{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

pset7-soln

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

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

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 λ 1 t x 1 + c 2 e λ 2 t x 2 where λ 1 , λ 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. λ 1 = 4 and λ 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 0 - 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 0 + 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 = 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 λ 2 - 5 λ - 4 = 0, so the eigenvalues are λ 1 = 1 2 (5 + 41) and λ 2 = 1 2 (5 - 41). 1

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

View Full Document
Another way to find the eigenvalues is to substitute y = e λt into the differential equation. We obtain λ 2 e λt = 5 λe λt + 4 e λt Dividing by e λt , we find the same relationship λ 2 - 5 λ - 4 = 0. Problem 3: a) Do problem 17 in section 6.3 (pg. 317). b) Do problem 24 in section 6.3 (pg. 318). Solution (5+5 points) a) The infinite series for e Bt is e Bt = I + tB + 1 2 t 2 B 2 + . . .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 6

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

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

View Full Document
Ask a homework question - tutors are online