= Solution
\b[Laplace identities.] The one-sided <Laplace transform> is
$$
\boxed{F(p)=\int_0^\infty e^{-pt}f(t)\,dt,}
$$
for $\operatorname{Re}p$ sufficiently large that the integral converges. For a piecewise smooth <function> of exponential order, <integration by parts> gives
$$
\mathcal L[f'](p)=[e^{-pt}f(t)]_0^\infty+p\int_0^\infty e^{-pt}f(t)\,dt=\boxed{pF(p)-f(0)}.
$$
Differentiating under the <integral> gives
$$
F'(p)=-\int_0^\infty t e^{-pt}f(t)\,dt,\qquad\boxed{\mathcal L[tf](p)=-F'(p)}.
$$
Iteration also yields $\mathcal L[f'']=p^2F-pf(0)-f'(0)$. For real $k$ and $p>0$, taking the imaginary part of $(p-ik)^{-1}$ gives
$$
\boxed{\mathcal L[\sin kt](p)=\frac k{p^2+k^2}.}
$$
\b[Transformed differential equation.] Let $G=\mathcal L[g]$, using the initial values $g(0)=1,g'(0)=0$. Then
$$
\begin{aligned}
\mathcal L[tg'']&=-\frac d{dp}(p^2G-p)=-2pG-p^2G'+1,\\
2\mathcal L[g']&=2pG-2,\\
\alpha^2\mathcal L[tg]&=-\alpha^2G'.
\end{aligned}
$$
Summing the three contributions gives
$$
\boxed{G'(p)=-\frac1{p^2+\alpha^2}.}
$$
For $\alpha\ne0$, the condition $G(p)\to0$ as $p\to\infty$ fixes the integration constant:
$$
G(p)=\int_p^\infty\frac{dq}{q^2+\alpha^2}=\frac1\alpha\arctan\frac\alpha p\qquad(p>0).
$$
It is simpler to invert the derivative: $\mathcal L[tg]=-G'=1/(p^2+\alpha^2)=\mathcal L[\sin(\alpha t)/\alpha]$. Thus
$$
\boxed{g(t)=\frac{\sin(\alpha t)}{\alpha t}\quad(t>0),\qquad g(0)=1.}
$$
This is the <sinc function>. Its <Taylor series> begins $1-\alpha^2t^2/6+O(t^4)$, verifying both initial values. For a direct differential check set $h=tg=\sin(\alpha t)/\alpha$. Then $h''+\alpha^2h=tg''+2g'+\alpha^2tg=0$. For $\alpha=0$, take the continuous limit $\boxed{g(t)=1}$, which also follows from $G=1/p$.
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