\begin{align*}
y^"+8y^{'}-9y=0 && {y(1)=1,y'(1)=0}\\
r^2+8r-9&=0\\
(r+9)(r-1)&=0\\
r=-9 \text{ or } r=1\\
y&=c_{1}e^{-9t}+c_{2}e^t\\
y(1)&=c_1e^{-9(1)}+c_{2}e^{1}\\
1&=c_{1}e^{-9t}+c_{2}e^t\\
\text{Differentiate y with respect to t, we get}\\
y^{'}&= -9c_{1}e^{-9t}+c_{2}e^{t}\\
y^{'}(1)&= -9c_{1}e^{-9(1)}+c_{2}e^{1}\\
0 &= -9c_{1}e^{-9}+c_{2}e\\
9c_{1}e^{-9}&=c_{2}e\\
c_{1}&=\frac{c_{2}e^{10}}{9}\\
\text{Substitute } c_{1}=\frac{c_{2}e^{10}}{9} \text{ in } 1=c_{1}e^{-9}+c_{2}e\\
1&=(\frac{c_{2}e^{10}}{9})e^{-9}+c_{2}e\\
1&=\frac{10}{9}c_{2}e\\
c_{2}&=\frac{9}{10e}\\
c_{1}&=\frac{1}{10}e^9\\
\text{Therefore, the general solution of the initial value problem(1) is}\\
y=\frac{1}{10}e^{9(1-t)}+\frac{9}{10}e^{t-1}
\end{align*}
