Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 326: 77

Answer

y is a solution

Work Step by Step

Find $y^{\prime} $ and $ y^{\prime\prime}$ $y=2(\ln x)+3$ $y^{\prime}=2\displaystyle \cdot\frac{1}{x}+0= \frac{2}{x}=(2x^{-1})$ $y^{\prime\prime}=(2x^{-1})^{\prime}=2(-x^{-2})=-\displaystyle \frac{2}{x^{2}}$ Substitute and see if the differential equation is satisfied $xy^{\prime\prime}+y^{\prime}=x(-\displaystyle \frac{2}{x^{2}})+\frac{2}{x}=-\frac{2}{x}+\frac{2}{x}=0$ It is, so, y is a solution.
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