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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

    Dick

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    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

    Dick

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    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
    4

    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

    Dick

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    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

    Dick

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    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)?
     
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