LaPlace Transformation DFQ Help [
I am struggling to move beyond the inverse step of taking the LaPlace general solution.
I know we take
$$\mathcal{L}\{y''\} + 9\mathcal{L}\{y\} = g(t)$$
I know that $\mathcal{L}\{y''\} = s^2$Y(s) - sf(0) - f'(0)$
So then from simple plug and chug, I get that $\mathcal{L}\{y''\} = s^2Y(s) - 1 - 0.$ This is just directly from the IVP problem.
Likewise I know that $\mathcal{L}\{y'\} = sY(s) - y(0)$. 
Combining these together into the given equation $y''+9y'= g(t)$
I solve for $Y(s)$ and get $(g(t)+1)/(s^2+9s).$
Up to this point, I am stuck.
 A: $$y''+9y'=g(t)$$
Take the Laplace transform of both sides
$$s^2F(s)-sy(0)-y'(0)+9(sF(s)-y(0))=\mathcal {L}(g(t))$$
$$s^2F(s)-1+9sF(s)=\mathcal {L}(g(t))$$
$$F(s)(s^2+9s)=1+\mathcal {L}(g(t))$$
You can use the heaviside function
$$g(t)=1(H(t)-H(t-1))+t(H(t-1))$$
$$g(t)=1H(t)+(t-1)(H(t-1))$$
You have also that the laplace transform of
$$\mathcal {L}(H(t-a)f(t-a))=e^{-as}F(s)$$
Therefore
$$\mathcal {L}(g(t))=\frac 1s+\frac {e^{-s}}{s^2}$$
you have now that
$$F(s)(s^2+9s)=1+\frac 1s+\frac {e^{-s}}{s^2}$$
$$F(s)=\frac 1 {s(s+9)}+\frac 1{s^2(s+9)}+\frac {e^{-s}}{s^3(s+9)}$$
You can use fractions decompositions ..

For the first fraction we decompose this way
$$h(s)=\frac 1 {s(s+9)}=\frac 19(\frac 1s-\frac 1{s+9})$$
then with the laplace transform table you see that
$$\mathcal{L}(h(s))=\frac 19 (1-e^{-9t})$$
Edit2
for the last fraction dont pay attention to the exponential put it outside
$$f(s)=\frac {e^{-s}}{s^3(s+9)}=e^{-s}\left (\frac {1}{s^3(s+9)} \right)$$
then perform fraction decomposition...the exponential is just a shift 
