Numerical Methods - Oridnary Differential Equations - 1

Numerical Methods
Ordinary Differential Equations - 1
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Technology & Science,
Rajkot (Gujarat) - INDIA
niravbvyas@gmail.com
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations -

Ordinary Differential Equations
Taylor’s Series Method:
Consider the ﬁrst order Diﬀerential Equation
dy
dx
= f(x, y), y(x0) = y0

Ordinary Differential Equations
Taylor’s Series Method:
Consider the ﬁrst order Diﬀerential Equation
dy
dx
= f(x, y), y(x0) = y0
The Taylor’s series is
y(x) = y(x0) +
(x − x0)
1!
y (x0) +
(x − x0)2
2!
y (x0) + . . .

Taylor’s Series Method
Ex:1
Solve y = x + y, y(0) = 1 by Taylor’s series
method. Hence ﬁnd values of y at x = 0.1 and
x = 0.2

Sol.: Here y = f(x, y) = x + y, x0 = 0 and y0 = 1

y = x + y

y = x + y ⇒ y (0) = 1

y = x + y ⇒ y (0) = 1
y = 1 + y

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2
y = y

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2
y = y ⇒ y (0) = y (0) = 2

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2
y = y ⇒ y (0) = y (0) = 2
yiv
= y

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2
y = y ⇒ y (0) = y (0) = 2
yiv
= y ⇒ yiv
(0) = y (0) = 2 . . .

y = x + y ⇒ y (0) = 1
y = 1 + y ⇒ y (0) = 1 + y (0) = 1 + 1 = 2
y = y ⇒ y (0) = y (0) = 2
yiv
= y ⇒ yiv
(0) = y (0) = 2 . . .
Taylor’s series is
y(x) = y(x0)+
(x − x0)
1!
y (x0)+
(x − x0)2
2!
y (x0)+. . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .
= 1 + x + x2
+
x3
3
+
x4
12
+ . . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .
= 1 + x + x2
+
x3
3
+
x4
12
+ . . .
y(0.1) = 1 + (0.1) + (0.1)2
+
(0.1)3
3
+
(0.1)4
12
+ . . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .
= 1 + x + x2
+
x3
3
+
x4
12
+ . . .
y(0.1) = 1 + (0.1) + (0.1)2
+
(0.1)3
3
+
(0.1)4
12
+ . . .
= 1.1103

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .
= 1 + x + x2
+
x3
3
+
x4
12
+ . . .
y(0.1) = 1 + (0.1) + (0.1)2
+
(0.1)3
3
+
(0.1)4
12
+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2
+
(0.2)3
3
+
(0.2)4
12
+ . . .

y(x) = 1 + xy (0) +
x2
2!
y (0) +
x3
3!
y (0) . . .
= 1 + x(1) +
x2
2
(2) +
x3
6
(2) +
x4
24
(2) + . . .
= 1 + x + x2
+
x3
3
+
x4
12
+ . . .
y(0.1) = 1 + (0.1) + (0.1)2
+
(0.1)3
3
+
(0.1)4
12
+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2
+
(0.2)3
3
+
(0.2)4
12
+ . . .
= 1.2428

Taylor’s Series
Ex Using Taylor’s series method, obtain the solution
of
dy
dx
= 3x + y2
, given that y(0) = 1. Find the
value of y for x = 0.1

Taylor’s Series
Sol. Here,y = f(x, y) = 3x + y2
, x0 = 0 and y0 = 1.

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy ⇒ y (x0) = 2(1)2
+ 2(5) = 12

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy ⇒ y (x0) = 2(1)2
+ 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy ⇒ y (x0) = 2(1)2
+ 2(5) = 12
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 1 + x +
x2
2!
(5) +
x3
3!
(12) + . . .

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy ⇒ y (x0) = 2(1)2
+ 2(5) = 12
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 1 + x +
x2
2!
(5) +
x3
3!
(12) + . . .
= 1 + x +
5x2
2!
+ 2x3
+ . . .

Taylor’s Series
, x0 = 0 and y0 = 1.
y = 3x + y2
⇒ y (x0) = 3(x0) + y2
0 = 3(0) + 1 = 1
y = 3 + 2yy ⇒ y (x0) = 3 + 2(1)(1) = 5
y = 2(y )2
+ 2yy ⇒ y (x0) = 2(1)2
+ 2(5) = 12
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 1 + x +
x2
2!
(5) +
x3
3!
(12) + . . .
= 1 + x +
5x2
2!
+ 2x3
+ . . .
y(0.1) =

Taylor’s Series
Ex Using Taylor’s series method, ﬁnd the solution of
dy
dx
= 2y + 3ex
, y(0) = 0,at x = 0.2

Taylor’s Series
Sol. Here,y = f(x, y) = 2y + 3ex
, x0 = 0 and y0 = 0.

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex
⇒ yiv
(x0) = 45

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex
⇒ yiv
(x0) = 45
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex
⇒ yiv
(x0) = 45
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 0 + x(3) +
x2
2!
(9) +
x3
3!
(21) +
x4
4!
(45) + . . .

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex
⇒ yiv
(x0) = 45
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 0 + x(3) +
x2
2!
(9) +
x3
3!
(21) +
x4
4!
(45) + . . .
= 3x +
9x2
2
+
7x3
2
+
15x4
8
+ . . .

Taylor’s Series
, x0 = 0 and y0 = 0.
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(0) + 3e0
= 3
y = 2y + 3ex
⇒ y (x0) = 2(y0) + 3ex0
= 2(3) + 3e0
= 9
y = 2y + 3ex
⇒ y (x0) = 2(y0 ) + 3ex0
= 2(9) + 3e0
= 21
yiv
= 2y + 3ex
⇒ yiv
(x0) = 45
y(x) = y0+(x−x0)y (x0)+
(x − x0)2
2!
y (x0)+
(x − x0)3
3!
y (x0)+. . .
= 0 + x(3) +
x2
2!
(9) +
x3
3!
(21) +
x4
4!
(45) + . . .
= 3x +
9x2
2
+
7x3
2
+
15x4
8
+ . . .
y(0.2) =

Ex:
Use Taylor’s series method to solve
dy
dx
= x2
+ y2
,
y(0) = 1. Find y(0.1) correct up to 4 decimal
places.

Ex:
Use Taylor’s series method to solve
dy
dx
= x2
y − 1,
y(0) = 1. Find y(0.03).

Picard’s Method
Picard’s Method:
Consider the ﬁrst order diﬀerential equation.

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)
subject to y(x0) = y0

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)
The equation (1) can be written as

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)
dy = f(x, y)dx

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)
dy = f(x, y)dx
Integrating between the limits for x and y, we get

Picard’s Method
Picard’s Method:
dy
dx
= f(x, y) − − − (1)
dy = f(x, y)dx
Integrating between the limits for x and y, we get
y
y0
dy =
x
x0
f(x, y)dx

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx
y = y0 +
x
x0
f(x, y)dx − − − (2)

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx
y = y0 +
x
x0
f(x, y)dx − − − (2)
Equation (2) is known as integral equation and
can be solved by successive approximation or
iteration.

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx
y = y0 +
x
x0
f(x, y)dx − − − (2)
iteration.
Now by Picard’s method, for 1st
approximation
y1

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx
y = y0 +
x
x0
f(x, y)dx − − − (2)
iteration.
approximation
y1
we replace y by y0 in f(x, y) in R.H.S of eq. (2),
we get

Picard’s Method
(y − y0) =
x
x0
f(x, y)dx
y = y0 +
x
x0
f(x, y)dx − − − (2)
iteration.
approximation
y1
we get
y1 = y0 +
x
x0
f(x, y0)dx

Picard’s Method
For 2nd
approximation y2,

Picard’s Method
For 2nd
approximation y2,
we get

Picard’s Method
For 2nd
approximation y2,
we get
y2 = y0 +
x
x0
f(x, y1)dx

Picard’s Method
For 2nd
approximation y2,
we get
y2 = y0 +
x
x0
f(x, y1)dx
In general,

Picard’s Method
For 2nd
approximation y2,
we get
y2 = y0 +
x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
x
x0
f(x, yn)dx for n = 0, 1, 2, . . .

Picard’s Method
For 2nd
approximation y2,
we get
y2 = y0 +
x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive values
of y are same up to desired accuracy.

Picard’s Method
Note:
This method is applicable to a limited class of
equations in which the successive integration can
be performed easily.

Picard’s Method
Ex Using Picard’s method solve
dy
dx
= 3 + 2xy where y(0) = 1 for x = 0.1.

Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
x
x0
f(x, yn) dx

Picard’s Method
Sol.:
x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy

Picard’s Method
Sol.:
x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st
approximation:

Picard’s Method
Sol.:
x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st
approximation:
put n = 0 and y0 = 1 in f(x, y)

Picard’s Method
Sol.:
x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st
approximation:
y1 = 1 +
x
0
(3 + 2x) dx

Picard’s Method
Sol.:
x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st
approximation:
y1 = 1 +
x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2

Picard’s Method
2nd
approximation:

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx
= 1 +
x
0
3 + 2x + 6x2
+ 2x3
) dx

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx
= 1 +
x
0
3 + 2x + 6x2
+ 2x3
) dx
∴ y2 = 1 + 3x + x2
+ 2x3
+
x4
2

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx
= 1 +
x
0
3 + 2x + 6x2
+ 2x3
) dx
∴ y2 = 1 + 3x + x2
+ 2x3
+
x4
2
which is approximate solution, putting x = 0.1

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx
= 1 +
x
0
3 + 2x + 6x2
+ 2x3
) dx
∴ y2 = 1 + 3x + x2
+ 2x3
+
x4
2
y(0.1) =

Picard’s Method
2nd
approximation:
put n = 1 and y1 = 1 + 3x + x2
in f(x, y)
y2 = 1 +
x
0
3 + 2x(1 + 3x + x2
) dx
= 1 +
x
0
3 + 2x + 6x2
+ 2x3
) dx
∴ y2 = 1 + 3x + x2
+ 2x3
+
x4
2
y(0.1) = 1.31205

Picard’s Method
Ex:
Using Picard’s method, obtain a solution upto
4th
approx of the equation
dy
dx
= y + x, y(0) = 1.

Picard’s Method
Sol.:
By Picard’s method y = y0 +
x
x0
f(x, y) dx

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st
approximation: put y = y0 = 1 in f(x, y)

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st
‘y1 = 1 +
x
0
(1 + x) dx

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st
‘y1 = 1 +
x
0
(1 + x) dx
∴ y1 = 1 + x +
x2
2

Picard’s Method
2nd
approximation:
put y = y1 = 1 + x +
x2
2
in f(x, y)

Picard’s Method
2nd
approximation:
put y = y1 = 1 + x +
x2
2
in f(x, y)
‘y2 = 1 +
x
0
1 + 2x +
x2
2
dx

Picard’s Method
2nd
approximation:
put y = y1 = 1 + x +
x2
2
in f(x, y)
‘y2 = 1 +
x
0
1 + 2x +
x2
2
dx
∴ y2 = 1 + x + x2
+
x3
6

Picard’s Method
3rd
approximation:
put y = 1 + x + x2
+
x3
6
in f(x, y)

Picard’s Method
3rd
approximation:
put y = 1 + x + x2
+
x3
6
in f(x, y)
‘y3 = 1 +
x
0
1 + 2x + x2
+
x3
6
dx

Picard’s Method
3rd
approximation:
put y = 1 + x + x2
+
x3
6
in f(x, y)
‘y3 = 1 +
x
0
1 + 2x + x2
+
x3
6
dx
∴ y3 = 1 + x + x2
+
x3
3
+
x4
24

Picard’s Method
4th
approximation:

Picard’s Method
4th
approximation:
put y = 1 + x + x2
+
x3
3
+
x4
24
in f(x, y)

Picard’s Method
4th
approximation:
put y = 1 + x + x2
+
x3
3
+
x4
24
in f(x, y)
‘y4 = 1 +
x
0
1 + 2x + x2
+
x3
3
+
x4
24
dx

Picard’s Method
4th
approximation:
put y = 1 + x + x2
+
x3
3
+
x4
24
in f(x, y)
‘y4 = 1 +
x
0
1 + 2x + x2
+
x3
3
+
x4
24
dx
∴ y4 = 1 + x + x2
+
x3
3
+
x4
12
+
x5
120

Picard’s Method
Ex:
Using Picard’s 2nd
approx. solution of the initial
value problem
dy
dx
= x2
+ y2
,for x = 0.4 correct to
4 decimal places given that y(0) = 0.

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2
+ y2

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2
+ y2
1st

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2
+ y2
1st
‘y1 = 0 +
x
0
(x2
+ 0) dx

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2
+ y2
1st
‘y1 = 0 +
x
0
(x2
+ 0) dx
∴ y1 =
x3
3

Picard’s Method
2nd
approximation:
put y = y1 =
x3
3
in f(x, y)

Picard’s Method
2nd
approximation:
put y = y1 =
x3
3
in f(x, y)
‘y2 = y0 +
x
0
x2
+
x3
3
2
dx

Picard’s Method
2nd
approximation:
put y = y1 =
x3
3
in f(x, y)
‘y2 = y0 +
x
0
x2
+
x3
3
2
dx
∴ y2 =
x3
3
+
x7
63

Picard’s Method
2nd
approximation:
put y = y1 =
x3
3
in f(x, y)
‘y2 = y0 +
x
0
x2
+
x3
3
2
dx
∴ y2 =
x3
3
+
x7
63
y(0.4) =

Picard’s Method
Ex:
Find the value of y for x = 0.1 by Picard’s
method given that
dy
dx
=
y − x
y + x
,y(0) = 1.

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =
y − x
y + x

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =
y − x
y + x
1st

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =
y − x
y + x
1st
y1 = 1 +
x
0
1 − x
1 + x
dx = 1 +
x
0
2 − (1 + x)
1 + x
dx

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =
y − x
y + x
1st
y1 = 1 +
x
0
1 − x
1 + x
dx = 1 +
x
0
2 − (1 + x)
1 + x
dx
= 1 +
x
0
2
1 + x
− 1 dx

Picard’s Method
Sol.:
x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =
y − x
y + x
1st
y1 = 1 +
x
0
1 − x
1 + x
dx = 1 +
x
0
2 − (1 + x)
1 + x
dx
= 1 +
x
0
2
1 + x
− 1 dx
∴ y1 = 1 − x + 2log(1 + x)

Picard’s Method
2nd
approximation:
put y = y1 = 1 + 2log(1 + x) − x in f(x, y)

Picard’s Method
2nd
approximation:
y2 = 1 +
x
0
1 − x + 2log(1 + x) − x
1 − x + 2log(1 + x) + x
dx

Picard’s Method
2nd
approximation:
y2 = 1 +
x
0
1 − x + 2log(1 + x) − x
1 − x + 2log(1 + x) + x
dx
= 1 +
x
0
1 − 2x + 2log(1 + x)
1 + 2log(1 + x)
dx

Picard’s Method
2nd
approximation:
y2 = 1 +
x
0
1 − x + 2log(1 + x) − x
1 − x + 2log(1 + x) + x
dx
= 1 +
x
0
1 − 2x + 2log(1 + x)
1 + 2log(1 + x)
dx
= 1 +
x
0
1 −
2x
1 + 2log(1 + x)
dx

Picard’s Method
= 1 + x −
x
0
2x
1 + 2log(1 + x)
dx

Picard’s Method
= 1 + x −
x
0
2x
1 + 2log(1 + x)
dx
which is diﬃcult to integrate therefore using 1st
approximation.

Picard’s Method
= 1 + x −
x
0
2x
1 + 2log(1 + x)
dx
which is diﬃcult to integrate therefore using 1st
approximation.
y(0.1) =

Numerical Methods - Oridnary Differential Equations - 1

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