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LINEAR ALGEBRA: Given a sequence as a basis for a solution of some O.D.E., find it

  1. Nov 16, 2006 #1
    Problem:

    The sequence (c, s, 1, [tex]e_1,\,e_{-1}[/tex]) is a basis for the solution space of some differential equation p(D)y = 0. Find this O.D.E.

    NOTE: c = cos(t) and s = sin(t)


    Work so far:

    I know that [tex]e_1[/tex] gives a (t - 1) and that the [tex]e_{-1}[/tex] gives a (t + 1), but how do I solve for the 1! I think that the c and s give ([tex]t^2[/tex] - 1).

    Also, can someone explain in detail or give a reference to what a Ker() is?

    thanks
     
  2. jcsd
  3. Nov 16, 2006 #2

    0rthodontist

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    Ker(x) means the kernel of the transformation x--the set of all values that x maps to the identity. In linear algebra that would be the set of all values that x maps to the zero vector. What exactly are [tex]e_1,\,e_{-1}[/tex]?
     
  4. Nov 17, 2006 #3

    HallsofIvy

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    I have no idea what you mean by this! Is it possible that [tex]e_1[/tex] and [tex]e_{-1}[/tex] were supposed to be [itex]e^1[/itex] and [tex]e^{-1}[/itex]? If that is the case then the roots of the characteristic equation are i, -i, 0, 1, and -1. From that information, you should be able to find the characteristic equation and from that the differential equation.

    The problem as given ("Find this O.D.E.") has no single solution. There exist an infinite number of differential equations having those functions as solutions. Why I am giving is the simplest linear, homogenous, differential equation.
     
  5. Nov 17, 2006 #4

    Office_Shredder

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    I think he meant [tex]e^t[/tex] and [tex]e^{-t}[/tex] since c, s, and 1 are functions (well, e and 1/e would be functions too, but for these purposes would be equivalent to 1)
     
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