University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 16 - Section 16.2 - First-Order Linear Equations - Exercises - Page 16-15: 30



Work Step by Step

Here, we have $y'-y=xy^2$ Compare the above equation with the Bernoulli equation with $n=2$ Plug $u=y^{1-2}=y^{-1}$ or, $y=u^{-1}$ Then $\dfrac{du}{dx}=(-1)y^{-2}\dfrac{dy}{dx}$ or, $\dfrac{du}{dx}+\dfrac{1}{y}=-x$ or, $\dfrac{du}{dx}+u=x$ The integrating factor is: $e^{\int (1)dx}=e^{x}$ In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides. $\int [ue^x]'=\int xe^x dx $ or, $-ue^x=e^{x}(x-1)+c$ Thus, $y=\dfrac{e^x}{e^x-xe^x+c}$
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