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Unformatted text preview: 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 + 2det 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 . Thus the eigenvalues are 2 1  2 1 . 2. Find the general form of the solution for each of the following differential equations. (a) (5 points) 3 y 00 y 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 + 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 + y = 0. Page 1 of 7 Name: Points earned: 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 = 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 ) 2cos( t ) sin( t ) + c 2 e 2 t sin( t ) 2sin( t ) + cos( t ) ....
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This note was uploaded on 01/26/2010 for the course MATH 20D 20D taught by Professor Eggers,john during the Fall '09 term at UCSD.
 Fall '09
 Eggers,John

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