Please help explain solution to population growth model

Hi, 

I'm taking Mathematical Modelling this semester and am having trouble making sense of a question given to us on a worksheet to complete. I've attached the question and the solution given to us in the hope that you may be able to help me make sense of the solution as I'm struggling to understand how the equilibrium was found and the conclusion was reached that the equilibrium is unstable. Thanks a million in advance.

Student X

Attachment:

An equilibrium value exists when the bird population neither grows nor shrinks, ie the net rate of change of population is simply zero (dP/dt =0). This value as computed happens to be b/a. 

Perhaps solving the differential equation might yield greater insight into what's exactly happening. 

Since dP/dt=aP-b, 

∫1/(aP-b) dP = ∫ dt 

1/a*∫ a/(aP-b) dP = ∫ dt 

1/a*ln |aP-b| = t + C   where C is a constant of integration  

ln|aP-b|= at+B (where B is another constant= aC)

 aP-b = e^(at+B)=e^B *e^(at) = Ae^(at)   (where A is another constant =e^B) 

P = 1/a*[Ae^(at)+ b] = De^(at) +(b/a)  (where D is another constant =A/a)  

Hence, at the onset when t=0, initial population Po = D + b/a 

CASE 1

If Po is greater than b/a, then D + b/a > b/a ===>  D > 0 

Since e^(at) is an exponential growth factor, as t approaches infinity in the long run, the component of P which is  De^(at) will be unbounded and grows infinitely huge as well, which therefore implies the same thing about P.  

CASE 2   

If Po is lesser than b/a, then D + b/a < b/a ===>  D < 0 

Since e^(at) is an exponential growth factor, as t approaches infinity in the long run, the component of P which is De^(at) will ultimately decay towards zero (due to the negative nature of D), which therefore implies P will head downwards as well.   

Hope this clarifies. Peace. 

Best Regards,

Mr Koh