Answer
(a) The set is a solution for the differential equation.
(b) $$ W=e^{-6x}.$$
Since $W$ is non zero then the set of solutions is linearly independent.
(c) The general solution is $$y=c_1e^{-3x}+c_2xe^{-3x}.$$
Work Step by Step
Given the ODE
$$y^{\prime \prime}+6 y^{\prime}+9 y=0.$$
(a) Let $y=e^{-3x}$ be given, then we have $$y^{\prime}=-3e^{-3x}, \quad y^{\prime \prime}=9e^{-3x}.$$
By substitution, we find that $y=e^{-3x}$ is a solution for the given linear differential equation. Also, for $y=xe^{-3x}$, we get
$$y^{\prime}=-3xe^{-3x}+e^{-3x}, \quad y^{\prime \prime}=9xe^{-3x}-6e^{-3x}.$$
By substitution into the ODE, we obtain that $y=xe^{-3x}$ is a solution.
(b) To test the linear independence we have to compute the Wronskian.
$$ W=\left| \begin {array}{cc} {{\rm e}^{-3\,x}}&x{{\rm e}^{-3\,x}}
\\ -3\,{{\rm e}^{-3\,x}}&{{\rm e}^{-3\,x}}-3\,x{
{\rm e}^{-3\,x}}\end {array} \right|=e^{-6x}.
$$
Since $W$ is non zero then the set of solutions is linearly independent.
(c) The general solution is $$y=c_1e^{-3x}+c_2xe^{-3x}.$$
