I took this Cauchy problem \begin{align} &y''-3y'+2y=4t+8,\\ &y(0) = 2,\\ &y'(0) =7 \end{align}
And transformed it through Laplace transform $$ Y(s)=\frac{4-8s}{s^2(s-1)(s-2)} + \frac{1+2s}{(s-1)(s-2)} $$
Simplified to $$ Y(s) =\frac{2}{s^2} + \frac{1}{s-1} + \frac{2}{s-2} $$
Inversed back to $$ y(t) = 2t + e^t + 2e^{2t} $$
However, the solution states that there should also be a constant of $-1$ to the function $y(t).$ I can not figure out where I lost a $1/s$ in my calculations. Transcript here.
Here, $s$ has a power $2$ in the denominator(in the first fraction). So besides $\dfrac{A}{s^2}$ you've to consider another term $\dfrac{A'}{s}$.
(for reference)
On solving, the first fraction becomes
$\dfrac{-1}{s} + \dfrac{2}{s^2} + \dfrac{4}{s-1} - \dfrac{3}{s-2}$.
Now you have that $-1$.