IE19

Convert to an integral equation: $y'' + y = \cos(x), y(0)=0, y'(0)=0\,$

$y'' = -y +\cos(x)\,$

$\int_0^x y''(\xi)d\xi = \int_0^x -y(\xi) + \cos(\xi) d\xi \,$

$y'(x) - y'(0) = \int_0^x -y(\xi) + \cos(\xi) d\xi\,$

$y'(x) = -\int_0^x y(\xi)d\xi + \left[ \sin(x) - \sin(0)\right]\,$

$\int_0^x y'(\xi) d\xi = -\int_0^x \!\!\! \int_0^s y(\xi) d\xi ds + \int_0^x \sin(\xi) d\xi\,$

$y(x) - y(0) = -\int_0^x y(\xi)(x-\xi)d\xi + \left[ -\cos(x)+\cos(0)\right]\,$

The solution is

$y(x) = -\int_0^x y(\xi) (x-\xi)d\xi + (1-\cos(x))\,$

To check the solution, start differentiating.

$y'(x) = -\frac{d}{dx} \int_0^x y(\xi)(x-\xi)d\xi + \sin(x)\,$

$y'(x) = -\int_0^x \frac{\partial}{\partial x} y(\xi)(x-\xi)d\xi + \sin(x)\,$

$y'(x) = -\int_0^x y(\xi) d\xi + \sin(x)\,$

$y''(x) = -\frac{d}{dx} \int_0^x y(\xi) d\xi + \cos(x)\,$

$y''(x) = -y(x) + \cos(x)\,$

$y''(x) + y(x) = \cos(x)\,$