1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Is the book incorrect here? Systems of diff eqs.

  1. Nov 19, 2012 #1


    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    Here is the question from my book along with their solution to the problem :

    Part 1 : http://gyazo.com/6467751155edcefec6cc583d164d2ae7
    Part 2 : http://gyazo.com/d91752f0a18a72af96f6afc74117f011
    Part 3 : http://gyazo.com/7a17dbca3e414d0ddc8458abce2eda9c

    2. Relevant equations

    I'll denote the coefficient matrix by the letter A.
    I denote my null space operator by N.
    t is the transpose of a vector.

    3. The attempt at a solution

    Now I'm almost positive the book is wrong here. Here's my solution.

    Assume that [itex]x = εe^{λt}[/itex] so our equation becomes :

    [itex]λεe^{λt} = Aεe^{λt} \Rightarrow (A - λI)ε = 0[/itex]

    So we seek eigenvector(s) ε such that the above equation is satisfied. So lets find the eigenvalues for A ( I'll skip some of the boring algebra here ) :

    [itex]det(A - λI) = (λ-3)(λ+1)[/itex] Thus λ1 = 3 and λ2 = -1 are eigenvalues for A.

    Now to find the eigenvectors for the eigenvalues, we seek the null space of A - λjI. So :

    [itex]ε(λ_1) = N(A - 3I) = span\left\{{[1/2 \space \space 1]^t}\right\} [/itex]

    [itex]ε(λ_2) = N(A + I) = span\left\{{[-1/2 \space \space 1]^t}\right\} [/itex]

    See ^ for some reason I get different eigenvectors than the book does. Everything else that follows after this is exactly the same as the book except my eigenvectors.

    Have I made an error somewhere? Or is the book wrong in this case? Some clarification on this would be greatly appreciated.

    Thanks in advance.
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, you are getting the same eigenvectors, but just scaled differently. Instead of <1/2,1> the book uses <1,2>, and instead of <-1/2,1> the book uses <1,-2>. Eigenvectors are, of course, not unique: any nonzero scalar multiple of an eigenvector is also an eigevector for the same eigenvalue.

  4. Nov 19, 2012 #3


    User Avatar
    Homework Helper

    Ahh yes, I remember now. So multiply my first one by 2 and my second one by -2 to get the same answer as they have.

    Thanks for clearing that up for me RGV.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook