= Solution
The one-sided <Laplace transform> is $\bar f(p)=\int_0^\infty e^{-pt}f(t)dt$. Its <Bromwich inversion formula> is $f(t)=(2\pi i)^{-1}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{pt}\bar f(p)dp$, with an upward vertical contour to the right of all singularities and within the appropriate convergence half-plane.
To derive the <convolution theorem for Laplace transforms>, transform $x(t)=\int_0^tf(\tau)g(t-\tau)d\tau$. With $u=t-\tau\geq0$, the triangular integration region becomes a quadrant. <Fubini's theorem> gives $\bar x(p)=\int_0^\infty\int_0^\infty e^{-p(u+\tau)}f(\tau)g(u)du\,d\tau=\bar f(p)\bar g(p)$, under the usual absolute convergence hypotheses. Inversion gives the required convolution formula.
For the differential equation, the transformed initial conditions give $(p^2-1)\bar y=\bar g+p+1$, hence $\bar y=\bar g/(p^2-1)+1/(p-1)$. Since $\mathcal L[\sinh t]=1/(p^2-1)$ for $\operatorname{Re}p>1$, the convolution yields
$$
\boxed{y(t)=e^t+\int_0^t\sinh(t-\tau)g(\tau)d\tau.}
$$
The integral and its first derivative vanish at zero, so the homogeneous term supplies both initial values. Differentiating twice gives the forcing through the unit derivative of the hyperbolic-sine kernel at zero.
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