Volterra equation of convolution type

Remarks

The integral is the convolution $(k * x)(t) = \int_0^t k(t - s) \, x(s) \, ds$ of functions on $[0, \infty)$, so the second kind is written $x = f + \lambda \, k * x$. Example: $x(t) = 1 + \int_0^t e^{-(t - s)} x(s) \, ds$, with $k(u) = e^{-u}$. The function $x(t) = 1 + t$ solves it: $\int_0^t e^{-(t - s)} (1 + s) \, ds = t$, by the substitution $u = t - s$ and $\int_0^t e^{-u}(1 + t - u) \, du = t$. Non-example: $x(t) = 1 + \int_0^t t s \, x(s) \, ds$ has kernel $K(t, s) = ts$, which is not a function of $t - s$ alone. The Laplace transform method solves convolution equations on $[0, \infty)$ when $k$ and $f$ have Laplace transforms.