x ˙ = x − y − x ( x 2 + y 2 ) y ˙ = y + x − y ( x 2 + y 2 )1. Determine the equilibrium points2. Show that this system has a periodic solution. Use the following substitution x = r cos ⁡ ( ϕ ) , y = r sin ⁡ ( ϕ ).3. Give the explicit solution for this system. Also determine the period of this solution.

Question

x    ˙    =  x  −  y  −  x  (      x    2    +      y    2    )      y    ˙    =  y  +  x  −  y  (      x    2    +      y    2    )1. Determine the equilibrium points2. Show that this system has a periodic solution. Use the following substitution   x  =  r  cos  ⁡  (  ϕ  )  ,  y  =  r  sin  ⁡  (  ϕ  ).3. Give the explicit solution for this system. Also determine the period of this solution.

Liberty Gates · Accepted Answer

x    2    +      y    2    =      r    2  Differentiating this gives you:  2  x      x    ′    +  2  y      y    ′    =  2  r      r    ′  Now you substitute       x    ′   and       y    ′  :  x  (  x  −  y  −  x  (      x    2    +      y    2    )  )  +  y  (  y  +  x  −  y  (      x    2    +      y    2    )  )  =  r      r    ′  This gives us:      x    2    −  x  y  −      x    2    (      x    2    +      y    2    )  +      y    2    +  x  y  −      y    2    (      x    2    +      y    2    )  =  r      r    ′        x    2    −  x  y  −      x    2        r    2    +      y    2    +  x  y  −      y    2        r    2    =  r      r    ′        x    2    −      r    2    (      x    2    +      y    2    )  +      y    2    =  r      r    ′        r    2    −      r    4    =  r      r    ′        r    2    −      r    4    =  r      r    ′    r  −      r    3    =      r    ′  To determine angle use:                    r        sin        ⁡        θ                    r        cos        ⁡        θ              =  tan  ⁡  θ  =            y      x      Using the quotient rule gives us:      θ    ′    =                    x                  y          ′                −        y                  x          ′                            r        2              =                    x        y        −                  y          2                −        x        y        (                  x          2                +                  y          2                )        −        x        y        −                  x          2                +        x        y        (                  x          2                +                  y          2                )                    r        2                  θ    ′    =  −  1      r    ′    =  r  −      r    3    =  r  (  1  −  r  )  (  1  +  r  )check:Polar coordinates differential equation                    r        sin        ⁡        θ                    r        cos        ⁡        θ              =  tan  ⁡  θ  =            y      x        tan  ⁡  θ  =            y      x        →  θ  =  a  r  c  t  a  n      y    x  Using the quotient rule gives us:      θ    ′    =  a  r  c  t  a  n  (      y    x    )  =                    (                  y          x                          )          ′                            1        +        (                  y          x                          )          2                      =                    x                  y          ′                −        y                  x          ′                            r        2

<mover> x ˙ </mover> = x − y −<!-- −

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