(1) d 2 θ d t = α ( θ − 1 ) + β ( θ − 1 ) 3 − γ d θ d t Where α , β , γ ∈ R.This is my solution in attempting to convert (1) into a system of first order ODE's: (2) d θ d t = ϕ (3) d ϕ d t = α ( θ − 1 ) + β ( θ − 1 ) 3 − γ ϕ Is the above system correct? Also in equation (3) is it okay to have the θ term?

Question

(1)                                                        d              2                        θ                                d            t                          =        α        (        θ        −        1        )        +        β        (        θ        −        1                  )          3                −        γ                              d            θ                                d            t                              Where   α  ,  β  ,  γ  ∈  R.This is my solution in attempting to convert (1) into a system of first order ODE&#039;s:                    (2)                                          d            θ                                d            t                                              =        ϕ                                          (3)                                          d            ϕ                                d            t                                              =        α        (        θ        −        1        )        +        β        (        θ        −        1                  )          3                −        γ        ϕ            Is the above system correct? Also in equation (3) is it okay to have the   θ term?

Makhi Lyons · Accepted Answer

We are given:                    (1)                                                        d              2                        θ                                d            t                          =        α        (        θ        −        1        )        +        β        (        θ        −        1                  )          3                −        γ                              d            θ                                d            t                              Where   α  ,  β  ,  γ  ∈  R.We can proceed as follows:-       x    1    =  θ    ⟹        x    1    ′    =      θ    ′    =      x    2  -       x    2    =      θ    ′      ⟹        x    2    ′    =      θ    ″    =  α  (  θ  −  1  )  +  β  (  θ  −  1      )    3    −  γ            d      θ              d      t        =  α  (      x    1    −  1  )  +  β  (      x    1    −  1      )    3    −  γ         x    2  So, our new system is:                              x          1          ′                                    =                  x          2                                              x          2          ′                                    =        α        (                  x          1                −        1        )        +        β        (                  x          1                −        1                  )          3                −        γ                           x          2                    Note, given initial conditions, similarly follow as:-   θ  (      t    0    )  =  a    ⟹        x    1    (      t    0    )  =  a-       θ    ′    (      t    0    )  =  b    ⟹        x    2    (      t    0    )  =  b

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