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

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

Find $\displaystyle y''(x) $.

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

$\displaystyle 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:

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

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

Differentiating again,

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

So now everything cancels?

I know the answer is $\displaystyle 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!