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22. x = 3x1+2x2+2x3, x2=- X3 = 5x1 +5x2 + 3x3 %3D %3D

22. x = 3x1+2x2+2x3, x2=- X3 = 5x1 +5x2 + 3x3 %3D %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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problem 22

26. Find the particular solution of the system 24. x,=2x1+ x2- X3, X=-4x1-3x2-X3, 25. x, = 5x1 + 5x2 + 2x3, x = -6x1 – 6x2 - 5x3, In Problems 17 through 25, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of each system. (28) 17. x = 4x1+x2+4x3, x =x1+7x2+x3, x = 4x1+x2+4x3 18. x = x1+2x2+2x3, x, =2x1+7x2+x3, x3 = 2x1+x2 +7x3 19. x=4x1+x2+x3, x,=x1++4x2+x3, x3=x1+x2+4x3 20. x = 5x1+x2 +3x3, x, =x1+7x2+x3 x3%=D3x1+x2+5x3 21. r. =5r1-6xa. x=2r1 –x2-2x2, x = 4x;-2x2-4x3 27. V1 = || 28. V = %D The amou %D Fig. 5.2.8 | 22. x = 3x1+2x2 +2x3, x, =-5x1-4x2 -2x3, dx x3= 5x1+5x2 +3x3 23. (29) d. %D 1=3X1+x2 + 13, X, =-5x1-3x2-X3, X3%3D5X1+5x2+3x3 ***= 2x1 +x2– x3, x, =-4x1-3x2-X3, X3= 4x1+4x2+2x3 where ki х1 (1) and (lb), and graphs of %3D %3D | %3D = 5x1 +5x2 + 2x3, x, = -6x1 – 6x2 - 5x3, X3=6x1 +6x2+5x3 %3D (30) (30) is at dx1 + X3, 3x1 01 dt dra
Transcribed Image Text:26. Find the particular solution of the system 24. x,=2x1+ x2- X3, X=-4x1-3x2-X3, 25. x, = 5x1 + 5x2 + 2x3, x = -6x1 – 6x2 - 5x3, In Problems 17 through 25, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of each system. (28) 17. x = 4x1+x2+4x3, x =x1+7x2+x3, x = 4x1+x2+4x3 18. x = x1+2x2+2x3, x, =2x1+7x2+x3, x3 = 2x1+x2 +7x3 19. x=4x1+x2+x3, x,=x1++4x2+x3, x3=x1+x2+4x3 20. x = 5x1+x2 +3x3, x, =x1+7x2+x3 x3%=D3x1+x2+5x3 21. r. =5r1-6xa. x=2r1 –x2-2x2, x = 4x;-2x2-4x3 27. V1 = || 28. V = %D The amou %D Fig. 5.2.8 | 22. x = 3x1+2x2 +2x3, x, =-5x1-4x2 -2x3, dx x3= 5x1+5x2 +3x3 23. (29) d. %D 1=3X1+x2 + 13, X, =-5x1-3x2-X3, X3%3D5X1+5x2+3x3 ***= 2x1 +x2– x3, x, =-4x1-3x2-X3, X3= 4x1+4x2+2x3 where ki х1 (1) and (lb), and graphs of %3D %3D | %3D = 5x1 +5x2 + 2x3, x, = -6x1 – 6x2 - 5x3, X3=6x1 +6x2+5x3 %3D (30) (30) is at dx1 + X3, 3x1 01 dt dra
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