Solution: Since one of the particular solution is $y_1(x)$, so we assume that the general solution is given by $y(x) = v(x) y_1(x)$, where $v(x)$ is the function that we need to determine. We have \begin{align*} y = xv & \implies y^\prime = v + xv^\prime \\ & \implies y^{\prime\prime} = v^\prime + v^\prime + xv^{\prime\prime} = 2v^\prime + xv^{\prime\prime} . \end{align*} Substituting the above in the given differential equation, \begin{align*} & x^2 \left( 2v^\prime + xv^{\prime\prime} \right) - x(x+2) \left( v+xv^\prime \right) +(x+2)xv = 0 \\ \implies & 2x^{2}v^\prime +x^3 v^{\prime\prime} - x^{2}v-x^{3}v^\prime -2xv-2x^{2}v^\prime + x^{2}v +2xv = 0 \\ \implies & x^{3}v^{\prime\prime} - x^3 v^\prime = 0 \\ \implies & v^{\prime\prime} -v^\prime =0. \end{align*} The solution of the above differential equation will be \[ v(x) = c_1 + c_2e^{x}, \] where $c_1$ and $c_2$ are arbitrary constants. Therefore, the general solution of the differential equation will be \[ y(x) = \left( c_1 + c_2 e^x \right)x = c_1x + c_2xe^x. \]