Answer
$y(x) = C_1e^{\frac{2x}{3}} + C_2xe^{\frac{2x}{3}}$
Work Step by Step
We assume a solution of the form $y=e^{mx}$,where $m$ Is a constant.
$y = e^{mx}$ $\implies$ $y' = m(e^{mx})$ and $y'' = m^2(e^{mx})$
$9y'' -12y' + 4y = 0$
$\implies$ $9m^2(e^{mx}) - 12m(e^{mx}) + 4(e^{mx}) = 0 $
$\implies$ $e^{mx}(9m^2-12m+4) = 0$
since $e^{mx}$ cant be 0
$\implies$ $9m^2 -12m +4 = 0$
$\implies$ $9m^2 - 6m - 6m + 4 = 0$
$\implies$ $(3m-2)(3m-2) = 0$
$\implies$ $(3m-2)^2 = 0$
$\implies$ $ m = \frac{2}{3}$(twice)
since we have equal roots,we have that :
$y(x) = C_1e^{\frac{2x}{3}} + C_2xe^{\frac{2x}{3}}$
