#### Answer

$r = -2$

#### Work Step by Step

Differentiating $y(x) = x^{r}$:
$y'(x) = rx^{r-1}$
Differentiating again:
$y"(x) = r(r - 1)x^{r-2}$
Substitute $y(x)$, $y'(x)$, and $y"(x)$ into the differential equation $x^{2}y" + 5xy' +4y = 0$
$x^{2}r(r - 1)x^{r-2} + 5xrx^{r-1} + 4x^{r} = 0$
-> $x^{r}(r(r - 1) + 5r + 4) = 0$
Since $x^{r} \ne 0$
$(r(r - 1) + 5r +4) = (r + 2)^{2} = 0$
Thus, $r = -2$.