d y d x = y 2 + 3 x 2 − 1 , y ( 1 ) = 1on D = | x − 1 | &lt;= 1 , | y − 1 | &lt;= 1 Find the second approximation to the solution and estimate the error term

Question

d      y              d      x        =      y    2    +  3      x    2    −  1  ,    y  (  1  )  =  1on   D  =            |        x    −    1          |        &amp;lt;=    1    ,          |        y    −    1          |        &amp;lt;=    1  Find the second approximation to the solution and estimate the error term

lodosr · Accepted Answer

We are given:                    (1)                                          d            y                                d            x                          =                  y          2                +        3                  x          2                −        1        ,        y        (        1        )        =        1        ,                           on                                  D        =                              |                    x          −          1                      |                    ≤          1          ,                      |                    y          −          1                      |                    ≤          1                    The Picard-Lindelöf iteration is given by:                    (2)                                          y            0                    (          x          )          =                      y            0                    ,                                            y                          n              +              1                                (          x          )          =                      y            0                    +                      ∫                                          x                0                                      x                    f          (          s          ,                      y            n                    (          s          )          )          d          s                    For   (  1  ), we have:   f  (  s  ,      y    n    (  s  )  )  =      y    n    2    +  3      s    2    −  1 and using   (  2  ), yields:            y      0        (    1    )    =          y      0        =    1              y      1        (    x    )    =          y      0        +          ∫                        x          0                    x        f    (    s    ,          y      0        (    s    )    )    d    s    =    1    +          ∫              1            x        (    (    1          )      2        +    3          s      2        −    1    )    d    s    =          x      3                  y      2        (    x    )    =          y      0        +          ∫                        x          0                    x        f    (    s    ,          y      1        (    s    )    )    d    s    =    1    +          ∫              1            x        (    (          s      3              )      2        +    3          s      2        −    1    )    d    s    =                            x          7                7              +          x      3        −    x    +                  6        7            If you need one more iteration, you would get:      y    3    (  x  )  =            x              15              735    +            2              x                  11                      77    −            2              x        9              63    +            3              x        8              98    +            x      7        7    −            2              x        5              5    +            3              x        4              7    +            4              x        3              3    −            6              x        2              7    −            13      x        49    +      4099    6930

d y </mrow> d x </mrow> </mfrac> =

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