How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation? y ′ = x 3 y 3 − 1

Name: How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation? y ′ = x 3 y 3 − 1
Uploaded: 2022-02-23T17:22:57+00:00
Description: How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation? y ′ = x 3 y 3 − 1

Question

How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation?      y    ′    =      x          3            y          3        −  1

trantegisis · Accepted Answer

Honestly, I don&#039;t think the solutions of your ODE can be written in elementary terms.Actually, any substitution of the type   u  =      y    α   won&#039;t simplify the ODE, because of that evil constant term -1.Neverthless, you could look for a power series solution of your ODE using the Frobenius method, that is:1. assume that a solution of your ODE can be expanded in a power series       ∑          n      =      0        ∞        a    n           x    n  2. evaluate the power series expansion of       y    3    (  x  ) and y&#039;(x) and plug them into the ODE,3. deduce from the ODE a recurrence relation for the coefficients       a    n  ,4. try to prove that the radius of convergence of       ∑          n      =      0        ∞        a    n           x    n   is   &amp;gt;  0then the sum   y  (  x  )  :=      ∑          n      =      0        ∞        a    n           x    n   will be an analytic solution of your ODE.It is easy to prove that if   y  (  x  )  =      ∑          n      =      0        ∞        a    n           x    n   then:1.       y    3    (  x  )  =      ∑          n      =      0        ∞        b    n           x    n  , where       b    n    :=      ∑          k      =      0        n        ∑          h      =      0              n      −      k            a    k           a    h           a          n      −      k      −      h       satisfies:      {                                        b            0                    =                      a            0            3                                                            b            n                    =                      1                          n                                             a                0                                                                     ∑                          k              =              1                                      n                                (          4          k          −          n          )                      a            k                                           b                          n              −              k                                              2.       y    ′    (  x  )  =      ∑          n      =      0        ∞    (  n  +  1  )         a          n      +      1               x    n  ,therefore plugging 1 and 2 into your ODE gives:      a    1    +  2      a    2       x  +  3      a    3           x    2    +      ∑          n      =      3        ∞    (  n  +  1  )         a          n      +      1               x    n    =  −  1  +      ∑          n      =      3        ∞        b          n      −      3            x    n      .Equating the coefficients of like powers of x, you obtain:      {                                        a            1                    =          −          1                                                  a            2                    =          0                                                  a            3                    =          0                                      (          n          +          1          )                                 a                          n              +              1                                =                      b                          n              −              3                                                , for           n          ≥          3                        i.e.:                    (1)                              {                                                                      a                  1                                =                −                1                                                                                      a                  2                                =                0                                                                                      a                  3                                =                0                                                                                      a                                      n                    +                    1                                                  =                                  1                                      n                    +                    1                                                                                     ∑                                      k                    =                    0                                                        n                    −                    3                                                                    ∑                                      h                    =                    0                                                        n                    −                    3                    −                    k                                                                    a                  k                                                                   a                  h                                                                   a                                      n                    −                    3                    −                    k                    −                    h                                                                              , for                 n                ≥                3.                                                                  Note that       a    0   cannot be determined using (1).Now there remains to be solved the problem of finding the radius of convergence of the power series whose coefficients are given by (1).

Jovany Clayton · Accepted Answer

You should highly notice that Bernoulli differental equation is of the form       y    ′    =  f  (  x  )      y    n    +  g  (  x  )  y rather than of the form       y    ′    =  f  (  x  )      y    n    +  g  (  x  )Approach 1:In fact       y    ′    =      x    3        y    3    −  1 belongs to an Abel equation of the first kind.Approach 2:Let u=xy,Then   y  =            u      x                          d        y                    d        x              =            1      x                          d        u                    d        x              −            u              x        2              ∴            1      x                          d        u                    d        x              −            u              x        2              =      u    3    −  1            1      x                          d        u                    d        x              =            u              x        2              +      u    3    −  1Let   v  =            1              x        2            Then                     d        u                    d        x              =                    d        u                    d        v                                d        v                    d        x              =  −            2              x        3                                d        u                    d        v               ∴  −            2              x        4                                d        u                    d        v              =            u              x        2              +      u    3    −  1  (  u  v  +      u    3    −  1  )                    d        v                    d        u              =  −  2      v    2  This belongs to an Abel equation of the second kind.

How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as

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