= Solution
Put $G(p)=\mathcal L[J_0](p)$, initially for real $p>0$. Besides the <Laplace transform of a derivative>, use multiplication by the original variable in the form
$$
\mathcal L[xg(x)](p)=-\frac d{dp}\mathcal L[g](p).
$$
Transforming the <Bessel differential equation>, with its given <initial conditions>, yields
$$
-\frac d{dp}\bigl(p^2G-p\bigr)+(pG-1)-G'=0.
$$
The constant terms cancel, leaving $(1+p^2)G'+pG=0$. This first-order <ordinary differential equation> integrates to
$$
G(p)=\frac C{\sqrt{1+p^2}}.
$$
The <initial value theorem for Laplace transforms> gives $\lim_{p\to\infty}pG(p)=J_0(0)=1$, so $C=1$. Thus the <Laplace transform of the Bessel function J0> is
$$
\boxed{\overline{J_0}(p)=\frac1{\sqrt{1+p^2}},\qquad p>0.}
$$
For complex $p$ with $\operatorname{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.
For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2)^{-3/2}$. On the other hand, the same multiplication rule gives
$$
\mathcal L[xJ_0(x)](p)=-G'(p)=\frac p{(1+p^2)^{3/2}}.
$$
Taking an <integral> from zero divides the <Laplace transform> by $p$. By <uniqueness of the Laplace transform>,
$$
\boxed{f(x)=\int_0^x tJ_0(t)\,dt.}
$$
A direct check also explains the cancellation. This expression has $f'=xJ_0$ and $f''=J_0+xJ_0'$. The <Bessel differential equation> says $(xJ_0')'=-xJ_0$, so integration from zero gives $xJ_0'=-\int_0^x tJ_0(t)\,dt=-f$. Hence $f''+f=J_0$, and both <initial conditions> hold.
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