Dependent differential equationsSo I have been working with these differential equations that are dependent on each other: d y d t = 0.1 y − 0.01 ( 10000 − a ) d a d t = 0.1 a − 10 y So the first equation represents the change rate of predatory fish while the other equation is a representation of small fish.My question is: is there a way to find a general solution for these equations, say in form of an exponential function? And what does that factor 0.01 in the first equation stand for, having 10000 as the population of the small fish at the beginning?

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Dependent differential equationsSo I have been working with these differential equations that are dependent on each other:                                          d            y                                d            t                                              =        0.1        y        −        0.01        (        10000        −        a        )                                                  d            a                                d            t                                              =        0.1        a        −        10        y            So the first equation represents the change rate of predatory fish while the other equation is a representation of small fish.My question is: is there a way to find a general solution for these equations, say in form of an exponential function? And what does that factor 0.01 in the first equation stand for, having 10000 as the population of the small fish at the beginning?

Alayna Phillips · Accepted Answer

Step 1      y    ′    =  0.1  y  −  0.01  (  10000  −  a  )     (  1  )      a    ′    =  0.1  a  −  10  y     (  2  )Differentiate (2) (since a and y differentiable):      a    ″    =  0.1      a    ′    −  10      y    ′      ⟹        y    ′    =            0.1              a        ′            −              a        ″              10  Step 2Also from (2) you get:  y  =            0.1      a      −              a        ′              10  Substitute those in (1) amd you get:            0.1              a        ′            −              a        ″              10    =  0.01  (  0.1  a  −      a    ′    )  −  0.001  (  10000  −  a  )    ⟹        a    ″    −  0.2      a    ′    +  0.011  a  =  10which is a non-homogeneous second order ODE. Once you find a you can find the solution y.

Scarlet Marshall · Accepted Answer

Step 1You can rewrite the system in matrix form, introducing the vector   p  :=  (  y  ,  a      )    T  . The system now reads      p    ′    =  M  p  +  qwhere M is a constant matrix and q is a constant vector. We can get rid of q by trying a contant solution       p    p  , such that      p    p    ′    =  0  =  M      p    p    +  q  .Then by subtraction,      p    ′    =  M  p  .Step 2Now assume a solution with an exponential form  p  =  c      e          λ      t        .We plug it in the equation and get  λ  c      e          λ      t        =  M  c      e          λ      t        ,or  M  c  =  λ  c  .This shows that   λ is an Eigenvalue of M and c an Eigenvector. The general solution will combine the two Eigenvalues of M.

Dependent differential equations. So I have been working with these differential equations that are dependent on each other: (dy)/(dt)=0.1y-0.01(10000-a), (da)/(dt)=0.1a-10y

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