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Luminous Blob
Jun10-04, 02:00 PM
I'm doing some revision on differential equations, and am getting stuck on what should be a simple problem.

The question is:

Solve: dP/dt = 0.2P(1000 - P)

Find particular solutions when (i) P(0) = 1000 and (ii) P(0) = 2000

The answers are supposed to be:

i) P(t) = 1000
ii) P(t) = 1000[1 - (1/2)exp(-200t)]^(-1)

I keep getting the wrong answers. The general solution I came up with (after doing the partial fraction decomposition and integrating) is:

t + C = 1/200[ ln|P| - ln|1000 - P| ]
exp(200t + C) = exp(ln|P| - ln|1000 - P|)
Aexp(200t) = P/(1000 - P) {where A = exp(C)}
P = 1000Aexp(200t) - (P)Aexp(200t)
P[ 1 + Aexp(200t) ] = 1000Aexp(200t)

P = 1000Aexp(200t)/[ 1 + Aexp(200t)]

But I don't get the right answers when I solve for A with the initial condition.

I'd appreciate it if anyone can tell me where I'm going wrong here.

arildno
Jun10-04, 02:24 PM
i) YOU HAVE PERFORMED DIVISION BY ZERO HERE!!
(Right when you started to solve the diff. eq.)

ii) Nothing wrong with your answer here, you'll get A=-2 (right?)

Remember:
\frac{1000Ae^{200t}}{1+Ae^{200t}}=\frac{1000}{1+\f rac{1}{A}e^{-200t}}, A\neq0

Luminous Blob
Jun10-04, 03:02 PM
i) YOU HAVE PERFORMED DIVISION BY ZERO HERE!!
(Right when you started to solve the diff. eq.)

ii) Nothing wrong with your answer here, you'll get A=-2 (right?)

Remember:
\frac{1000Ae^{200t}}{1+Ae^{200t}}=\frac{1000}{1+\f rac{1}{A}e^{-200t}}, A\neq0

Aaahh...of course. Thanks, that clears things up a bit.