Given the system of differential equations x ′ = 2 x + y 3 and y ′ = − y i found the flow ϕ t ( x , y ) = ( ( x 0 + 1 / 5 y 0 3 ) e 2 t − 1 / 5 y 0 3 e − 3 t , y 0 e − t )

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Given the system of differential equations       x    ′    =  2  x  +      y    3   and       y    ′    =  −  y i found the flow      ϕ    t    (  x  ,  y  )  =  (  (      x    0    +  1  /  5      y    0    3    )      e          2      t        −  1  /  5      y    0    3        e          −      3      t        ,      y    0        e          −      t        )

pheniankang · Accepted Answer

Since   y  =      y    0        e          −      t      , there are no periodic solutions. If a solution had period   T  &amp;gt;  0, then       y    0          −      (      t      +      T      )        =      y    0        e          −      t      , clearly impossible unless       y    0    =  0. If       y    0    =  0,   x  =      x    0        e          2      t       and a similar argument applies.

Given the system of differential equations x ′ = 2 x + y 3 and...

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