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: 24

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 $r - 1 = 0$
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