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Homework Help: Determining whether two functions are linear independent via wronskian

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine which of the following pairs of functions are linearly independent.

    (a) [itex]f(t)=3t,g(t)=|t|[/itex]

    (b) [itex]f(x)=x^{2},g(x)=4|x|^{2}[/itex]

    2. Relevant equations

    the Wronskian is defined as,


    if {f(u),g(u)} are linearly dependent, W=0

    if W=/=0, {f(u),g(u)} are linearly independent

    3. The attempt at a solution

    The assumed interval for the independent variables t,x are [itex] x,t \in (-\infty,\infty) [/itex]

    for (a),

    I determined [itex] W(t)=3t-3|t| [/itex], which for x>0 is [itex] W(t)=3t-3t=0 [/itex].
    for x<0 we have [itex] W(t)=-3t-3t=-6t [/itex]. Since for some value of [itex] t \in (-\infty,\infty)[/itex] I found a [itex]W(t) \neq 0[/itex] I can conclude that the functions f(t) and g(t) are linearly independent.

    Now for (b),

    Similarly to (a), I find a [itex] W(x)=8x^{2}|x|-8|x|^{2}x[/itex]

    for [itex] x>0 : W(x)=8x^{3}-8x^{3}=0[/itex]
    for [itex] x<0 : W(x)=8(-x)^{2}|-x|-8|-x|^{2}(-x)=8x^{3}+8x^{3}=16x^{3}\neq 0[/itex]

    Similarly I conclude that f(x) and g(x) are linearly independent since I found values of x which make the wronskian not equal to 0.

    However, while my conclusion is correct for (a), (b) is supposedly linearly dependent.

    Is my method correct? if so what mistake did I make in concluding that the functions of (b) were linearly independent?

  2. jcsd
  3. Mar 4, 2012 #2


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    |x|^2 is equal to x^2.
  4. Mar 4, 2012 #3

    I use that fact, but I still find that for
    [itex] x<0 : W(x)=8(-x)^{2}|-x|-8(-x)^{3}=8x^{3}+8x^{3}=16x^{3}[/itex]

    Am I just completely missing something here?
  5. Mar 4, 2012 #4


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    Your calculation of [itex]W(x)=8x^{2}|x|-8|x|^{2}x[/itex] is wrong. Just use that |x|^2=x^2 from the beginning, so g(x)=4x^2.
  6. Mar 4, 2012 #5


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    The specific place where your calculations are incorrect is$$
    \frac d {dx}|x|^2 = 2|x|$$is false.
  7. Mar 4, 2012 #6
    Ah, I see. Thank you
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