> 
> 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.