find general soln for dP/dt = [bkPe^(-kt)] / [1 - be^(-kt)]

Assume that a population grows at a rate given by:

. . .dP/dt = [bkPe^(-kt)] / [1 - be^(-kt)]

...where b and k are positive constants, and the initial population is P_o. Find the general solution for the differential equation.

I'm not sure where to begin. I'm assuming I have to separate the equations, but I'm not sure how.

almostacasper said:

I'm assuming I have to separate the equations, but I'm not sure how.

Click to expand...

Try dividing through by "P", and "multiplying" the "dt" over to the right-hand side. Then try a u-substitution on the right-hand side.

Eliz.

Re: find general soln for dP/dt = [bkPe^(-kt)] / [1 - be^(-k

Hello, almostacasper!

Assume that a population grows at a rate given by:

. . .\(\displaystyle \L\frac{dP}{dt} \:= \:\frac{bkPe^{-kt}}{1\,-\,be^{-kt}}\)

...where \(\displaystyle b\) and \(\displaystyle k\) are positive constants, and the initial population is \(\displaystyle P_o.\)
Find the general solution for the differential equation.

I'm assuming I have to separate the equations. . Right!

Click to expand...

Eliz. Stapel is absolutely correct . . . multiply by \(\displaystyle \L\frac{dt}{P}\)

. . \(\displaystyle \L\frac{\not{dt}}{P}\,\cdot\,\frac{dP}{\not{dt}}\;=\;\frac{dt}{\not{P}}\,\cdot\,\frac{bk\not{P}e^{-kt}}{1\,-\,be^{-lt}}\)

. . . . . \(\displaystyle \L\frac{dP}{P} \;=\;bk\,\cdot\,\frac{e^{-kt}}{1\,-\,be^{-kt}}\)


Then: \(\displaystyle \L\;\int\frac{dP}{P} \;=\;bk\int \frac{e^{-kt}}{1\,-\,be^{-kt}}\,dt\)


On the right, let \(\displaystyle \L\,u\:=\:1\,-\,be^{-kt}\) . . . etc.

Ok, so:

dP/P = bk [e^(-kt)]/[1-be^(-kt)]

ln(p) = bk*ln[1-be^(-kt)] + c

p = e^( bk*ln[1-be^(-kt)]+c)

p = e^(bk*ln[1-be^(-kt)]) * e^c

p = ce^(bk*ln[1-be^(-kt)])

I think i can simplify it more, but im not exactly sure how.
I'm thinking since b and k are constants then i could make e^(bk) into c,
and e^(ln[1-be^(-kt)]) into 1-be^(-kt).
But i'm not totally sure since they are multiplied in the power instead of added, like the c.

Hint:

\(\displaystyle \L\,e^{k*ln(p)} = e^{ln(p)*k} = (e^{ln(p)})^{k} = p^{k}\)