pset7-soln

pset7-soln - 18.06 Problem Set 7 Due Wednesday, 16 April...

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Unformatted text preview: 18.06 Problem Set 7 Due Wednesday, 16 April 2008 at 4 pm in 2-106. Problem 1: Do problem 1 in section 6.3 (pg. 315) in the book. Solution (10 points) We solve a linear system of differential equations by taking u ( t ) = c 1 e 1 t x 1 + c 2 e 2 t x 2 where 1 , 2 are the eigenvalues, x 1 , x 2 are the eigenvectors, and c 1 , c 2 are constants that satisfy c 1 x 1 + c 2 x 2 = u (0). To write down the matrix exponential explicitly, we must find the eigenvalues and eigenvectors of A . Since this A is diagonal, its eigenvalues are just the diagonal entries, i.e. 1 = 4 and 2 = 1. The eigenvectors are x 1 = (1 , 0) and x 2 = (1 ,- 1). Finally, if u (0) = (5 ,- 2) we must find how to write u (0) as a linear combination of x 1 and x 2 . We do this by solving the equation 1 1- 1 c 1 c 2 = 5- 2 We get c 2 = 2 and c 1 = 3. So, the final equation is u ( t ) = 3 e 4 t 1 + 2 e t 1- 1 Problem 2: Do problem 3 in section 6.3 (pg. 315). Solution (10 points) To linearize this system, we identify u with the vector [ y, y ] T , so that we have two equations dy/dt = y and dy /dt = y 00 = 4 y + 5 y . That is, we can decouple the differential equation by adding y as a new variable, to obtain the system dy/dt dy /dt = 0 1 4 5 y y We call the coefficient matrix A as usual. The eigenvalues of A satisfy the equation 2- 5 - 4 = 0, so the eigenvalues are 1 = 1 2 (5 + 41) and 2 = 1 2 (5- 41). 1 Another way to find the eigenvalues is to substitute y = e t into the differential equation. We obtain 2 e t = 5 e t + 4 e t Dividing by e t , we find the same relationship 2- 5 - 4 = 0....
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.

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pset7-soln - 18.06 Problem Set 7 Due Wednesday, 16 April...

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