Exercise 5.4 [Milne's Predictor and Corrector method]

EXERCISE 5.4 [Milne's Predictor and Corrector method]

Using Milne's method solve the following problems

1. Given y' = 1+ y², y (0) = 0, y (0.2) = 0.2027, y (0.4) = 0.4228, y (0.6) = 0.6841, estimate y (0.8), y (1) using Milne's method upto three decimals.

[Ans. 1.029, 1.555]

2. Solve, y = x - y², y (0) 1 to obtain y (0.4) by Milne's method. Obtain the data you require by any method of your liking.

[Ans. 0.7796]

3. Compute, y (0.6) by Milne's method given y' = x + y, y (0) with h = 0.2. Obtain the required data by Taylor's series method.

[Ans. 2.0442]

4. Given y' = 1/x+y, y (0) = 2, y (0.2)= 2.0933, y (0.4) = 2.1755, y(0.6) =  2.2493, find y (0.8) by Milne's predictor corrector method [A.U N/D 2013]

[Ans. 2.3162, 2.3164]

5. Given y' = x (x² + y²) e^-x, y (0) = 1, find y at x = 0.1, 0.2 and 0.3 by Taylor's method; compute y (0.4) by Milne's method.

[Ans. 1.0047, 1.01813, 1.03975, y(0.4) = 1.0709]

6. Given dy/dx = 1/ 2 (1 + x²) y² and y (0) = 1, y (0.1) = 1.06, y (0.2) = 1.12, y (0.3) = 1.21 find y (0.4), using Milne's predictor corrector formula.

[Ans. y_4,p = 1.2772, y_4,c = 1.2797]

7. Given dy/dx + y + y - x² = 0, y (0.1) = 0.9052, y (0.2) = 0.8213 find correct to four places of decimals y (0.4) and y (0.5), using Milne's Predictor corrector method.

 [Ans. y (0.4)p = 0.6897, y (0.4) c = 0.6897, y(0.5)p = 0.6435, y(0.5), = 0.6435]

8. Given dy/dx = 1/x+y' y(0) = 2. Taking the value of y(0.2) = 2.0933,y(0.4) = 2.1755, y (0.6) = 2.2493 find y (0.8) using Milne's method.

[Ans. yp (0.8) = 2.3162, yc (0.8) = 2.3164]

9. The differential equation dy/dx = 1 / 2 (1 + x)y², y (0) = 1 is given.

Solve it by Taylor's series method for x = 0.0, (0.2), 0.6. Advance the solution, to x = 1.0 by Milne's predictor corrector formula.

[Ans. y (0.2)= 1.123, y (0.4) = 1.3135, y (0.6) = 1.6312, y (0.8) = 2.2376, yp (1) = 3.6028, yc (1) = 3.84]

10. Using Taylor's series method (of fourth order) to solve dy / dx = x² + y2 - 2, y (0) = 1 at x = ± 0.1, + 0.2. Continue the solution of the problem at x = 0.3 using Milne's method.

[Ans. y (0.1)= 1.0047, y (-0.1)= 1.09, y (0.2) = 0.76049, y3,p = 0.61463, y3,c = 0.6148]

11. The differential equation dy/dx  = x² + y/2 satisfied by y (1) = 2, y (1.1) = 2.2156, y (1.2)= 2.4649, y (1.3)= 2.7514. Using Milne's method, find y (1.4) correct to 4 decimal places.

[Ans. y_p (1.4) = 3.079, y_c (1.4) = 3.0794]

12. The differential equation 10 dy/dx = x² + y² is satisfied by  y (0) = 1, y (0.1) = 1.0110, y (0.2) = 1.0207, y (0.3) = 1.0318.

Determine y (0.4) and y (0.5) correct to four decimal places, using Milne's method.

[Ans. y (0.4) = 1.0438, y (0.5) = 1.05577]

13. Solve y' = x² + y², y (0) = 1. Compute y (0.4) by Milne's predictor and corrector method. Obtain the starting values by Taylor's series method with terms containing h⁴.  [A.U. May 1999] [A.U M/J 2016 R-13]

14. Compute y at x = 0.3 by Milne's method if y' x² + y², y (-0.1) = 0.9088, y (0) = 1 = 1.1115, y (0.2) = 1.2530.  [A.U. May 1999]