Solve non-linear ode system as a function of t. { x ˙ = y y ˙ = − x + x 2 = x ( x − 1 ) To solve it as a function x ( y ) or y ( x ) is trivial, but I need the solution as a function of time: ( x ( t ) , y ( t ) ) .The system has a first integral: H ( x , y ) = 1 2 ( x 2 + y 2 ) − 1 3 x 3 . In particular, if no general solution is possible, I'm searching for a solution for H = 1 6 .

Question

Solve non-linear ode system as a function of   t.      {                                                x            ˙                    =          y                                                          y            ˙                    =          −          x          +                      x            2                    =          x          (          x          −          1          )                        To solve it as a function   x  (  y  ) or   y  (  x  ) is trivial, but I need the solution as a function of time:       (    x    (    t    )    ,    y    (    t    )    )  .The system has a first integral:   H  (  x  ,  y  )  =      1    2        (          x      2        +          y      2        )    −      1    3        x    3  . In particular, if no general solution is possible, I&#039;m searching for a solution for   H  =      1    6  .

luisjoseblash2 · Accepted Answer

Multiplying the 2nd order equation       x    ¨    =  x  (  x  −  1  ) by       x    ˙   on both sides and integrating with respect to time, we get a nonlinear 1st order equation for   x:      x    ¨    =  x  (  x  −  1  )        x    ¨        x    ˙    =  x  (  x  −  1  )      x    ˙          ∫    0      t        x    ˙    (  τ  )      x    ¨    (  τ  )  d  τ  =      ∫    0      t    x  (  τ  )  (  x  (  τ  )  −  1  )      x    ˙    (  τ  )  d  τ        ∫                        x        ˙            0                          x        ˙            (      t      )            x    ˙      d      x    ˙    =      ∫            x      0              x      (      t      )        x  (  x  −  1  )    d  x      1    2    (              x      ˙        2    −              x      ˙        0    2    )  =      1    3    (  x  −  1      )    3    +      1    2    (  x  −  1      )    2    −      1    3    (      x    0    −  1      )    3    −      1    2    (      x    0    −  1      )    2                  x      ˙        2    =              x      ˙        0    2    +      2    3    (  x  −  1      )    3    +  (  x  −  1      )    2    −      2    3    (      x    0    −  1      )    3    −  (      x    0    −  1      )    2    =  f  (  x  ;      x    0    ,              x      ˙        0    )      1    2    (              x      ˙        2    −              x      ˙        0    2    )  =      1    3    (  x  −  1      )    3    +      1    2    (  x  −  1      )    2    −      1    3    (      x    0    −  1      )    3    −      1    2    (      x    0    −  1      )    2                  x      ˙        2    =              x      ˙        0    2    +      2    3    (  x  −  1      )    3    +  (  x  −  1      )    2    −      2    3    (      x    0    −  1      )    3    −  (      x    0    −  1      )    2    =  f  (  x  ;      x    0    ,              x      ˙        0    )Once we have               x      ˙        2    =  f  (  x  ;      x    0    ,              x      ˙        0    ), we can at least find an implicit equation for   x:            d      x              d      t        =      f    (    x    ;          x      0        ,                      x        ˙            0        )          ∫            x      0        x              d      χ              f      (      χ      ;              x        0            ,                              x          ˙                0            )        =      ∫    0    t    d  τ  =  t  .The final integral will in general ony be computable in terms of elliptic integrals.

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