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Simple integral question

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data

    solve the integral equation:

    [tex]\int_0^2 t y(t)dt = 3[/tex]

    2. Relevant equations



    3. The attempt at a solution

    If we differentiate we get:
    0 = xy(x)
    which means y=0.
    but this is the wrong answer and i think this is the wrong approach.
     
  2. jcsd
  3. Jan 31, 2010 #2
    Your proposed calculation is not correct (as you know of course), because both sides of the integral equation are simply constant numbers, i.e. they don't depend on any variable with respect to which you can differentiate.

    It would have been correct if the upper integration limit had been t, but here it is simply 2.

    So you must think a bit differently on this one.

    Torquil
     
  4. Jan 31, 2010 #3
    i cant think of a different method
    :(
     
  5. Jan 31, 2010 #4
    Something about that problem statement appears to be suspicious... Solving an integral equation would normally mean solving for the general form of y(x), and applying boundary conditions to get it into the exact form you want. Here, there aren't enough constraints to uniquely specify y; in fact, there aren't even enough constraints to specify the FORM of y! You could write y = C, y = Ct, y = Csin(t), y = Ce^t, etc, and then solve for C in each of those by doing the integral to get different expressions for y. In short, this problem makes no sense.
     
  6. Jan 31, 2010 #5
    Ok, but it's not really "a method".

    Hint: There is an infinite number of different functions y(t) that will be able to solve the equation. In fact, almost any integrable function will be able to do it, if it is correctly normalized.

    The problem only says to "solve it", so any solution will do.

    Torquil
     
  7. Jan 31, 2010 #6
    oh right, i see.
    like y=3/2 can be a solution for an example.

    so there's infinitely many solutions.
     
  8. Jan 31, 2010 #7
    Almost...

    EDIT: Yes! :-)
     
  9. Jan 31, 2010 #8
    thanks
    :)
     
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