How to solve DE that relate values of derivatives at different points?

I try to solve for the specific function $f(x) = \frac{2-2a}{x-1} \int_0^{x-1} f(y) dy + af(x-1)$

It looks similar to the function used to find the Renyi's parking constant because it came out from a simple generalization of that problem.

The skill I have gained in my differential class can't even solve $f(x) = f'(x-1)$

I'm not looking for anyone to solve it. I just want to know the techniques for solving DE where functions and it's derivatives are evaluated at different points.(What's the terminology for this kind of DE?)

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– kennytm Sep 4 '10 at 17:50
Can't $f(y + 1) = f'(y)$ be solved by just substituting $y = e^{ry}$ so you get $e^r = r$ (You need a special function to solve that). So then you appeal to some uniqueness and existence theorem. So the function that maps $y(t)$ to $y(t + 1)$ must be continuous and so must its derivative be. – Jonas Teuwen Sep 4 '10 at 17:54
Differential-difference equation. (Google it.) – whuber Sep 4 '10 at 18:51
The first one, naturally is a delay integro-differential equation (DIDE). ;) – J. M. Sep 4 '10 at 22:07
whuber: My understanding of differential-difference equations is that they look something like this formula for the derivative of a Bessel function: $\frac{\mathrm{d}}{\mathrm{d}x}C_n(x)=\frac12(C_{n-1}(x)-C_{n+1}(x))$, and it can be shown that only the two solutions to the Bessel DE are the solutions to this differential-difference equation. – J. M. Sep 4 '10 at 22:43

Let $u=x-1$ ,

Then $f(u+1)=\dfrac{2(1-a)}{u}\int_0^uf(y)~dy+af(u)$

$\dfrac{2(a-1)}{u}\int_0^uf(y)~dy=af(u)-f(u+1)$

$2(a-1)\int_0^uf(y)~dy=auf(u)-uf(u+1)$

$2(a-1)f(u)=auf'(u)+af(u)-uf'(u+1)-f(u+1)$

$uf'(u+1)+f(u+1)-auf'(u)+(a-2)f(u)=0$

Let $f(u)=\int_Ce^{us}K(s)~ds$ ,

Then $u\int_Cse^{(u+1)s}K(s)~ds+\int_Ce^{(u+1)s}K(s)~ds-au\int_Cse^{us}K(s)~ds+(a-2)\int_Ce^{us}K(s)~ds=0$

$\int_Cs(e^s-a)e^{us}K(s)~d(us)+\int_C(e^s+a-2)e^{us}K(s)~ds=0$

$\int_Cs(e^s-a)K(s)~d(e^{us})+\int_C(e^s+a-2)e^{us}K(s)~ds=0$

$[s(e^s-a)e^{us}K(s)]_C-\int_Ce^{us}~d(s(e^s-a)K(s))+\int_C(e^s+a-2)e^{us}K(s)~ds=0$

$[s(e^s-a)e^{us}K(s)]_C-\int_C(s(e^s-a)K'(s)+((s+1)e^s-a)K(s))e^{us}~ds+\int_C(e^s+a-2)e^{us}K(s)~ds=0$

$[s(e^s-a)e^{us}K(s)]_C-\int_C(s(e^s-a)K'(s)+(se^s-2(a-1))K(s))e^{us}~ds=0$

$\therefore s(e^s-a)K'(s)+(se^s-2(a-1))K(s)=0$

$s(e^s-a)K'(s)=-(se^s-2(a-1))K(s)$

$\dfrac{K'(s)}{K(s)}=-\dfrac{e^s}{e^s-a}+\dfrac{2(a-1)}{s(e^s-a)}$

$\int\dfrac{K'(s)}{K(s)}ds=\int\left(-\dfrac{e^s}{e^s-a}+\dfrac{2(a-1)}{s(e^s-a)}\right)ds$

$\ln K(s)=-\ln(e^s-a)+\int_k^s\dfrac{2(a-1)}{r(e^r-a)}dr+c_1$

$K(s)=\dfrac{ce^{\int_k^s\dfrac{2(a-1)}{r(e^r-a)}dr}}{e^s-a}$

$\therefore f(u)=\int_C\dfrac{ce^{us+\int_k^s\dfrac{2(a-1)}{r(e^r-a)}dr}}{e^s-a}ds$

$f(x)=\int_C\dfrac{ce^{xs+\int_k^s\dfrac{2(a-1)}{r(e^r-a)}dr}}{e^s-a}ds$

But since the above procedure in fact suitable for any complex number $s$,

$\therefore f_n(x)=\int_{a_n}^{b_n}\dfrac{c_ne^{x(p_n+q_ni)t+\int_{k_n}^{(p_n+q_ni)t}\dfrac{2(a-1)}{r(e^r-a)}dr}}{e^{(p_n+q_ni)t}-a}d((p_n+q_ni)t)$

For some $x$-independent real number choices of $a_n$ , $b_n$ , $p_n$ , $q_n$ and $k_n$ such that:

$\displaystyle\lim_{t\to a_n}(p_n+q_ni)te^{x(p_n+q_ni)t+\int_{k_n}^{(p_n+q_ni)t}\dfrac{2(a-1)}{r(e^r-a)}dr}=\lim_{t\to b_n}(p_n+q_ni)te^{x(p_n+q_ni)t+\int_{k_n}^{(p_n+q_ni)t}\dfrac{2(a-1)}{r(e^r-a)}dr}$

$\int_{a_n}^{b_n}\dfrac{e^{x(p_n+q_ni)t+\int_{k_n}^{(p_n+q_ni)t}\dfrac{2(a-1)}{r(e^r-a)}dr}}{e^{(p_n+q_ni)t}-a}d((p_n+q_ni)t)$ converges

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