= Solution
A sufficient condition for existence of the <Laplace transform> is that $y$ be piecewise continuous on each bounded interval and have <exponential order>: $|y(t)|\le Ce^{at}$ for sufficiently large $t$. The defining integral then converges absolutely for real $p>a$, and more generally defines a <holomorphic function> for complex $p$ with $\operatorname{Re}p>a$. The boundary $p=a$ may or may not be included, depending on $y$.
Write $Y(p)=\mathcal L[y](p)$. To solve the <differential equation> by <Laplace transform>, initially seek a sufficiently regular solution with $y(0)=1$ and finite $y'(0)=b$. The <Laplace transform of a derivative> gives
$$
\mathcal L[y']=pY-1,\qquad \mathcal L[y'']=p^2Y-p-b.
$$
Using $\mathcal L[ty]=-Y'$ from part (i), the transformed <differential equation> becomes
$$
-(2pY+p^2Y'-1)-Y-pY'-(pY-1)-Y=0,
$$
or
$$
p(p+1)Y'+(3p+2)Y=2.
$$
After division by $p(p+1)$, an <integrating factor> is $p^2(p+1)$, since $(3p+2)/(p(p+1))=2/p+1/(p+1)$. Consequently
$$
\frac d{dp}\bigl[p^2(p+1)Y\bigr]=2p,\qquad
Y=\frac1{p+1}+\frac{C}{p^2(p+1)}.
$$
For a bounded $y$, $|Y(p)|\le\|y\|_\infty/p$ for $p>0$. Letting $p\downarrow0$ forces $C=0$, an instance of how <boundedness selects a Laplace transform at a singular endpoint>. By the exponential formula in part (iii), this yields
$$
\boxed{y(t)=e^{-t}.}
$$
Direct substitution checks the <differential equation>, boundedness and the required limit.
To complete the uniqueness argument, no unproved endpoint regularity is needed. The two functions $e^{-t}$ and $t-1$ solve the homogeneous <linear differential equation> on $(0,\infty)$, and their <Wronskian> is $te^{-t}\ne0$. They form a <fundamental system of solutions> on that interval, so every classical solution is $Ae^{-t}+B(t-1)$. Boundedness forces $B=0$ and the limit at zero forces $A=1$. Thus \b[the bounded solution with the specified limit is unique]. The limit alone would not suffice: $e^{-t}+C(t-1+e^{-t})$ has the same limit for every $C$, but is unbounded when $C\ne0$.
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