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Linear Indepdant or not? the Wronskin != 0, so whats going on!

  1. Feb 18, 2006 #1
    Hello everyone, can you tell me what i'm doing wrong here? The question says:
    Determine which of the following pairs of functions are linearly independent.

    1. f(x) = x^2\quad, g(x) = 4|x|^2

    2. f(x) = x^3\quad , g(x)=|x|^3

    3. [​IMG]

    4. f(t) = 2t^2+14t, g(t)=2t^2-14t

    Here is my work:
    [​IMG]


    Once i put it in the form of y1y2'-y1'y2, i would plug in 0 for t, and if th4e answer was != 0, I thought it was L.I.
     
    Last edited: Feb 18, 2006
  2. jcsd
  3. Feb 18, 2006 #2

    0rthodontist

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    Which question specifically are you confused on? I see you have a problem with #2 since |x| is not the same as x. Also what is this with plugging in 0 for t? You don't do that.
     
  4. Feb 18, 2006 #3
    so taking the derivative of |x|^3 isn't t he same as x^3?
    Once you put the equations into y1y2'-y1'y2 What do you to to determine if they are linearly indepdant or not?
     
  5. Feb 18, 2006 #4

    0rthodontist

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    if y1y2' - y1'y2 is 0, then the functions are L.D. Otherwise the functions are L.I.

    The two functions are L.I. if one is a multiple of the other, that is if y1/y2 = k where k is a constant (or if y2/y1). Differentiating both sides and multiplying through by y2^2 gives you y1y2' - y1'y2 = 0 since the derivative of a constant is 0.

    In summary if y1y2' - y1'y2 = 0, then either y1/y2 or y2/y1 is a constant and the functions are l.d., and if y1y2' - y1'y2 != 0, then y1/y2 is not a constant and the functions are l.i.

    Yes, the derivative of |x|^3 is not the same as the derivative of x^3. If x > 0 then |x|^3 = x^3, but if x < 0 then |x|^3 = -(x^3).
     
  6. Feb 19, 2006 #5
    thank u for the explanation!
     
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