= Solution
{c}
Write $F_j(p)=\mathcal Lf_j(p)$, using the square-root branch positive on real $p>0$. Integrating the differential equation and using boundedness at zero gives
$$
xf_1(x)=\int_0^x tf_0(t)\,dt.
$$
The integration constant is zero since $xf_1(x)\to0$. Continuity of $f_0$ then shows $f_1(x)=O(x)$, so $f_1$ extends continuously with $f_1(0)=0$.
The <Laplace transform of a derivative> and <differentiation under the integral sign> give
$$
-pF_1'(p)=-F_0'(p),\qquad F_1'(p)=\frac{F_0'(p)}p=-\frac1{(p^2+1)^{3/2}}.
$$
Integrating and using $F_1(p)\to0$ as real $p\to\infty$ fixes the constant:
$$
\boxed{F_1(p)=1-\frac p{\sqrt{p^2+1}}.}
$$
The <initial value theorem for Laplace transforms> gives $f_0(0)=\lim_{p\to\infty}pF_0(p)=1$. It is tempting to apply the derivative rule immediately to $f_0$, but its stated regularity is only continuity. To justify that step, define $h(x)=1-\int_0^x f_1(t)\,dt$. Then
$$
\mathcal Lh(p)=\frac{1-F_1(p)}p=\frac1{\sqrt{p^2+1}}=F_0(p).
$$
Uniqueness of the <Laplace transform> gives $h=f_0$, and hence $f_0$ is continuously differentiable and
$$
\boxed{f_1=-f_0'.}
$$
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