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HW6SOL

# HW6SOL - HW#6 Chapter.7.1#2 u Let x u x u then x t x t...

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Unformatted text preview: HW#6. Chapter .7.1 #2. u Let x u, x u then x t x t Therefore, x t x t 0 2 x t 1 0.5 x t 0 3sin t u t u t x t 3 sin t 0.5u t 2u t 3 sin t 0.5x t 2x t 0.5u 2u 3 sin t #6. u Let x u, x p t u q t u g t , u 0 u , u 0 u u then x t x t x 0 x 0 u t u t u 0 u 0 x t g t p t u t q t u t g t p t x t q t x t u u Therefore, x t x t 0 p t 1 q t x t x t x 0 0 , g t x 0 u u #7. x a) From the first equation, x Substitute these into the second equation, x x The characteristic equation is r Therefore, 4r 3 0 2x 4x x 3x 2x 4x x 2x , and x x 2x 2x x , x x 2x 0 r , and x x b) The initial conditions x 0 c e x 2x c e 1, r 3 c e 3c e 2c e 2c e c e c e 2 , x 0 3 Substitute the initial conditions into the general solution, x x Therefore, c x x c) 5 , 2 5 e 2 5 e 2 c 1 e 2 1 e 2 1 2 c e c e c e c e c c c c 2 3 #21. L dI dt R I V, C dV dt I V R , , , , , , , , Let I , I , I andI be the current through the resistors, R , R , inductor, L, and capacitor, C, respectively. Assign V , V , V andV as the respective voltage drops. Based on Kirchhoff's second law, the net voltage drops, around each loop, satisfy V V V 0 , V V V 0, and V V 0. Applying Kirchhoff's first law to each node, I That is, V V 0, V V V 0, I I I 0 I I 0 , I I 0, I I I 0 Using the current-voltage relations, V Combining these equations, R I Set I Then, L dI dt R I V and C dV dt I V R I, V V LI V 0 and CV I V R R I , V R I , LI V , CV I Chapter .7.2 #2. 1 3 2 1 2 2 , 2 3 2 a) 2 2 1 3 2 1 3 2 1 2 1 2 2 2 2 3 2 3 2 1 2 1 1 2 7 2 2 3 6 b) 3 3 c) 1 3 2 3 2 5 1 2 7 7 2 2 3 2 1 3 2 2 2 2 1 3 3 2 3 1 2 2 2 2 3 2 2 3 4 11 6 6 5 d) 5 2 1 2 1 3 3 2 2 3 2 1 1 2 2 3 2 2 2 2 8 6 3 2 7 4 4 1 3 2 1 2 2 2 4 4 #4. 3 2 2 a) T T 1 2 T 3 3 2 2 1 2 3 1 2 T 3 2 1 2 2 3 b) 3 2 2 c) T 3 3 2 2 1 2 3 3 2 1 2 2 3 #9 x a) T T 1 2 2 , y 2 3 1 2 T 1 2 2 T 2 3 1 2 3 1 2 1 2 2 2 2 4 3 1 2 4 5 3 2 4 3 1 2 4 5 3 Therefore, T T b) , , , T 1 2 2 2 3 1 2 2 1 2 2 2 4 3 1 2 4 3 5 , Therefore, T 2 3 1 2 4 3 1 2 4 3 5 3 5 , , #13. 1 2 1 The augmented matrix is 1 2 1 1 2 1 1 0 0 1 1 1 1 3 0 1 1 0 0 1 0 1 0 row2 2 0 0 1 row3 1 1 1 1 1 0 0 1 0 1 0 2 0 0 1 1 0 0 1 3 0 1 1 0 0 3 2 1 0 3 1 0 1 0 0 1/3 0 0 1/3 1/3 0 1/3 1/3 0 1/3 1 1 1 1 1 2 2 row1 row1 1 1 0 0 3 2 1 0 row2/ 3 row3/3 3 1 0 1 1 1 1 2/3 1/3 1 0 0 row1 1/3 0 row2 0 1/3 row2 row3 1 1 0 1 0 0 1 1 2/3 1 1/3 1 1 1 0 1 0 0 Therefore, 1 0 0 1/3 0 1 0 1/3 0 0 1 1/3 1 2 1 1 1 1 1 1 2 1/3 1/3 1/3 1/3 0 1/3 1/3 0 1/3 #25. 1 4 1 2 , e 4e e e 3e 12e 1 4 1 2 1 4 1 2 e 4e e e e 4e 2e 2e 4e 8e e 4e e 2e 3e 12e 2e 2e Chapter .7.3 #4. x 2x x 2x x x x x 2x 0 0 0 This system of equation can be written as 1 2 1 The augmented matrix is 1 2 1 1 2 1 1 0 0 2 1 1 2 1 1 1 0 1 0 2 0 2 1 1 1 1 2 x x x 0 0 0 1 0 1 2 1 0 3 3 0 1 0 row2 2 row1 0 0 3 3 0 2 0 row3 row1 2 1 0 1 2 1 0 0 3 3 0 3 3 0 0 0 0 0 3 3 0 row3 row2 x 2x x x 0 0 . x This system of equations does not have unique solutions. Hence, set x Then, x x x 1 1 1 #5. x x 3x x x x x 2x 0 0 0 This system of equation can be written as 1 0 3 1 1 1 The augmented matrix is 1 1 2 x x x 0 0 0 1 0 3 1 1 1 1 0 3 1 1 1 1 0 1 0 2 0 1 0 0 1 0 1 1 0 0 1 0 0 1 0 4 0 1 0 1 0 4 0 3 0 1 0 1 0 row2 3 row1 2 0 row3 row1 1 0 1 0 0 1 4 0 0 1 1 0 row3 row2 x x x 4x 3x 0 0 0 Therefore, x x x 0 0 0 #13. e , 2e By inspection, 2 , e ,e , 3e , 0 #14. 2 sin t , sin t , sin t , 2sin t They are linearly dependent if and only if c , where c is an arbitrary constant However, such a constant does not exist. Therefore, they are linearly independent. #25. 3 2 4 2 0 2 4 2 3 3 2 4 2 4 2 2 3 8 4 4 0 4 1 5 15 8 8 3 3 4 2 6 2 For 1, 4 2 4 2 1 2 4 2 4 The matrix is singular. Therefore, set x Then, x Or if set x Then, x Therefore eigenvectors corresponding 1 are 1, 8 x x x 0 0 0 1, x 1 4 1, x 1 0 1 4 , 1 1 0 1 Either of them can be replaced with any linear combination of those two vectors. For 8, 5 2 4 The augmented matrix is 5 2 4 5 2 4 2 8 2 4 0 2 0 2 row2 5 0 5 row3 5x 2x 18x 2 8 2 4 0 2 0 5 0 5 0 0 2 4 0 18 9 0 18 9 0 2 8 2 4 2 5 x x x 0 0 0 row3 4 row1 4x 9x 0 0 This system of equations does not have unique solutions. Hence, set x Then, 1 1/2 1 1. Chapter .7.4 #9. a) Let t be a solution of t is also a solution. Now let t t . t It follows that c t c t c t t , . Then the collection of vectors t , t , , t , t Constitutes n+1 vectors with n components. Based on the assertion in Problem 12, Section 7.3, these vectors are necessarily linearly dependent. That is, there are n+1 constants b , b , such that b From Problem 8, we have b for all t , . Now b t b t b t b t 0 t b t b t b t 0 ,b ,b (not all zero) 0 , otherwise that would contradict the fact that the first n vectors are linearly independent. Hence y t and the assertion is true. b) Consider t c t c t t Based on the assumption, t The collection of vectors t , is linearly independent on t , , c t 0 , for 1,2, , n. k c t k c t k c t k c t t , and suppose that we also have k t k t 1 b b t b t b t . It follows that k Chapter .7.5 #5. 2 1 a) Let e . Then 1 2 r e 2 1 1 2 r 1 2 e e 0 0 2 r 1 To have non-trivial solution 2 r 1 and r for r 1, 1 1 which gives 1 2 r r 4r 3 0 1, 3 1 1 0 0 1 1 for r 3 1 1 1 1 which gives 1 1 Therefore, the general solution is c 1 e 1 c 1 e 1 0 0 b) #7. 4 8 a) Let e . Then r 4 8 To have non-trivial solution 4 8 and r for r 0, 4 8 which gives 3 6 r 6 3 r r 6 e 3 r 4 8 3 6 3 6 e e 0 0 r 2r 0 2, 0 0 0 3 4 for r 2, 6 8 which gives 1 2 Therefore, the general solution is c b) 3 4 c 1 e 2 3 4 0 0 Chapter .7.6 #5. 1 5 a) Let e . Then r 1 5 To have non-trivial solution 1 5 and r for r 1 , 2 5 2 1 0 0 1 r 3 1 r r 2r 2 0 r 3 e 1 r 1 5 1 3 e e 0 0 1 3 which gives 2 Hence, 2 for r 1 , 2 which gives 1 2 Hence, 1 2 e e cos t sin t 2 cos t sin t 5 2 1 0 0 1 e e 2 cos t sin t cos t sin t 1 Therefore, the general solution is c e b) cos t 2cos t sin t c e sin t cos t 2sin t #7. 1 0 2 1 3 2 Let e . Then r 1 2 3 r e 0 2 1 0 2 1 3 2 r 0 2 0 2 1 e e 0 0 0 0 2 1 1 1 r To have non-trivial solution, 1 2 3 r 1 0 2 r 0 2 0 r 7r 5 r 1 r 2r 5 0 1 r 0 0 2 1 r 2 3 2 1 r For r 1, r 1 r 1 4 r r 1, 1 3r 2 0 0 2 0 3 2 Set 2. Then 0 2 0 e 0 0 0 and 2 3 2 2 3 e 2 For r 1 2 , 2 2 3 The augmented matrix is 2 2 3 2 2 3 2 0 0 Set 1. Then and 0 1 0 e cos 2t cos 2t sin 2t sin 2t e 0 1 0 2 2 0 2 2 0 0 row2 2 0 2 0 row3 3 0 0 2 0 2 0 row3 0 2 2 0 0 2 0 2 0 2 0 0 2 0 0 0 2 0 0 2 2 0 0 2 0 2 0 0 0 2 0 0 0 0 2 2 0 2 2 e 0 0 0 row1 row1/2 row2 For r 1 2 , 2 2 3 0 2 2 0 2 2 e 0 0 0 The augmented matrix is 2 2 3 2 2 3 0 2 2 2 0 0 Set 1. Then and 0 1 0 0 row2 2 0 2 0 row3 3 0 2 2 0 0 2 0 2 0 row3 0 2 2 0 0 2 0 2 0 row1 row1/2 2 0 0 2 0 0 0 2 0 0 2 2 0 0 2 0 2 0 0 0 2 0 0 0 row2 0 1 By linear combination of and e 0 cos 2t sin 2t cos 2t sin 2t e 1 2 1 2 1 2 1 2 0 cos 2t sin 2t 0 sin 2t cos 2t e Therefore, the general solution is c or c 2 3 e 2 2 3 e 2 e c 0 cos 2t sin 2t e c 0 sin 2t cos 2t e c 0 cos 2t sin 2t e c 0 sin 2t cos 2t e #9. 1 1 Let e . Then r 1 1 To have non-trivial solution 1 1 and r for r 1 , 2 1 2 5 0 0 1 r 3 5 r r 2r 2 0 r 3 e 5 r 1 1 5 3 e e 0 0 5 3 , 0 1 1 which gives 5 2 Hence, 5 2 for r 1 , e e 5 cos t 5sin t 2 cos t sin t 2 which gives 1 2 5 0 0 2 Hence, 2 5 e e 5 2 5cos t 5sin t cos t sin t Therefore, the general solution is c e Apply the initial condition c c Therefore, e cos t cos t 3sin t sin t 5 2 c 1 ,c 5 0 1 3 5 1 1 5cos t 2cos t sin t c e 5sin t cos t 2sin t ...
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