Solve $(y* \sin)(t)=t^2$ 
Solve $(y* \sin)(t)=t^2$

What I did: We know that
$$\mathscr{L} \{ y(t) \}=Y(s)\quad\mathscr{L} \{ \sin(t) \}=\frac{1}{s^2+1} \quad\mathscr{L} \{ t^2 \}=\frac{1}{s^3} \quad\mathscr{L} \{ f(t)*g(t) \}=F(s) \ G(s)$$
hence
$$Y(s)=\frac{s^2+1}{s^3}=\frac{A}{s^3}+\frac{B}{s^2}+\frac{C}{s}$$
with
$$A=\lim_{s\rightarrow 0}\ s^2+1=1$$
$$B=\lim_{s\rightarrow 0} \ D(s^2+1)=\lim_{s\rightarrow 0} \ 2s=0$$
$$C=\frac{1}{2} \lim_{s\rightarrow 0} \ D^2(s^2+1)=\frac{1}{2} \lim_{s\rightarrow 0} \ 2=1$$
hence
$$Y(s)=\frac{1}{s^3}+\frac{1}{s}$$
and
$$y(t)=t^2+1$$

Is this correct?

 A: Well, you used some properties:


*

*The Laplace transform of a function $\text{y}\left(t\right)$, looks like:
$$\text{Y}\left(\text{s}\right)=\mathcal{L}_t\left[\text{y}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{y}\left(t\right)e^{-\text{s}t}\space\text{d}t\tag1$$

*The concolution theorem:
$$\mathcal{L}_t\left[\text{f}\left(t\right)*\text{g}\left(t\right)\right]_{\left(\text{s}\right)}=\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_t\left[\text{g}\left(t\right)\right]_{\left(\text{s}\right)}:=$$
$$\left\{\int_0^\infty\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t\right\}\cdot\left\{\int_0^\infty\text{g}\left(t\right)e^{-\text{s}t}\space\text{d}t\right\}=\text{F}\left(\text{s}\right)\cdot\text{G}\left(\text{s}\right)\tag2$$

*The Laplace transform of $\sin\left(t\right)$, looks like:
$$\mathcal{L}_t\left[\sin\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\sin\left(t\right)e^{-\text{s}t}\space\text{d}t=\frac{1}{1+\text{s}^2}\tag3$$
When $\Re\left(\text{s}\right)>0$.

*The Laplace transform of $t^2$, looks like:
$$\mathcal{L}_t\left[t^2\right]_{\left(\text{s}\right)}:=\int_0^\infty t^2e^{-\text{s}t}\space\text{d}t=\frac{2}{\text{s}^3}\tag4$$
When $\Re\left(\text{s}\right)>0$.


So, when we've your problem for the LHS:
$$\mathcal{L}_t\left[\text{y}\left(t\right)*\sin\left(t\right)\right]_{\left(\text{s}\right)}=\mathcal{L}_t\left[\text{y}\left(t\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_t\left[\sin\left(t\right)\right]_{\left(\text{s}\right)}=\text{Y}\left(\text{s}\right)\cdot\frac{1}{1+\text{s}^2}\tag5$$
And for the RHS:
$$\mathcal{L}_t\left[t^2\right]_{\left(\text{s}\right)}=\frac{2}{\text{s}^3}\tag6$$
So, set the LHS and RHS equal and solve for $\text{Y}\left(\text{s}\right)$:
$$\text{Y}\left(\text{s}\right)\cdot\frac{1}{1+\text{s}^2}=\frac{2}{\text{s}^3}\space\Longleftrightarrow\space\text{Y}\left(\text{s}\right)=\frac{\frac{2}{\text{s}^3}}{\frac{1}{1+\text{s}^2}}=\frac{2\cdot\left(1+\text{s}^2\right)}{\text{s}^3}=2\cdot\left\{\frac{1}{\text{s}^3}+\frac{1}{\text{s}}\right\}\tag7$$
Now, using inverse Laplace transform we get:
$$\text{y}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\text{Y}\left(\text{s}\right)\right]_{\left(t\right)}=2\cdot\left\{\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}^3}\right]_{\left(t\right)}+\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\right]_{\left(t\right)}\right\}=2\cdot\left(\frac{t^2}{2}+1\right)=2+t^2\tag8$$
