= Solution
Differentiating the transform with respect to $p$ gives
$$
\boxed{\mathcal L[ty](p)=-\frac d{dp}\overline y(p).}
$$
Applying this to $\mathcal L[1]=1/p$ also gives $\boxed{\mathcal L[t]=1/p^2}$ for $\operatorname{Re}p>0$.
Now put $Y=\overline y$, $y_0=y(0)$ and $v_0=\dot y(0)$. Transforming each term of the differential equation, and using the <derivative> identities, gives
$$
-\frac d{dp}(p^2Y-py_0-v_0)-\frac d{dp}(pY-y_0)+(pY-y_0)+Y=\frac1{p^2}.
$$
The initial-value terms cancel, leaving
$$
-p(p+1)Y'-pY=\frac1{p^2},\qquad
\frac d{dp}\{(p+1)Y\}=-\frac1{p^3}.
$$
Integrating once yields $(p+1)Y=C+1/(2p^2)$. Partial fractions give
$$
Y=\frac C{p+1}+\frac1{2p^2(p+1)}
=\frac{C+1/2}{p+1}-\frac1{2p}+\frac1{2p^2}.
$$
The transform pairs therefore imply, on writing $A=C+1/2$,
$$
\boxed{y(t)=Ae^{-t}+\frac{t-1}{2}.}
$$
For the final parameter count, the differential equation is the <derivative> of $t(\dot y+y)$: it says $[t(\dot y+y)]'=t$. Integrating gives $t(\dot y+y)=t^2/2+B$. A regular solution at $t=0$ has finite $y$ and $\dot y$, forcing $B=0$. Thus $\dot y+y=t/2$ is first order and leaves only one constant. Equivalently, the original equation at zero requires $\dot y(0)+y(0)=0$, so the two usual initial values cannot be chosen independently. The displayed solution satisfies this constraint and verifies the original equation directly.
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