Find two linearly independent solutions of the differential equation $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$ I want to find two linearly independent solutions of the differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
Previously I have seen that the following holds  for the differential equation $y''+ \frac{1}{x}y'-\frac{1}{x^2}y=0, x>0$:

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*We are looking for solutions of the differential equation of the form $x^r$. Then the function $x^r$ is a solution of the differential equation at $(0,+\infty)$ if:

$$r(r-1)x^{r-2}+ \frac{1}{x} r x^{r-1}- \frac{1}{x^2}x^r=0 \forall x >0 \Rightarrow r=1 \text{ or } r=-1$$

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*So, the functions $y_1(x)=x, y_2= \frac{1}{x}$ are solutions of the differential equation and it also holds that they are linearly indepedent since $W(y_1, y_2) \neq 0$
For this differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
I thought the following:
$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3} \Rightarrow  y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$

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*We are looking for solutions of the differential equation of the form $\left( x- \frac{1}{3}\right)^r$.
Then the function $\left( x- \frac{1}{3}\right)^r$ is a solution of the differential equation at $( \frac{1}{3},+\infty)$ if:
$$r(r-1) \left( x- \frac{1}{3}\right)^{r-2}+ \frac{1}{ \frac{3x-1}{3}} r \left( \frac{3x-1}{3}\right)^{r-1}- \frac{9}{(3x-1)^2} \left( x- \frac{1}{3}\right)=0 \Rightarrow \dots \Rightarrow r= \pm 1$$
Therefore, the functions $z_1(x)=x- \frac{1}{3}, z_2(x)=\frac{1}{x- \frac{1}{3}}$ are solutions of the differential equation at $\left( \frac{1}{3}, +\infty\right)$.
$$z_1(x) z_2'(x)-z_1'(x) z_2(x)=\frac{-2}{x- \frac{1}{3}} \neq 0$$
So, $z_1, z_2$ are linearly independent solutions of the differential equation.
Thus, the general solution of $y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$ is of the form:
$$c_1 \left( x- \frac{1}{3} \right)+ c_2 \left( \frac{1}{x- \frac{1}{3}}\right) | c_1, c_2 \in \mathbb{R}, x> \frac{1}{3}$$
EDIT:
We set $t=x-\frac{1}{3}$ and we have:
$$\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}=\frac{dy}{dx}$$
$$\frac{d^2y}{dt^2}=\frac{d}{dt} \left( \frac{dy}{dt}\right)=\frac{d}{dt} \left( \frac{dy}{dx} \right)=\frac{dx}{dt} \frac{d}{dx} \left( \frac{dy}{dx} \right)=\frac{d^2y}{dx^2}$$
$$y''(x)+ \frac{1}{x-\frac{1}{3}}y'(x)-\frac{1}{\left( x-\frac{1}{3}\right)^2}y(x)=0 \\ \Rightarrow y''(t)+\frac{1}{t}y'(t)-\frac{1}{t^2}y(t)$$
Two linearly independent solutions are $y_1(t)=t$ and $y_2(t)=\frac{1}{t}, y \in (0,+\infty)$.
Thus, two linearly independent solutions of $y''+\frac{1}{x-\frac{1}{3}}y'-\frac{1}{\left( x-\frac{1}{3} \right)^2}y=0, x> \frac{1}{3}$
are $y_1(x)=x-\frac{1}{3}, y_2(x)=\frac{1}{x-\frac{1}{3}}$
Is it right or have I done something wrong?

Hint
I do not know how much this could help you; so, please forgive me if this is off-topic.
If you define first $y=(3x-1)u$, $$(3x-1)^2 y''+(9x-3)y'-9y=0$$ rewrite $$(3 x-1)^2 \Big((3 x-1) u''+9 u'\Big)=0$$ for which order can be reduced and integration seems simple.
