Solve the differential equation with the laplace transformation. $$y''-4y'+9y=9\quad,\quad y(0)=0 \quad ,\quad y'(0)=-8$$
I will solve this question to this state, but I cannot continue.
\begin{align} s^2 y(s)-4sy(s)+9y(s)=\frac{9}{s} \end{align}
\begin{align} y(s)&=\frac{9}{s}(\frac{1}{s^2}-4s+9) \\ &=\frac{1}{s}-\frac{s-4}{(s-2)^2}+5 \\ &=\frac{1}{s}-\frac{s-2}{(s-2)^2}-5 + \frac{2}{(s-2)^2}+5 \end{align} I did this so far, but I cannot transform last term.
You should first know that, if
$$\hat{y}(s) = \int_0^{\infty} dt \: y(t) e^{-s t}$$
is the LT of $y$, then the LT of the first and second derivative of $y$ is
$$\hat{y'}(s) = -y(0) + s \hat{y}(s) = s \hat{y}(s)$$ $$\hat{y''}(s) = -y'(0) - s y(0) + s^2 \hat{y}(s) = 8 + s^2 \hat{y}(s)$$
which may be shown by integration by parts. I think you erred here in not including the initial conditions. The differential equation becomes
$$[8 + s^2 \hat{y}(s)] -4 [ s \hat{y}(s)] + 9\hat{y}(s) = 9 \int_0^{\infty} dt \: e^{-s t} = \frac{9}{s} $$
or
$$(s^2-4 s+9) \hat{y}(s) = \frac{9}{s}-8 \implies \hat{y}(s) = \frac{9-8 s}{s (s^2-4 s+9)}$$
To find the inverse transform, use a partial fraction decomposition. Note that
$$s^2-4 s+9=0 \implies s_{\pm} = 2 \pm i \sqrt{5}$$
I will spare you further algebra; the decomposition is
$$\frac{9-8 s}{s (s^2-4 s+9)} = \frac{1}{s} + \frac{B}{s-s_+} + \frac{C}{s-s_-}$$
where $B = (4+s_+)/(s_+-s_-) = -\frac{1}{2} - i \frac{3}{\sqrt{5}}$ and $C= -(4+s_-)/(s_+-s_-) = -\frac{1}{2} + i \frac{3}{\sqrt{5}}$.
At this point, you should be able to do the inverse transform by lookup. After some simplification, the result is
$$y(t) = 1 + e^{2 t} \left ( -\cos{\sqrt{5} t} + \frac{6}{\sqrt{5}} \sin{\sqrt{5} t} \right )$$