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Differential equations - algebraic properties of solutions

  1. Mar 6, 2007 #1
    Show that the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear. (Hint: Use integration by parts.)



    Attempt at the solution:
    A[y](t) = integ(s(y(s))ds) [integ from t to a]

    Using integration by parts:
    Let u = s
    du = ds

    let dv = y(s)
    v = 1/2 y^2(s)

    so,
    A[y](t) = s(1/2y^2(s)) - integ(ds 1/2y^2(s)) [i'm not sure how to change the integration limits]
    A[y](t) = 1/2s y^2(s) - [1/2 (1/3 y^3(s))] (from ___ to ___)
    and just sub in the integrating limits and simplify?
    am i doing something wrong.
     
  2. jcsd
  3. Mar 6, 2007 #2

    cristo

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    Well, firstly, what is the definition of a linear operator?

    Does this mean s times the function y(s)? i.e. the integral [tex]\int_t^asy(s)ds[/tex].

    y is a function, and so you can't integrate it like this! I would let dv=s ds and u=y(s). But then, I don't know the definition of a linear operator that you are using, and so do not know what you're trying to prove!
     
  4. Mar 6, 2007 #3

    HallsofIvy

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    What in the world are you doing? No one asked you to actually integrate!

    A transformation, T, is linear if and only if T(ay1+ by2)= aT(y1)+ bT(y2).

    What can you do with [itex]\int s(ay_1+ by_2)ds[/itex]?
     
  5. Mar 7, 2007 #4
    i integrated since the hint was to use integration of parts
    i have no idea what i am doing
     
  6. Mar 7, 2007 #5
    i'm looking at the integral you posted hallsofivy
    you can multiply the s in and split it into two integrals???
     
  7. Mar 7, 2007 #6

    radou

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    You can, since linearity is a property of intergals.
     
  8. Mar 7, 2007 #7
    can i integrate it then???
    or is that not the point.
    im not quite sure what i'm doing, ive read the class notes. i dont really understand
     
  9. Mar 10, 2007 #8
    anybody know what i can do
    this is relaly causing me trouble
     
  10. Mar 10, 2007 #9

    HallsofIvy

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    First go back and check the problem! If you are really to "Show that the operator A defined by A[y](t) = integ(s(y(s))ds) [integration from t to a] is linear" then do as Cristo suggested originally- check the definition of "linear" and show that it works here. I can not imagine how one would "use integration by parts" to do that.
     
  11. Mar 11, 2007 #10
    definition of a linear operator means it must satisfy A[cy] = cA[y] and A[y1+y2]=A[y1] + A[y2]

    i would probably know how to do this if it wasn't an integral and i had to show it is linear, but i really do not know where to start.
     
  12. Mar 11, 2007 #11
    i think i have an idea how to do it now, without the integration by parts.. but my prof said that we are suppose to do it using that, i have no idea how.

    but this is what im doing right now (it seems right to me but correct me if i'm wrong):
    A[y] = integ(s(y(s))ds)

    A[cy] = integ(s(cy) ds)
    A[cy] = c integ(s y(s) ds)
    A[cy] = c A[y]
    so i satisfied property 1

    A[y1 + y2] = integ( s(y1+y2) ds)
    A[y1 + y2] = integ( sy1 ds) + integ( sy2 ds)
    A[y1 + y2] = A[y1] + A[y2]
    so i satisfied property 2

    so it is a linear operator?
    but since i need integration by parts, i've no idea what to do.
     
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