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

Solving Coupled Differential Equations
I have the following differential equations, for modeling predator-prey relationships:
$\frac{dx}{dt}=Ax-Bxy$
$\frac{dy}{dt}=Cxy-Dy$
Where 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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

vermieterbx
Step 1
Generally 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 2
First find the null-clines setting: $0=Ax-Bxy={f}_{1}\left(x,y\right)$ and $Cxy-Dy={f}_{2}\left(x,y\right)=0$ then you get $x=0,y=A/B$ for ${f}_{1}\left(x,y\right)$ and $y=0,x=D/C$ for ${f}_{2}\left(x,y\right)$. 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 ${f}_{1}\left(x,y\right),{f}_{2}\left(x,y\right)$ computed in the two fixed points you can understand which is the stable one and where the system ends up.
###### Not exactly what you’re looking for?
Kailey Vargas
Step 1
While complete analytic solution is not likely, it's curious that we can reduce this system to a couple of nonlinear first order ODEs for x(t) and y(t).
Let's do this for y only, since the same procedure can be applied to x as well.
First, we write down some useful relationships:
$\stackrel{˙}{x}=x\left(A-By\right)$
$Cx-D=\frac{\stackrel{˙}{y}}{y}$
$Cxy=\stackrel{˙}{y}+Dy$
Now we differentiate the second equation w.r.t. t:
$\stackrel{¨}{y}=\left(Cx-D\right)\stackrel{˙}{y}+Cy\stackrel{˙}{x}=\frac{{\stackrel{˙}{y}}^{2}}{y}+\left(A-By\right)\left(\stackrel{˙}{y}+Dy\right)$
Since the equation doesn't containt t explicitly, we can reduce the order by the usual substitution:
$\stackrel{˙}{y}=u\left(y\right),\phantom{\rule{2em}{0ex}}\stackrel{¨}{y}=u\cdot {u}^{\prime }$
We obtain:
$u{u}^{\prime }=\frac{{u}^{2}}{y}+\left(A-By\right)\left(u+Dy\right)$
Now we introduce another function:
$u=y\cdot v\left(y\right),\phantom{\rule{2em}{0ex}}{u}^{\prime }=v+y{v}^{\prime }$
We get:
$yv\left(v+y{v}^{\prime }\right)=y{v}^{2}+\left(A-By\right)\left(v+D\right)y$
Step 2
Simplifying, we obtain:
$yv{v}^{\prime }=\left(A-By\right)\left(v+D\right)$
But this is a separable equation. So:
$\frac{vdv}{v+D}=\left(\frac{A}{y}-B\right)dy$
$v-D\mathrm{ln}\left(v+D\right)=A\mathrm{ln}y-By+{c}_{1}$
Getting back to the original function, we have:
$\frac{\stackrel{˙}{y}}{y}-D\mathrm{ln}\left(\frac{\stackrel{˙}{y}}{y}+D\right)=A\mathrm{ln}y-By+{c}_{1}$
Or:
$\begin{array}{}\text{(1)}& \stackrel{˙}{y}-Dy\mathrm{ln}\left(\stackrel{˙}{y}+Dy\right)=\left(A-D\right)y\mathrm{ln}y-B{y}^{2}+{c}_{1}y\end{array}$
This doesn'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't be here.
Step 2
We can actually resolve (1) for the derivative using Lambert W (product logarithm) function. Transforming the equation:
$\left(\stackrel{˙}{y}+Dy{\right)}^{-Dy}{e}^{\stackrel{˙}{y}}={y}^{\left(A-D\right)y}{e}^{-B{y}^{2}+{c}_{1}y}$
$\frac{\stackrel{˙}{y}+Dy}{Dy}\mathrm{exp}\left(-\frac{\stackrel{˙}{y}+Dy}{Dy}\right)=\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)$
This has a solution:
$\frac{\stackrel{˙}{y}+Dy}{Dy}=-W\left(-\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)\right)$
$\begin{array}{}\text{(2)}& \stackrel{˙}{y}=-Dy\left(1+W\left(-\frac{{e}^{-1-{c}_{1}/D}}{D}{y}^{-A/D}\mathrm{exp}\left(\frac{B}{D}y\right)\right)\right)\end{array}$
###### Not exactly what you’re looking for?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee