Explanation Given information: The provided first order linear initial value problem is x d y d x − 2 y = x 3 . The above equation passes through ( 2 , − 1 ) and the graphical solution is, Concept used: The first order linear initial value problem is of the form, d y d x + p ( x ) y = q ( x ) The general solution is given by, y = 1 I ( x ) [ ∫ I ( x ) q ( x ) d x + C ] ; where I ( x ) is the integrating factor and equals to I ( x ) = e ∫ p ( x ) d x . Consider the first order linear initial value problem x d y d x − 2 y = x 3 . The provided equation can be written as, x d y d x − 2 y = x 3 d y d x − 2 y x = x 2 The integrating factor is, I ( x ) = e ∫ ( − 2 x ) d x = e − 2 ln x = 1 x 2

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 14.1, Problem 11PS

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The solution of the first order linear initial value problem xdydx−2y=x3 and compare the result with the provided graphical solution.

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Chapter 14.1, Problem 10PS

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Chapter 14 Solutions

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Ch. 14.1 - Prob. 1PS Ch. 14.1 - Prob. 2PS Ch. 14.1 - Prob. 3PS Ch. 14.1 - Prob. 4PS Ch. 14.1 - Prob. 5PS Ch. 14.1 - Prob. 6PS Ch. 14.1 - Prob. 7PS Ch. 14.1 - Prob. 8PS Ch. 14.1 - Prob. 9PS Ch. 14.1 - Prob. 10PS

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The solution of the first order linear initial value problem x d y d x − 2 y = x 3 and compare the result with the provided graphical solution.

Solution Summary: The author compares the solution of the first order linear initial value problem xdy dx to the provided graphical solution.

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The solution of the first order linear initial value problem xdydx−2y=x3 and compare the result with the provided graphical solution.

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Chapter 14 Solutions