Differential equations notations confusionGiven these differential equations: d 2 x d t 2 = 2 Ω d y d t sin ⁡ ( λ ) − g L x d 2 y d t 2 = − 2 Ω d x d t sin ⁡ ( λ ) − g L yNow making the following substitutions: d x d t = u and d y d t = v we have the following differential equations d u d t = 2 Ω v sin ⁡ ( λ ) − g L x d v d t = − 2 Ω u sin ⁡ ( λ ) − g L yNow here is where I get confused, note that we can write d u d t = f ( t , u ) and likewise for the other equation. But if that is so, then what would be my unknown variable for the first equation? Or is it wrong to write d u d t = f ( t , u )?

Question

Differential equations notations confusionGiven these differential equations:                    d        2            x              d              t        2              =  2  Ω            d      y              d      t        sin  ⁡  (  λ  )  −      g    L    x                    d        2            y              d              t        2              =  −  2  Ω            d      x              d      t        sin  ⁡  (  λ  )  −      g    L    yNow making the following substitutions:            d      x              d      t        =  u and             d      y              d      t        =  v we have the following differential equations            d      u              d      t        =  2  Ω  v  sin  ⁡  (  λ  )  −      g    L    x            d      v              d      t        =  −  2  Ω  u  sin  ⁡  (  λ  )  −      g    L    yNow here is where I get confused, note that we can write             d      u              d      t        =  f  (  t  ,  u  ) and likewise for the other equation. But if that is so, then what would be my unknown variable for the first equation? Or is it wrong to write             d      u              d      t        =  f  (  t  ,  u  )?

Julianne Mccoy · Accepted Answer

Step 1I think if you write the system as follows, it gets simpler, although it takes time.      D    2    x  =  a  D  y  −  b  x  ,                     d      x              d      t        =  D  x        D    2    y  =  −  a  D  x  −  b  y  ,                     d      y              d      t        =  D  ywherein   a  =  2  Ω  sin  ⁡  (  λ  )  ,  b  =      g    L  . So we get:  (      D    2    +  b  )  x  −  a  D  y  =  0      (      D    2    +  b  )  y  +  a  D  x  =  0Step 2By any method you know for solving abbove system, one gets:  [  (      D    2    +  b      )    2    +      a    2        D    2    ]  y  =  0and  [  (      D    2    +  b      )    2    +      a    2        D    2    ]  x  =  0Each of the above can converetd to standard form as, for example for y(t),:  [  (      D    2    +  b      )    2    +      a    2        D    2    ]  y  =  0  →      y          (      4      )        +  (  2  b  +      a    2    )      y    ″    +      b    2    y  =  0which can be solved easily.

Differential equations notations confusion. Given these differential equations: (d^2x)/(dt^2) = 2Omega (dy)/(dt) sin(lambda) - g/L x

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