Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 17



Work Step by Step

Here, we have $y'+\dfrac{2y}{x+1}=\dfrac{x}{x+1}$ The integrating factor is: $I=e^{\int (\dfrac{2}{x+1}) dx}=(x+1)^2$ Now, $(x+1)^2[y'+\dfrac{2}{x+1} y]=(x+1)^2(\dfrac{x}{x+1})$ $\int [(x+1)^2y]' =\int x(x+1) dx$ This implies that $(x+1)^2 (y)=\int (x^2+x) dx$ $\implies y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+c)$ Further, apply the initial conditions. $y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+c) \implies y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+1)$
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