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Higher order linear equations- ODEs

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Verify that the differential operator defined by
    L[y] = y(n) + p1(t)y(n−1) +· · ·+ pn(t)y

    is a linear differential operator. That is, show that
    L[c1y1+ c2 y2] = c1L[y1] + c2L[y2],

    where y1 and y2 are n times differentiable functions and c1 and c2 are arbitrary constants.
    Hence, show that if y1, y2, . . . , yn are solutions of L[y] = 0, then the linear combination c1y1+· · ·+cnyn is also a solution of L[y] = 0.


    2. Relevant equations



    3. The attempt at a solution

    I think I don't understand what I need to find.
    What's the question here ?
    First I need to verify that it's a linear differential operator.
    Next, I need to show that
    L[c1y1+ c2 y2] = c1L[y1] + c2L[y2]

    and then, there is another part ?
    does this question have three parts ?

    I don't really understand how to approach this question.

    Would appreciate any help.

    Thanks,
    Roni.

    EDIT:

    ohh, to show that it's a linear differential operator, I just need to show that this one is true:
    L[c1y1+ c2 y2] = c1L[y1] + c2L[y2]

    correct ?

    now I'm trying to understand the second part of the question
     
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2

    Mark44

    Staff: Mentor

    Re: higher order linear equations

    What have you tried?
     
  4. Feb 24, 2010 #3
    Re: higher order linear equations

    Hello,

    I edited my post....
    I am still trying to understand the question...
     
  5. Feb 24, 2010 #4

    Mark44

    Staff: Mentor

    Re: higher order linear equations

    There are two parts.
    Show that L is a linear differential operator.
    Show that if y1, y2, ..., yn are solutions of L[y] = 0, then any linear combination of the yi's is also a solution of L[y] = 0. IOW, show that L[c1y1 + c2y2 + ... + cnyn] = 0.

    They're both pretty straightforward, especially if you keep in mind some fo the basic principles of differentiation; namely, (f + g)'(x) = f'(x)+ g'(x) and (cf)'(x) = cf'(x).
     
  6. Feb 24, 2010 #5
    Re: higher order linear equations

    oh right, I just had to read the definition :\

    thank you very much for your help, I appreciate it.
     
  7. Feb 24, 2010 #6

    Mark44

    Staff: Mentor

    Sure, you're welcome! A thank you goes a long way!
     
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