University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 16 - Practice Exercises - Page 16-34: 19

Answer

$y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$

Work Step by Step

The standard form of the given equation is: $y'+3x^2 y=x^2$ ....(1) The integrating factor is: $e^{\int (3x^2) dx}=e^{x^3}$ In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides. $\int [e^{x^3}y]' =\int x^2 e^{x^3}dx $ or, $e^{x^3}y=(1/3)e^{x^3}+c$ Applying the initial conditions, we get $c=\dfrac{-4}{3}$ Thus, we have $e^{x^3}y=(1/3)e^{x^3}-\dfrac{4}{3}$ Hence, $y=\dfrac{1}{3}-\dfrac{4}{3}e^{-x^{3}}$
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