SolutionLastHW - Solutions to 7.4: 2, 3, 4, 7, 8 2. Note...

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Unformatted text preview: Solutions to 7.4: 2, 3, 4, 7, 8 2. Note that det( A- I ) = 0 - - 4 4- = 0 2 + 16 = 0 = 4 i or =- 4 i. Eigenvectors for = 4 i : Here we have A- 4 iI =- 4 i- 4 4- 4 i , and so we have an eigenvector corresponding to nullspace( A- 4 iI ) as follows: v = 1- i . Hence, we obtain the solution x ( t ) = e 4 it 1- i = (cos4 t + i sin4 t ) 1 + i- 1 = cos4 t 1- sin4 t- 1 + i cos4 t- 1 + sin4 t 1 . Therefore, we obtain the two real-valued solutions x 1 ( t ) = cos4 t 1- sin4 t- 1 and x 2 ( t ) = cos4 t- 1 + sin4 t 1 . Putting the two solutions x 1 ( t ) and x 2 ( t ) together, we obtain the general solution to this linear system: x ( t ) = c 1 cos4 t 1- sin4 t- 1 + c 2 cos4 t- 1 + sin4 t 1 . Equivalently, this can be expressed as x ( t ) = c 1 cos4 t sin4 t + c 2 sin4 t- cos4 t . Note that one can also obtain the solution by working with the eigenvalue =- 4 i . The calculations are similar. 1 3. Note that det( A- I ) = 0 1- - 2 5- 5- 2 + 4 + 5 = 0 =- 2 i. Eigenvectors for =- 2 + i : Here we have A- (- 2 + i ) I = 3- i- 2 5- 3- i . The two rows of this matrix must be proportional, since we must be able to find a nonzero vector in its nullspace corresponding to =- 2 + i , so concentrating on the first row, we can determine an eigenvector corresponding to =- 2 + i as follows: v = 2 3- i . Note that other choices of v are possible here, such as 3 + i 5 , obtained by concentrating on the second row instead of the first row. Proceeding with our choice of v , we obtain the solution x ( t ) = e (- 2+ i ) t 2 3- i = e- 2 t (cos t + i sin t ) 2 3 + i- 1 = e- 2 t cos t 2 3- sin t- 1 + i cos t- 1 + sin t 2 3 . Therefore, we obtain the two real-valued solutions x 1 ( t ) = e- 2 t cos t 2 3- sin t- 1 and x 2 ( t ) = e- 2 t cos t- 1 + sin t 2 3 . Putting the two solutions x 1 ( t ) and x 2 ( t ) together, we obtain the general solution to this linear system: x ( t ) = c 1 e- 2 t cos t 2 3- sin t- 1 + c 2 e- 2 t cos t- 1 + sin t 2 3 . Equivalently, this can be expressed as x ( t ) = c 1 e- 2 t 2cos t 3cos t + sin t + c 2 e- 2 t 2sin t- cos t + 3sin t . Note that one can also obtain the solution by working with the eigenvalue =- 2- i . The calculations are similar. 2 4. Note that det( A- I ) = 0 - 1- - 2 2- 1- = 0 2 + 2 + 5 = 0 =- 1 2 i. Eigenvectors for =- 1 + 2 i : Here we have A- (- 1 + 2 i ) I =- 2 i 2- 2- 2 i . One choice of eigenvector, found by examining the nullspace of this matrix, is v = 1 i . We obtain the solution x ( t ) = e (- 1+2 i ) t 1 i = e- t (cos2 t + i sin2 t ) 1 + i 1 = e- t cos2 t 1- sin2 t 1 + i cos2 t 1 + sin2 t 1 . Therefore, we obtain the two real-valued solutions x 1 ( t ) = e- t cos2 t 1- sin2 t 1 and x 2 ( t ) = e- t cos2 t 1 + sin2 t 1 ....
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This note was uploaded on 10/03/2011 for the course CHEM 31 taught by Professor Roberts during the Fall '10 term at Lehigh University .

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SolutionLastHW - Solutions to 7.4: 2, 3, 4, 7, 8 2. Note...

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