Showing that solutions to $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ are linearly independent 
Find solutions for $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ which have the form $y(x)=x^r$ and then show that they are linearly independent.

So my method for solving this was to let $y(x)=x^r$ and then substituting that into the original equation:
$x^3\cdot r(r-1)(r-2)x^{r-3}+2x^2r(r-1)x^{r-2}-4xrx^{r-1}+4x^r=0$
Then solving $r(r-1)(r-2)+2\cdot r(r-1)-4r+4=0$ I guessed the solution $r=1$
Then used polynomial long division to get $r(r-1)(r-2)=0$
And so have the general solution $y(x)=c_1+c_2x+c_3x^2$
So I believe my $3$ linearly independent solutions should be $x^0, x^1, x^2$ however when I try these solutions, the only one that seems to work is $y(x)=x$. 
So my first question is, is my general solution correct?
And then are solutions supposed to be any linear combination of $x^0,x^1,x^2$ or do only some linear combination actually solve the equation?
As for showing they are linearly independent I was going to use a theorem that polynomials of different degrees are linearly independent.
 A: Then solving $ r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$ I guessed the solution $r=1$
You need to solve completely your equation :
$r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$
$r(r−1)(r−2)+2⋅r(r−1)−4(r-1)=0$
$(r-1)(r(r-2)+2r-4)$
$(r-1)(r^2-4)=0 \implies (r-1)=0 \text{ or } (r^2-4)=0$
$r^2-4 =0 \implies r=2,r=-2$ 
$r-1 =0 \implies r=1$
$$S_r=\{1,2,-2\}$$ 
So the general solution is :
$$y=c_1x+c_2x^2+\frac {c_3}{x^2}$$
A: Make the ansatz $$y=x^\lambda$$ then you will get
$$\lambda=-2,\lambda=1,\lambda=2$$ so our solutions are
$$y=\frac{c_1}{x^2},y=c_2x,y=c_3x^2$$
A: As @Aryadeva has shown the general solution is
$ \boxed{y(x) = c_1 x  +c_2 x^2 + c_3 \frac1{x^2}} \quad(ODE)$
To show there solutions are linearly independent you can use The Wronskian Determinant
$y_1(x) = x, y_2(x) = x^2, y_3(x)=x^{-2}$
$
W(y_1,y_2)(t)  = 
\left| \begin{array}{ccc}
y_1 & y_2  & y_3\\
y^{'}_1  &  y^{'}_2  &y^{'}_3 \\
y^{''}_1 & y^{''}_2 & y^{''}_3 
 \end{array} \right| = 
\left| \begin{array}{ccc}
x & x^2 & x^{-2} \\
1 & 2x  & -2 x^{-3}\\
0 & 2& 6 x^{-4} 
 \end{array} \right| = u(x) \neq 0 
$
Therefore the solutions are linearly independent.
