Tag Archives: solving differential equations

June 12, 2024 · 7:02 am

The Logistic Equation

My year 12 Specialist students are working on logistic growth at the moment. An example might be helpful.

A new viral disease was found to spread according to the equation \frac{dN}{dt}=kn(M-N), where M is the susceptible population, N is the number of people infected at time t months and k=1.5\times 10^{-9}. In March 2010, it was thought only 100 people out of a population of 18 million were infected. Use the logistic model to find the number infected in:

(a) March 2011

(b) June 2012

(c) January 2017

Specialist 12 – Nelson Senior Maths

\frac{dN}{dt}=kN(M-N)

\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^{6}-N)

\frac{dN}{N(18\times10^{6}-N)}=1.5\times10^{-9}dt

Use partial fractions to separate the denominator \frac{dN}{N(18\times10^{6}-N)}

\frac{1}{N(18\times10^{6}-N)}=\frac{A}{N}+\frac{B}{18\times10^{6}-N}

1=A(18\times10^{6}-N)+BN

When N=0

1=A(18\times10^{6})

A=\frac{1}{18\times10^{6}}

When N=18\times10^{6}

1=B(18\times10^{6})

B=\frac{1}{18\times10^{6}}

\frac{1}{18\times10^{6}}(\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN)=1.5\times10^{-9}dt

\int\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN=\int27\times10^{-3}dt

\ln|N|-\ln|18\times10^{6}-N|=(27\times10^{-3})t+c

\ln|\frac{N}{18\times10^{6}-N}|=(27\times10^{-3})t+c

\frac{N}{18\times10^{6}-N}=e^{(27\times10^{-3})t+c}

Let A=e^{c} and rearrange to make N the subject.

N=\frac{(18\times10^{6})Ae^{(27\times10^{-3})t}}{1+Ae^{(27\times10^{-3})t}}

Divide by Ae^{(27\times10^{-3})t}

N=\frac{(18\times10^{6})}{\frac{1}{A}e^{-(27\times10^{-3})t}+1}

Initially 100 people were infected.

100=\frac{(18\times10^{6})}{\frac{1}{A}+1}

A=\frac{1}{179999}

N=\frac{(18\times10^{6})}{179999e^{-(27\times10^{-3})t}+1}

(a) t=12, N=138.3, hence 138

(b) t=27, N=207.3, hence 207

(c) t=82, N=915.2, hence 915.

It is not necessary to solve the differential equation, you can use the formula

\frac{dP}{dt}=rP(k-P)\leftrightarrowP=\frac{kP_0}{P_0+(k-P_0)e^{-rkt}}

This formula is on the Year 12 Mathematics Specialist formula sheet for Western Australia.

For our question,

\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^6-N)

So, P=\frac{18\times10^{6}\times100}{100+(18\times10^6-100)e^{-(1.5\times10^{-9})(18\times10^6)t}}

And you can substitute values for t.

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