I have a linear ODE that has terms that are shifted, for example
$$\frac{d^k f(x-n)}{dx^k}$$
from a general equation
$$f(x)=\sum_{k=0}^{n} a_k \frac{d^k f(x-n)}{dx^k}$$
where $a_k$ are constants. Since the ODE was linear i assumed a Laplace transform would be easy to apply to this, but a laplace transform for something like $\frac{df(t-n)}{dt}$ gives
$$\int_{0}^{\infty} \frac{df(t-n)}{dt} e^{-st}dt=\int_{n}^{\infty} \frac{df(t)}{dt} e^{-s(t+n)}dt$$
$$=e^{-ns}[\int_{n}^{\infty} \frac{d}{dt}(f(t) e^{-st})dt - s\int_{n}^{\infty}f(t)e^{-st}]$$
$$=e^{-ns}[-f(n^{-})e^{-ns}-s(F(s)-\int_{0}^{n} f(t)e^{-st}dt)]$$
Is this correct? What about the last term, is there an evaluation for it or is it computer only? Is there an easier way to solve the above ODE?
The Laplace transforms applies to functions that are assumed to be zero on $t < 0$. The shifted function $f(t-t_0)$ would be zero on $t < t_0$. The time-shifting property says
$$ \int_0^\infty f^{(n)}(t-t_0)e^{-st}dt = \int_{t_0}^\infty f^{(n)}(t)e^{-s(t+t_0)}dt = e^{-t_0s} \int_0^\infty f^{(n)}(t)e^{-st}dt $$
where the last integral is just the Laplace transform of $f^{(n)}(t)$