Explanation Procedure used: Find the critical points by equating the differential equation to zero. The point of inflection of a function the points is the point where the function’s second derivative is zero. Obtain the nature of curve in the regions obtained from function’s second derivative and sketch the curve. Calculation: Consider the autonomous differential equation. d y d x = y 2 − y − 6 Equate the above equation to zero. y 2 − y − 6 = 0 ( y − 3 ) ( y + 2 ) = 0 From above equation, the roots are, y = 3 and y = − 2 So, 3 and − 2 are critical a points. d 2 y d x 2 = ( 2 y − 1 ) d y d x = ( 2 y − 1 ) ( y − 3 ) ( y + 2 ) So the only possible point of inflection is at y = 1 2 as the other two are critical points. The concavity of solutions can be different on either side y = − 2 and y = 3 . For y > 3 and − 2 < y < 1 3 , y ″ ( x ) > 0

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 2.1, Problem 36E

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To find: The intervals for which the curves are concave down, information about the location of points of inflection of a solution curve and sketch the solution curve for which y(0)=−1 and y(2)=2.

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Chapter 2.1, Problem 35E

Chapter 2.1, Problem 37E

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - dydx=1xy (a) y(0) = 0 (b) y(1) = 0 (c) y(2) = 2...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - In Problems 512 use computer software to obtain a...Ch. 2.1 - In Problems 512 use computer software to obtain a...Ch. 2.1 - In Problems 512 use computer software to obtain a...Ch. 2.1 - In Problems 512 use computer software to obtain a...Ch. 2.1 - In Problems 512 use computer software to obtain a...Ch. 2.1 - In Problems 512 use computer software to obtain a...

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The intervals for which the curves are concave down, information about the location of points of inflection of a solution curve and sketch the solution curve for which y ( 0 ) = − 1 and y ( 2 ) = 2 .

Solution Summary: The author explains the procedure used to find the critical points by equating the differential equation to zero.

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To find: The intervals for which the curves are concave down, information about the location of points of inflection of a solution curve and sketch the solution curve for which y(0)=−1 and y(2)=2.

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