Given a second order differential equation y ″ + f ( y ) y ′ + g ( y ) = 0 ,write an equivalent system of first order equations with transformations x 1 = y , x 2 = y ′ + ∫ 0 y f ( s ) d s .This is what I did: x 1 ′ = y ′ = x 2 − ∫ 0 y f ( s ) d s = x 2 − ∫ 0 x 1 f ( s ) d s x 2 ′ = y ″ + f ( y ) = − f ( y ) y ′ − g ( y ) + f ( y ) = f ( y ) ( 1 − y ′ ) − g ( y ) = f ( x 1 ) ( 1 − x 2 + ∫ 0 x 1 f ( s ) d s ) − g ( x 1 )I feel like this answer is wrong though, because I am not sure if I'm doing the standard procedure.

Question

Given a second order differential equation      y    ″    +  f  (  y  )      y    ′    +  g  (  y  )  =  0  ,write an equivalent system of first order equations with transformations      x          1        =  y  ,      x          2        =      y    ′    +      ∫          0              y        f  (  s  )  d  s  .This is what I did:      x          1        ′    =      y    ′    =      x          2        −      ∫          0              y        f  (  s  )  d  s  =      x          2        −      ∫          0                      x                  1                      f  (  s  )  d  s      x          2        ′    =      y    ″    +  f  (  y  )  =  −  f  (  y  )      y    ′    −  g  (  y  )  +  f  (  y  )  =  f  (  y  )  (  1  −      y    ′    )  −  g  (  y  )  =  f  (      x          1        )  (  1  −      x          2        +      ∫          0                      x                  1                      f  (  s  )  d  s  )  −  g  (      x          1        )I feel like this answer is wrong though, because I am not sure if I&#039;m doing the standard procedure.

Angelo Murray · Accepted Answer

Given a second-order, generally non-linear, ordinary differential equaition of the form                    (1)                              y          ″                +        f        (        y        )                  y          ′                +        g        (        y        )        =        0        ,            and the specified transformation of variables                    (2)                              x          1                =        y        ,                                (3)                              x          2                =                  y          ′                +                              ∫            0            y                    f          (          s          )                    d          s          ,                    we need to find       x    1    ′   and       x    2    ′  ; we start with the obvious                    (4)                              x          1          ′                =                  y          ′                ,            and writing (3) in the form                    (5)                              y          ′                =                  x          2                −                              ∫            0            y                    f          (          s          )                    d          s          ,                    we further transform (4) to                    (6)                              x          1          ′                =                  x          2                −                              ∫            0            y                    f          (          s          )                    d          s          =                      x            2                    −                                    ∫              0                                                x                  1                                                      f            (            s            )                        d            s            ;                              turning now to       x    2    ′  , we differentiate (3), applying the Leibniz integral rule to                    (7)                                          ∫            0            y                    f          (          s          )                    d          s          ,                    and find                    (8)                              x          2          ′                =                  y          ″                +        f        (        y        )                  y          ′                ;            we then write (1) in the form                    (9)                              y          ″                =        −        f        (        y        )                  y          ′                −        g        (        y        )            and substitute this into (8) to obtain      x    2    ′    =      y    ″    +  f  (  y  )      y    ′  then the system (6), (10) is the transformed version of (1).Going the other way:we start with (6),                    (9)                              x          1          ′                =                  x          2                −                              ∫            0                                          x                1                                              f          (          s          )                    d          s          ,                    and (10)                    (10)                              x          2          ′                =        −        g        (                  x          1                )        ;            using (2) and (3), and th Leibniz rule again,      y    ″    =      x    1    ″    =      x    2    ′    −  f  (      x    1    )      x    1    ′  that is,                    (12)                              y          ″                +        f        (        y        )                  y          ′                +        g        (        y        )        =        0        ,            our original equation (1).

Given a second order differential equation y ″ </msup> + f ( y

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