Given:

\(y''-4y'+9y=0\)

\(y(0)=0\)

\(y'(0)=-8\)

The auxiliary equation of the givendifferential equation is

\(m^2-4m+9=0\)

\(m=\frac{-(-4)+-\sqrt{(-4)^2-(4)(1)(9)}}{2}\)

\(=\frac{4\pm\sqrt{16-36}}{2}\)

\(=\frac{(4\pm i2\sqrt5)}{2}\)

\(m=2\pm i\sqrt5\)

Hence the general solution is

\(y=e^{2x}(A\cos\sqrt5x+B\sin\sqrt5x)\)

\(y(0)=0\)

\(\Rightarrow0=1(A(1)+B(0))\)

\(\Rightarrow A=0\)

\(y=e^{2x}(B\sin\sqrt5x)\)

\(y'(0)=-8\)

\(y'=2e^x(B\sin\sqrt5x)+\sqrt5e^x(B\cos\sqrt5x)\)

\(-8=0+\sqrt5(B)\)

\(B=\frac{-8}{\sqrt5}\)

Hence, the solution is

\(y=-e^{2x}(\frac{8}{\sqrt5}\sin\sqrt5x)\)