KVL on series RCL circuit
Whatever you did is correct. You have got an expression for current. Solution manual has used a different approach and they have obtained expression for voltage. 

Finally what you want is voltage. Let's start from where you have stopped. 

$$i(t)=Ae^{-9t}+Be^{-t}$$

so voltage across capacitor:
$$\begin{align*}
v(t)&amp;=\frac{1}{C}\int i(t) dt\\
&amp;=-27\left[\frac{A}{9}e^{-9t}+Be^{-t}\right]
\end{align*}$$

Now we have to evaluate the constants \$A\$ and \$B\$. From the initial conditions, you know that \$i(0^-)=0\$ and current through an inductor can not vary instantly. Which gives:
$$0=A+B\tag1$$

Now we need one more condition. We know that voltage across capacitor can not vary instantly, and \$v(0^-)=16\$.

$$16=-3A-27B\tag2$$

From (1) and (2) you can solve for \$A\$ and \$B\$. You will get the same answer as given in solution manual once you submit these values to voltage expression.