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Calculus/Modelling with differential equations

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Question
It is on modelling with differential equations.

Mould is spreading through a piece of cheese of mass 1kg in such a way that at a time t days, x kg of the cheese has become mouldy where: -

dx/dt = 2x(1-x)

Given that initially 0.01 kg of the cheese is mouldy.

a) Show that x = [e^(2t)] / [99+e^(2t)]

So far I've got dx/(2x(1-x) = dt
Then partial fractions (1/(2x) - 1/(2(1-x) ) dx = dt
The integration (1/2) (lnx + ln(1-x)) = t +c

At which point I can't get where I'm supposed to end up.  Have I integrated incorrectly?  If not can you please supply the next couple of steps for me

Many thanks

Answer
Hello Maggie,

What you have done is fine, except for the sign in between the two terms in your partial fraction expansion.  It should be: (1/(2x) + 1/(2(1-x)) dx = dt and thus, the integral is:
(1/2) (lnx - ln(1-x)) = t + c...now it is just the algebra to solve for x in terms of t.

First, let's multiply both sides of (1/2) (lnx - ln(1-x)) = t + c by 2 to get
ln(x) - ln(1-x) = 2t + C ==> ln[x/(1-x)]=2t + C ==> x(1-x)=e^(2t+C)
==> x/(1-x)=e*(2t)*e^C, but e^C=K, another constant.
==> x/(1-x)=Ke^(2t)/1, cross multiplying gives: x=(1-x)*Ke^(2t)...solving for x gives:
x=Ke^(2t)/[1+Ke^(2t)], now impose the initial condition x(0)=0.01:
0.01=K/(1+K)...so K=1/99...thus,
x(t)=(1/99)e^(2t)/[1+(1/99)e^(2t)], multiplying by 99/99 gives
x(t)=e^(2t)/[99+e^(2t)]

OK?

Abe

Calculus

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Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

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Over 15 years teaching at the college level.

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NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

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B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

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