Answer
$r + 1 = 0$ or $r - 1 = 0$
Work Step by Step
Differentiating $y(x)=x^{yr}$,
$y'(x)=rx^{r-1}$
Differentiating again,
$y"(x)=r(r-1)x^{r-2}$
Substitute $y(x)$, $y'(x)$ and $y"(x)$ into differential equation $x^{2}y" + xy' - y = 0$:
$x^{2}r(r-1)x^{r-2} + xrx^{r-1} - x^{r}$
$x^{r}(r(r-1) + r -1) = 0$
Since $x^{y} \ne 0$
$(r(r-1) +r -1) = 0$
$(r(r-1) +r -1) = (r-1)(r+1) = 0$
-> $r + 1 = 0$ or $r - 1 = 0$
