Explanation Given: The plane autonomous system is x ′ = x ( 10 − x − y 2 ) and y ′ = y ( 16 − y − x ) . Concept used: Stable node: The condition for a critical point to be a stable node is that τ 2 − 4 Δ > 0 , τ < 0 and Δ > 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. 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. Unstable node: The condition for a critical point to be a unstable node is that τ 2 − 4 Δ > 0 , τ > 0 and Δ > 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Calculation: The plane autonomous system is given as follows: x ′ = x ( 10 − x − y 2 ) y ′ = y ( 16 − y − x ) Consider that x ′ = P ( x , y ) and y ′ = Q ( x , y ) . The above equations are written as follows: P ( x , y ) = x ( 10 − x − y 2 ) Q ( x , y ) = y ( 16 − y − x ) The critical point ( x , y ) is the point for which x ′ = 0 and y ′ = 0 . Solve the equation y ′ = 0 as follows: y ′ = Q ( x , y ) y ( 16 − y − x ) = 0 From the above equation it is observed that either the value of y is 0 or the value of 16 − y − x is 0 . Solve the equation x ′ = 0 as follows: x ′ = P ( x , y ) x ( 10 − x − y 2 ) = 0 From the above equation it is observed that either the value of x is 0 or the value of 10 − x − y 2 is 0 . Substitute 0 for x in 16 − y − x = 0 . 16 − y − 0 = 0 16 − y = 0 y = 16 Substitute 0 for y in 10 − x − y 2 = 0 . 10 − x − 0 = 0 10 − x = 0 x = 10 Solve the equation 10 − x − y 2 = 16 − y − x . 10 − x − y 2 = 16 − y − x y − y 2 = 16 − 10 y 2 = 6 y = 12 Substitute 12 for y in 10 − x − y 2 = 0 . 10 − x − 12 2 = 0 x = 10 − 6 x = 4 The critical points for the given autonomous system are ( 0 , 0 ) , ( 10 , 0 ) , ( 0 , 16 ) and ( 4 , 12 ) . Differentiate P ( x , y ) partially with respect to x . ∂ P ( x , y ) ∂ x = ∂ ( x ( 10 − x − y 2 ) ) ∂ x = 10 − 2 x − y 2 Differentiate P ( x , y ) partially with respect to y . ∂ P ( x , y ) ∂ y = ∂ ( y ( 16 − y − x ) ) ∂ y = 16 − 2 y − x Differentiate Q ( x , y ) partially with respect to x . ∂ Q ( x , y ) ∂ x = ∂ ( x ( 10 − x − y 2 ) ) ∂ x = 10 − 2 x − y 2 Differentiate Q ( x , y ) partially with respect to y . ∂ Q ( x , y ) ∂ y = ∂ ( x ( 10 − x − y 2 ) ) ∂ y = − x 2 The Jacobian matrix formed for the given system is as follows: g ′ ( x , y ) = ( ∂ P ∂ x ∂ P ∂ y ∂ Q ∂ x ∂ Q ∂ y ) = ( 10 − 2 x − y 2 − 1 2 x − y 16 − x − 2 y ) For the critical point ( 0 , 0 ) the Jacobian matrix is as follows: g ′ ( 0 , 0 ) = ( 10 0 0 16 ) Consider the trace of the above matrix as τ

Bundle: Differential Equations with Boundary-Value Problems, 9th + WebAssign Printed Access Card for Zill's Differential Equations with Boundary-Value Problems, 9th Edition, Single-Term

9th Edition

ISBN: 9781337604918

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 10.3, Problem 19E

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The category of the critical point for the plane autonomous system x′=x(10−x−y2) and y′=y(16−y−x).

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Chapter 10.3, Problem 18E

Chapter 10.3, Problem 20E

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Bundle: Differential Equations with Boundary-Value Problems, 9th + WebAssign Printed Access Card for Zill's Differential Equations with Boundary-Value Problems, 9th Edition, Single-Term

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The category of the critical point for the plane autonomous system x ′ = x ( 10 − x − y 2 ) and y ′ = y ( 16 − y − x ) .

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The category of the critical point for the plane autonomous system x′=x(10−x−y2) and y′=y(16−y−x).

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