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: 46

Answer

See below.

Work Step by Step

Take the derivatives of the function. $$y(x)=c_1x^4+c_2x^{-2}$$ $$y'(x)=4c_1x^3-2c_2x^{-3}$$ $$y''(x)=12c_1x^2+6c_2x^{-4}$$ Substitute these functions into the differential equation to get $$x^2y''-xy'-8y=0$$ $$x^2(12c_1x^2+6c_2x^{-4})-x(4c_1x^3-2c_2x^{-3})-8(c_1x^4+c_2x^{-2})=0$$ $$12c_1x^4+6c_2x^{-2}-4c_1x^4+2c_2x^{-2}-8c_1x^4-8c_2x^{-2}=0$$ $$0=0$$ This equation is always true when $x>0$, so the equation is a solution to the differential equation.
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