Question

Im using the logistic equation to work out a rabbit population question and just want to confirm my working out...

dy/dt = ay(b - y)

Separate variables:
dy / ay(b-y) = dt

Separate fractions:
dy [1/aby + 1/ab(b-y)] = dt

Multiply both sides by ab:
dy [1/y + 1/(b - y)] = ab dt

Integrate:
ln(y) - ln(b-y) = abt + C

Logarithm property:
ln(y / b - y) = abt + C

Algebra to solve for y:
y / (b - y) = Ce^(abt) [positive C]
-1 + b/b-y = Ce^(abt)
b/b-y = Ce^(abt)+ 1
b - y = b / (Ce^(abt) + 1)
y = b - b/(Ce^(abt) + 1)

is this right?

Answer 1

How did you separate the fractions?
that part looks fishy

Answer 2

...and if a is a constant I would leave it on the other side

Answer 3

\[{dy\over y(b-y)}=adt\]then do partial fractions

Answer 4

A rabbit population satisfies the logic equation
\[dy/dt=2\times10^{-7}y(10^6-y)\]
where t is the time measured in months. The population is suddenly reduced to 40% of its steady state size by myxamatiosis. If the myxamatosis then has no further effect, how large is the population 8 months later? How long will it take for the population to build up again to 90% of its steady state size?

Answer 5

ok you just did it differently than me
it looks good as far as I can tell

Answer 6

so \[a=2\times10^-7\]and \[b=10^6\]
so using the formula...\[dy/dt=ay(b-y)\] I need to take dt over to the RHS of the equation and bring everything else to the left?

Answer 7

yeah but it seems to work out the same...\[{dy\over y(b-y)}\implies\frac1b(\frac1y-\frac1{b-y})=adt\implies \frac1y-\frac1{b-y}=abdt\]\[\ln {y\over(1-y)}=abt+C\]\[{y\over(b-y)}=Ce^{abt}=-1+{b\over b-y}=Ce^{abt}\]I just kept going for a morning exercise, but clearly you and I are going to get the same answer :)

Answer 8

\[{b\over b-y}=Ce^{abt}+1\implies y=b-{b\over Ce^{abt}+1}\]yep :)

Answer 9

awesome... thanks!!

Answer 10

very welcome, thanks for the wake-up :)