Given the system x ′ = x e y − 3 y ′ = 2 sin ⁡ ( x ) + 3 − yFind 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 }.

Question

Given the system      x    ′    =  x      e          y      −      3                                                         y    ′    =  2  sin  ⁡  (  x  )  +  3  −  yFind 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    }.

Karson French · Accepted Answer

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 · Accepted Answer

How you&#039;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.

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

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Answer & Explanation

New Questions in Algebra I