Getting a differential equation from a parametric function family x = c e − t − 2 t + 2 y = c ( 1 + t ) e − t − t 2 + 2

Question

Getting a differential equation from a parametric function family  x  =  c      e          −      t        −  2  t  +  2       y  =  c  (  1  +  t  )      e          −      t        −      t    2    +  2

Rubi Boyle · Accepted Answer

Step 1This is not an answer but a comment too long to be edited in the comments section.                    (1)                    x        =        c                  e                      −            t                          −        2        t        +        2                                (2)                    y        =        c        (        1        +        t        )                  e                      −            t                          −                  t          2                +        2            I agree with the Ted Shifrin&#039;s comment. Supposing that the problem is to eliminate c and t in order to get a differential equation involving only   x  ,  y  ,            d      y              d      x        ,                    d        2            y              d              x        2              ,  .  .  . they are an infinity of possible answers depending on the kind of differential equation and the order expected.For example, for a first order nonlinear differential equation :Take   c      e          −      t       from (1) and put it into (2)                    (3)                    y        =        (        1        +        t        )        (        x        +        2        t        −        2        )        −                  t          2                +        2            Differentiate (1) and (2)                    (4)                              {                                                                                          d                    x                                                        d                    t                                                  =                −                c                                  e                                      −                    t                                                  −                2                                                                                                          d                    y                                                        d                    t                                                  =                −                c                                  e                                      −                    t                                                  t                −                2                t                                                                              ⟹                                              d            y                                d            x                          =        t            Putting   t  =      y    ′    =            d      y              d      x       from (4) into (3) gives the ODE :  y  =  (  1  +      y    ′    )  (  x  +  2      y    ′    −  2  )  −      y          ′      2        +  2But probably this is not what the question is asking for. The question should be clarified.

Getting a differential equation from a parametric function family x = c e <

Answered question

Answer & Explanation

New Questions in Integral Calculus