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math20d stephen young fall 09 final

# math20d stephen young fall 09 final - Math 20D 1(10 points...

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Math 20D Final Exam 8 December 2009 1. (10 points) Show that the matrix 4 - 2 2 1 1 2 - 1 2 1 has eigenvalues 3 , 3 , 0 and find the eigenvectors associated with 3. Solution: det 4 - λ - 2 2 1 1 - λ 2 - 1 2 1 - λ = (4 - λ ) det 1 - λ 2 2 1 - λ - ( - 2) det 1 2 - 1 1 - λ + 2 det 1 1 - λ - 1 2 = (4 - λ )((1 - λ ) 2 - 4) + 2(1 - λ + 2) + 2(2 + 1 - λ ) = (4 - λ )( - 3 - 2 λ + λ 2 ) + 12 - 4 λ = - 12 - 8 λ + 4 λ 2 + 3 λ + 2 λ 2 - λ 3 + 12 - 4 λ = - λ 3 + 6 λ 2 + 9 λ = - λ ( λ - 3) 2 . Thus the eigenvalues are 3, 3, and 0. Perform Guassian elimination on A - 3 I 1 - 2 2 1 - 2 2 - 1 2 - 2 1 - 2 2 0 0 0 0 0 0 . Thus the eigenvalues are 2 1 0 - 2 0 1 . 2. Find the general form of the solution for each of the following differential equations. (a) (5 points) 3 y 00 - y 0 - 2 y = 0 Solution: The characteristic polynomial is 3 r 2 - r - 2 = 0 which has roots 1 ± 1+24 6 = 1 ± 5 6 . Thus the general form of the solution is c 1 e t + c 2 e - 2 3 t . (b) (5 points) y 00 + 4 y 0 + 13 y = 0 Solution: The characteristic polynomial is r 2 +4 r +13 = 0 which has roots - 4 ± 16 - 52 2 = - 4 ± 6 i 2 = - 2 ± 3 i . Thus the general form of the solution is e 3 t ( c 1 cos(2 t ) + c 2 sin(2 t )). (c) (5 points) 4 y 00 + 4 y 0 + y = 0. Page 1 of 7 Name: Points earned:

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Math 20D Final Exam 8 December 2009 Solution: The characteristic polynomial is 4 r 2 +4 r +1 = 0 which has roots - 4 ± 16 - 16 8 = - 1 2 . Thus the general form of the solution is c 1 e - t 2 + c 2 te - t 2 . 3. Find the general form of the solution for each of the following systems of differential equations of the form x 0 = Ax . (a) (15 points) A = - 1 1 - 2 - 3 Solution: det - 1 - λ 1 - 2 - 3 - λ = (1 + λ )(3 + λ ) + 2 = λ 2 + 4 λ + 5 . Thus the eigenvalues are - 2 ± i and the eigenvectors are 1 2 i . Thus the general form of the solution is c 1 e - 2 t cos( t ) 2 cos( t ) - sin( t ) + c 2 e - 2 t sin( t ) 2 sin( t ) + cos( t ) . (b) (15 points) A = 1 3 4 2 Solution: det 1 - λ 3 4 2 - λ = (1 - λ )(2 - λ ) - 12 = λ 2 - 3 λ - 10 = ( λ - 5)( λ + 2) .
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