find the solution to the problem y ′ = ( 1 1 0 1 ) y , y ( 0 ) = ( 4 0 ) I know i have to find the eigenvalues and eigenvectors of the matrix A = ( 1 1 0 1 ) there's only one eigenvalue which is 1 and only "one" eigenvector and we can choose ( 1 0 ) but now I dont know what to do.what comes next?

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find the solution to the problem       y    ′    =      (                            1                          1                                      0                          1                      )    y  ,  y  (  0  )  =      (                            4                                      0                      )  I know i have to find the eigenvalues and eigenvectors of the matrix   A  =      (                            1                          1                                      0                          1                      )  there&#039;s only one eigenvalue which is 1 and only &quot;one&quot; eigenvector and we can choose       (                            1                                      0                      )  but now I dont know what to do.what comes next?

talhekh · Accepted Answer

Note: this method is not a general one, but is probably easier for the particular example provided.In this case, you do not need to diagonalize the matrix. Indeed, if we write   y  =  (      y    1    ,      y    2    ), then the system is simply:      {                                        y            1            ′                                              =                      y            1                    +                      y            2                                                            y            2            ′                                              =                      y            2                                  with initial conditions       y    1    (  0  )  =  4, and       y    2    (  0  )  =  0. Notice that the second equation is easy to solve:      y    2    (  x  )  =  C  exp  ⁡  (  x  )  .Since       y    2    (  x  )  =  0, we get       y    2    =  0, thus the first equation reduces to       y    1    ′    =      y    1  . Finally,  y  (  x  )  =      (                            4          exp          ⁡          (          x          )                                      0                      )    .

Audrina Jackson · Accepted Answer

If you know how to find an exponential of the matrix, then you no longer need to find eigenvectors and generalised eigenvectors, because your solution is precisely   y  (  t  )  =  e  x  p  (  A  t  )  y  (  0  ); in your case it is       y    1    (  t  )  =  4      e    t    ,        y    2    (  t  )  =  0.

find the solution to the problem y ′ </msup> = ( <mtable r

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