Consider the following linear system of differential equations: { x ˙ = − 4 y y ˙ = x where x ( t ) and y ( t ) are unknown real functions.One can simply verify that the general solution is ( x ( t ) y ( t ) ) = c 1 ( cos ⁡ ( 2 t ) 1 2 sin ⁡ ( 2 t ) ) + c 2 ( − 2 sin ⁡ ( 2 t ) cos ⁡ ( 2 t ) ) where c 1 and c 2 are real parameters.Question: Which is the exponential form of this expression? It should by something like ( x ( t ) y ( t ) ) = k 1 e i 2 t ( P 11 P 21 ) + k 2 e − i 2 t ( P 12 P 22 ) where k 1 and k 2 should (?) be complex parameters.

Question

Consider the following linear system of differential equations:      {                                                      x              ˙                                =          −          4          y                                                                y              ˙                                =          x                        where   x  (  t  ) and   y  (  t  ) are unknown real functions.One can simply verify that the general solution is      (                            x          (          t          )                                      y          (          t          )                      )    =      c    1        (                            cos          ⁡          (          2          t          )                                                  1            2                    sin          ⁡          (          2          t          )                      )    +      c    2        (                            −          2          sin          ⁡          (          2          t          )                                      cos          ⁡          (          2          t          )                      )  where       c    1   and       c    2   are real parameters.Question: Which is the exponential form of this expression? It should by something like      (                            x          (          t          )                                      y          (          t          )                      )    =      k    1        e          i      2      t            (                                        P                          11                                                                        P                          21                                            )    +      k    2        e          −      i      2      t            (                                        P                          12                                                                        P                          22                                            )  where       k    1   and       k    2   should (?) be complex parameters.

heilaritikermx · Accepted Answer

I think the key here is to rewrite the linear system as the following matrix equation                                          (                                                            0                                                  −                  4                                                                              1                                                  0                                                      )                                (                                                            x                                                                              y                                                      )                    =                      (                                                                                                      x                      ˙                                                                                                                                                              y                      ˙                                                                                            )                          ⏟                    A                        x          →                    =                                                  x              →                                ˙                      .The characteristic equation is       λ    2    +  4  =  0  , so you&#039;re correct in using   cos  ⁡  (  2  t  ) and   sin  ⁡  (  2  t  )  . Now, to get at a complex exponential solution, you need to find eigenvectors of   A with eigenvalues   −  2  i  ,  2  i  . These are      (                            2          i                                      1                      )      and        (                            −          2          i                                      1                      )  respectively.Consequently, you can write      (                            x          (          t          )                                      y          (          t          )                      )    =      c    1        (                            2          i                                      1                      )        e          2      i      t        +      c    2        (                            −          2          i                                      1                      )        e          −      2      i      t        .It is worth noticing that the eigenvectors, are attached to their associated eigenfunctions, and that the eigenvalue of the derivative operator is identical to that of the eigenvalue under the action of   A  ..

kwisangqaquqw3 · Accepted Answer

Well, defining   k  ≡      c    2    −  i      c    1        /    2, you get a real      (                            x          (          t          )                                      y          (          t          )                      )    =  k         e          i      2      t            (                            i                                      1                      /                    2                      )    +      k    ∗           e          −      i      2      t            (                            −          i                                      1                      /                    2                      )    .You did not specify anything about the form of the Ps, but you have sufficient freedom to recast these into something of your choice.Note the problem simplifies to a triviality if you defined   z  ≡  x  +  i  2  y, so             z      ˙        =  i  2     z... take the c.c. ... do you see the point?

Consider the following linear system of differential equations: { <mtable columnalign="

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