1. ## 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!

2. Originally Posted by hunkydory19
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!
$y'(x) = \int^x_0 y(t) \ dt + \int^1_x -y(t) \ dt$

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

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

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