Answer
$${y}=\frac{9(e^{x-1}-e^{-9(x-1)})}{8}$$
Work Step by Step
$$y''+8y'-9y=0,\quad{y}(1)=1,\quad{y'}(1)=0$$Let $y=e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$${\lambda}^2+8{\lambda}-9=0$$
$$(\lambda-1)(\lambda+9)=0$$
$$\lambda_{1,2}=1,-9$$
The general solution is ${y}=c_{1}e^{x}+c_{2}e^{-9x}$. Substituting in the constraints, we obtain
$c_{1}e+c_{2}e^{-9}=1$ and $c_{1}e-9c_{2}e^{-9}=0\Rightarrow-8c_{2}e^{-9}=1\Rightarrow{c}_{2}=-\frac{1}{8e^{-9}}\Rightarrow{c}_{1}=\frac{9}{8e}$
$$\therefore{y}=\frac{9(e^{x-1}-e^{-9(x-1)})}{8}$$
