Given the system x′=xe^(y−3), y′=2sin(x)+3−y Find the equilibrium points and decide if they are stable.

peckishnz

peckishnz

Answered question

2022-09-08

Given the system
x = x e y 3
                              y = 2 sin ( x ) + 3 y
Find the equilibrium points and decide if they are stable.
The equilibrium points of the system are the solutions X such that X = 0, which means x = 0 = y ′. By this condition, I get the points X = ( x , y ) { { ( 0 , 3 ) , ( k π , 3 ) } , k Z }.

Answer & Explanation

Karson French

Karson French

Beginner2022-09-09Added 15 answers

Given equations become
(1) x ˙ = x e y 3
and
(2) y ˙ = 2 sin x + 3 y .
At an equilibrium point, we have x ˙ = y ˙ = 0, so
(3) x e y 3 = 0
and
(4) 2 sin x + 3 y = 0.
Since e y 3 0 for all y R , ( 3 ) yields x = 0; then by ( 4 ), y = 3; there are no other equilibria. ( x , y ) = ( 3 , k π ), k Z , is not an equilibrium as direct substitution of these values into ( 1 ) ( 4 ) will reveal.
The Jacobian J ( x , y ) of the system ( 1 ) ( 2 ) at the point ( x , y ) is given by the matrix
(5) J ( x , y ) = [ x ˙ x x ˙ y y ˙ x y ˙ y ] = [ e y 3 x e y 3 2 cos x 1 ] ;
at ( 0 , 3 ) we have
(6) J ( 0 , 3 ) = [ 1 0 2 1 ] .
It is easy to see that the characteristic polynomial of the matrix J ( 0 , 3 ) is λ 2 1; the eigenvalues are thus ± 1, as may also easily be seen by inspection of the triangular matrix J ( 0 , 3 ). Since J ( 0 , 3 ) has one positive and one negative eigenvalue, the point ( 0 , 3 ) is a saddle; hence, it is an unstable equilibrium.
Terry Briggs

Terry Briggs

Beginner2022-09-10Added 1 answers

How you're getting the ( 3 , k π ). You want x e y 3 = 0, which says x = 0, and 2 sin ( x ) + 3 y = 0, which when x = 0 says y = 3.
For stability, you want to find the Jacobian matrix and evaluate it at the equilibrium point, then see what are the eigenvalues.

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