= Solution
{c}
Taking the <Laplace transform>, using <linearity> and the preceding three formulas, gives
$$
p\bar x(p)-x(0)-\bar x(p)=\frac2{p^2}-\frac2{p^3}
=\frac{2(p-1)}{p^3}.
$$
Since $x(0)=0$, division by $p-1$ gives $\bar x(p)=2/p^3$ for $p\ne1$ in the domain of convergence. There is no genuine singularity at $p=1$: <continuity> of the <Laplace transform> there supplies the same value. One can also first work with $\operatorname{Re}p>1$, where division is harmless.
The formula in part (iii) identifies $2/p^3$ as the <Laplace transform> of $t^2$. By <uniqueness of the Laplace transform>,
$$
\boxed{\bar x(p)=\frac2{p^3},\qquad x(t)=t^2\quad(t\geq0).}
$$
As a direct check, $\dot x-x=2t-t^2$ and $x(0)=0$. Any two solutions of this <initial value problem> differ by $Ce^t$ and have the same initial value, forcing $C=0$. This also verifies uniqueness independently of transform inversion.
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