Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 136: 43

Answer

$$ n=4,\ \ \ n= -1 $$

Work Step by Step

Given $$x^{2} y^{\prime \prime}-2 x y^{\prime}=4 y$$ Since \begin{align*} y&=x^n \\ y'&=nx^{n-1}\\ y''&=n(n-1)x^{n-2} \end{align*} Then \begin{align*} x^{2} y^{\prime \prime}-2 x y^{\prime}&=4 y\\ n(n-1)x^{n-2}x^2-2 xnx^{n-1}&=4x^n \\ [n(n-1) -2n-4]x^{n} &=0\\ (n-4)(n+1)&=0 \end{align*} Hence $$ n=4,\ \ \ n= -1 $$
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