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Second order differential equation question

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Verify that [itex]y_1=x^3[/itex] and [itex]y_2=|x|^3[/itex] are linearly independent
    solutions of the differential equation [itex]x^2y''-4xy'+6y=0 [/itex] on the interval
    [itex](-\infty,\infty)[/itex]. Show that [itex]W(y_1,y_2)=0[/itex] for every real number
    x, where W is the wronskian.

    2. Relevant equations
    theorems on differential equations

    3. The attempt at a solution

    First I need to check that y1 and y2 are the solutions of the
    given diff. equation. y1 is easy. To prove that y2 is the solution,
    I divided the whole interval [itex](-\infty,\infty)[/itex], in three parts [itex]x>0\; ,x=0\;,\;x>0[/itex]. And then I showed that the diff. equation is satisfied on all the different
    parts. So that means , y2 is the solution of the diff. equation

    Now, to check the linear independence, lets consider the equation
    [tex]c_1 x^3+c_2 |x|^3=0[/tex]

    Now here I am stuck. How do I prove that [itex]c_1=c_2=0[/itex] for all values of x in

  2. jcsd
  3. May 1, 2012 #2


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    Pick a couple of particular values of x. How about x=1 and x=(-1)?
  4. May 1, 2012 #3
    Ok, if I plug those values of x, I get [itex]c_1+c_2=0[/itex]. But this just means that
    c1 can be expressed in terms of c2...

    Edit: Oh mistake.. another equation is [itex]-c_1 +c_2 = 0[/itex] which gives me
    [itex]c_1=0\;\;, c_2=0[/itex] . so [itex]y_1\;,y_2[/itex] are linearly independent.

    Now for the remaining part, I have to show that [itex]W(y_1,y_2)=0[/itex] for every real
    number. So should I split the interval in 3 parts, since[itex]|x|^3[/itex] is not differentiable on [itex](-\infty,\infty)[/itex].
    Last edited: May 1, 2012
  5. May 1, 2012 #4


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    It's true that ##|x|## isn't differentiable at ##x=0##. But you are asserting ##|x|^3## isn't. Are you sure about that? Have you graphed it? Have you checked its right and left hand derivatives?
  6. May 2, 2012 #5
    Ok I see that [itex]|x|^3[/itex] is differentiable on [itex](-\infty,\infty) [/itex]. But [itex] y_2 '[/itex] is a piecewise function now. So can wronskian be evaluated on different parts of the
    interval [itex](-\infty,\infty) [/itex] ? The examples I have seen with the wronskian don't involve piecewise functions.
  7. May 2, 2012 #6


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    Just evaluate the wronskian on each piece. Being piecewise defined doesn't really change things.
  8. May 2, 2012 #7
    Ok thats what I thought. So its valid to evaluate wronskian on different parts of the interval. Can you give the link to such examples.... my book on diff. equations (dennis g zill) doesn't have any ...
  9. May 2, 2012 #8


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    I'm not sure why you'd need one. Just try it. If x>=0, y2=x^3 and if x<=0, y2=(-x^3). Just split it up like that.
  10. May 2, 2012 #9
    By writing |x| as √(x2) and using the chain rule, you can show that [itex]\frac{d}{dx}|x| = \frac{|x|}{x} = \frac{x}{|x|}[/itex] without having to split it up into different cases. Then you can easily show that the Wronskian of x3 and |x|3, after some rewriting, is equal to 0.
  11. May 2, 2012 #10
    Bohrok, if we write derivative like that, should not we worry about the case when x=0 . That where things get ugly..
  12. May 2, 2012 #11


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    There's nothing ugly about x=0 in the original problem. I don't think Bohrok's hint really simplifies it. Just do it directly.
  13. May 3, 2012 #12
    thanks..... Dick..
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