System of differential equations x ′ = 4 x − 2 y y ′ = 3 x − y − 2 e 3 t Initial conditions are x ( 0 ) = y ( 0 ) = 0

Question

System of differential equations      x    ′    =  4  x  −  2  y      y    ′    =  3  x  −  y  −  2      e          3      t      Initial conditions are   x  (  0  )  =  y  (  0  )  =  0

altaryjny94 · Accepted Answer

Step 1            x        ′    +      [                            −          4                          2                                      −          3                          1                      ]        x    =      [                            0                                      −          2                      e                          3              t                                            ]        e          A      t                  x        ′    +  A      e          A      t            x    =      e          A      t        g  (  t  )  A  =      P          −      1        D  P  A  =      [                            1                          2                                      1                          3                      ]        [                            −          2                          0                                      0                          −          1                      ]        [                            3                          −          2                                      −          1                          1                      ]          e          A      t        =      P          −      1            e          D      t        P      x    =       e          −      A      t        C  +      e          −      A      t        ∫      e          A      t        g  (  t  )  d  t        e          −      A      t        C  +      P          −      1            e          −      D      t        ∫      e          D      t        P  g  (  t  )  d  tStep 2      e          D      t        P  g  (  t  )  =      [                            4                      e            t                                                −          2                      e                          2              t                                            ]        x    =      e          −      A      t        C  +      P          −      1            [                            4                                      −          1                      ]        e          3      t          P      x    (  0  )  =  C  +      [                            4                                      −          1                      ]      C  =      [                            −          4                                      1                      ]        x    =      [                            −          4                      e                          2              t                                +          2                      e                          t                                +          2                      e                          3              t                                                            −          4                      e                          2              t                                +          3                      e            t                    +                      e                          3              t                                            ]

Keenan Conway · Accepted Answer

Step 1Subtracting both equations, you get      x    ′    −      y    ′    =  x  −  y  +  2      e          −      3      t      which you can solve for   x  −  y, giving  x  −  y  =  C      e    t    −            e              −        3        t              2    ,where   C  =      1    2  Step 2Then solve      y    ′    =  3      (    y    +                            e          t                −                  e                      −            3            t                              2        )    −  y  +  2      e          −      3      t        .

System of differential equations. x'=4x-2y, y'=3x-y-2e^{3t}

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