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 - Page 21: 16

Answer

$y=C_{1}x^2\cos(3\ln x)$ is solution of ($\alpha$).

Work Step by Step

$x^2y''-3xy'+13y=0$ _____($\alpha$) $y=C_{1}x^2\cos(3\ln x)$ ____(1) Differentiate (1) with respect to $x$ $y'=C_{1}[2x\cos(3\ln x)-\frac{3x^2}{x}\sin(3\ln x)]$ $y'=C_{1}x[2\cos(3\ln x)-3\sin(3\ln x)]$ ___(2) Differentiate (2) with respect to $x$ $y''=C_{1}[2\cos(3\ln x)-3\sin(3\ln x)]+C_{1}x[\frac{-6}{x}\sin(3\ln x)-\frac{9}{x}\cos(3\ln x)]$ $y''=C_{1}[-7\cos(3\ln x)-9\sin(3\ln x)]$ __(3) $x^2y''=C_{1}[-7x^2\cos(3\ln x)-9x^2\sin(3\ln x)]$ ___(4) $-3xy'=C_{1}[-6x^2\cos(3\ln x)+9x^2\sin(3\ln x)]$ ____(5) $13y=13C_{1}x^2\cos(3\ln x)$ ___(6) Adding (4),(5) and (6) $x^2y''-3xy'+13y=C_{1}[-13x^2\cos(3\ln x)-9x^2\sin(3\ln x)+13x^2\cos(3\ln x)+9x^2\sin(3\ln x)]=0$ Hence, $y=C_{1}x^2\cos(3\ln x)$ is solution of ($\alpha$).
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