System of equations: ( y ˙ 1 y ˙ 2 ) = ( 2 0 4 − 1 ) ( y 1 y 2 ) Looking at matrix A I can see a value not on the diagonal equal to zero. So the eigenvalues here are 2 and − 1, meaning I have a saddle node. Next I look at the eigenvectors: λ = 2 , [ 0 0 0 4 − 3 0 ] λ = − 1 , [ 3 0 0 4 0 0 ] y = A e 2 x ( 3 4 ) + B e − x ( 0 1 ) So I have an attractive trajectory on the y-axis, and a repulsive trajectory on the line y = 3 4 xWith the slope at origin being y ˙ 2 y ˙ 1 = 4 y 1 − y 2 2 y 1 = 0 0 = ± ∞Does all of this seem correct?

Question

System of equations:      (                                                                      y                ˙                                      1                                                                                          y                ˙                                      2                                )    =      (                            2                          0                                      4                          −          1                      )        (                                        y            1                                                            y            2                                )  Looking at matrix   A I can see a value not on the diagonal equal to zero. So the eigenvalues here are   2 and   −  1, meaning I have a saddle node. Next I look at the eigenvectors:  λ  =  2  ,      [                            0                          0                          0                                      4                          −          3                          0                      ]    λ  =  −  1  ,      [                            3                          0                          0                                      4                          0                          0                      ]    y  =  A      e          2      x            (                            3                                      4                      )    +  B      e          −      x            (                            0                                      1                      )  So I have an attractive trajectory on the   y-axis, and a repulsive trajectory on the line   y  =      3    4    xWith the slope at origin being                              y          ˙                    2                                y          ˙                    1        =            4              y        1            −              y        2                    2              y        1              =      0    0    =  ±  ∞Does all of this seem correct?

domwaheights0m · Accepted Answer

It is indeed correct, but the slope you computed is the slope of the orbit on the phase space, because each coordinate       y    1   and       y    2   have an exponencial growth/decay rate. also the origin is not a node, but a saddle equilibria

System of equations: (y_1 y_2)=(2 0, 4 −1)(y_1 y_2)

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