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Math Help - integral equation

  1. #1
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    integral equation

    This is my first attempt at an integral equation so I don't really know what's going on.

    Suppose that y = y(x) satisfies the integral equation

     y(x) = 1 + \int^1_0 |x - t| y(t) dt

    Find  y''(x) .

    I've split the integral at t = x to get:

     y(x) = 1 + \int^x_0 (x - t) y(t) dt + \int^1_x (t - x) y(t) dt

    Using Leibnitz's integral rule, differentiating once gives:

     y'(x) = \int^x_0 y(t) dt + y(x) + \int^1_x -y(t) dt - y(x)

    So  y'(x) = \int^x_0 y(t) \ dt + \int^1_x -y(t) \ dt

    Differentiating again,

    y''(x) = \int^x_0 y(x) \ dt + \int^1_x -y(x) \ dt

    So now everything cancels?

    I know the answer is y''(x) = 2y(x) but I just can't see how to get there from here...

    Thanks in advance for any help, I've literally been working on this for days!
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  2. #2
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    Quote Originally Posted by hunkydory19 View Post
    This is my first attempt at an integral equation so I don't really know what's going on.

    Suppose that y = y(x) satisfies the integral equation

     y(x) = 1 + \int^1_0 |x - t| y(t) dt

    Find  y''(x) .

    I've split the integral at t = x to get:

     y(x) = 1 + \int^x_0 (x - t) y(t) dt + \int^1_x (t - x) y(t) dt

    Using Leibnitz's integral rule, differentiating once gives:

     y'(x) = \int^x_0 y(t) dt + y(x) + \int^1_x -y(t) dt - y(x)

    So  y'(x) = \int^x_0 y(t) \ dt + \int^1_x -y(t) \ dt

    Differentiating again,

    y''(x) = \int^x_0 y(x) \ dt + \int^1_x -y(x) \ dt

    So now everything cancels?

    I know the answer is y''(x) = 2y(x) but I just can't see how to get there from here...

    Thanks in advance for any help, I've literally been working on this for days!
     y'(x) = \int^x_0 y(t) \ dt + \int^1_x -y(t) \ dt

    = \int^x_0 y(t) \ dt - \int^1_x y(t) \ dt

    = \int^x_0 y(t) \ dt + \int^x_1 y(t) \ dt

     \Rightarrow y''(x) = y(x)  + y(x)
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