## Differential Equations and Linear Algebra (4th Edition)

$r = -2$
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, $r = -2$.