Answer
Plugging $y$ into the differential equation verifies that it is indeed a solution. The particular solution that satisfies the initial conditions is $y = 4e^{2x/3}(1 - x/3)$.
Work Step by Step
If $y = e^{2x/3}(C_1 + C_2x)$, then its first derivative is
$y' = \frac{2}{3}e^{2x/3}\cdot(C_1 + C_2x) + e^{2x/3}\cdot (C_2) = e^{2x/3}(\frac{2}{3}C_1 +C_2 + \frac{2}{3}C_2x)$
and its second derivative is
$y''=\frac{2}{3}e^{2x/3}\cdot(\frac{2}{3}C_1 +C_2 + \frac{2}{3}C_2x) + e^{2x/3}\cdot(\frac{2}{3}C_2) = e^{2x/3}(\frac{4}{9}C_1 + \frac{4}{3}C_2 + \frac{4}{9}C_2x)$
So plugging these into the left-hand side of $9y''-12y'+4y=0$ yields
$e^{2x/3}[9(\frac{4}{9}C_1 + \frac{4}{3}C_2 + \frac{4}{9}C_2x)-12(\frac{2}{3}C_1 +C_2 + \frac{2}{3}C_2x)+4(C_1 + C_2x)]$
$= e^{2x/3}[4C_1 + 12C_2 + 4C_2x-8C_1-12C_2-8C_2x+4C_1+4C_2x] = e^{2x/3}[(4-8+4)C_1+(12-12)C_2 + (4 - 8 + 4)C_2x] = 0$
So the general solution does indeed satisfy the differential equation. The initial conditions imply:
$4 = e^{2\cdot 0/3}(C_1 + C_2\cdot 0)=C_1\implies C_1=4$
$0 = e^{2}(C_1 + 3C_2)\implies C_2 = -\frac{1}{3}C_1 = -\frac{4}{3}$
So the particular solution that satisfies the initial conditions is
$y = 4e^{2x/3}(1 - x/3)$
