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: Diff Eq linear systems help

  1. Dec 15, 2008 #1
    i am trying to study for a DE test and was wondering if you could help with some problems

    dY = -2 1
    dt 0 2

    i found the eigen values to be lamda = 2

    i need some help on how to find the eigen vectors is this a reapeted root case and if it is how do i start off

    would i go AV = (lamda)V then solve form there and in this case there would only be one eigen vector. I aplolgize on my lack of knowledge on typing in math symbols.
  2. jcsd
  3. Dec 15, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    The eigenvalues aren't both 2. There are two linearly independent eigenvectors. Try that again.
  4. Dec 15, 2008 #3
    i will give it another go
  5. Dec 15, 2008 #4
    if i subtract the identiy matrix multiplied by lamda from

    -2 1
    0 2

    i get the new matrix

    ( to make this simpler im going to use L for lamda)

    ( -2 - L ) 1
    0 (2 - L)

    then if i take the determinate of this matrix and set it equal to zero i get

    (-2 - L)(2 - L) - 1(0) = 0

    L^2 - 4 = 0

    L^2 = 4

    L = 2

    i must be making a mistake somewhere can some point out where thanks for the help
  6. Dec 15, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    Stop with (L-2)*(-L-2)=0. It's already factored. The roots are +2 and -2. Likewise for L^2=4.
  7. Dec 15, 2008 #6
    ok thank you so then +2 and -2 are my eigen values thank you for the help
  8. Dec 15, 2008 #7
    using those values i calculated the following eigenvectors.

    AV = LV

    from this i get the eqautions

    -2x + y = 2x and 2y = 2y then i get

    y = 4x and y= y

    if i let x = 1 i get

    V = 1

    for my first eigen vector

    then for the second i get

    AV = LV then i get the eqns

    -2x + y = -2x solving i get y = 0

    and the other eqn is 2y = -2y im a little confused on how i can get my eigen vector from these two eqns. y = 0 but how can i get the x val
  9. Dec 15, 2008 #8


    User Avatar
    Science Advisor
    Homework Helper

    You've got that y=0 and x=anything. There's no equation for x. Pick an 'anything'. like x=1. That's works doesn't it?
  10. Dec 15, 2008 #9
    ok i was kinda thinking that but was a little unclear but thank you. the next part i am working on is for each eigen value pick an associated eigen vector V and determine the solution Y(t) with Y(0) = V

    could you please help me start this one off im confused on that an what it meanse to pick an eigen vector cant i just use the ones from the prior part.
  11. Dec 15, 2008 #10


    User Avatar
    Science Advisor
    Homework Helper

    Ok, pick (1,4) (I'll write the vectors that way instead of in column form). Assume the solution has the form (1,4)*w(t) (where w(t) is an ordinary function, not a vector). Substitute that into the DE. Can you solve for w(t)?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook