 Solution

Write $F_j(p)=\mathcal{L}{f}_j(p)$, using the square-root branch positive on real $p>0$. Integrating the differential equation and using boundedness at zero givesThe integration constant is zero since $x{f}_1(x)\to 0$. 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 giveIntegrating and using $F_1(p)\to 0$ as real $p\to \infty$ fixes the constant:The initial value theorem for Laplace transforms gives $f_0(0)={\lim }_{p\to \infty }p{F}_0(p)=1$. It is tempting to apply the derivative rule immediately to $f_0$, but its stated regularity is only continuity. To justify that step, define $h(x)=1-{\int }_0^x{f}_1(t)dt$. ThenUniqueness of the Laplace transform gives $h=f_0$, and hence $f_0$ is continuously differentiable and