Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.2 Basic Ideas and Terminology - Problems: 25

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, $r = -2$.
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