Explanation Given: The plane autonomous system is x ′ = − 2 x y and y ′ = y − x + x y − y 3 . Concept used: Saddle point: The condition for a critical point to be a saddle point is that τ 2 − 4 Δ > 0 and Δ < 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Calculation: Consider that x ′ = P ( x , y ) and y ′ = Q ( x , y ) . The above equations are written as follows: P ( x , y ) = − 2 x y Q ( x , y ) = y − x + x y − y 3 The critical point ( x , y ) is the point for which x ′ = 0 and y ′ = 0 . Solve the equation x ′ = 0 as follows: x ′ = P ( x , y ) − 2 x y = 0 From the above equation it is observed that either the value of x is 0 or the value of y is 0 . Solve the equation y ′ = 0 for x = 0 as follows: y ′ = Q ( x , y ) y − x + x y − y 3 = 0 y ( 1 − y 2 ) − x + x y = 0 Substitute 0 for x in the above equation. y ( 1 − y 2 ) = 0 From the above equation it is observed that either the value of y is 0 or the value of 1 − y 2 is 0 . Solve the equation 1 − y 2 = 0 . 1 − y 2 = 0 y 2 = 0 y = ± 1 Therefore, the critical points for the given plane autonomous system is ( 0 , 0 ) , ( 0 , − 1 ) and ( 0 , 1 ) . Differentiate P ( x , y ) partially with respect to x as follows: ∂ P ( x , y ) ∂ x = ∂ ( − 2 x y ) ∂ x = − 2 y Differentiate P ( x , y ) partially with respect to y as follows: ∂ P ( x , y ) ∂ y = ∂ ( − 2 x y ) ∂ y = − 2 x Differentiate Q ( x , y ) partially with respect to x as follows: ∂ Q ( x , y ) ∂ x = ∂ ( y − x + x y − y 3 ) ∂ x = − 1 + y Differentiate Q ( x , y ) partially with respect to y as follows: ∂ Q ( x , y ) ∂ y = ∂ ( y − x + x y − y 3 ) ∂ y = 1 + x − 3 y 2 The Jacobian matrix formed for the given system is as follows: g ′ ( x , y ) = ( ∂ P ∂ x ∂ P ∂ y ∂ Q ∂ x ∂ Q ∂ y ) = ( − 2 y − 2 x − 1 + y 1 + x − 3 y 2 ) For the critical point ( 0 , 0 ) the Jacobian matrix is as follows: g ′ ( 0 , 0 ) = ( 0 0 − 1 1 ) Consider the trace of the above matrix as τ

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 10.3, Problem 17E

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The category of the critical point for the plane autonomous system x′=−2xy and y′=y−x+xy−y3.

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Chapter 10.3, Problem 16E

Chapter 10.3, Problem 18E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The category of the critical point for the plane autonomous system x ′ = − 2 x y and y ′ = y − x + x y − y 3 .

Solution Summary: The author explains the critical point for the plane autonomous system xprime, which is a saddle point, and the determinant of the matrix.

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The category of the critical point for the plane autonomous system x′=−2xy and y′=y−x+xy−y3.

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