1. ## Prey/Predator Model (need help asap please)

Hey I need help with a question regarding a prey/predator model. The question is in 2 parts (paraphrasing them). (i) if there are no predators, what is the solution (e.g will the population exponentially grow/decline). (ii) if there are no prey, what is the solution (e.g will the population exponentially grow/decline).

dx/dt = 0.3x - xy/100 <--- prey

dy/dt = 15y(1-y/15) +25xy <--- predator

My solutions.

(i) dx/dt = 0.3x - xy/100
dx/dt = 0.3x (y=0)

dx/x = 0.3 dt

integrating

ln(x) = 0.3t + c
x= ke^0.3t (k = e^c)

therefore the population is increasing exponentially.

(ii) dy/dt = 15y(1-y/15) +25xy
dy/dt = 15y(1-y/15) (x=0)
dy/dt = 15y - y^2
dy/dt = -y(-15 + y)
dy/y(-15+y) = -1 dt

integrating from this point is where I'm finding difficulty. I've attempted it, but I'm pretty sure I'm wrong.

(1/y - 1/(-15+y))dy = -1 dt
ln(y) - ln(-15+y) = -t + c
ln(y/(-15+y)) = -t + c
y/(-15+y) = ke^-t (k = e^c)
y = (ke^-t)(-15 + y)

the last line doesn't make sense to me, because I was trying to get y = something that doesn't include a 'y' in it... can someone please help me?

2. Also I'd like to ask another question related to this. If you've got this equation:

dR/dt = 0.4R

and you integrate it.

dR/R = 0.4 dt

ln(R) = 0.4t + c

R(t) = ke^0.4t (k = e^c)

So dR/dt = 0.4(8)
= 3.2

(1) R = 8 + 3.2 = 11.2

Now if I plug the same values into R(t):

k = 8
t = 0

R(t) = 8e^0.4(0) = 8

I get 8 which seems correct.

now If I put t = 1, I get:

R(t) = 8e^0.4(1) = 11.9346 (4 d.p)

Now according to my (1) I should get 11.2 not 11.9346. Why is this?

Also I'd like to know why R(t) = k(1.4)^t works?

From what I understand, R(t) = ke^0.4t is the solution, not R(t) = k(1.4)^t.