# Math Help - tricky degree standard differential equation and calculus question! need some advice

1. ## tricky degree standard differential equation and calculus question! need some advice

Hi,

We've been given this tricky question and I have a few ideas jotted down but when I've gone to try them they haven't been making sense and I've been going round in circles and it's truly bugging me now so I need somebodies help please!! The question is:

Suppose we have two species interacting on the same environment. After observations we find that the size of their populations obey the following set of equations:

dW/dt = αW−βV
dV/dt = −γV+δW

where α,β,γ,δ are positive constants therefore solve the system by proposing
W (t) = A sin(ωt) + B cos(ωt)
V (t) = C sin(ωt) + D cos(ωt)

Find the values of the constants A,B,C and D ONLY in terms of the parameters α,β,γ,δ ω

So basically I was thinking maybe it involves subbing them into the original system, differentiating them and comparing coefficients and then solving a system of equations (however this doesn't seem to work as you can only get the constants A,B,C and D in terms of a MIX of constants and parameters and we solely what parameters and plus you only obtain 4 homogenous equations implying that at least 1 of the constants must equal zero so that's got to incorrect or is it?

Now I was thinking perhaps converting them into harmonic form and then playing around it with but then I come to the same conclusion as above pretty much where I get a mix of constants and parameters when I only want parameters.

So maybe involving an idea to do with these set of differential equations representing a second order equation or some sort (A second order equation and its solutions)

but I'm just going round in circles any help please guys!

Best Wishes,

Charlotte

2. ## Re: tricky degree standard differential equation and calculus question! need some adv

You'll end up with four equations if you also consider the homogenous solutions to the two differential equations. The homogenous solutions give you $\alpha A = \beta C$ and $\alpha B = \beta D$

Substitute them into the two other equations:
$A\omega \cos{\omega t}-B\omega \sin{\omega t}=\alpha (A \sin{\omega t}+B \cos{\omega t})-\beta (C \sin{\omega t}+D \cos{\omega t})$
$C\omega \cos{\omega t}-D\omega \sin{\omega t}=\delta (A \sin{\omega t}+B \cos{\omega t})-\gamma (C \sin{\omega t}+D \cos{\omega t})$