Numerical Methods - Oridnary Differential Equations - 1

Numerical MethodsOrdinary Differential Equations - 1

Dr. N. B. Vyas

Department of Mathematics,Atmiya Institute of Technology & Science,

Rajkot (Gujarat) - [email protected]

Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1

Ordinary Differential Equations

Taylor’s Series Method:

Consider the first order Differential Equationdy

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) + . . .


Ordinary Differential Equations

Taylor’s Series Method:

Consider the first order Differential Equationdy

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 seriesmethod. Hence find values of y at x = 0.1 andx = 0.2



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

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 + 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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(x0) +(x− x0)

1!y′(x0) +

(x− x0)2

2!y′′(x0) + . . .




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(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) . . .

= 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



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



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



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



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



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



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



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



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

ofdy

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.

y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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)+. . .

= 1 + x +x2

2!(5) +

x3

3!(12) + . . .

= 1 + x +5x2

2!+ 2x3 + . . .

y(0.1) =


Taylor’s Series


y′ = 3x + y2

⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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


y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 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 ofdy

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.

y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex

⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex

⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex

⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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) =


Taylor’s Series


y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3

y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9

y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 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 solvedy

dx= x2 + y2,

y(0) = 1. Find y(0.1) correct up to 4 decimalplaces.



Ex:

Use Taylor’s series method to solvedy

dx= x2y− 1,

y(0) = 1. Find y(0.03).


Picard’s Method

Picard’s Method:

Consider the first order differential equation.

dy

dx= f(x, y)−−− (1)

subject to y(x0) = y0

The equation (1) can be written as

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

Picard’s Method:

Consider the first order differential equation.dy

dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ x

x0

f(x, y)dx


Picard’s Method

Picard’s Method:


dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ x

x0

f(x, y)dx


Picard’s Method

Picard’s Method:


dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ x

x0

f(x, y)dx


Picard’s Method

Picard’s Method:


dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ x

x0

f(x, y)dx


Picard’s Method

Picard’s Method:


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

Picard’s Method:


dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ x

x0

f(x, y)dx


Picard’s Method

Picard’s Method:


dx= f(x, y)−−− (1)



dy = f(x, y)dx


y0

dy =

∫ 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)

Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.

Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get

y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

(y − y0) =

∫ x

x0

f(x, y)dx

y = y0 +

∫ x

x0

f(x, y)dx−−− (2)



y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

(y − y0) =

∫ x

x0

f(x, y)dx

y = y0 +

∫ x

x0

f(x, y)dx−−− (2)



y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

(y − y0) =

∫ x

x0

f(x, y)dx

y = y0 +

∫ x

x0

f(x, y)dx−−− (2)


Now by Picard’s method, for 1st approximationy1

we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get

y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

(y − y0) =

∫ x

x0

f(x, y)dx

y = y0 +

∫ x

x0

f(x, y)dx−−− (2)



y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

(y − y0) =

∫ x

x0

f(x, y)dx

y = y0 +

∫ x

x0

f(x, y)dx−−− (2)



y1 = y0 +

∫ x

x0

f(x, y0)dx


Picard’s Method

For 2nd approximation y2,


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 valuesof y are same up to desired accuracy.


Picard’s Method



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



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



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



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



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

Note:

This method is applicable to a limited class ofequations in which the successive integration canbe performed easily.


Picard’s Method

Ex Using Picard’s method solvedy

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

Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy

1st approximation:

put n = 0 and y0 = 1 in f(x, y)

y1 = 1 +

∫ x

0

(3 + 2x) dx

∴ y1 = 1 + 3x + x2


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

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

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

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

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:

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

2which is approximate solution, putting x = 0.1

y(0.1) = 1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) = 1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) = 1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) = 1.31205


Picard’s Method

2nd approximation:


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

y(0.1) = 1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) = 1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) =

1.31205


Picard’s Method

2nd approximation:


y2 = 1 +

∫ x

0

[3 + 2x(1 + 3x + x2)

]dx

= 1 +

∫ x

0

[3 + 2x + 6x2 + 2x3)

]dx

∴ y2 = 1 + 3x + x2 + 2x3 +x4


y(0.1) = 1.31205


Picard’s Method

Ex:

Using Picard’s method, obtain a solution upto

4th approx of the equationdy

dx= y + x, y(0) = 1.


Picard’s Method

Sol.:

By Picard’s method y = y0 +

∫ 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)

‘y1 = 1 +

∫ x

0

(1 + x) dx

∴ y1 = 1 + x +x2

2


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

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


‘y1 = 1 +

∫ x

0

(1 + x) dx

∴ y1 = 1 + x +x2

2


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

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


‘y1 = 1 +

∫ x

0

(1 + x) dx

∴ y1 = 1 + x +x2

2


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

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


‘y1 = 1 +

∫ x

0

(1 + x) dx

∴ y1 = 1 + x +x2

2


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

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


‘y1 = 1 +

∫ x

0

(1 + x) dx

∴ y1 = 1 + x +x2

2


Picard’s Method

2nd approximation:

put y = y1 = 1 + x +x2

2in f(x, y)

‘y2 = 1 +

∫ x

0

(1 + 2x +

x2

2

)dx

∴ y2 = 1 + x + x2 +x3

6


Picard’s Method

2nd approximation:

put y = y1 = 1 + x +x2

2in f(x, y)

‘y2 = 1 +

∫ x

0

(1 + 2x +

x2

2

)dx

∴ y2 = 1 + x + x2 +x3

6


Picard’s Method

2nd approximation:

put y = y1 = 1 + x +x2

2in f(x, y)

‘y2 = 1 +

∫ x

0

(1 + 2x +

x2

2

)dx

∴ y2 = 1 + x + x2 +x3

6


Picard’s Method

2nd approximation:

put y = y1 = 1 + x +x2

2in 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

6in f(x, y)

‘y3 = 1 +

∫ x

0

(1 + 2x + x2 +

x3

6

)dx

∴ y3 = 1 + x + x2 +x3

3+

x4

24


Picard’s Method

3rd approximation:

put y = 1 + x + x2 +x3

6in f(x, y)

‘y3 = 1 +

∫ x

0

(1 + 2x + x2 +

x3

6

)dx

∴ y3 = 1 + x + x2 +x3

3+

x4

24


Picard’s Method

3rd approximation:

put y = 1 + x + x2 +x3

6in 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:

put y = 1 + x + x2 +x3

3+

x4

24in 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

4th approximation:

put y = 1 + x + x2 +x3

3+

x4

24in 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

4th approximation:

put y = 1 + x + x2 +x3

3+

x4

24in 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

4th approximation:

put y = 1 + x + x2 +x3

3+

x4

24in 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 problemdy

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


‘y1 = 0 +

∫ x

0

(x2 + 0) dx

∴ y1 =x3

3


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 0, f(x, y) = x2 + y2


‘y1 = 0 +

∫ x

0

(x2 + 0) dx

∴ y1 =x3

3


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 0, f(x, y) = x2 + y2


‘y1 = 0 +

∫ x

0

(x2 + 0) dx

∴ y1 =x3

3


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 0, f(x, y) = x2 + y2


‘y1 = 0 +

∫ x

0

(x2 + 0) dx

∴ y1 =x3

3


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 0, f(x, y) = x2 + y2


‘y1 = 0 +

∫ x

0

(x2 + 0) dx

∴ y1 =x3

3


Picard’s Method

2nd approximation:

put y = y1 =x3

3in f(x, y)

‘y2 = y0 +

∫ x

0

[x2 +

(x3

3

)2]dx

∴ y2 =x3

3+

x7

63y(0.4) =


Picard’s Method

2nd approximation:

put y = y1 =x3

3in f(x, y)

‘y2 = y0 +

∫ x

0

[x2 +

(x3

3

)2]dx

∴ y2 =x3

3+

x7

63y(0.4) =


Picard’s Method

2nd approximation:

put y = y1 =x3

3in f(x, y)

‘y2 = y0 +

∫ x

0

[x2 +

(x3

3

)2]dx

∴ y2 =x3

3+

x7

63y(0.4) =


Picard’s Method

2nd approximation:

put y = y1 =x3

3in f(x, y)

‘y2 = y0 +

∫ x

0

[x2 +

(x3

3

)2]dx

∴ y2 =x3

3+

x7

63

y(0.4) =


Picard’s Method

2nd approximation:

put y = y1 =x3

3in f(x, y)

‘y2 = y0 +

∫ x

0

[x2 +

(x3

3

)2]dx

∴ y2 =x3

3+

x7

63y(0.4) =


Picard’s Method

Ex:

Find the value of y for x = 0.1 by Picard’s

method given thatdy

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


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 1 +

∫ x

0

[2

1 + x− 1

]dx

∴ y1 = 1− x + 2log(1 + x)


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 1, f(x, y) =y − x

y + x


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 1 +

∫ x

0

[2

1 + x− 1

]dx

∴ y1 = 1− x + 2log(1 + x)


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 1, f(x, y) =y − x

y + x


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 1 +

∫ x

0

[2

1 + x− 1

]dx

∴ y1 = 1− x + 2log(1 + x)


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 1, f(x, y) =y − x

y + x


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 1 +

∫ x

0

[2

1 + x− 1

]dx

∴ y1 = 1− x + 2log(1 + x)


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 1, f(x, y) =y − x

y + x


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 1 +

∫ x

0

[2

1 + x− 1

]dx

∴ y1 = 1− x + 2log(1 + x)


Picard’s Method

Sol.:


∫ x

x0

f(x, y) dx

Here x0 = 0, y0 = 1, f(x, y) =y − x

y + x


y1 = 1 +

∫ x

0

1− x

1 + xdx = 1 +

∫ x

0

2− (1 + x)

1 + xdx

= 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)

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

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

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

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

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

which is difficult to integrate therefore using 1stapproximation.

y(0.1) =


Picard’s Method

= 1 + x−∫ x

0

2x

1 + 2log(1 + x)dx


y(0.1) =


Picard’s Method

= 1 + x−∫ x

0

2x

1 + 2log(1 + x)dx


y(0.1) =


Date post:	09-Jan-2017
Category:	Education
Upload:	dr-nirav-vyas
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Numerical Methods - Oridnary Differential Equations - 1

Education