Solving Coupled Differential EquationsI have the following differential equations, for modeling predator-prey relationships: d x d t = A x − B x y d y d t = C x y − D yWhere A, B, C, and D are constants. How could I go about solving this? I've only really worked with basic first order differential equations before, and I've found little to help me figure this out. Any help would be much appreciated.

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Solving Coupled Differential EquationsI have the following differential equations, for modeling predator-prey relationships:            d      x              d      t        =  A  x  −  B  x  y            d      y              d      t        =  C  x  y  −  D  yWhere A, B, C, and D are constants. How could I go about solving this? I&#039;ve only really worked with basic first order differential equations before, and I&#039;ve found little to help me figure this out. Any help would be much appreciated.

vermieterbx · Accepted Answer

Step 1Generally for this type of model one would start analysing what happens at equilibrium, i.e. whether there is a fixed point, if they are stable or not by looking at the Jacobian of the system.Step 2First find the null-clines setting:   0  =  A  x  −  B  x  y  =      f    1    (  x  ,  y  ) and   C  x  y  −  D  y  =      f    2    (  x  ,  y  )  =  0 then you get   x  =  0  ,  y  =  A      /    B for       f    1    (  x  ,  y  ) and   y  =  0  ,  x  =  D      /    C for       f    2    (  x  ,  y  ). Finding the intersection of the null-clines you get two fixed points (0,0) (both population are extinct) and (D/C,A,B) (both populations are non-zero). Then if you look at the Jacobian (matrix of partial derivatives of       f    1    (  x  ,  y  )  ,      f    2    (  x  ,  y  ) computed in the two fixed points you can understand which is the stable one and where the system ends up.

Kailey Vargas · Accepted Answer

Step 1While complete analytic solution is not likely, it&#039;s curious that we can reduce this system to a couple of nonlinear first order ODEs for x(t) and y(t).Let&#039;s do this for y only, since the same procedure can be applied to x as well.First, we write down some useful relationships:            x      ˙        =  x  (  A  −  B  y  )  C  x  −  D  =                    y        ˙              y    C  x  y  =            y      ˙        +  D  yNow we differentiate the second equation w.r.t. t:            y      ¨        =  (  C  x  −  D  )            y      ˙        +  C  y            x      ˙        =                              y          ˙                    2        y    +  (  A  −  B  y  )  (            y      ˙        +  D  y  )Since the equation doesn&#039;t containt t explicitly, we can reduce the order by the usual substitution:            y      ˙        =  u  (  y  )  ,              y      ¨        =  u  ⋅      u    ′  We obtain:  u      u    ′    =            u      2        y    +  (  A  −  B  y  )  (  u  +  D  y  )Now we introduce another function:  u  =  y  ⋅  v  (  y  )  ,        u    ′    =  v  +  y      v    ′  We get:  y  v  (  v  +  y      v    ′    )  =  y      v    2    +  (  A  −  B  y  )  (  v  +  D  )  yStep 2Simplifying, we obtain:  y  v      v    ′    =  (  A  −  B  y  )  (  v  +  D  )But this is a separable equation. So:            v      d      v              v      +      D        =      (          A      y        −    B    )    d  y  v  −  D  ln  ⁡  (  v  +  D  )  =  A  ln  ⁡  y  −  B  y  +      c    1  Getting back to the original function, we have:                    y        ˙              y    −  D  ln  ⁡      (                            y          ˙                    y        +    D    )    =  A  ln  ⁡  y  −  B  y  +      c    1  Or:                    (1)                                          y            ˙                          −        D        y        ln        ⁡        (                              y            ˙                          +        D        y        )        =        (        A        −        D        )        y        ln        ⁡        y        −        B                  y          2                +                  c          1                y            This doesn&#039;t look like anything solvable, but it is indeed a 1st order ODE for y only.We also need to determine c1 from the original system somehow, because the extra constant shouldn&#039;t be here.Step 2We can actually resolve (1) for the derivative using Lambert W (product logarithm) function. Transforming the equation:  (            y      ˙        +  D  y      )          −      D      y            e                            y          ˙                      =      y          (      A      −      D      )      y            e          −      B              y        2            +              c        1            y                                    y          ˙                    +      D      y              D      y        exp  ⁡      (    −                                        y            ˙                          +        D        y                    D        y              )    =            e              −        1        −                  c          1                          /                D              D        y          −      A              /            D        exp  ⁡      (          B      D        y    )  This has a solution:                              y          ˙                    +      D      y              D      y        =  −  W      (    −                  e                  −          1          −                      c            1                                /                    D                    D              y              −        A                  /                D              exp    ⁡          (              B        D            y      )        )                      (2)                                          y            ˙                          =        −        D        y                  (          1          +          W                      (            −                                          e                                  −                  1                  −                                      c                    1                                                        /                                    D                                            D                                      y                              −                A                                  /                                D                                      exp            ⁡                          (                              B                D                            y              )                        )                    )

Solving Coupled Differential Equations. I have the following differential equations, for modeling predator-prey relationships: dx/dt=Ax-Bxy

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