Explanation Given information: Given initial value problem is y ' = 6 x y 2 , y ( 1 ) = 4 Calculation: On solving the given initial value problem using separation of variable, y ' = 6 x y 2 y ' y 2 = 6 x ∫ 1 y 2 d y = 6 ∫ x d x ( y − 2 + 1 − 2 + 1 ) = 6 ( x 2 2 ) + c − 1 y = 3 x 2 + c y = − 1 3 x 2 + c

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 9 Solutions

Calculus: Early Transcendentals

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 4PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E

Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Prob. 32E Ch. 9.1 - Prob. 33E Ch. 9.1 - Prob. 34E Ch. 9.1 - Prob. 35E Ch. 9.1 - Prob. 36E Ch. 9.1 - Prob. 37E Ch. 9.1 - Prob. 38E Ch. 9.1 - Prob. 39E Ch. 9.1 - Prob. 40E Ch. 9.1 - Prob. 41E Ch. 9.1 - Prob. 42E Ch. 9.1 - Prob. 43E Ch. 9.1 - Prob. 44E Ch. 9.1 - Prob. 45E Ch. 9.1 - Prob. 46E Ch. 9.1 - Prob. 47E Ch. 9.1 - Prob. 48E Ch. 9.1 - Prob. 49E Ch. 9.1 - Prob. 50E Ch. 9.1 - Prob. 51E Ch. 9.1 - Prob. 52E Ch. 9.1 - Prob. 53E Ch. 9.1 - Prob. 54E Ch. 9.1 - Prob. 55E Ch. 9.1 - Prob. 56E Ch. 9.1 - 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