math20d stephen young fall 09 final

math20d stephen young fall 09 final - Math 20D Final Exam 8...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ) ....
View Full Document

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.

Page1 / 7

math20d stephen young fall 09 final - Math 20D Final Exam 8...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online