Solve the initial value problem.
\begin{align*}
y'' -6y' +9y = 0\\
y(0) = 0, y'(0) = 2
\end{align*}
\begin{align}
r^2 - 6r +9 = 0 \\
(r-3)^2 = 0\\
r_1 = r_2 = 3\\\
y(t) = c_1e^{3t}+c_2te^{3t}\\
y(0)=c_1e^0+0=0\\
c_1=0
\end{align}
\begin{align}
y'(0)&=3c_1e^{3t}+c_2(e^{3t}+3te^{3t})\\
&=3c_1e^{3t}+c_2e^{3t}+3c_2 te^{3t} \\
&=3c_1e^0+c_2e^0+0\\
&=3c_1+c_2=2 \\
c_2&=2-3c_1\\
&=2-0=2\\
y(t)&=2te^{3t}
\end{align}
