For this linear system: d x d t = x + y − 2 d y d t = x − y − 4Classify ( 1 , 1 )

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For this linear system:                    d        x                    d        t              =  x  +  y  −  2                    d        y                    d        t              =  x  −  y  −  4Classify   (  1  ,  1  )

cydostwng6c · Accepted Answer

d      y              d      x        =                              d          y                          d          t                                      d          x                          d          t                      =                    x        −        y        −        4                    x        +        y        −        2            At   (  1  ,  1  )  ,                      d        y                    d        x              ≠  0,  (  1  ,  1  )  ,                      d        y                    d        x              ≠  0. Indeed, at   (  1  ,  1  ),                     d        y                    d        x             is not defined.                    d        y                    d        x              =  0 when   x  −  y  −  4  =  0But this doesn&#039;t tell us too much: only that   x  −  y  =  4. (Two variables and one equation. Infinitely many solutions.)On the other hand, if you find the Jacobian and evaluate the eigenvalues, you&#039;ll find that your critical point is   (  3  ,  −  1  ), and is a saddle point.

For this linear system:dx/(dt)=x+y−2 dy/(dt)=x−y−4 Classify (1,1)

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