IEVS1

$u(x) = f(x) + \int_0^x k(x-y)u(y)\,dy$

The integral is of convolution type. Take the Laplace transform of both sides.

$\mathcal{L}u = \mathcal{L}f + \mathcal{L}k\mathcal{L}u\,$

$\mathcal{L}u = \frac{\mathcal{L}f}{1-\mathcal{L}k}\,$

So the answer is the inverse Laplace transform of the right hand side.

$u(x) = \mathcal{L}^{-1}\left(\frac{\mathcal{L}f}{1-\mathcal{L}k}\right)\,$