Help solving system of differential equations

[frank1234]

I am doing a project for my ODE class that involves modeling a predator-prey relationship with a harvesting factor. I have not yet learned how to solve a system of equations. It would be greatly appreciated if somebody could point me in the right direction on how to go about solving this. The two equations I have are as follows:



[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT]
[FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]γ[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT]

[FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]γ[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]2[/FONT] are all positive constants

x and y are function of t

 

[Ishuda]

I am doing a project for my ODE class that involves modeling a predator-prey relationship with a harvesting factor. I have not yet learned how to solve a system of equations. It would be greatly appreciated if somebody could point me in the right direction on how to go about solving this. The two equations I have are as follows:



[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT]
[FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]′[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]γ[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT]

[FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]γ[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main]2[/FONT] are all positive constants

x and y are function of t

As in ordinary differential equations, different techniques are used depending on the particular equations. In this particular example, that x'/x = a-E1 - [FONT=MathJax_Math-italic]α[/FONT][FONT=MathJax_Math-italic]y looks inviting but, looking at the second equation, it also looks like I might do a substitution of f(t) =\(\displaystyle \int^t y(s) ds\)[/FONT] so that f' = y, etc. One could then solve for x in terms of f and t, plug that into the second equation and plow through the necessities [I haven't gone all the way to the end but I think it is just tedious rather than difficult or impossible].

 

[HallsofIvy]

These can be written as \(\displaystyle x'= (a- E_1)x- \alpha xy\) and \(\displaystyle y'= -(c+ E_2)y- \gamma xy\). Because of that "xy" these are non-linear equations- and there generally is no "easy" way to solve non-linear equations!

One way of approaching this is to differentiate the first equation again: \(\displaystyle x''= (a- E_1)x'- \alpha y- \alpha xy'\). We can replace that y' with the second equation: \(\displaystyle y'= -(c+ E_2)y- \gamma xy\) so that \(\displaystyle x''= (a- E_1)x'- \alpha y- \alpha x((c+ E_2)y- \gamma xy)= (a- E_1)x'- \alpha(1+ x(c+ E_2)- \gamma x^2)y\).

Now, from the first equation, \(\displaystyle y= \frac{(a- E_1)x- x'}{\alpha x}\). Replace y by that in the previous equation to get a single (non-linear) equation for x.