Show that $y_1(x) = \int_c^x f(x-t) R(t) dt$ is a particular solution of $L(y) = R$ This is problem 14 from Chapter 6.15 of Apostol Calculus, Volume 2 (p. 167):

If $L(y) = y'' + ay' + by$, where $a$ and $b$ are constants, let $f$ be that particular solution of $L(y) = 0$ satisfying the conditions $f(0) = 0$ and $f'(0) = 1$.  Show that a particular solution of $L(y) = R$ is given by the formula
  $$ y_1 (x) = \int_c^x f(x-t) R(t) \, dt  $$
  for any choice of $c$.

I'm not quite sure how to prove this.  Certainly we can evaluate 
$$f'' + af' + bf = 0$$
at $0$ to find $f''(0) = -a$, but that doesn't seem of much use.  We also know solutions of this equation are all of the form
$$f(x) = e^{-ax/2} (c_1 u_1(x) + c_2 u_2(x))$$
where $u_1$ and $u_2$ will depend on the value of $d = a^2 - 4b$.  Without constraints on $a$ and $b$, I'm not sure what to do with this.
 A: suppose $$Lf = R,\quad f(0) = 0,\quad f'(0) = 1.$$ look at $$y=\int_c^xf(x-t)R(t)\,dt$$
then by leibniz rule of differentiaton, we have 
$$\begin{align}y' &= \int_c^xf'(x-t)R(t)dt + f(0)R(x)\\
   &= \int_c^xf'(x-t)R(t)dt\\
y'' &= \int_c^xf''(x-t)R(t)dt+f'(0)R(x) \\
&= \int_c^xf''(x-t)R(t)dt+R(x) \\
Ly &= \int_c^x\left(af''(x-t) + bf'(x-t) + cf(x-t)\right)R(t)\, dt + R(x) \\
&= R(x).\end{align}$$
A: You can prove 
$$\tag{1} y' =   \int_c^xf'(x-t)R(t)\,dt + f(0)R(x)$$
Without use Leibniz rule of differentiation, since Apostol doesn't mention that theorem in the book.
Well, we have
$$y = \int_c^x f(x-t) R(t) \, dt $$
Make the following substitution:
$$u(x-c) = x-t \implies dt = -du(x-c)$$ 
$$\implies y(x) = \int_0^1 f(u(x-c))R(x-u(x-c))(x-c)du $$
$$ y'(x) = \frac{d}{dx}\int_0^1 f(u(x-c))R(x-u(x-c))(x-c)du $$
Then, the operator $\frac{d}{dx}$ can get into the integral. Thus,
$$y'(x) = \int_0^1 f'(u(x-c))uR(x-u(x-c))(x-c) + f(u(x-c))R'(x-u(x-c))(1-u)(x-c) + f(u(x-c))R(x-u(x-c))\,du$$
We return to the original variable of integration $t$:
$$\tag{2}y'(x) = \int_c^xf'(x-t)R(t)\frac{x-t}{x-c} + f(x-t)R'(t)\left(1-\frac{x-t}{x-c}\right)+ \frac{f(x-t)R(t)}{x-c} \,dt$$
We can use integration by parts in the second term of the right side:
$$ \int_c^x f(x-t)R'(t)\frac{t-c}{x-c}dt = $$
$$f(x-t)R(t)\frac{t-c}{x-c}\Big\vert_c^x - \int_c^xR(t)\left(-f'(x-t)\frac{t-c}{x-c}+ \frac{f(x-t)}{x-c}\right)dt =$$
$$f(0)R(x) + \int_c^xR(t)\left(f'(x-t)\frac{t-c}{x-c}- \frac{f(x-t)}{x-c}\right)dt  $$
Substituting this result in $(2)$, equation $(2)$ becomes $(1)$
$$ y'(x) = \int_c^xf'(x-t)R(t)\,dt + f(0)R(x) $$
With $f(0)=0$, we have
$$\tag{3} y'(x) = \int_c^xf'(x-t)R(t)\,dt$$
Since equation $(3)$ has the same form that $(1)$, it's straightforward that
$$y''(x) = \int_c^xf''(x-t)R(t)dt+f'(0)R(x) $$
After that, you can continue with abel's solution.
