## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 2 - The Derivative - 2.3 Introduction To Techniques Of Differentiation - Exercises Set 2.3 - Page 141: 48

#### Answer

If $x \neq 0$, then $y = \frac{1}{x}$ satisfies $x^3y'' + x^2y' - xy =0$.

#### Work Step by Step

If $y = \frac{1}{x} = x^{-1}$, then \begin{align} y' &= (-1)x^{-1-1} \\ &= -x^{-2} \\ y'' &= (-2)(-x^{-2-1}) \\ &= 2x^{-3} \\ \end{align} Substitution $y, y'$, and $y''$ into the differential equation, we obtain $x^3(\frac{2}{x^3}) + x^2(-\frac{1}{x^2}) - x(\frac{1}{x}) = 2 -1 - 1 = 0$

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