we consider the following model: { x ′ = x ( 4 − x − y ) y ′ = y ( 2 + 2 α − y − α x ) a) Find the critical point P does not depend on α and having coordinates strictly positive.-&gt; The critical points are: ( 0 , 0 ) , ( 0 , 2 + 2 α ) , ( 2 , 2 ) , ( 4 , 0 ) then P = ( 2 , 2 )b) Assuming d y d x ,solve the system for the value α 0 of α and such that P is not hyperbolic,Draw the phase portrait for α = α 0 c) what is the nature of P for 0 &lt; α &lt; α 0 and α 0 &lt; α, Draw the shape of the phase portraits for these values of αCan someone tell me how to solve b), why they use d y d x ?

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we consider the following model:      {                                        x            ′                                    =          x          (          4          −          x          −          y          )                                                  y            ′                                    =          y          (          2          +          2          α          −          y          −          α          x          )                        a) Find the critical point   P does not depend on   α and having coordinates strictly positive.-&amp;gt; The critical points are:   (  0  ,  0  )  ,  (  0  ,  2  +  2  α  )  ,  (  2  ,  2  )  ,  (  4  ,  0  ) then   P  =  (  2  ,  2  )b) Assuming                     d        y                    d        x            ,solve the system for the value       α    0   of   α and such that   P is not hyperbolic,Draw the phase portrait for   α  =      α    0  c) what is the nature of   P for   0  &amp;lt;  α  &amp;lt;      α    0   and       α    0    &amp;lt;  α, Draw the shape of the phase portraits for these values ​​of   αCan someone tell me how to solve b), why they use                     d        y                    d        x            ?

Shawn Castaneda · Accepted Answer

For system of the shape      {                                        x            ′                    =          f          (          x          ,          y          )                                                  y            ′                    =          g          (          x          ,          y          )                        If one had a solution such one could write   y  (  t  )  =      y    ~    (  x  (  t  )  ) then one would have that            d      y              d      t        =            d                  y        ~                    d      x                  d      x              d      t      So, with an abuse of notation one writes            d      y              d      x        =                    d        y                    d        t                            d        x                    d        t              =            f      (      x      ,      y      )              g      (      x      ,      y      )      So assumming             d      y              d      x       could simply mean that   g  (  x  ,  y  )  ≠  0 for a solution, and thus the previous is well defined. With this trick you turn a PDE into a ODE (provided the above works). Now you have a fraction of polynomial, which maybe you can integrate and get something nice, hence they say&#039;&#039;solve&#039;&#039;.

we consider the following model: { <mtable columnalign="left left" rowspacing=".2em"

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