Explanation Given: The plane autonomous system is x ′ = x y − 3 y − 4 and y ′ = y 2 − x 2 . Concept used: Unstable spiral point: The condition for a critical point to be a 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. Stable spiral point: The condition for a critical point to be a stable 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: Consider that x ′ = P ( x , y ) and y ′ = Q ( x , y ) . The above equations are written as follows: P ( x , y ) = x y − 3 y − 4 Q ( x , y ) = y 2 − x 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 ) y 2 − x 2 = 0 ( y + x ) ( y − x ) = 0 The values of y is y = x and y = − x . Solve the equation x ′ = 0 as follows: x ′ = P ( x , y ) x y − 3 y − 4 = 0 Substitute x for y in the above equation. x 2 − 3 x − 4 = 0 ( x + 1 ) ( x − 4 ) = 0 The values of x are x = − 1 and x = 4 . Substitute − x for y in the x y − 3 y − 4 = 0 . − x 2 + 3 x − 4 = 0 x 2 − 3 x + 4 = 0 No real solution exists for the above quadratic equation. Therefore, the critical points for the given plane autonomous system is ( 4 , 4 ) and ( − 1 , − 1 ) . Differentiate P ( x , y ) partially with respect to x as follows: ∂ P ( x , y ) ∂ x = ∂ ( x y − 3 y − 4 ) ∂ x = y Differentiate P ( x , y ) partially with respect to y as follows: ∂ P ( x , y ) ∂ y = ∂ ( x y − 3 y − 4 ) ∂ y = x − 3 Differentiate Q ( x , y ) partially with respect to x as follows: ∂ Q ( x , y ) ∂ x = ∂ ( y 2 − x 2 ) ∂ x = − 2 x Differentiate Q ( x , y ) partially with respect to y as follows: ∂ Q ( x , y ) ∂ y = ∂ ( y 2 − x 2 ) ∂ y = 2 y For the critical point ( − 1 , − 1 ) , the Jacobian matrix is as follows: g ′ ( − 1 , − 1 ) = ( − 1 − 4 2 − 2 ) Consider the trace of the above matrix as τ

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Chapter 10.3, Problem 16E

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The category of the critical point for the plane autonomous system x′=xy−3y−4 and y′=y2−x2.

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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 ′ = x y − 3 y − 4 and y ′ = y 2 − x 2 .

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