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: 14

Answer

See below.

Work Step by Step

Take the derivatives of the function. $$y(x)=c_1x^{-3}+c_2x^{-1}$$ $$y'(x)=-3c_1x^{-4}-c_2x^{-2}$$ $$y''(x)=12c_1x^{-5}+2c_2x^{-3}$$ Substitute these functions into the differential equation to get $$x^2y''+5xy'+3y=0$$ $$x^2(12c_1x^{-5}+2c_2x^{-3})+5x(-3c_1x^{-4}-c_2x^{-2})+3(c_1x^{-3}+c_2x^{-1})=0$$ $$12c_1x^{-3}+2c_2x^{-1}+-15c_1x^{-3}-5c_2x^{-1}+3c_1x^{-3}+3c_2x^{-1}=0$$ $$0=0$$ This means that the solution is valid for all real values. The interval linked to this solution is $(-\infty,\infty)$.
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