How could we show the 2ODE: $$ f''(x) + \frac{2}{x}\, f'(x) + f(x) = 0 \tag{eq.(A)} $$
with $f(0) = 1$, $f'(0)=0$, $f''(0) = -\frac{1}{3}$ (the latter induced by L'Hopital), can be written in its integral form:
$$f(x) = 1 + \int^x_0 \left(\frac{s^2}{x} - s\right)f(s)ds \tag{eq.(B)}$$
It is clear to me that at $x = 0$ this is just $1$ satisfying the initial conditions of eq.(A), but how can we manipulate the integrator term when taking the derivatives for $f'(x)$ and $f''(x)$ ? How can we show the equivalence between $eq.(A)$ and $eq.(B)$. I am trying to understand this before using this scheme for computation.
Notice that $$ (xf(x))''=xf''(x)+2f'(x)=-xf(x) $$ so that for $g(x)=xf(x)$ the first order Taylor formula with integral remainder term reads as $$ g(x)=g(0)+g'(0)x+\int_0^x(x-s)g''(s)\,ds $$ Now insert the ODE and apply one partial integration $$ xf(x)=x-\int_0^x(x-s)sf(s)\,ds. $$ This is, after some minor rearrangements, the claimed integral formula.
One could of course also directly read off the solution for $g$ and find that $f(x)=\frac{\sin x}{x}$.