Solve the initial value problem using Laplace Transforms.
$\displaystyle \ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0 $
Apply the Laplace transform to both sides of the equation to get
$\displaystyle s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}$
Apply the initial conditions and simplify to get
$\displaystyle s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6$ $\displaystyle \implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-3)^2}$
Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.
Spoiler: