Dynamical system equilibrium point increment d x 1 d x ( t ) = x 2 ( t ) d x 2 d x ( t ) = − k m x 1 ( t ) + f ( t ) m Then we find the equilibrium point by setting 0 = x 2 ( t ) 0 = − k m x 1 ( t ) + f ( t ) m Which gives as result x e q = [ x 1 x 2 ] = [ f / k 0 ] Finally, the book defines the increment with respect to the equilibrium point Δ x ( t ) = x ( t ) − x e q . Substracting equation ( 1 , 3 ) and ( 2 , 4 ) the result is d Δ x 1 d x ( t ) = Δ x 2 ( t ) d Δ x 2 d x ( t ) = − k m Δ x 1 ( t ) So far so good, but for the next step they get "the general solution, parametrized by the initial state" as Δ x 1 ( t ) = Δ x 1 ( 0 ) c o s ( ω t ) + Δ x 2 ( 0 ) ω s i n ( ω t ) Δ x 2 ( t ) = − Δ x 1 ( 0 ) ω s i n ( ω t ) + Δ x 2 ( 0 ) c o s ( ω t ) This result I don't understand where does it comes from, could someone give me a hint?

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Dynamical system equilibrium point increment                    d                  x          1                            d        x              (    t    )    =          x      2        (    t    )                      d                  x          2                            d        x              (    t    )    =                  −        k            m              x      1        (    t    )    +                  f        (        t        )            m      Then we find the equilibrium point by setting      0    =          x      2        (    t    )        0    =                  −        k            m              x      1        (    t    )    +                  f        (        t        )            m      Which gives as result      x          e      q        =      [                                        x                      1                                                            x                      2                                ]    =      [                            f                  /          k                                      0                      ]  Finally, the book defines the increment with respect to the equilibrium point   Δ  x  (  t  )  =  x  (  t  )  −      x          e      q      . Substracting equation   (  1  ,  3  ) and   (  2  ,  4  ) the result is                    d        Δ                  x          1                            d        x              (    t    )    =    Δ          x      2        (    t    )                      d        Δ                  x          2                            d        x              (    t    )    =                  −        k            m        Δ          x      1        (    t    )  So far so good, but for the next step they get &quot;the general solution, parametrized by the initial state&quot; as      Δ          x      1        (    t    )    =    Δ          x      1        (    0    )    c    o    s    (    ω    t    )    +                  Δ                  x          2                (        0        )            ω        s    i    n    (    ω    t    )        Δ          x      2        (    t    )    =    −    Δ          x      1        (    0    )    ω    s    i    n    (    ω    t    )    +    Δ          x      2        (    0    )    c    o    s    (    ω    t    )  This result I don&#039;t understand where does it comes from, could someone give me a hint?

aqueritztv · Accepted Answer

This is ODE business, the solution of:                                          d                          x              1                                            d            t                                              =                              x          2                                                          d                          x              2                                            d            t                                              =                    −                  k          m                          x          1                    is of the form you wrote. Let       ω    2    =      k    m   for simplification.A way to prove it is to differentiate the first equation and substitute the second in:                    d        2                    x        1                    d              t        2              =            d              x        2                    d      t        =  −      ω    2        x    1  So the solution for       x    1   is of the form:      x    1    (  t  )  =      A    1    cos  ⁡  (      ω    t  )  +      A    2    sin  ⁡  (      ω    t  )Differentiate it to find the solution to       x    2    (  t  ).Then substituting the initial conditions you find       A    1    =      x    1    (  0  ) and       A    2    =                    x        2            (      0      )        ω  .

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