += /////
y1y +=
.....y
3!
h
y
2!
h
y
1!
h
y)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!
0.1
0f(1.1)y
32
1 ++++==
0.1103y(1.1)y1 ==

5. Taylor’s and Picard’s methods 6
Dr. V. Ramachandra Murthy
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 =?
.....y
3!
h
y
2!
h
y
1!
h
y)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!
0.1
0.1103y
32
2 ++++=
0.2427y(1.2)y2 ==
3
1
/
xyy =
.....///
ny
3!
3h//
ny
2!
2h/
ny
1!
h
ny)1nf(x1ny ++++=+=+
3
1
xy
dx
dy
=
3
1
/
xyy = /3
2
3
1
//
.y.y
3
1
x.yy
−
+= ( ) 










 −
++=
−−
−
2/3
5
//3
2
/3
2
///
yy
3
2
yy
3
x
yy
3
2
y

6. Taylor’s and Picard’s methods 7
Dr. V. Ramachandra Murthy
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
∴
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
1
3
//
1
2
/
1122 ++++==
( )( ) 111yxy 3
1
3
1
00
/
0 ===
1.3333(1)
3
1
(1)(1).y.y
3
1
.xyy
3
2
3
1
/
0
3
2
00
3
1
0
//
0 =





+=+=
−
−
( ) 8888.0yy
3
2
yy
3
x
yy
3
2
y
2/
0
3
5
0
//
0
3
2
0
0/
0
3
2
0
///
0 =



−+=
−−−
( ) ( ) .....(0.8888)
3!
0.1
(1.3333)
2!
0.1
(1)
1!
0.1
1y
32
1 ++++=
1.1068y(1.1)y1 ==
( ) ( ) ( ) 4242.11.13781.1068(1.1)
3
1
1.1068.y.yx
3
1
yy 3
2-
3
1/
1
3
2-
11
3
1
1
//
1 =+=+=
( ) 1.13781.1068(1.1)yxy 3
1
3
1
11
/
1 ===
( ) 8438.0yy
3
2
yy
3
x
yy
3
2
y
2/
1
3
5
1
//
1
3
2
1
1/
1
3
2
1
///
1 =



−+=
−−−
( ) ( ) (0.8438)
3!
0.1
(1.4242)
2!
0.1
(1.1378)
1!
0.1
1.1068y
32
2 +++=
1.2278y(1.2)y2 ==

7. Taylor’s and Picard’s methods 8
Dr. V. Ramachandra Murthy
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)
12102
0030 =+=+=′ 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 +=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
2/
yx3y += ////
2yy3y +=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
2
y3x
dx
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!
0.1
1y
432
1 =++++=

8. Taylor’s and Picard’s methods 9
Dr. V. Ramachandra Murthy
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/
+=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
xy1
dx
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
/
0y =+=+=
3(3)(1)(0)(2)3yyxy //
0
///
00
IV
0 =+=+=
( ) ( ) ( ) ........(3)
4!
40.2
(2)
3!
30.2
(1)
2!
20.2
(1)
1!
0.2
11y +++++=
1.2228y(0.2)y1 ==

9. Taylor’s and Picard’s methods 10
Dr. V. Ramachandra Murthy
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/
=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
1-yx
dx
dy 2
=
/2//
yx2xyy +=
//2///2/////
yx4xy2yyx2xy2xy2yy ++=+++=
///2//////2//////IV
yx6xy6yyx2xy4xy4y2yy ++=++++=
1-yxy 2/
=
0(0)(-1)2(0)(1)yxy2xy /
0
2
000
//
0 =+=+=
2(0)(0)4(0)(-1)2(1)yxy4x2yy //
0
2
0
/
000
///
0 =++=++=
11(0)(1)1-yxy 0
2
0
/
0 −=−==
-6(0)(2)6(0)(0)6(-1)yxy6x6yy ///
0
2
0
//
00
/
0
IV
0 =++=++=
( ) ( ) ( ) 0.90030...(-6)
4!
40.1
(2)
3!
30.1
(0)
2!
20.1
(-1)
1!
0.1
11y =+++++=

10. Taylor’s and Picard’s methods 11
Dr. V. Ramachandra Murthy
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 =?
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
1
3
//
1
2
/
1122 ++++==
1.40590=
0.990991(0.90030)1(0.1)-yxy 2
1
2
1
/
1 −=−=
( )( ) ( ) ( ) 17015.099099.01.09003.01.02yxy2xy
2/
1
2
111
//
1 =−+=+=
(0.17015)(0.1)99099)4(0.1)(-0.2(0.9003)yxy4x2yy 2//
1
2
1
/
111
///
1 ++=++=
( ) ( )
( ) (-5.82979)
4!
0.1
(1.40590)
3!
0.1
(0.17015)
2!
0.1
(-0.99099)
1!
0.1
0.9003y
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 =

11. Taylor’s and Picard’s methods 12
Dr. V. Ramachandra Murthy
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
∴
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
1-xy
dx
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!
0.02
21y +++++=
02000.2y(1.02)1y ==

12. Taylor’s and Picard’s methods 13
Dr. V. Ramachandra Murthy
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 =?
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
x
3e2y
dx
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
//
0
0
=+=+=
213e(2)(9)3e2yy 0x//
0
///
0
0
=+=+=
453e(2)(21)3e2yy 0x///
0
IV
0
0
=+=+=

13. Taylor’s and Picard’s methods 14
Dr. V. Ramachandra Murthy
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!
0.2
01y +++++=
0.811y(0.2)y1 ==
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
3
xy
dx
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 =+=+=

14. Taylor’s and Picard’s methods 15
Dr. V. Ramachandra Murthy
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!
h
1y)2f(x2y ++++==
( ) ( ) ( ) 1.2269....(17)
4!
40.1
(11)
3!
30.1
(5)
2!
20.1
(2)
1!
0.1
11y =+++++=
2.5579(1.1)(1.2269)xyy 33
11
/
1 =+=+=
6.18793(1.1)(2.5579)3xyy 22
1
/
1
//
1 =+=+=
12.78796(1.1)(6.1879)6xyy 1
//
1
///
1 =+=+=
18.7879612.78796yy ///
1
IV
1 =+=+=
( ) ( ) ( )
5158.1
...(18.7879)
4!
40.1
(12.7879)
3!
30.1
(6.1879)
2!
20.1
(2.5579)
1!
0.1
2269.12y
=
++++=
2xy-1
dx
dy
=
0xy
dx
dy
=+

15. Taylor’s and Picard’s methods 16
Dr. V. Ramachandra Murthy
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
x
0
0
y)dxf(x,yy
∫+=
x
x
001
0
)dxyf(x,yy
∫+=
x
x
102
0
)dxyf(x,yy
∫+=
x
x
203
0
)dxyf(x,yy
∫+=
x
x
1-n0n
0
)dxyf(x,yy
1-ny ny
2
y1
dx
dy
+=
0y;0x;y1y)f(x, 00
2
==+=
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( ) ( )∫∫∫ +=++=
x
0
2
1-n
x
0
x
0
2
1-nn dxy1.dxdxy10y
( )∫+=
x
0
2
1-nn dxyxy

16. Taylor’s and Picard’s methods 17
Dr. V. Ramachandra Murthy
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)=?
( ) ( ) xdx0xdxyxy
x
0
x
0
2
01 =+=+= ∫∫
( ) ( ) 3
x
xdxxxdxyxy
3
x
0
2
x
0
2
12 +=+=+= ∫∫
( )
63
x
15
2x
3
x
dx
3
x
xxdxyxy
753
x
0
23
x
0
2
23 ++=





++=+= ∫∫
2
yx
dx
dy
+=
2
yxy)f(x, +=
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( )∫ ++=
x
x
2
1-n0n
0
dxyxyy
( ) ( ) 0.0050
2
x
dx0x0dxyxyy
0.1
0
20.1
0
x
x
2
001
0
=





=++=++= ∫∫

17. Taylor’s and Picard’s methods 18
Dr. V. Ramachandra Murthy
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)=?
( ) ( )( ) 0.0050
2
x
dx0.0050x0dxyxyy
0.1
0
20.1
0
2x
x
2
102
0
=





=++=++= ∫∫
( ) ( )( ) 0.0050
2
x
dx0050.0x0dxyxyy
0.1
0
2
0.1
0
2x
x
2
203
0
=





=++=++= ∫∫
( ) ( )( ) 02.0dx0.0050x0050.0dxyx0050.0y
0.2
0.1
20.2
0.1
2
01 =++=++= ∫∫
1.0x0 = 0.0050y0 =
( )∫ ++=
0.2
0.1
2
1-nn dxyx0.0050y
( )( ) ( )( )
( )
0.0200
x0.02
2
x
0.0050
dx0.02x0.0050dxyx0.0050y
0.2
0.1
2
2
0.2
0.1
20.2
0.1
2
12
=






++=
++=++= ∫∫
( )( ) 0.0200dx0.02x0.0050y
0.2
0.1
2
3 =++= ∫
yx
dx
dy 2
−=
yxy)f(x, 2
−=

18. Taylor’s and Picard’s methods 19
Dr. V. Ramachandra Murthy
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
x
1-n0n
0
)dxyf(x,yy
( )∫ −+=
0.2
0
1-n
2
n dxyx1y
( ) ( ) 0.8026dx1x1dxyx1y
0.2
0
2
0.2
0
0
2
1 =−+=−+= ∫∫
( ) ( ) 0.8421dx0.8026x1dxyx1y
0.2
0
2
0.2
0
1
2
2 =−+=−+= ∫∫
( ) ( ) 0.8342dx0.8421x1dxyx1y
0.2
0
2
0.2
0
2
2
3 =−+=−+= ∫∫
( ) ( ) 0.8358dx0.8342x1dxyx1y
0.2
0
2
0.2
0
3
2
4 =−+=−+= ∫∫
8355.0y5 = 8355.0y6 =
2xyx
dx
dy 2
+=
0y;0x;2xyxy)f(x, 00
2
==+=
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( )∫ −++=
x
0
1n
2
n dx2xyx0y

19. Taylor’s and Picard’s methods 20
Dr. V. Ramachandra Murthy
--------(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
x
2
3
x
dx
3
x
2xxdx2xyxy
53
x
0
3
2
x
0
1
2
2 +=





+=+= ∫∫
( ) ( )
3
x
dxxdx2xyxy
3
x
0
2
x
0
0
2
1 ∫∫ ==+=
∫ ++=∫ +=





















x
0
dx
15
5
x
2
3
3
x
2x
2
x
x
0
dx22xy
2
x3y
7x
105
45x
15
2
3
3xx
0
dx6x
15
44x
3
22x3y ++=∫ ++= 







xy1
dx
dy
+=
xy1y)f(x, +=
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( )∫ −++=
0.1
0
1nn dxxy11y
( ) ( ) 1.105
2
x
x1x11dxxy11y
0.1
0
0.1
0
2
0.1
0
01 =





++=++=++= ∫∫

20. Taylor’s and Picard’s methods 21
Dr. V. Ramachandra Murthy
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)=?
yx
dx
dy
−=
y-xy)f(x, =
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( )∫ −+=
0.2
0
1nn dxy-x1y
( ) ( ) 0.82x
2
x
11-x1dxy-x1y
0.2
0
0.2
0
20.2
0
01 =





−+=+=+= ∫∫
( ) ( )
0.8560.82x-
2
x
1
dx0.82-x1dxy-x1y
0.2
0
2
0.2
0
0.2
0
12
=





+=
+=+= ∫∫
8500.0y7,n
8500.0y6,n
8499.0y5,n
8502.0y4,n
8488.0y3,nfor
7
6
5
4
3
==
==
==
==
==

21. Taylor’s and Picard’s methods 22
Dr. V. Ramachandra Murthy
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
x
1-n0n
0
)dxyf(x,yy
xy
xy
dx
dy
+
−
=
xy
x-y
y)f(x,
+
=
∫ 





+
−
+=
−
−
0.1
0
1n
1n
n dx
xy
xy
1y
dx
1x
2
11dx
1x
2
1x
1x
1dx
1x
21x
1
dx
x1
1x
1dx
x1
x1
1dx
xy
xy
1y
0.1
0
0.1
0
0.1
0
0.1
0
0.1
0
0.1
0
0
0
1
∫∫∫
∫∫∫






+
−−=





+
−
+
+
−=





+
−+
−=






+
−
−+=





+
−
+=





+
−
+=
( ) ( )[ ] 090.11xlog2x1y
0.1
0
0.1
01 =++−=
( )[ ] ( )x1.090xlog2.181dx1
1.090x
2.18
1
dx
1.090x
1.090)(x-2(1.090)
1dx
x1.090
1.0901.090x-1.090
1
dx
x1.090
x1.090
1dx
xy
xy
1y
0.1
0
0.1
0
0.1
0
0.1
0
0.1
0
0.1
0
0.1
0
1
1
2
−++=





−
+
+=






+
+
+=





+
+−
+=






+
−
+=





+
−
+=
∫
∫∫
∫∫

22. Taylor’s and Picard’s methods 23
Dr. V. Ramachandra Murthy
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
0.1
0
x0.1
0
1.091xlog2.1821
dx
0.1
0
1
1.091x
2.182
1
dx
0.1
0
1.091x
1.091)(x-2(1.091)
1
dx
0.1
0
x1.091
1.0911.091x-1.091
1
dx
0.1
0
x1.091
x1.091
1
0.1
0
dx
x2y
x2y
13y
=
−++=






−
+
+=






+
+
+=






+
+−
+=






+
−
+=





+
−
+=
∫
∫
∫
∫∫
2
xy
dx
dy
−=
2
xyy)f(x, −=
∫+=
x
x
1-n0n
0
)dxyf(x,yy
( )∫ −+= −
x
0
2
1nn dxxy1y
( ) ( )
3
x
x1
3
x
x1dxx11dxxy1y
3x
0
3
x
0
2
x
0
2
01 −+=





−+=−+=−+= ∫∫

23. Taylor’s and Picard’s methods 24
Dr. V. Ramachandra Murthy
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
x
x1
dx
12
x
3
x
2
x
x11
dxx
3
x
12
x
2
x
x11dxxy1y
5432
x
0
432
x
0
2
342x
0
2
23
−−−++=






−−−++=






−−−+++=−+=
∫
∫∫
2
y3x
dx
dy
+=
22
yx
dx
dy
+=

24. Taylor’s and Picard’s methods 25
Dr. V. Ramachandra Murthy
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,2x
dx
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,nxhfny
0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
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,nxhfny
0
1n
ywhere
,..2,1,0
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
=→+=
+
=









++++=
+
+
r
0.051x = 0.12x =
10y = ?1y = ?2y =

25. Taylor’s and Picard’s methods 26
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( ) ______(4)0y,0xhf0y
0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
[ ] 1.05100.0510y2
0
x0.051
0y,0xhf0y
0
1
y
=++=



 ++=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( )
( ) ( )
( ) ( ) 0513.11.0520.05120
2
0.05
1
(0)
1
y2
1x0y2
0x
2
0.05
1
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=



 ++++=



 ++++=









++=
Second approximation:
Put r=1 in Eqn(3)
( ) ( )
( ) ( )
( ) ( ) 0513.11.051320.05120
2
0.05
1
(1)
1
y2
1x0y2
0x
2
0.05
1
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=



 ++++=



 ++++=









++=
Since
( )1
1
y and
( )2
1
y are the same correct to four decimal places
y(0.05)=1.0513
Step-(2): (To find y(0.1))2y(x2y == )
Put n=1 in Equations (1) and (2)

26. Taylor’s and Picard’s methods 27
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( ) ______(6)1y,1xhf1y0
2
ywhere
_______(5)
(r)
2
y,2xf1y,1xf
2
h
y1
1r
2
y
+=









++=
+
Initial approximation:
From Eqn(6)
( ) ( )
( ) 1.10391.051320.050.051.0513
1y2
1
x0.051.05131y,1xhf1y0
2
y
=



 ++=




 ++=+=
First approximation:
Put r=0 in Eqn(5)
( ) ( )
( ) ( )
( ) ( )
1.1054
1.103920.11.051320.05
2
0.05
1.0513
(0)
2
y2
1x1y2
1x
2
0.05
1.0513
(0)
2
y,2xf1y,1xf
2
h
1y
1
2
y
=



 ++++=



 ++++=









++=
Second approximation:
Put r=1 in Eqn(5)
( ) ( )
( ) ( )
( ) ( ) 1055.11.105420.11.051320.05
2
0.05
1.0513
(1)
2
y2
1x1y2
1x
2
0.05
1.0513
(1)
2
y,2xf1y,1xf
2
h
1y
2
2
y
=



 ++++=



 ++++=









++=
Similarly
( ) 1055.1
3
2
y =
Since
( ) ( )3
2
y&
2
2
y are the same correct to four decimal places
y(0.1)=1.1055
Problem(2):
Obtain the solution of the equation yx
dx
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

27. Taylor’s and Picard’s methods 28
Dr. V. Ramachandra Murthy
decimal places.
Soln: Given data: ( ) yxyx,f += , h=0.3
00x =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+
Step-(1): (To find y(0.3))1y(x1y == )
Put n=0 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
[ ] [ ] 1.3100.310y0x0.31
0y,0xhf0y
0
1
y
=++=++=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( )
( ) ( ) 





++++=









++=
(0)
1
y1x0y0x
2
0.3
1
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
( ) ( )[ ] 3660.11.30.310
2
0.3
1 =++++=
Second approximation:
Put r=1 in Eqn(3)
0.31x = 0.62x =
10y = ?1y = ?2y =

28. Taylor’s and Picard’s methods 29
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( )
( ) ( ) ( )[ ] 3703.11.36600.310
2
0.3
1
2
1
y
(1)
1
y1x0y0x
2
0.3
1
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=++++=






++++=









++=
Similarly
( ) 3703.1
3
1
y =
Since
( ) ( )3
1
y&
2
1
y are the same correct to four decimal places
y(0.3)=1.3703
Step-(2): (To find y(0.6))2y(x2y == )
Put n=1 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(6)1y,1xhf1y0
2
ywhere
_______(5)
(r)
2
y,2xf1y,1xf
2
h
0y
1r
2
y
+=









++=
+
Initial approximation:
From Eqn(6)
( ) ( ) ( )
( )[ ] 1.81141.37030.30.31.3703
1y1x0.31.37031y,1xhf1y0
2
y
=++=
++=+=
First approximation:
Put r=0 in Eqn(5)
( ) ( )
( ) ( ) 





++++=









++=
(0)
2
y2x1y1x
2
0.3
3703.1
(0)
2
y,2xf1y,1xf
2
h
1y
1
2
y
( ) ( )[ ] 8827.11.81140.61.37030.3
2
0.3
3703.1 =++++=
Similarly
( ) ( ) ( ) 8869.1
4
2
y,8869.1
3
2
y,8667.1
2
2
y ===
∴∴∴∴ y(0.6)=1.8869

29. Taylor’s and Picard’s methods 30
Dr. V. Ramachandra Murthy
Problem(3):
Using Modified Euler’s method find y(0.2) given that yx
dx
dy
+= ; y(0)=1
correct to four decimal places.
Soln: Given data: ( ) yxyx,f += , h=0.2
00x = 0.21x =
10y = ?1y =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+
To find y(0.2))1y(x1y ==
Put n=0 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
[ ] [ ] 1.2100.210y0x0.21
0y,0xhf0y
0
1
y
=++=++=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( )
[ ] 24.12.12.010
2
0.2
1
(0)
1
y1x0y0x
2
0.2
1
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=++++=



 ++++=









++=
Second approximation:
Put r=1 in Eqn(3)

30. Taylor’s and Picard’s methods 31
Dr. V. Ramachandra Murthy
( ) ( )
( ) [ ] 244.124.12.010
2
0.2
1
2
1
y
(1)
1
y1x0y0x
2
0.2
1
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=++++=



 ++++=









++=
Similarly
( ) 2444.1
3
1
y = &
( ) 2444.1
4
1
y =
Since
( ) ( )4
1
y&
3
1
y are the same correct to four decimal places
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 2xy
dx
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 =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+
To find y(1.1))1y(x1y ==
Put n=0 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)

31. Taylor’s and Picard’s methods 32
Dr. V. Ramachandra Murthy
( ) ( )
[ ] [ ] 1.2)1)(1)(2(0.110y02x0.11
0y,0xhf0y
0
1
y
=+=+=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( )
[ ] 1.232)2(1.1)(1.22(1)(1)
2
0.1
1
(0)
1
y12x0y02x
2
0.1
1
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=++=



 ++=









++=
Second approximation:
Put r=1 in Eqn(3)
( ) ( )
( ) [ ] 1.235532)2(1.1)(1.22(1)(1)
2
0.1
1
2
1
y
(1)
1
y12x0y02x
2
0.1
1
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
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
3x
x
2y
dx
dy
+= , y(1)=0.5 correct to three decimal places.
Soln: Given data: ( ) 3x
x
2y
yx,f += , h=0.2
10x = 1.21x = 1.41x =
0.50y = ?1y = ?1y =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+

32. Taylor’s and Picard’s methods 33
Dr. V. Ramachandra Murthy
Step(1): To find y(1.2))1y(x1y ==
Put n=0 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
( )
( )
0.9
31
1
2(0.5)
0.20.5
3
0x
0x
02y
0.20.5
0y,0xhf0y
0
1
y
=




++=






++=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
0227.1
31.2
1.2
0.9231
1
0.52
2
0.2
0.5
3
1x
1x
(0)
1
2y3
0x
0x
02y
2
0.2
0.5
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=




++++=










++++=









++=
Second approximation:
Put r=1 in Eqn(3)

33. Taylor’s and Picard’s methods 34
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
043.1
31.2
1.2
1.0227231
1
0.52
2
0.2
0.5
3
1x
1x
(1)
1
2y3
0x
0x
02y
2
0.2
0.5
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=




++++=










++++=









++=
Similarly
( ) 046.1
3
1
y = and
( ) 046.1
4
1
y =
Since
( ) ( )4
1
y&
3
1
y are the same correct to four decimal places
y(1.2)=1.2444
Step(2): To find y(1.4))2y(x2y ==
Put n=1 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(6)1y,1xhf1y0
2
ywhere
5)_______(
(r)
2
y,2xf1y,1xf
2
h
1y
1r
2
y
+=









++=
+
Initial approximation:
From Eqn(6)
( ) ( )
( )
( )
1.740
31.2
1.2
2(1.046)
0.2046.1
3
1x
1x
12y
0.2046.1
1y,1xhf1y
0
2
y
=




++=






++=
+=
First approximation:
Put r=0 in Eqn(5)

34. Taylor’s and Picard’s methods 35
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
916.1
31.4
1.4
1.74231.2
1.2
1.0462
2
0.2
046.1
3
2x
2x
(0)
2
2y3
1x
1x
12y
2
0.2
046.1
(0)
2
y,2xf1y,1xf
2
h
1y
1
2
y
=




++++=










++++=









++=
Similarly
( ) 941.1
2
2
y =
( ) 944.1
3
2
y =
,
( ) 945.1
4
2
y =
,
( ) 945.1
5
2
y =
Since
( ) ( )5
2
y&
4
2
y are the same correct to three decimal places
y(1.4)=1.945
Problem(6):
solve y1
dx
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 =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+
To find y(0.1))1y(x1y ==
Put n=0 in Equations (1) and (2)

35. Taylor’s and Picard’s methods 36
Dr. V. Ramachandra Murthy
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
[ ] 1.0010.10
]01[00y,0xhf0y
0
1
y
=−+=
−+=+= yhy
First approximation:
Put r=0 in Eqn(3)
( ) ( )
[ ] 095.01.0101
2
0.1
0
(0)
1
y101
2
h
0
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=−+−+=



 ++−+=








++= yy
Second approximation:
Put r=1 in Eqn(3)
( ) ( )
0952.0]095.0101[
2
0.1
0
(1)
1
y10y-1
2
h
0y
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=−+−+=



 −++=









++=
Similarly
( ) 0952.0
3
1
y = ,
Since
( ) ( )4
1
y&
3
1
y are the same correct to four decimal places
y(0.1)=0.0952
Problem(7):
Use Modified Euler’s method to solve the differential equation
2yx
dx
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 =

36. Taylor’s and Picard’s methods 37
Dr. V. Ramachandra Murthy
10y = 1.11741y = 1.27622y =
Modified Euler’s Formula is given by
( ) ( )
( ) ( ) ______(2)ny,nxhfny0
1n
ywhere
1)_______(
(r)
1n
y,1nxfny,nxf
2
h
ny
1r
1n
y
+=
+










++++=
+
+
Step(1): To find y(0.1))1y(x1y ==
Put n=0 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(4)0y,0xhf0y0
1
ywhere
3)_______(
(r)
1
y,1xf0y,0xf
2
h
0y
1r
1
y
+=









++=
+
Initial approximation:
From Eqn(4)
( ) ( )
( ) ( ) 1.12100.112
0y0x0.11
0y,0xhf0y
0
1
y
=



 ++=



 ++=
+=
First approximation:
Put r=0 in Eqn(3)
( ) ( ) ( )
( ) ( ) 1155.121.11.0210
2
0.1
1
2
(0)
1
y1x2
0y0x
2
h
0y
(0)
1
y,1xf0y,0xf
2
h
0y
1
1
y
=



 ++++=













++++=








++=
Second approximation:
Put r=1 in Eqn(3)
( ) ( )
( )
1172.1]2)1155.1(1.02)1(0[
2
0.1
1
2
(1)
1
y1x2
0y0x
2
h
0y
(1)
1
y,1xf0y,0xf
2
h
0y
2
1
y
=++++=













++++=









++=

37. Taylor’s and Picard’s methods 38
Dr. V. Ramachandra Murthy
Similarly
( ) 1174.1
3
1
y = ,
( ) 1174.1
4
1
y =
Clearly
( ) ( )4
1
y&
3
1
y are same correct to four decimal places.
∴∴∴∴ y(0.1)=1.1174
Step-(2): (To find y(0.2))2y(x2y == )
Put n=1 in Equations (1) and (2)
( ) ( )
( ) ( ) ______(6)1y,1xhf1y0
2
ywhere
_______(5)
(r)
2
y,2xf1y,1xf
2
h
1y
1r
2
y
+=









++=
+
Initial approximation:
From Eqn(6)
( ) ( ) ( )
( ) 2522.121.11741.00.11.1174
2
1y1xh1y1y,1xhf1y
0
2
y
=



 ++=




 ++=+=
First approximation:
Put r=0 in Eqn(5)
( ) ( )
( )
( ) ( ) 1.273221.25220.221.11740.1
2
0.1
1.1174
2
(0)
2
y2x2
1y1x
2
h
1y
(0)
2
y,2xf1y,1xf
2
h
1y
1
2
y
=



 ++++=













++++=









++=
Similarly
( ) ( ) ( ) 2762.1
4
2
y,2762.1
3
2
y,2758.1
2
2
y ===
Since
( )3
2
y and
( )4
2
y are the same correct to four decimal places
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}

38. Taylor’s and Picard’s methods 39
Dr. V. Ramachandra Murthy
Problem(9):
Solve y2x
dx
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
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
++=






++=






++=
=
++++=+
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
y
3xyx,f += , h=0.1
00x = 0.11x = 0.21x =
10y = 1.06641y = 1670.12y =
The Runge-Kutta method of 4th
order is given by

39. Taylor’s and Picard’s methods 40
Dr. V. Ramachandra Murthy
[ ]
( )
( ) 












++=






++=






++=
=
++++=+
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
-------------------- (1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= --------------------------- (2)
( )
( )









++=






++=






++=
=
3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
( )
0.05
2
1
3(0)0.1
2
0y
03xh
0y,0xhf1k
=



+=





+=
=
0.0662
2
0.05
1
2
1
2
0.1
030.1
2
1k
0y
2
1
2
h
0x3h
2
1k
0y,
2
h
0xhf2k
=











++





+=












++





+=






++=
0.0666
2
0.0662
1
2
1
2
0.1
030.1
2
2k
0y
2
1
2
h
0x3h
2
2k
0y,
2
h
0xhf3k
=











++





+=












++





+=






++=

40. Taylor’s and Picard’s methods 41
Dr. V. Ramachandra Murthy
( )
( ) ( )
( ) ( ) 0.08330.06661
2
1
0.1030.1
3k0y
2
1
h0x3h
3k0yh,0xhf4k
=



+++=




+++=
++=
Substituting all these values in Eqn(2), we get
[ ] 1.06640.08332(0.0666)2(0.0662)0.05
6
1
11y =++++=
Put n=1 in Eqn(1)
[ ]4k32k22k1k
6
1
1y2y ++++= ---------------------- (3)
Where
( )
( )3k1yh,1xhf4k
2
2k
1y,
2
h
1xhf3k
2
1k
1y,
2
h
1xhf2k
1y,1xhf1k
++=






++=






++=
=
( )
0.0833
2
1.0664
3(0.1)0.1
2
1y
13xh
1y,1xhf1k
=



+=





+=
=
0.1004
2
0.0833
1.0664
2
1
2
0.1
0.130.1
2
1k
1y
2
1
2
h
1x3h
2
1k
1y,
2
h
1xhf2k
=











++





+=












++





+=






++=
0.1008
2
0.1004
1.0644
2
1
2
0.1
0.130.1
2
2k
1y
2
1
2
h
1x3h
2
2k
1y,
2
h
1xhf3k
=











++





+=












++





+=






++=

41. Taylor’s and Picard’s methods 42
Dr. V. Ramachandra Murthy
( )
( ) ( )
( ) ( ) 0.11830.10081.0664
2
1
0.10.130.1
3k1y
2
1
h1x3h
3k1yh,1xhf4k
=



+++=




+++=
++=
Substituting all these values in Eqn(3), we get
[ ]
1.1670
0.11832(0.1008)2(0.1004)0.0833
6
1
1.06642y
=
++++=
Problem(2):
Apply Runge-Kutta method of fourth order to find an approximate value
of y(0.1) and y(0.2) of 2yx
dx
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 =
The Runge-Kutta method of 4th
order is given by
[ ]
( )
( ) 












++=






++=






++=
=
++++=+
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
-------------------------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= --------------------------(2)

42. Taylor’s and Picard’s methods 43
Dr. V. Ramachandra Murthy
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
( ) ( ) 0.12100.12
0y0xh
0y,0xhf1k
=



 +=



 +=
=
0.1152
2
2
0.1
1
2
0.1
00.1
2
2
1k
0y
2
h
0xh
2
1k
0y,
2
h
0xhf2k
=














++





+=














++





+=






++=
0.1168
2
2
0.1152
1
2
0.1
00.1
2
2
2k
0y
2
h
0xh
2
2k
0y,
2
h
0xhf3k
=














++





+=














++





+=






++=
( )
( ) ( )
( ) ( ) 0.134720.116810.100.1
2
3k0yh0xh
3k0yh,0xhf4k
=



 +++=



 +++=
++=
Substituting all these values in Eqn(2), we get
[ ] 1.11640.13472(0.1168)2(0.1152)0.1
6
1
11y =++++=
Put n=1 in Eqn(1) [ ]4k32k22k1k
6
1
1y2y ++++= -----------(3)
Where

43. Taylor’s and Picard’s methods 44
Dr. V. Ramachandra Murthy
( )
( )3k1yh,1xhf4k
2
2k
1y,
2
h
1xhf3k
2
1k
1y,
2
h
1xhf2k
1y,1xhf1k
++=






++=






++=
=
( )
( ) ( ) 0.134621164.1(0.1)0.12
1y1xh
1y,1xhf1k
=



 +=



 +=
=














++





+=






++=
2
2
1k
1y
2
h
1xh
2
1k
1y,
2
h
1xhf2k
0.1551
2
2
0.1346
1.1164
2
0.1
0.10.1 =














++





+=
0.1575
2
2
0.1551
1.1164
2
0.1
0.10.1
2
2
2k
1y
2
h
1xh
2
2k
1y,
2
h
1xhf3k
=














++





+=














++





+=






++=
( )
( ) ( )
( ) ( ) 0.182220.15751.11640.10.10.1
2
3k1yh1xh
3k1yh,1xhf4k
=



 +++=



 +++=
++=
Substituting all these values in Eqn(3), we get
[ ]
27341.
0.18222(0.1575)2(0.1551)0.1346
6
1
1.11642y
=
++++=

44. Taylor’s and Picard’s methods 45
Dr. V. Ramachandra Murthy
Problem(3):
Use Runge-Kutta method of fourth order to approximate y when x=0.1,
given that y=1 when x=0 and yx
dx
dy
+=
Soln: Given data: ( ) yxyx,f += , h=0.1
00x = 0.11x =
10y = ?1y =
The Runge-Kutta method of 4th
order is given by
[ ]
( )
( )3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
++=






++=






++=
=
++++=+
------------------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= -------------------(2)
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
( )[ ] [ ] 0.1100.10y0xh
0y,0xhf1k
=+=+=
=

45. Taylor’s and Picard’s methods 46
Dr. V. Ramachandra Murthy
0.11
2
0.1
1
2
0.1
00.1
2
1k
0y
2
h
0xh
2
1k
0y,
2
h
0xhf2k
=











++





+=












++





+=






++=
1105.0
2
0.11
1
2
0.1
00.1
2
2k
0y
2
h
0xh
2
2k
0y,
2
h
0xhf3k
=











++





+=












++





+=






++=
( )
( ) ( )[ ]
( ) ( )[ ] 0.12100.110510.100.1
3k0yh0xh
3k0yh,0xhf4k
=+++=
+++=
++=
Substituting all these values in Eqn(2), we get
[ ]
11031.
0.12102(0.1105)2(0.11)0.1
6
1
11y
=
++++=
Problem(4):
Use Runge-Kutta method of fourth order to obtain an approximation to
y(1.5) for the solution of 2xy
dx
dy
= ;y(1)=1 correct to four decimal places.
Soln: Given data: ( ) 2xyyx,f = , h=0.5
10x = 1.51x =
10y = ?1y =
The Runge-Kutta method of 4th
order is given by

46. Taylor’s and Picard’s methods 47
Dr. V. Ramachandra Murthy
[ ]
( )
( ) 












++=






++=






++=
=
++++=+
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
------------------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= --------------------------(2)
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
[ ] [ ] 12(1)(1)0.50y02xh
0y,0xhf1k
===
=
1.875
2
1
1
2
0.5
120.5
2
1k
0y
2
h
0x2h
2
1k
0y,
2
h
0xhf2k
=











+





+=












+





+=






++=
4218.2
2
1.875
1
2
0.5
120.5
2
2k
0y
2
h
0x2h
2
2k
0y,
2
h
0xhf3k
=











+





+=












+





+=






++=

47. Taylor’s and Picard’s methods 48
Dr. V. Ramachandra Murthy
( )
( )( )[ ]
( )( )[ ] 5.13272.421810.5120.5
3k0yh0x2
3k0yh,0xhf4k
=++=
++=
++=
h
Substituting all these values in Eqn(2), we get
[ ]
4543.3
5.13272(2.4218)2(1.875)1
6
1
11y
=
++++=
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 =
The Runge-Kutta method of 4th
order is given by
[ ]
( )
( ) 












++=






++=






++=
=
++++=+
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
-----------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= -------------(2)
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=

48. Taylor’s and Picard’s methods 49
Dr. V. Ramachandra Murthy
( )
[ ] [ ] -0.11-0.10y-h
0y,0xhf1k
===
=
095.0
2
0.1
-10.1
2
1k
0yh
2
1k
0y,
2
h
0xhf2k
−=











−=












+−=






++=
0952.0
2
0.095
10.1
2
2k
0yh
2
2k
0y,
2
h
0xhf3k
−=











−−=












+−=






++=
( )
( )[ ]
( )[ ] -0.09040.0952-10.1
3k0yh
3k0yh,0xhf4k
=−=
+−=
++=
Substituting all these values in Eqn(2), we get
[ ] 9048.00.0904-2(-0.0952)2(-0.095)0.1-
6
1
11y =+++=
Put n=1 in Eqn(1) [ ]4k32k22k1k
6
1
1y2y ++++= -------------(3)
( )
( )3k1yh,1xhf4k
2
2k
1y,
2
h
1xhf3k
2
1k
1y,
2
h
1xhf2k
1y,1xhf1k
++=






++=






++=
=
( )
( )[ ] [ ] -0.09049048.00.11y-h
1y,1xhf1k
=−==
=

49. Taylor’s and Picard’s methods 50
Dr. V. Ramachandra Murthy
-0.0859
2
0.0904
0.90480.1
2
1k
1yh
2
1k
1y,
2
h
1xhf2k
=



−−=












+−=






++=
-0.0861
2
0.0859
0.90480.1
2
2k
1yh
2
2k
1y,
2
h
1xhf3k
=











−−=












+−=






++=
( )
( )[ ]
( )[ ] 0.08180.0861-0.90480.1
3k1yh
3k1yh,1xhf4k
−=−=
+−=
++=
Substituting all these values in Eqn(3), we get
[ ]
8187.0
0.0818-2(-0.0861)2(-0.0859)0.0904-
6
1
0.90482y
=
+++=
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 =
The Runge-Kutta method of 4th
order is given by

50. Taylor’s and Picard’s methods 51
Dr. V. Ramachandra Murthy
[ ]
( )
( ) 












++=






++=






++=
=
++++=+
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
---------------------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= -----------------------(2)
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
0.5
10
1
0.5
0y0x
1
h
0y,0xhf1k
=



+
=





+
=
=
3333.0
2
0.5
1
2
5.0
0
1
0.5
2
1k
0y
2
h
0x
1
h
2
1k
0y,
2
h
0xhf2k
=










+++
=










+++
=






++=
3529.0
2
0.3333
1
2
0.5
0
1
0.5
2
2k
0y
2
h
0x
1
h
2
2k
0y,
2
h
0xhf3k
=










+++
=










+++
=






++=

51. Taylor’s and Picard’s methods 52
Dr. V. Ramachandra Murthy
( )
0.2698
0.352910.50
1
0.5
3k0yh0x
1
h
3k0yh,0xhf4k
=




+++
=





+++
=
++=
Substituting all these values in Eqn(2), we get
[ ] 3570.10.26982(0.3529)2(0.3333)0.5
6
1
11y =++++=
Put n=1 in Eqn(1) [ ]4k32k22k1k
6
1
1y2y ++++= --------------(3)
( )
( )3k1yh,1xhf4k
2
2k
1y,
2
h
1xhf3k
2
1k
1y,
2
h
1xhf2k
1y,1xhf1k
++=






++=






++=
=
( )
0.2692
1.3570.5
1
0.5
1y1x
1
h
1y,1xhf1k
=



+
=





+
=
=
2230.0
2
0.2692
1.3570
2
5.0
0.5
1
0.5
2
1k
1y
2
h
1x
1
h
2
1k
1y,
2
h
1xhf2k
=










+++
=










+++
=






++=
2253.0
2
0.2230
1.357
2
0.5
0.5
1
0.5
2
2k
1y
2
h
1x
1
h
2
2k
1y,
2
h
1xhf3k
=










+++
=










+++
=






++=

52. Taylor’s and Picard’s methods 53
Dr. V. Ramachandra Murthy
( )
0.1936
0.22531.3570.50.5
1
0.5
3k1yh1x
1
h
3k1yh,1xhf4k
=




+++
=





+++
=
++=
Substituting all these values in Eqn(3), we get
[ ]
5835.1
0.19362(0.2253)2(0.2230)0.2692
6
1
1.35702y
=
++++=
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
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny
++=






++=






++=
=
++++=+
---------------(1)
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= -------------------(2)

53. Taylor’s and Picard’s methods 54
Dr. V. Ramachandra Murthy
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
0.32(0)03e0.1
02yx3eh
0y,0xhf1k
0
=



 +=



 +=
=
0.345
2
0.3
02
)
2
0.1
(0
3e0.5
2
1k
0y2
)
2
h
(x
3eh
2
1k
0y,
2
h
0xhf2k
0
=
















++
+
=
















++
+
=






++=
0.349
2
0.345
02
)
2
0.1
(0
3e0.1
2
2k
0y2
)
2
h
(x
3eh
2
2k
0y,
2
h
0xhf3k
0
=
















++
+
=
















++
+
=






++=
( )
( ) ( )
401.0
349.002)1.0(03e1.03k0y2h)(x3eh
3k0yh,0xhf4k
00
=



 +++=



 +++=
++=
Substituting all these values in Eqn(2), we get
[ ]
348.0
0.4012(0.349)2(0.345)0.3
6
1
01y
=
++++=
Problem(8):
Obtain the value of y at x=0.2 using Runge-Kutta method of 4th
order for

54. Taylor’s and Picard’s methods 55
Dr. V. Ramachandra Murthy
the differential equation
xy
x-y/y
+
= ; y(0)=1 correct to four decimal
places.
Soln: Given data: ( )
xy
x-y
yx,f
+
= , h=0.2
00x = 0.21x =
10y = ?1y =
The Runge-Kutta method of 4th
order is given by
[ ]
( )
( )
)1.(..............................
3knyh,nxhf4k
2
2k
ny,
2
h
nxhf3k
2
1k
ny,
2
h
nxhf2k
ny,nxhf1kwhere
4k32k22k1k
6
1
ny1ny













++=






++=






++=
=
++++=+
Put n=0 in Eqn(1)
[ ]4k32k22k1k
6
1
0y1y ++++= -----------------(2)
( )
( )3k0yh,0xhf4k
2
2k
0y,
2
h
0xhf3k
2
1k
0y,
2
h
0xhf2k
0y,0xhf1kwhere
++=






++=






++=
=
( )
0.2
01
0-1
0.2
0x0y
0x0y
h
0y,0xhf1k
=



+
=





+
−
=
=

55. Taylor’s and Picard’s methods 56
Dr. V. Ramachandra Murthy
0.1666
2
0.2
0
2
0.2
1
2
0.2
0
2
0.2
1
0.2
2
h
0x
2
1k
0y
2
h
0x
2
1k
0y
h
2
1k
0y,
2
h
0xhf2k
=


















++





+






+−





+
=


















++





+






+−





+
=






++=
0.1661
2
0.2
0
2
0.1666
1
2
0.2
0
2
0.1666
1
0.2
2
h
0x
2
2k
0y
2
h
0x
2
2k
0y
h
2
2k
0y,
2
h
0xhf3k
=


















++





+






+−





+
=


















++





+






+−





+
=






++=
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
1414.0
2.001661.01
2.001661.01
0.2
h0x3k0y
h0x3k0y
0.2
3k0yh,0xhf4k
=






+++
+−+
=





+++
+−+
=
++=
Substituting all these values in Eqn(2), we get
[ ]
1678.1
0.14142(0.1661)2(0.1666)0.2
6
1
11y
=
++++=
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}

56. Taylor’s and Picard’s methods 57
Dr. V. Ramachandra Murthy
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-ny(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,1xf1fwhere
formulaCorrector(2)
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
formulaPredictor(1)32f2f12f
3
4h
0yp4,y
≠==




=





====
→−−−−−




 +++=
+
→−−−−−+−+=
2
yx
dx
dy +
=
2
yx
y)f(x,
+
=
( ) (1)2ff2f
3
4h
yy 3210p4, −−−−−+−+=

57. Taylor’s and Picard’s methods 58
Dr. V. Ramachandra Murthy
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 =
568.1
2
2.6360.5
2
yx
f 11
1 =
+
=
+
=
2975.2
2
3.5951
2
yx
f 22
2 =
+
=
+
=
234.3
2
4.9681.5
2
yx
f 33
3 =
+
=
+
=
{ } 871.6)234.3(22975.2)568.1(2
3
4(0.5)
2y p4, =+−+=
( ) ( )(r)
44
(r)
4
(r)
4322
1)(r
c4, y,xffwhere)2(f4ff
3
h
yy =−−−−−+++=+
( )
( ) ( )
( ) 6.87314.43554(3.234)2.2975
3
0.5
3.595y
4.4355
2
6.8712
2
yx
y,xfy,xff
Where,f4ff
3
h
yy
(1)
c4,
p4,4
p4,4
(0)
44
(0)
4
(0)
4322
(1)
c4,
=+++=∴
=
+
=
+
===
+++=
( )
( ) ( )
( )4.43654(3.234)2.2975
3
0.5
3.595y
4.4365
2
6.87312
2
yx
y,xfy,xff
where
f4ff
3
h
yy
(2)
c4,
(1)
c4,4(1)
c4,4
(1)
44
(1)
4
(1)
4322
(2)
c4,
+++=∴
=
+
=
+
===
+++=

58. Taylor’s and Picard’s methods 59
Dr. V. Ramachandra Murthy
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
yx
dx
dy
+=
( )
( ) ( )
( ) 6.87334.43664(3.234)2.2975
3
0.5
3.595y
4.4366
2
6.87332
2
yx
y,xfy,xff
where
f4ff
3
h
yy
(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 =
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
n
3
//
n
2
/
nn1n1n ++++== ++
( )( )
( ) ( )///////////////IV
2//////
///
22/
y3yyy2y2yyyyy2y
yyy22y
2yy2xy
yxyGiven
+=++=
++=
+=∴
+=
.....y
3!
h
y
2!
h
y
1!
h
y)f(xy
///
0
3
//
0
2
/
0011 ++++==
( )( )
( ) 28yy3yy2y
8yyy22y
2y2y2xy
1yxy
//
0
/
0
///
00
IV
0
2/
0
//
00
///
0
/
000
//
0
2
0
2
0
/
0
=+=
=++=
=+=
=+=

59. Taylor’s and Picard’s methods 60
Dr. V. Ramachandra Murthy
Substituting all these values in Eqn(2), we get
∴ y(0.1)=1.1114
similarly y(-0.1)=0.9087, y(0.2)=1.2529
Thus
Milne’s Predictor formula is given by
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!
0.1
1y
432
1
=
+++++=
0.1x0 −=
1y1 =
0x1 = 0.1x2 = 0.2x3 = 0.3x4 =
9087.0y0 = 1114.1y2 = 2529.1y3 = ?y4 =
( ) (3)32f2f12f
3
4h
0yp4,y −−−−−+−+=
( ) ( )2
i
2
iiii yx)y,f(xf +==ix iy
0x1 = 1y1 = ( ) ( ) ( ) ( ) 110yxf
222
1
2
11 =+=+=
0.1x2 = 1114.1y2 = ( ) ( ) 2452.1yxf
2
2
2
22 =+=
0.2x3 = 2529.1y3 = ( ) ( ) 6097.1yxf
2
3
2
33 =+=
{ } 4385.1)6097.1(22452.1)1(2
3
4(0.1)
0.9087y p4, =+−+=
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f
(r)
4
y,4xf
(r)
4
fwhere)4(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
≠==





=−−−−−




 +++=
+
rand

60. Taylor’s and Picard’s methods 61
Dr. V. Ramachandra Murthy
( ) ( ) ( ) ( )
( ) 4395.11592.24(1.6097)1.2452
3
0.1
1.1114
(1)
c4,
y
1592.221.438520.32
p4,y2
4x
(0)
4
f
(0)
4
y,4xf
(0)
4
fWhere,
(0)
4
f34f2f
3
h
2y
(1)
c4,
y
=+++=∴
=+=+=∴





=




 +++=
Second improvement: Put r=1 in eqn(4)
( )
( ) ( )
( )
4396.1
(3)
c4,
ysimilarly
1.43961621.24(1.6097)1.2452
3
0.1
1.1114
(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
f34f2f
3
h
2y
(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
2y2
y)f(x,:dataGiven =
−
=
Milne’s Predictor formula is given by
( ) (1)32f2f12f
3
4h
0yp4,y −−−−−+−+=
40x =
0049.11y =
4.11x = 4.22x = 4.33x = 4.44x =
10y = 0097.12y = 0143.13y = ?4y =

61. Taylor’s and Picard’s methods 62
Dr. V. Ramachandra Murthy
Substituting all the values in eqn(1) we get,
{ } 0816.1)0451.0(20466.0)0483.0(2
3
4(0.1)
1p4,y =+−+=
Milne’s Corrector formula is given by





=−−−−−




 +++=
+ (r)
4
y,4xf
(r)
4
fwhere)2(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f ≠== rand
First improvement: Put r=0 in eqn(2)
( )
i5x
2
iy2
)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
2
1y2
1f =
−
=
−
=
( ) ( ) 0.0466
5(4.2)
21.00972
25x
2
2y2
2f =
−
=
−
=
( ) ( ) 0.0451
5(4.3)
21.01432
35x
2
3y2
3f =
−
=
−
=

62. Taylor’s and Picard’s methods 63
Dr. V. Ramachandra Murthy
( ) ( ) ( )
( ) 0187.10.04374(0.0451)0.0466
3
0.1
1.0097
(1)
c4,
y
0437.0
5(4.4)
21.01862
45x
2
p4,y2
p4,y,4xf
(0)
4
y,4xf
(0)
4
f
Where,
(0)
4
f34f2f
3
h
2y
(1)
c4,
y
=+++=∴
=
−
=
−
==




=





 +++=
Second improvement: Put r=1 in eqn(2)
( )
( ) 0187.10437.04(0.0451)0.0466
3
0.1
1.0097
(2)
c4,
y
0437.0
5(4.4)
21.01872
45x
2
(1)
c4,
y2
(1)
c4,
y,4xf
(1)
4
y,4xf
(1)
4
f
where
(1)
4
f34f2f
3
h
2y
(2)
c4,
y
=+++=∴
=
−
=





−
=




=




=





 +++=
Since
(2)
c4,
y&
(1)
c4,
y are the same up to four decimal places
y(4.4)=1.0187
Problem(4):
Given 2y2x1
2
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,2y2x1
2
1
y)f(x,:dataGiven =



 +=
Milne’s Predictor formula is given by
00x =
06.11y =
0.11x = 0.22x = 0.33x =
10y = 12.12y = 21.13y = ?4y =
0.44x =

63. Taylor’s and Picard’s methods 64
Dr. V. Ramachandra Murthy
( ) (1)32f2f12f
3
4h
0yp4,y −−−−−+−+=
Substituting all the values in eqn(1) we get,
{ } 2771.1)7979.0(26522.0)5674.0(2
3
4(0.1)
1p4,y =+−+=
Milne’s Corrector formula is given by
)2(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y −−−−−




 +++=
+
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f,
(r)
4
y,4xf
(r)
4
fewher ≠==




= rand
First improvement: Put r=0 in eqn(2)
2
i
y2
i
x1
2
1
)iy,if(xif 



 +==
ix iy
0.11x =
0.22x =
0.33x =
06.11y =
12.12y =
21.13y =
5674.02
1
y2
1
x1
2
1
1f =



 +=
6522.02
2
y2
2
x1
2
1
2f =



 +=
7979.02
3
y2
3
x1
2
1
3f =



 +=

64. Taylor’s and Picard’s methods 65
Dr. V. Ramachandra Murthy
( ) ( )
( ) 2796.10.94594(0.7979)0.6522
3
0.1
1.12
(1)
c4,
y
9459.022771.124.01
2
12
p4,
y2
4
x1
2
1(0)
4
y,4xf
(0)
4
f
Where,
(0)
4
f34f2f
3
h
2y
(1)
c4,
y
=+++=∴
=



 +=



 +=




=





 +++=
Second improvement: Put r=1 in eqn(2)
( ) ( )
( )
1.2797
(3)
c4,
ySimilarly
1.27970.94964(0.7979)0.6522
3
0.1
1.12
(2)
c4,
y
0.949621.279620.41
2
1(1)
c4,
y2
4
x1
2
1(1)
4
y,4xf
(1)
4
f
Where,
(1)
4
f34f2f
3
h
2y
(2)
c4,
y
=
=+++=∴
=



 +=



 +=




=





 +++=
Since
(3)
c4,
y&
(2)
c4,
y are the same up to four decimal places
y(0.4)=1.2797
Problem(5):
Using Milne’s predictor-corrector method solve 2y2y
dx
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,:dataGiven =−=
Milne’s Predictor formula is given by
( ) (1)32f2f12f
3
4h
0yp4,y −−−−−+−+=
00x =
0499.11y =
0.051x = 0.12x = 0.153x = 0.44x =
10y = 0996.12y = 1488.13y = ?4y =
2
i
y-i2y)iy,if(xif ==ix iy
0.051x = 0499.11y = 9975.02
1
y-12y1f ==

65. Taylor’s and Picard’s methods 66
Dr. V. Ramachandra Murthy
9960.02
2
y-22y2f ==
Substituting all the values in eqn(1) we get,
{ } 1969.1)9778.0(29960.0)9975.0(2
3
4(0.05)
1p4,y =+−+=
Milne’s Corrector formula is given by
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f,
(r)
4
y,4xf
(r)
4
fwhere
)2(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
≠==




=
−−−−−




 +++=
+
rand
First improvement: Put r=0 in eqn(2)
( ) ( ) ( )
( ) 1.19740.96124(0.9778)0.9960
3
0.05
1.0996
(1)
c4,
y
0.961221.19691.196922
p4,yp4,2y
(0)
4
y,4xf
(0)
4
f
Where,
(0)
4
f34f2f
3
h
2y
(1)
c4,
y
=+++=∴
=−=−=




=





 +++=
Second improvement: Put r=1 in eqn(4)
( ) ( )
( )
1.1974
0.96104(0.9778)0.9960
3
0.05
1.0996
(2)
c4,
y
0.961021.19741.19742
2
(1)
c4,
y
(1)
c4,
2y
(1)
4
y,4xf
(1)
4
f
Where,
(1)
4
f34f2f
3
h
2y
(2)
c4,
y
=
+++=∴
=−=




−=




=





 +++=
Since
(2)
c4,
y&
(1)
c4,
y are the same up to four decimal places
y(0.2)=1.1974
0.12x =
0.153x =
0996.12y =
1488.13y = 9778.02
3
y-32y3f ==

66. Taylor’s and Picard’s methods 67
Dr. V. Ramachandra Murthy
Problem(6):
Solve the initial value problem 1y(0);2xy1
dx
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,:dataGiven =+= x
Milne’s Predictor formula is given by
( ) (1)32f2f12f
3
4h
0yp4,y −−−−−+−+=
( )( ) 122.12105.11.012
1
y1x11f =+=+=
( )( ) 299.12223.12.012
2
y2x12f =+=+=
( )( ) 1.55021.3550.312
3
y3x13f =+=+=
Substituting all the values in eqn(1) we get,
{ } 526.1)550.1(2299.1)122.1(2
3
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 =
2
i
yix1)iy,if(xif +==ix iy
0.11x =
0.22x =
0.33x =
105.11y =
223.12y =
355.13y =

67. Taylor’s and Picard’s methods 68
Dr. V. Ramachandra Murthy
Milne’s Corrector formula is given by
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f,
(r)
4
y,4xf
(r)
4
fwhere
)2(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
≠==




=
−−−−−




 +++=
+
rand
First improvement: Put r=0 in eqn(2)
( ) ( )( )
( ) 1.5371.9314(1.550)1.299
3
0.1
1.223
(1)
c4,
y
1.93121.5260.412
p4,y4x1
(0)
4
y,4xf
(0)
4
f
Where,
(0)
4
f34f2f
3
h
2y
(1)
c4,
y
=+++=∴
=+=+=




=





 +++=
Second improvement: Put r=1 in eqn(2)
( )( )
( )
1.537
1.9444(1.550)1.299
3
0.1
1.223
(2)
c4,
y
944.121.5370.41
2
(1)
c4,
y4x1
(1)
4
y,4xf
(1)
4
f
Where,
(1)
4
f34f2f
3
h
2y
(2)
c4,
y
=
+++=∴
=+=




+=




=





 +++=
Since
(2)
c4,
y&
(1)
c4,
y are the same up to four decimal places
y(0.4)=1.537
Problem(7):
Part of a Numerical solution of ( ) ( )y0.1x0.2
dx
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

68. Taylor’s and Picard’s methods 69
Dr. V. Ramachandra Murthy
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,:dataGiven =+=
Milne’s Predictor formula is given by
( ) (1)32f2f12f
3
4h
0yp4,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 =+=
Substituting all the values in eqn(1) we get,
{ } 0444.2)2332.0(22221.0)2110.0(2
3
4(0.05)
1p4,y =+−+=
Milne’s Corrector formula is given by
0,
(r)
c4,
y
(r)
4
fp4,y
(0)
4
f,
(r)
4
y,4xf
(r)
4
fwhere
)2(
(r)
4
f34f2f
3
h
2y
1)(r
c4,
y
≠==




=
−−−−−




 +++=
+
rand
First improvement: Put r=0 in eqn(2)
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 =

69. Taylor’s and Picard’s methods 70
Dr. V. Ramachandra Murthy
( )
( )
( ) 0444.20.24444(0.2332)0.2221
3
0.05
2.0211
(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
f34f2f
3
h
2y
(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
=∴
===

70. Taylor’s and Picard’s methods 71
Dr. V. Ramachandra Murthy
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)
4y&p4,y
(0)
4ywhere
(0)
4y,4xf
(0)
4f:Note
(r)
4
y,4xf(r)
4
f,3y,3xf3f,2y,2xf2f,1y,1xf1fwhere
formulaCorrector1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y
formulaPredictor09f137f259f355f
24
h
3yp4,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
yx
y)f(x,:dataGiven =
+
=
00x = 0.51x = 1.02x = 1.53x = 2.04x =
636.21y = 595.32y = 968.43y = ?4y =
Adam’s Predictor formula is given by
( ) -(1)-------09f137f259f355f
24
h
3yp4,y −+−+=
20y =

71. Taylor’s and Picard’s methods 72
Dr. V. Ramachandra Murthy
ix iy 2
iyix
)iy,if(xif
+
==
00x = 20y =
1
2
20
2
0y0x
)0y,0f(x0f =
+
=
+
==
0.51x = 636.21y =
568.1
2
636.25.0
2
1y1x
)1y,1f(x1f
=
+
=
+
==
12x = 595.32y = 2975.2
2
595.30.1
2
2y2x
)2y,2f(x2f
=
+
=
+
==
1.53x = 968.43y = 234.3
2
968.45.1
2
3y3x
)3y,3f(x3f
=
+
=
+
==
( ) ( ) ( ) ( )( )
8707.6
191.568372.2975593.23455
24
0.5
4.968p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by
(2)1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y −−−−−−−




 +−++=
+
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)

72. Taylor’s and Picard’s methods 73
Dr. V. Ramachandra Murthy
( )
( ) ( ) ( )( )
6.8730
1.5682.297553.234194.43539
24
0.5
4.968
(1)
c4,
y
4.4353
2
6.87072
2
p4,y4x
p4,y,4xf
(0)
4
y,4xf
(0)
4
fwhere
1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=
+−++=∴
=
+
=
+
==




=





 +−++=
Second improvement: Put r=1 in eqn(2)
( ) ( ) ( )( )
6.8733
1.5682.297553.234194.43659
24
0.5
4.968
(2)
c4,
y
4.4365
2
6.87302
2
(1)
c4,
y4x
(1)
c4,
y,4xf
(1)
4
y,4xf
(1)
4
fwhere
1f25f319f
(1)
4
9f
24
h
3y
(2)
c4,
y
=
+−++=∴
=
+
=
+
=




=




=





 +−++=
Third improvement: Put r=2 in eqn(2)
( ) ( ) ( )( )
6.8733
1.5682.297553.234194.43669
24
0.5
4.968
(3)
c4,
y
4.4366
2
6.87332
2
(2)
c4,
y4x
(2)
c4,
y,4xf
(2)
4
y,4xf
(2)
4
f
1f25f319f
(2)
4
9f
24
h
3y
(3)
c4,
y
=
+−++=∴
=
+
=
+
=




=




=





 +−++=
Since
(3)
c4,
y&
(2)
c4,
y are the same up to four decimal places
y(2)=6.8733

73. Taylor’s and Picard’s methods 74
Dr. V. Ramachandra Murthy
Problem(2):
Obtain the solution of the initial value problem 2xy2x
dx
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!
h
ny)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!
h
0y)1f(x1y ++++== ------
(2)
661x186x1x66x2///
0
y2
0
x//
0
y06x/
0
6yIV
0
y
181x61)2(14x1x2//
0
y2
0
x)0y2(1/
0
y04x///
0
y
61x21)2(1)(1/
0
y2
0
x)0y(102x//
0
y
21)1(1)0y(12
0
x/
0
y
=++=++=
=+++=+++=
=++=++=
=+=+=
Substituting all these values in Eqn(2), we get
( ) ( ) ( )
1.2332
.......(66)
24
40.1
(18)
3!
30.1
(6)
2!
20.1
(2)
1!
0.1
11y
=
+++++=
Put n=1 in eqn(1) .....///
1y
3!
3h//
1y
2!
2h/
1y
1!
h
1y2y ++++= ----------(3)

74. Taylor’s and Picard’s methods 75
Dr. V. Ramachandra Murthy
( )
( )
99.5907///
1
y2
1
x//
1
y16x/
1
6yIV
1
y
25.8968//
1
y2
1
x)1y2(1/
1
y14x///
1
y
7.8853
2.702121.11.2332)2(1.1)(1/
1
y2
1
x)1y(112x//
1
y
2.70211.2332)(121.1)1y(12
1
x/
1
y
=++=
=+++=
=
++=++=
=+=+=
Substituting all these values in Eqn(3), we get
( ) ( )
( )
1.5475
.......(99.5907)
4!
40.1
(25.8968)
3!
30.1
(7.8853)
2!
20.1
(2.7021)
1!
0.1
1.23322y
=
++
+++=
similarly 9785.13y =
Thus
5475.12y = 9785.13y =
Adam Predictor formula is given by
( )09f137f259f355f
24
h
3yp4,y −+−+= ---------(4)
( ) ( )iy12
ix)iy,if(xif +==
( ) ( ) ( ) ( ) 211210y12
0x0f =+=+=
( ) ( ) ( ) ( )
7021.2
1.2332121.11y12
1x1f
=
+=+=
( ) ( ) ( ) ( )
6684.3
1.5475121.22y12
2x2f
=
+=+=
( ) ( ) ( ) ( )
0336.5
1.9785121.33y12
3x3f
=
+=+=
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 =

75. Taylor’s and Picard’s methods 76
Dr. V. Ramachandra Murthy
( ) ( ) ( ) ( )( )
8707.6
191.568372.2975593.23455
24
0.5
4.968p4,y
becomesEqn(4)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y ---------(5)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(5)
( ) ( ) ( ) ( ) ( )
7.0005
2.5717121.4p4,y12
4xp4,y,4xf0
4
y,4xf
(0)
4
f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=
+=+==



=





 +−++=
( ) ( ) ( )( )
2.5743
2.70213.668455.0336197.00059
24
0.1
1.9785
(1)
c4,
y
=
+−++=∴
Second improvement: Put r=1 in eqn(5)
( ) ( ) ( )
2.5745
(3)
c4,
ySimilarly2.5745
(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
9f
24
h
3y
(2)
c4,
y
==∴
=
+=




 +=




=




=





 +−++=
Since
(3)
c4,
y&
(2)
c4,
y are the same up to four decimal places
y(1.4)=2.5745

76. Taylor’s and Picard’s methods 77
Dr. V. Ramachandra Murthy
Problem(3):
Solve for y(0.4) given that ;2yx
dx
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: =−=
Adam Predictor formula is given by
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
( )2
iy-ix)iy,if(xif ==
( ) ( ) 121-02
0y-0x0f −===
( ) ( )
7311.0
20.9117-0.12
1y-1x1f
−=
==
( ) ( )
0.5214
20.8494-0.22
2y-2x2f
−=
==
( ) ( )
0.3497
20.8061-0.32
3y-3x3f
−=
==
( ) ( ) ( ) ( )( )
7789.0
1-90.7311-370.5214-590.3497-55
24
0.1
0.8061p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by
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

77. Taylor’s and Picard’s methods 78
Dr. V. Ramachandra Murthy





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y ---------(2)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)
( ) ( ) ( )
( ) ( ) ( )
0.7784
0.7311
0.5214-50.3497-190.2066-9
24
0.1
0.8061
(1)
c4,
y
0.206620.77890.42
p4,y4xp4,y,4xf
(0)
4
y,4xf
(0)
4
f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=






−
−+
+=∴
−=−=−==




=





 +−++=
Second improvement: Put r=1 in eqn(2)
( )
( ) ( ) ( )( )
0.7785
(3)
c4,
ySimilarly
0.7785
0.73110.5214-50.3497-190.2059-9
24
0.1
0.8061
(2)
c4,
y
0.205920.77840.4
2
(1)
c4,
y4x
(1)
c4,
y,4xf
(1)
4
f
where1f25f319f
(1)
4
9f
24
h
3y
(2)
c4,
y
=
=
−−++=∴
−=−=




−=




=





 +−++=
Since
(3)
c4,
y&
(2)
c4,
y are the same up to four decimal places
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.

78. Taylor’s and Picard’s methods 79
Dr. V. Ramachandra Murthy
Soln: 0.1h,
2
xy
y)f(x,:dataGiven ==
Adam Predictor formula is given by
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
ix iy
2
iyix
)iy,if(xif ==
00x = 10y =
0
2
0(1)
2
0y0x
0f ===
0.11x = 01.11y =
0505.0
2
0.1(1.01)
2
1y1x
1f ===
0.22x = 022.12y =
1022.0
2
0.2(1.022)
2
2y2x
2f ===
0.33x = 023.13y =
1534.0
2
0.3(1.023)
2
3y3x
3f ===
( ) ( ) ( ) ( )( )
0408.1
090.0505370.1022590.153455
24
0.1
1.023p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
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 =

79. Taylor’s and Picard’s methods 80
Dr. V. Ramachandra Murthy
First improvement: Put r=0 in eqn(2)
( )
( ) ( ) ( )
1.0410
0505.0
0.102250.1534190.20819
24
0.1
023.1
(1)
c4,
y
2081.0
2
)0.4(1.0408
2
p4,y4x
p4,y,4xf0
4
y,4xf
(0)
4
f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=






+
−+
+=∴
====



=





 +−++=
Second improvement: Put r=1 in eqn(2)
( ) ( ) ( )
1.0410
0505.0
0.102250.1534190.20829
24
0.1
023.1
(2)
c4,
y
2082.0
2
)0.4(1.0410
2
(1)
c4,
y4x
(1)
c4,
y,4xf
(1)
4
f
where1f25f319f
(1)
4
9f
24
h
3y
(2)
c4,
y
=






+
−+
+=∴
===




=





 +−++=
Since
(2)
c4,
y&
(1)
c4,
y are the same up to four decimal places
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
1
y)f(x,:dataGiven =
+
=
2492.23y =
Adam- Predictor formula is given by
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
00x =
0932.21y =
0.21x = 0.42x =
20y = 1754.22y =
0.33x = 0.44x =
?4y =

80. Taylor’s and Picard’s methods 81
Dr. V. Ramachandra Murthy
ix iy iyix
1
)iy,if(xif
+
==
00x = 20y =
5.0
20
1
0y0x
1
0f =
+
=
+
=
0.21x = 0932.21y =
4360.0
0932.22.0
1
1y1x
1
1f =
+
=
+
=
0.42x = 1754.22y =
3882.0
1754.24.0
1
2y2x
1
2f =
+
=
+
=
0.63x =
2492.23y = 3509.0
2492.26.0
1
3y3x
1
3f =
+
=
+
=
( ) ( ) ( ) ( )( )
2.3160
0.590.4360370.3882590.350955
24
0.2
2.2492p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y ---------(2)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)
( )
( ) ( ) ( )( )
2.3162
0.43600.388250.3509190.32099
24
0.2
2.2492
(1)
c4,
y
0.3209
2.31600.8
1
p4,y4x
1
p4,y,4xf0
4
y,4xf
(0)
4
f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=
+−++=∴
=
+
=
+
==



=





 +−++=

81. Taylor’s and Picard’s methods 82
Dr. V. Ramachandra Murthy
Second improvement: Put r=1 in eqn(2)
( ) ( ) ( )( )
2.3162
0.43600.388250.3509190.32099
24
0.2
2.2492
(2)
c4,
y
0.3209
2.31620.8
1
(1)
c4,
y4x
1(1)
c4,
y,4xf
(1)
4
f
where1f25f319f
(1)
4
9f
24
h
3y
(2)
c4,
y
=
+−++=∴
=
+
=
+
=




=





 +−++=
Since
(2)
c4,
y&
(1)
c4,
y are the same up to four decimal places
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
xy1
x
y
2x
1y)f(x,:dataGiven =
−
=−=
Adam-Bashforth Predictor formula is given by
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
10x =
996.01y =
1.11x = 1.22x =
10y = 986.02y =
1.33x = 1.44x =
972.03y = ?4y =

82. Taylor’s and Picard’s methods 83
Dr. V. Ramachandra Murthy
ix iy ( )2
ix
iyix-1
)iy,if(xif ==
00x = 10y = ( ) ( )
0
21
1x1-1
2
0x
0y0x-1
0f ===
.111x = 996.01y = ( ) ( )
0.079
21.1
1.1x0.996-1
2
1x
1y1x-1
1f
−=
==
1.22x = 986.02y = ( ) ( )
0.127
21.2
1.2x0.986-1
2
2x
2y2x-1
2f
−=
==
1.33x = 972.03y = ( ) ( )
155.0
21.3
1.3x0.972-1
2
3x
3y3x-1
3f
−=
==
( ) ( ) ( ) ( )( )
955.0
090.079-370.127-590.155-55
24
0.1
0.972p4,y
becomesEqn(1)
=
−+−+=
∴
Adam’s-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y --------(2)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)
( )
( ) ( )
( ) ( ) ( )( ) 955.00.0790.171-50.155-190.171-9
24
0.1
0.972
(1)
c4,
y
-0.171
21.4
)1.4)(0.955-1
2
4x
p4,y4x-1
p4,y,4xf)
(0)
4
y,4(
(0)
4
f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
=−−++=∴
=====





 +−++=
xf
Since
(1)
c4,
y&p4,y are the same up to four decimal places
y(1.4)=0.955

83. Taylor’s and Picard’s methods 84
Dr. V. Ramachandra Murthy
Problem(7):
Using Adam-Bashforth Predictor Corrector method obtain the solution of
y2x
dx
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
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
ix iy
( ) iy-2
ix)iy,if(xif ==
00x = 10y =
( ) ( ) 11-200y-2
0x0f −===
.101x = 9051.01y =
( ) ( ) 8951.09051.0-20.11y-2
1x1f −===
0.22x = 8212.02y =
( ) ( ) 7812.08212.0-20.22y-2
2x2f −===
0.33x = 7491.03y =
( ) ( ) 6591.07491.0-20.33y-2
3x3f −===
00x =
9051.01y =
0.11x = 0.22x =
10y = 8212.02y =
0.33x = 0.44x =
7491.03y = ?4y =

84. Taylor’s and Picard’s methods 85
Dr. V. Ramachandra Murthy
( ) ( ) ( ) ( )( )
6896.0
1-90.8951-370.7812-590.6591-55
24
0.1
0.7491p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y --------(2)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)
( ) ( ) ( )
( ) ( ) ( )( )
.68960
0.89510.7812-50.6591-190.5296-9
24
0.1
0.7491(1)
c4,
y
-0.52960.6896-20.4p4,y-2
4xp4,y,4xf
(0)
4f
where1f25f319f
(0)
4
9f
24
h
3y
(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
yx2e
dx
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
( )09f137f259f355f
24
h
3yp4,y −+−+= --------(1)
00x =
010.21y =
0.11x = 0.22x =
20y = 040.22y =
0.33x = 0.44x =
090.23y = ?4y =

85. Taylor’s and Picard’s methods 86
Dr. V. Ramachandra Murthy
ix iy
iyx2e)iy,if(xif i
−==
00x = 20y =
0202e0yx2e0f 0i
=−=−=
.101x = 010.21y =
0.20032.0100.12e1yx2e1f i
=−=−=
0.22x = 040.22y =
0.40282.0400.22e2yx2e2f 2
=−=−=
0.33x = 090.23y =
0.60972.0900.32e3yx2e3f 3
=−=−=
( ) ( ) ( ) ( )( )
2.1615
090.2003370.4028590.609755
24
0.1
2.090p4,y
becomesEqn(1)
=
−+−+=
∴
Adam-Bashforth Corrector formula is given by





 +−++=
+
1f25f319f
(r)
4
9f
24
h
3y
1)(r
c4,
y --------(2)
0rfor
(r)
c4,
y
(r)
4
y&p4,y
(0)
4
ywhere
(0)
4
y,4xf
(0)
4
f ≠==




=
First improvement: Put r=0 in eqn(2)
( )
( ) ( ) ( )( )
2.1615
0.20030.402850.6097190.82219
24
0.1
2.090(1)
c4,
y
2.16150.42ep4,yx2ep4,y,4xf)
(0)
4y,4(
(0)
4f
where1f25f319f
(0)
4
9f
24
h
3y
(1)
c4,
y
4
=
+−++=∴
−=−===





 +−++=
xf
Since
(1)
c4,
y&p4,y are the same up to four decimal places
y(0.4)=2.1615

86. Taylor’s and Picard’s methods 87
Dr. V. Ramachandra Murthy
Problem(8):
Using Adam-Bashforth Predictor Corrector method obtain the solution of
2yx
dx
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}