Solve $y'(t)=t+1$ using Laplace transform I'm studying how to use Laplace transform to solve ODEs. 
I have thought to use this very simple example: $$y'(t)=t+1 \qquad y(0)=0$$
I can use integration to find $y(t)$: $$y(t)=\int (t+1) \ \  dt=\frac{1}{2} t^2+t+C$$
$C \in \mathbb{R} $, $C=0$ for the initial condition, so:
$$y(t)=\frac{1}{2} t^2+t$$

I consider $F(s)$ the Laplace transform of $f(t)=t+1$:
$$F(s)=\frac{1}{s}+\frac{1}{s^2}=\frac{1+s}{s^2}$$
I consider, now, the coefficient of the linear ODE:
$$H(s)=\frac{1}{s}$$
So: $$Y(s)=H(s) \ F(s)=\frac{1+s}{s^3}$$
Partial Fraction Decomposition of $Y(s)$:
$$Y(s)=\frac{1}{s^2}+\frac{1}{s^3}$$
Antitransform:
$$y(t)=t^2+t \ne \frac{1}{2} t^2+t $$
Where is the mistake?
Thanks!
 A: Your inverse Laplace transform of $\frac1{s^3}$  is wrong, it should be $\frac{t^2}2$ as you expect.
$$\mathcal L^{-1} \left\{ \frac1{s^{n+1}} \right\} = \frac{t^n}{\color{red}{n!}} $$
A: $$\mathscr{L}^{-1}\left\{\frac1{s^2}+\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\mathscr{L}^{-1}\left\{\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\color{blue}{\frac{1}{2}}\mathscr{L}^{-1}\left\{\frac{\color{blue}{2}}{s^3}\right\}=t+\frac12t^2$$
A: $$y'(t)=t+1\Longleftrightarrow$$
$$\mathcal{L}_t\left[y'(t)\right]_{(s)}=\mathcal{L}_t\left[t+1\right]_{(s)}\Longleftrightarrow$$
$$sy(s)-y(0)=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$

Use $y(0)=0$:

$$sy(s)-0=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$
$$sy(s)=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$
$$y(s)=\frac{\frac{1}{s^2}+\frac{1}{s}}{s}\Longleftrightarrow$$
$$y(s)=\frac{1+s}{s^3}\Longleftrightarrow$$
$$y(s)=\frac{1}{s^3}+\frac{1}{s^2}\Longleftrightarrow$$

Use $$\mathcal{L}_s^{-1}\left[\frac{1}{s^n}\right]_{(t)}=\frac{t^{n-1}}{\Gamma(n)}$$

$$y(s)=\frac{1}{s^3}+\frac{1}{s^2}\Longleftrightarrow$$
$$\mathcal{L}_s^{-1}\left[y(s)\right]_{(t)}=\mathcal{L}_s^{-1}\left[\frac{1}{s^3}+\frac{1}{s^2}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=\frac{t^2}{2}+t$$
