Solve $y'(t)=t+1$ using Laplace transform

Question

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!

Answer 1

Your inverse Laplace transform of $\frac1{s^3}$ is wrong, it should be $\frac{t^2}2$ as you expect.

$$\mathcal L^{-1} \left\{ \frac1{s^{n+1}} \right\} = \frac{t^n}{\color{red}{n!}} $$

Answer 2

$$\mathscr{L}^{-1}\left\{\frac1{s^2}+\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\mathscr{L}^{-1}\left\{\frac1{s^3}\right\}=\mathscr{L}^{-1}\left\{\frac1{s^2}\right\}+\color{blue}{\frac{1}{2}}\mathscr{L}^{-1}\left\{\frac{\color{blue}{2}}{s^3}\right\}=t+\frac12t^2$$

Answer 3

$$y'(t)=t+1\Longleftrightarrow$$ $$\mathcal{L}_t\left[y'(t)\right]_{(s)}=\mathcal{L}_t\left[t+1\right]_{(s)}\Longleftrightarrow$$ $$sy(s)-y(0)=\frac{1}{s^2}+\frac{1}{s}\Longleftrightarrow$$

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Use $y(0)=0$:

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

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

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