The system is as follows: i 1 = i 2 + i 3 , 50 sin ⁡ t = 6 i 1 + i 2 ′ + 5 i 2 , 50 sin ⁡ t = 6 i 1 + i 3 ′ , Find i 2 , i 3 .

Question

The system is as follows:                              i          1                                    =                  i          2                +                  i          3                ,                            50        sin        ⁡        t                            =        6                  i          1                +                  i          2          ′                +        5                  i          2                ,                            50        sin        ⁡        t                            =        6                  i          1                +                  i          3          ′                ,            Find       i    2    ,      i    3  .

scranna0o · Accepted Answer

Notice that the equations could be re-written as                                                        i              1                                ′                =                                            i              2                                ′                +                                            i              3                                ′                                                                        i              2                                ′                =        50        sin        ⁡        t        −        6                              i                          1              ,                                                H                                                                    −        5                              i                          2              ,                                                H                                                                                                                            i              3                                ′                =        50        sin        ⁡        t        −        6                              i                          1              ,                                                H                                                                        now replacing from the second and third equations into the first one, after simplification, we could write      d          d      t            [                                                                                        i                  1                                                                                                                                          i                  2                                                                                                                                          i                  3                                                                          ]    =      [                                                      −              12                                                          −              5                                            0                                                              −              6                                                          −              5                                            0                                                              −              6                                            0                                0                                ]        [                                                                                        i                  1                                                                                                                                          i                  2                                                                                                                                          i                  3                                                                          ]    +      [                                                      100              sin              ⁡              t                                                                          50              sin              ⁡              t                                                                          50              sin              ⁡              t                                            ]  now for the homogeneous part      d          d      t                                    i                                      H                      =      [                                                      −              12                                                          −              5                                            0                                                              −              6                                                          −              5                                            0                                                              −              6                                            0                                0                                ]                                i                                      H                    Decomposition of the above matrix yields      [                                                      −              12                                                          −              5                                            0                                                              −              6                                                          −              5                                            0                                                              −              6                                            0                                0                                ]    =      [                                        5                                1                                0                                                3                                              −              2                                            0                                                2                                3                                1                                ]        [                                                      −              15                                            0                                0                                                0                                              −              2                                            0                                                0                                0                                0                                ]                      [                                                            5                                            1                                            0                                                                    3                                                              −                  2                                                            0                                                                    2                                            3                                            1                                                    ]                    −        1            so that by defining                              j                                      H                      =                    [                                                            5                                            1                                            0                                                                    3                                                              −                  2                                                            0                                                                    2                                            3                                            1                                                    ]                    −        1                                          i                                      H                    we get      d          d      t                                    j                                      H                      =      [                                                      −              15                                            0                                0                                                0                                              −              2                                            0                                                0                                0                                0                                ]                                j                                      H                    which are three uncoupled first order ODEs which yield                              j                                      H                      =      [                                                                                        c                  1                                                                              e                                      −                    15                    t                                                                                                                                                            c                  2                                                                              e                                      −                    2                    t                                                                                                                                                            c                  3                                                                          ]  Also, for the particular part, according to the inhomogeneous part, we should have                              i                                      P                      =  sin  ⁡  t      [                                                                                        A                  1                                                                                                                                          A                  2                                                                                                                                          A                  3                                                                          ]    +  cos  ⁡  t      [                                                                                        B                  1                                                                                                                                          B                  2                                                                                                                                          B                  3                                                                          ]  Replacing the particular solution form into the original equation gives  sin  ⁡  t      (                  [                                                                              −                  12                                                                              −                  5                                                            0                                                                                      −                  6                                                                              −                  5                                                            0                                                                                      −                  6                                                            0                                            0                                                    ]                    [                                                                                                                        A                      1                                                                                                                                                                                      A                      2                                                                                                                                                                                      A                      3                                                                                                          ]            +              [                                                                                                                        B                      1                                                                                                                                                                                      B                      2                                                                                                                                                                                      B                      3                                                                                                          ]            +              [                                                                              100                                                                                                      50                                                                                                      50                                                                    ]              )    +  cos  ⁡  t      (                  [                                                                                                                        A                      1                                                                                                                                                                                      A                      2                                                                                                                                                                                      A                      3                                                                                                          ]            −              [                                                                              −                  12                                                                              −                  5                                                            0                                                                                      −                  6                                                                              −                  5                                                            0                                                                                      −                  6                                                            0                                            0                                                    ]                    [                                                                                                                        B                      1                                                                                                                                                                                      B                      2                                                                                                                                                                                      B                      3                                                                                                          ]              )    =      [                                        0                                                0                                                0                                ]  Now you should only summarize these steps to get the final result.

omgespit9q · Accepted Answer

Now an alternative (non-matrix based) solution. After eliminating 𝑖1 from the first equation we get                                                        i              2                                ′                =        −        11                              i            2                          −        6                              i            3                          +        50        sin        ⁡        t                                                                i              3                                ′                =        −        6                              i            2                          −        6                              i            3                          +        50        sin        ⁡        t            Decomposing into homogeneous and particular parts for the unknowns, for the first part we have                                                        i                              2                ,                                                      H                                                                                ′                =        −        11                              i                          2              ,                                                H                                                                    −        6                              i                          3              ,                                                H                                                                                                                            i                              3                ,                                                      H                                                                                ′                =        −        6                              i                          2              ,                                                H                                                                    −        6                              i                          3              ,                                                H                                                                        Now we shall assume the solutions to have an exponential behavior. That is                                          i                          2              ,                                                H                                                                    =                              A            2                                                e                          r              t                                                                                i                          3              ,                                                H                                                                    =                              A            3                                                e                          r              t                                          Replacing into the equations, we get                    (        11        +        r        )                              A            2                          +        6                              A            3                          =        0                            6                              A            2                          +        (        6        +        r        )                              A            3                          =        0

The system is as follows: i_1=i_2+i_3, 50sin t=6i_1+i'_2+5i'_2, 50sin t=6i_1+i'_3, Find i_2,i_3.

Answered question

Answer & Explanation

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