Explanation Consider the differential equation d p d t = t 2 p − p + t 2 − 1 d p d t = p ( t 2 − 1 ) + ( t 2 − 1 ) d p d t = ( t 2 − 1 ) ( p + 1 ) Separate the variables, d p ( p + 1 ) = ( t 2 − 1 ) d t Integrate on bothsides, ∫ 1 p + 1 d p = ∫ ( t 2 − 1 ) d t ∫ 1 p + 1 d p = ∫ t 2 d t − ∫ d t ln | p + 1 | = t 3 3 − t + C <

Single Variable Calculus: Early Transcendentals

8th Edition

ISBN: 9781305270336

Author: James Stewart

Publisher: Cengage Learning

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Chapter 9.3, Problem 9E

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To solve: The differential equation dpdt=t2p−p+t2−1

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Single Variable Calculus: Early Transcendentals

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The differential equation d p d t = t 2 p − p + t 2 − 1

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To solve: The differential equation dpdt=t2p−p+t2−1

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Chapter 9 Solutions