Solving $u(t)=f(t)+a\int_0^t u(s)ds$ by assuming $f$ is differentiable Original

Let $a\in\mathbb R$ and $f\colon [0,1]\to\mathbb R$ a continuous
  function. Solve the integral equation $$u(t)=f(t)+a\int_0^t
 u(s)\mathrm ds,\quad t\geq 0$$ and find an explicit formula for the
  solution.

First assume $f$ is differentiable. Then
$$u'(t)=f'(t)+au(t)$$
$$\to u'(t)-au(t)=f'(t)$$
$$\to u'(t)e^{-at}-au(t)e^{-at}=f'(t)e^{-at}$$
$$\to (u(t)e^{-at})'=f'(t)e^{-at}$$
$$\to (u(t)e^{-at})=\int f'(t)e^{-at} dt$$
$$\to u(t)=e^{at}\int f'(t)e^{-at} dt$$
$$\to u(t)=e^{at}[e^{-at}f(t)+a\int f e^{-at} dt]$$
$$\to u(t)=f(t)+ae^{at}\int f e^{-at} dt$$
Second, use the above as an Ansatz to show that the solution for when $f$ is differentiable is the same as when $f$ isn't.
Is this merely coincidence? Or is there some rule like in solving integral equations with conditions of blah blah blah, the Ansatz is of when $f$ is differentiable under the conditions of blah blah blah, is right?
 A: Here's a solution which does not assume $f(t)$ is differentiable; it answers the question as originally asked.
The key idea of the ansatz here is to multiply through by $e^{-at}$, which serves as an integrating factor and leads to a primitive $e^{-at}\int_0^t u(s) \; ds$ which occurs on the left of equation (3) below.  This approach is very similar to the standard method for solving the differential equation
$u'(t) = au(t) + g(t), \tag 0$
which occurs in many elementary texts on ordinary differential equations.
Starting with
$u(t) = f(t) + a \displaystyle \int_0^t u(s) \; ds, \tag 1$
we have
$u(t) - a \displaystyle \int_0^t u(s) \; ds = f(t); \tag 2$
observe that
$\left (e^{-at} \displaystyle \int_0^t u(s) \; ds \right )' = e^{-at} u(t) - a e^{-at} \displaystyle \int_0^t u(s) \; ds$
$= e^{-at} \left ( u(t) - a \displaystyle \int_0^t u(s) \; ds \right ) = e^{-at} f(t); \tag 3$
therefore,
$e^{-at} \displaystyle \int_0^t u(s) \; ds  = e^{-at} \displaystyle \int_0^t u(s) \; ds -  e^{-a\cdot 0} \displaystyle \int_0^0 u(s) \; ds$
$\displaystyle \int_0^t \left (e^{-aw} \displaystyle \int_0^w u(s) \; ds \right )' \; dw = \displaystyle \int_0^t e^{-as}f(s) \; ds, \tag 4$
whence
$\displaystyle \int_0^t u(s) \; ds  = e^{at} \displaystyle \int_0^t e^{-as}f(s) \; ds, \tag 4$
we now substitute (4) into (1) to obtain
$u(t) = f(t) + a e^{at} \displaystyle \int_0^t e^{-as}f(s) \; ds, \tag 5$
as per request.
A: You could do it like this: Let 
$$ v(t) = \int_0^t u(s)\ ds $$
Then by the FTOC, $v'(t) = u(t)$, so you have
$$ v'(t) - av(t) = f(t) $$
Using the integrating factor
$$ (e^{-at}v(t))' = e^{-at}f(t) $$
Integrate both sides, knowing $v(0) = 0$
$$ v(t) = e^{at}\int_0^t e^{-as}f(s)\ ds $$
Finally, differentiate again
$$ u(t) = v'(t) = f(t) + ae^{at}\int_0^te^{-as}f(s)\ ds $$
