Is there a way to solve for $f(x)$ when
$$ g(x)=\int_0^x dx' W(x,x') f(x') $$
If it weren't for the x-dependence in $W(x,x')$, I could write for example, $$ f(x)=\frac{1}{W(x)}\frac{\partial g(x)}{\partial x} $$
The best I can do is to split the (known) $W$ function such that I now have,
$$ g(x)=\int_0^x dx' P(x') f(x')+R(x)\int_0^x dx' Q(x') f(x') $$
where $P(x)$, $Q(x)$ and $R(x)$ are known functions, and again, I'd like to get an expression for $f(x)$ in terms of $g(x)$, $P(x)$, $Q(x)$, $R(x)$ and their derivatives.
In order to avoid confusion with the meaning of the apostrophes, $x'$ is remplaced by $t$ : $$ g(x)=\int_0^x P(t) f(t)dt+R(x)\int_0^x Q(t) f(t)dt $$
where $P(x)$, $Q(x)$ and $R(x)$ are known functions.
Now, the character apostrophe means derivative : $$ g'(x)= P(x) f(x) +R(x)Q(x)f(x) +R'(x)\int_0^x Q(t) f(t)dt $$ $$ \frac{g'}{R'}= \frac{P+RQ}{R'}f +\int_0^x Q(t) f(t)dt $$ $$ \left(\frac{g'}{R'}\right)'= \left(\frac{P+RQ}{R'}f\right)' +Q(x) f(x) $$ $$ \left(\frac{g'}{R'}\right)'= \left(\frac{P+RQ}{R'}\right)f'+\left(\frac{P+RQ}{R'}\right)'f +Q f $$
Let : $ \begin{cases} u(x)=\left(\frac{P+RQ}{R'}\right)'+Q \\ v(x)=\frac{P+RQ}{R'} \\ w(x)=\left(\frac{g'}{R'}\right)' \end{cases} $
Since $P(x)$, $Q(x)$ and $R(x)$ are known functions. $u(x)$ , $v(x)$ and $w(x)$ are known functions. $$ u(x)f(x)+v(x)f'(x)=w(x) $$ Solving this first order linear ODE leads to $f(x)$.