My Problem is this given System of differential equations. y 1 ′ = 5 y 1 + 2 y 2 y 2 ′ = − 2 y 1 + y 2 My Approach was: again, i analyze, it must be a ordinary, linear System of equations, with both being of first-order. Than i built the corresponding Matrix as follows: ( y 1 ′ y 2 ′ ) ⏟ y → ′ = ( 5 2 − 2 1 ) ⏟ A ( y 1 y 2 ) ⏟ y → that's why: y → ′ = ( 5 2 − 2 1 ) y → Then I determined the eigenvalues:they are r 1 = 3 and r 2 = 3Knowing them, I can build the corresponding eigenvectors:they are v → 1 = ( − 1 + 1 ) and v → 2 = ( 0 0 ) Now i plug into the equation: x → = c 1 e r 1 t v 1 → + c 2 e r 2 t v 2 → x → = c 1 e 3 t ( − 1 1 ) + c 2 e 3 t ( 0 0 ) this lead to my result: y 1 = − c 1 e 3 t + 0 c 2 e 3 t y 2 = c 1 e 3 t + 0 c 2 e 3 t y 1 = − c 1 e 3 t y 2 = c 1 e 3 t But I doubt it's correct. My suspect are the eigenvectors, I really don't know if they are correct. And this could have lead to a wrong solution.

Question

My Problem is this given System of differential equations.      y      1      ′    =  5      y      1    +  2      y      2          y      2      ′    =  −  2      y      1    +      y      2  My Approach was: again, i analyze, it must be a ordinary, linear System of equations, with both being of first-order. Than i built the corresponding Matrix as follows:                      (                                                            y                1                              ′                                                                                        y                2                              ′                                                    )            ⏟                                                y          →                      ′              =                      (                                            5                                      2                                                          −              2                                      1                                      )            ⏟                        A                            (                                                            y                1                                                                                        y                2                                                    )            ⏟                                y        →            that&#039;s why:              y      →          ′    =      (                            5                          2                                      −          2                          1                      )        y    →  Then I determined the eigenvalues:they are       r    1    =  3 and       r    2    =  3Knowing them, I can build the corresponding eigenvectors:they are               v      →        1    =      (                            −          1                                      +          1                      )   and               v      →        2    =      (                            0                                      0                      )  Now i plug into the equation:      x    →    =      c    1        e                  r        1            t                  v      1        →    +      c    2        e                  r        2            t                  v      2        →          x    →    =      c    1        e          3      t            (                            −          1                                      1                      )    +      c    2        e          3      t            (                            0                                      0                      )  this lead to my result:      y    1    =  −      c    1        e          3      t        +  0      c    2        e          3      t              y    2    =      c    1        e          3      t        +  0      c    2        e          3      t                y    1    =  −      c    1        e          3      t              y    2    =      c    1        e          3      t      But I doubt it&#039;s correct. My suspect are the eigenvectors, I really don&#039;t know if they are correct. And this could have lead to a wrong solution.

Mateo Barajas · Accepted Answer

Hint:      X    ′    (  t  )  =  A  X  (  t  )if you have repeated eigen value like   c and   v is eigen vector correspond to   c then general solution is            X      (      t      )      =              e                  c          t                    v      and here solution is:            X      (      t      )      =      c              e                  3          t                            (                                            −              1                                                          1                                      )

My Problem is this given System of differential equations. <msubsup> y 1 <mi cla

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