Page 335 - Advanced engineering mathematics

10.3 Solution of X = AX + G 315

This is

z 1 = 20 z 1 + −1/4 1/4 8
z 2 06 z 2 1/4 3/4 4e 3t
3t
2z 1 − 2 + e
= 3t .
6z 2 + 2 + 3e
This is an uncoupled system, consisting of one differential equation for just z 1 , and a second
differential equation for just z 2 . Solve each of these first-order linear differential equations to
obtain
3t
2t
z 1 (t) = c 1 e + e + 1
1
6t
3t
z 2 (t) = c 2 e − e − .
3
Then
2t 3t
−3 1 c 1 e + e + 1
X(t) = PZ(t) =
1 1 c 2 e − e − 1/3
6t
3t

3t
2t
3t
−3c 1 e + c 2 e − 4e − 10/3 −3e 2t e 6t −4e − 10/3
6t
= = C + .
c 1 e + c 2 e + 2/3 e 2t e 6t 2/3
6t
2t
This is the general solution in the form (t)C+ p , which is a sum of the general solution of the
associated homogeneous equation and a particular solution of the nonhomogeneous equation.
SECTION 10.3 PROBLEMS
⎛ ⎞ ⎛ −2t ⎞ ⎛ ⎞
In each of Problems 1 through 9, use variation of param- 1 −3 0 te 6
eters to find the general solution, with A and G given. If 9. ⎝ 3 −5 0 ⎠ , ⎝ te −2t ⎠ ; ⎝ 2 ⎠
2 −2t
initial conditions are given, also satisfy the initial value 4 7 −2 t e 3
problem
In each of Problems 10 through 19, find a general solu-
t tion of the system. If initial values are given, also solve the
5 2 −3e
1. ,
−2 1 e 3t initial value problem.

2 −4 1 10. x =−2x 1 + x 2 , x =−4x 1 + 3x 2 + 10cos(t)
2. , 1 2
1 −2 3t 11. x = 3x 1 + 3x 2 + 8, x = x 1 + 5x 2 + 4e 3t

6t 1 2
7 −1 2e 12. x = x 1 + x 2 + 6e , x = x 1 + x 2 + 4
3t

3. , 1 2
1 5 6te 6t

13. x = 6x 1 + 5x 2 − 4cos(3t), x = x 1 + 2x 2 + 8
⎛ ⎞ ⎛ 2t ⎞ 1 2
2 0 0 e cos(3t) 14. x = 3x 1 − 2x 2 + 3e , x = 9x 1 − 3x 2 + e 2t
2t

4. ⎝ 0 6 −4 ⎠ , ⎝ −2 ⎠ 1 2
2t
2t

0 4 −2 −2 15. x = x 1 + x 2 + 6e , x = x 1 + x 2 + 2e ;
1
2
x 1 (0) = 6, x 2 (0) = 0
1 0 0 0 0
⎛ ⎞ ⎛ ⎞

4 3 0 0 −2e t ⎟ 16. x = x 1 − 2x 2 + 2t, x =−x 1 + 2x 2 + 5;
1
2
⎜ ⎟ ⎜
,
5. ⎜ ⎟ ⎜ ⎟
⎝ 0 0 3 0 ⎠ ⎝ 0 ⎠ x 1 (0) = 13, x 2 (0) = 12
−1 2 9 1 e t 17. x = 2x 1 − 5x 2 + 5sin(t), x = x 1 − 2x 2 ;

1 2
x 1 (0) = 10, x 2 (0) = 5
2 0 2 0
6. , ;
5 2 10t 3 18. x = 5x 1 − 4x 2 + 4x 3 − 3e −3t , x = 12x 1 − 11x 2 +

1 2
12x 3 + t, x = 4x 1 − 4x 2 + 5x 3 ; x 1 (0) = 1,

t
5 −4 2e −1 3
7. , ; x 2 (0) =−1, x 3 (0) = 2
4 −3 2e t 3
19. x = 3x 1 − x 2 − x 3 , x = x 1 + x 2 − x 3 + t,

⎛ ⎞ ⎛ 2t ⎞ ⎛ ⎞ 1 2
2 −3 1 10e 5 x = x 1 − x 2 + x 3 + 2e ; x 1 (0) = 1,
t

2t
3
8. ⎝ 0 2 4 ⎠ , ⎝ 6e ⎠ ; ⎝ 11 ⎠ x 2 (0) = 2, x 3 (0) =−2
0 0 1 −e 2t −2
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October 14, 2010 20:32 THM/NEIL Page-315 27410_10_ch10_p295-342