## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 109

#### Answer

$y=x^3-4x^2+5$

#### Work Step by Step

We need to find the anti-derivative for $\dfrac{d^3y}{dx^3}=6$ We have: $\dfrac{d^2y}{dx^2}=6x+c$; $\dfrac{dy}{dx}=3x^2+cx+c'$, $y=x^3+\dfrac{cx^2}{2}+c'x+c''$ Apply the initial conditions $y''(0)=-8 ; y'(0)=0,y(0)=5$ Thus, $c=-8$ , $c'=2$ and $c''=5$ Hence, $y=x^3-4x^2+5$

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