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Can someone explain this to me please?

  1. Nov 2, 2011 #1
    I'm asking this integral equation (I'm not sure if it's an integral equation or not by it's a problem in my ODE book and because it has an integral in it I called it that way). anyways, this is the problem:

    [tex]y=\int^{x}_{1}ty(t)dt[/tex]

    I differentiated y with respect to x and I turned that equation into this ODE: [tex]y'=xy[/tex]
    Solving this ODE yields [tex]y=Ce^{x^2/2}[/tex]

    But from the definition of y, it is clear that y(1)=0 while my solution suggests that y=e1/2.

    Then I substituted y(t)=Cex2 in the original equation and I obtained:
    [tex]y=\int^{x}_{1}tCe^{t^2/2}dt → y=C(e^{t^2/2})|^{x}_{1}→y=C(e^{x^2/}-e^{1/2})[/tex]
    And in this case y(1) is indeed equal to 0.

    Would someone explain why the y that is obtained from the ODE solution tells me that y(1)≠0? What's wrong in my solution?
     
  2. jcsd
  3. Nov 3, 2011 #2
    Any ideas on this matter will be appreciated. Are my questions really that hard that they usually get no responses back on physics forum or there's a conspiracy against me? lol.
     
  4. Nov 5, 2011 #3
    The condition y(1)=0 is given in order to find constant.THE PURPOSE OF GIVING BOUNDARY CONDITIONS IS TO OBTAIN CONSTANTS AFTER INTEGRATION.
     
  5. Nov 5, 2011 #4
    "In order to solve this you need to know what value of t... x and 1 correspond to and then you can go from there. If you keep it in terms of t, you get the same function, but t^2, x^2.

    Knowing that x=t, at y=?, and t=0, y=?, you can transform accordingly .

    This an ODe, you you don't have what y=, when x=t, t=0, etc.

    YS
     
  6. Nov 5, 2011 #5
    I guess I've already found an explanation to this.
    C must be zero. probably that's the only answer this integral equation can have. other answers would lead to contradiction.
     
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