University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 105

Answer

$y=x^2-x^3+4x+1$

Work Step by Step

We need to find the anti-derivative for $\dfrac{d^2y}{dx^2}=2-6x$ We have: $\dfrac{dy}{dx}=2x-3x^2+c$; $y=x^2-x^3+cx+c'$ Apply the initial conditions $y'(0)=4 ; y(0)=1$ Thus, $0-0+c=4$ and $0+c'=1 \implies c'=1$ Hence, $y=x^2-x^3+4x+1$
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