Page 278 - Schaum's Outline of Differential Equations

CHAP. 26] SOLUTIONS BY MATRIX METHODS 261

Thus,

If we define two new arbitrary constants k^ = (3k 1 + k 2)/6 and k 4 = (3k± - k 2)/6, then

Supplementary Problems

A(
Solve each of the following systems by matrix methods. Note that e for the first five problems is found in Problem 16.2,
A(
while e for Problems 26.15 through 26.17 is given in Problem 16.3.
26.9. x + 2x - Sx = 0; x(l) = 1, i(l) = 0 26. 10. x + 2x - Sx = 4; x(0) = 0, jfc(O) = 0

26.11. x + 2x-Sx = 4;x(l) = 0, i(l) = 0 26.12. x + 2x-8x = 4;x(G) = 1, i(0) = 2
26.13. x + 2x-8x = 9e-'; x(0) = 0, i(0) = 0 26.14. The system of Problem 26.4, using Eq. (26.2)

26.15. x + x = 0 26. 16. x + x = 0; x(2) = 0, x(2) = 0

26.17. x + x = t;x(l) = 0,x(l) = l 26.18. y-y-2y = 0

3
26.19. y-y-2y = 0; y(0) = 2, /(O) = 1 26.20. y-y-2y = e ';y(0) = 2, /(O) = 1

26.21. y - y - 2y = e ' 3 ; y(0) = 1, /(O) = 2 26.22. z + 9z + Uz = -sint; z(0) = 0, z(0) = - 1
9
26.23. x = - 4x + 6y 26.24. x + 5x - 12y = 0
y = -3x + 5y; y + 2x - 5y = 0;
x(0)=3,y(0) = 2 ^(0) = 8, X°)= 3

26.25. x-2y = 3 26.26. x = x + 2y
2
y + y-X = -t ; y = 4x + 3y
x(0) = 0,y(0) = -l

26.27. x = 6t; x(0) = 0, x(0) = 0 , x(0) = 12 26.28. x + y = 0
y + x = 2e~' ;
x(0) = 0, x(0) = - 2, y(0) = 0

26.29. x = 2x + 5y + 3,
y = ~ x - 2y;
x(0) = 0,x(0) = 0,y(0) = l