Explanation Given: The plane autonomous system is x ′ = − 3 x + y 2 + 2 and y ′ = x 2 − y 2 . 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 spiral point: The condition for a critical point to be an unstable spiral point is that τ ≠ 0 , τ 2 − 4 Δ < 0 and α > 0 where, τ is the trace of the matrix, α is the real part of the eigenvalue and Δ is the determinant of the matrix. Calculation: The plane autonomous system is given as follows: x ′ = − 3 x + y 2 + 2 y ′ = x 2 − y 2 Consider that x ′ = P ( x , y ) and y ′ = Q ( x , y ) . The above equations are written as follows: P ( x , y ) = − 3 x + y 2 + 2 Q ( x , y ) = x 2 − y 2 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 ) x 2 − y 2 = 0 x 2 = y 2 x = ± y Solve the equation x ′ = 0 as follows: x ′ = P ( x , y ) − 3 x + y 2 + 2 = 0 Substitute x for y in the above equation. x 2 − 3 x + 2 = 0 ( x − 1 ) ( x − 2 ) = 0 The value of x obtained from above equations are x = 1 and x = 2 . Therefore, the critical points for the given plane autonomous system is ( 1 , 1 ) , ( 1 , − 1 ) , ( 2 , 2 ) and ( 2 , − 2 ) . Differentiate P ( x , y ) partially with respect to x as follows: ∂ P ( x , y ) ∂ x = ∂ ( − 3 x + y 2 + 2 ) ∂ x = − 3 Differentiate P ( x , y ) partially with respect to y as follows: ∂ P ( x , y ) ∂ y = ∂ ( − 3 x + y 2 + 2 ) ∂ y = 2 y Differentiate Q ( x , y ) partially with respect to x as follows: ∂ Q ( x , y ) ∂ x = ∂ ( x 2 − y 2 ) ∂ x = 2 x Differentiate Q ( x , y ) partially with respect to y as follows: ∂ Q ( x , y ) ∂ x = ∂ ( x 2 − y 2 ) ∂ x = − 2 y The Jacobian matrix formed for the given system is as follows: g ′ ( x , y ) = ( ∂ P ∂ x ∂ P ∂ y ∂ Q ∂ x ∂ Q ∂ y ) = ( − 3 2 y 2 x − 2 y ) For the critical point ( 2 , − 2 ) , the Jacobian matrix is as follows: g ′ ( 2 , − 2 ) = ( − 3 2 ⋅ ( − 2 ) 2 ⋅ ( 2 ) − 2 ⋅ ( − 2 ) ) = ( − 3 − 4 4 4 ) Consider the trace of the above matrix as τ . Trace of a matrix is the sum of all the elements of the main diagonal. Trace of the above matrix is calculated as follows: τ = − 3 + 4 = 1 The determinant of the matrix is Jacobian matrix is calculated as follows: Δ = | − 3 − 4 4 4 | = − 12 + 16 = 4 The value of τ 2 − 4 Δ is calculated as follows: τ 2 − 4 Δ = ( 1 ) 2 − ( 4 ⋅ 4 ) = 1 − 16 = − 15 The eigenvalues are calculated as follows: λ = τ ± τ 2 − 4 Δ 2 = 1 ± − 15 2 = 1 2 ± i 15 2 The real part ( α ) of the eigenvalue obtain as above is positive

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

9th Edition

ISBN: 9781337604901

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 10.3, Problem 15E

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The category of the critical point for the plane autonomous system x′=−3x+y2+2 and y′=x2−y2.

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Chapter 10.3, Problem 14E

Chapter 10.3, Problem 16E

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

Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 2E Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 4E Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - In Problems 16 write the given nonlinear...Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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

Solution Summary: The author describes the critical points for the plane autonomous system: stable node, saddle point, and unstable spiral point.

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The category of the critical point for the plane autonomous system x′=−3x+y2+2 and y′=x2−y2.

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