Two-Point Boundary Value Problems - Numerical Methods in Engineering with MATLAB - page 308

The results are:

>> x y1 y2 y3 y4
0.0000e+000 0.0000e+000 0.0000e+000 -9.7607e-001 9.7131e-001
1.0000e-001 -4.7184e-003 -9.2750e-002 -8.7893e-001 9.7131e-001
3.9576e-001 -6.6403e-002 -3.1022e-001 -5.9165e-001 9.7152e-001
7.0683e-001 -1.8666e-001 -4.4722e-001 -2.8896e-001 9.7627e-001
9.8885e-001 -3.2061e-001 -4.8968e-001 -1.1144e-002 9.9848e-001
1.0000e+000 -3.2607e-001 -4.8975e-001 6.4879e-016 1.0000e+000

0.000

-0.050

-0.100

-0.150

-0.200

-0.250

-0.300

-0.350

0.00 0.20 0.40 0.60 0.80 1.00

x

By good fortune, our initial estimates y (0) 1 and y (0) =− = 1were very close to the

final values.

PROBLEM SET 8.1

1. Numerical integration of the initial value problem

y + y − y (0) y = 0 y (0) = 0 = 1

741028. What is the valueof y (0) that wouldresult in y (1) yielded y (1) = 0 . = 1,

assuming that y (0) is unchanged?

2. The solution of the differentialequation

y + y + 2 y = 6

2, y (0) 0 and y (0) with the initialconditions y (0) = = = 1, yielded y (1) =

03765. When the solutionwas repeatedwith y (0) 3 . = 0 (the other conditions

= .

72318. Determine the valueof y (0) so being unchanged), the result was y (1) 2

that y (1) = 0.