Solving $x^2y''-xy'=0$ and Showing Linear Independence

Question

For the equation $x^2y''-xy'=0$, find two solutions, show that they are linearly independent, and find the general solution.

Using $y=x^r$ gives $$y'=rx^{r-1}$$ and $$y''=(r-1)rx^{r-2}$$ To solve for $r$ (assuming that $x ≥ 0$), substitute to rewrite $x^2y''-xy'=0$ as $$x^2(r-1)rx^{r-2}-xrx^{r-1} = 0$$ Dividing by $\frac {1}{xr}$ gives $$x(r-1)x^{r-2}-x^{r-1}=0$$ But past this point, I am unsure of how to solve for r, and more generally of how to show linear independence of whichever two solutions arise.

Answer 1

$$x^2(r-1)rx^{r-2}-xrx^{r-1} = 0$$ $$x^r(r(r-1)-r)=0$$

This should be true for any x so that you have $$r^2-2r=0 \implies r=0,r=2$$ $$y=c_1+c_2x^2$$

Another approach

$$x^2y''-xy'=0$$ $$xy''-y'=0$$ $$\left (\frac {y'}x\right )'=0$$ Integrate $$\frac {y'}x=C_1$$ $$y'=C_1x$$ integrate again $$y=C_1x^2+C_2$$

Answer 2

$$ r(r-1) x^r -rx^r =0$$

$$ r(r-2)x^r=0$$

$$r=0 \text { or } r=2$$ $$ y=c_1 x^0 + c_2 x^2 = c_1 +c_2 x^2$$