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Wize AP Calculus (BC) Textbook > Differential Equations

Applications of DE

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Exponential Growth and Decay

Any equation that involves an unknown function yy and its derivatives dy/dtdy/dt is called a differential equation.

Differential Equation for Exponential Growth


This equation states that the rate of change dy/dtdy/dt of a quantity is proportional to the current value of that quantity.
dydt=ky\boxed{\frac{dy}{dt}=ky}
where kk is a constant called the proportionality constant (k > 0 for growth, k < 0 for decay).\text{(k > 0 for growth, k < 0 for decay).} If y(0)y(0) is the value of the function yy at time t=0,t=0, the solution becomes
y(t)=y(0)ekt\boxed{y(t)=y(0)e^{kt}}

Wize Tip
The half-life of a radioactive substance is defined as the time t1/2t_{1/2} it takes the substance to decay to half of its original size.
t1/2=ln⁡2k\displaystyle t_{1/2}=\frac{\ln2}{k}

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Logistic Growth Model


A variant on the Exponential Growth Model is the Logistic Growth Model. This model's growth rate decreases as the population approaches the carrying capacity.
dPdt=kP(M−P)\displaystyle \boxed{\frac{dP}{dt}=kP\left(M-P\right)}
Where PPis the population at time tt, MMis the carrying capacity, and kk is the proportionality constant.
P(t)=M1+Ae(−Mk)t\boxed{P(t)=\frac{M}{1+Ae^{\left(-Mk\right)t}}}
Where AA is determined by the initial condition P(0)P(0).

Wize Tip
The population growth rate is fastest when the population is equal to M2\displaystyle \frac{M}{2}.

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Example: Radioactive Decay

A radioactive substance weighs initially 30 grams and decays at a half-life of 10 days. Find the amount of substance left after 20 days.



y(0)=30and y(10)=y(0)2=15y(0)=30 \quad \text{and} \quad\, y(10)=\dfrac{y(0)}{2}=15
dydt=ky\frac{dy}{dt}=ky

y(t)=y(0)ekt=30ekty(t)=y(0)e^{kt}=30e^{kt}
With 2nd condition, we have:

y(10)=15=30e10ky(10)=15=30e^{10k}

⇒10k=ln⁡(1/2)\Rightarrow10k=\ln(1/2)

⇒k=110ln⁡(1/2)\Rightarrow k=\frac{1}{10}\ln(1/2)

⇒y(t)=30e(1/10)ln⁡(1/2)t\Rightarrow y(t) = 30e^{(1/10)\ln(1/2)t}

y(20)=30e2ln⁡(1/2)=30eln⁡(12)2=30(12)2=7.5 grams\displaystyle y(20) = 30e^{2\ln(1/2)}=30e^{\ln{(\frac{1}{2})^2}}=30\Big(\frac{1}{2}\Big)^2=7.5 \text{ grams}


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Example: Logistic Growth Model

The growth of a population is given by the logistic differential equation

dPdt=0.02P(58−P)\displaystyle \frac{dP}{dt}=0.02P\left(58-P\right)

with time t measured in years.

a) What is the solution to this differential equation if P(0)=1P(0)=1?

b) At what population is the fastest growth rate?

a) P(t)=M1+Ae(−Mk)t=581+Ae−1.16tP(t)=\displaystyle\frac{M}{1+Ae^{\left(-Mk\right)t}}=\frac{58}{1+Ae^{-1.16t}}

P(0)=1P(0)=1

⇒1=581+Ae−1.16(0)\displaystyle \Rightarrow1=\frac{58}{1+Ae^{-1.16(0)}}

⇒1+A=58\Rightarrow 1+A=58

⇒A=57\Rightarrow A=57

P(t)=581+57e−1.16t\boxed{\displaystyle P(t)=\frac{58}{1+57e^{-1.16t}}}

b) Find the critical points of the differential equation.

dPdt=0.02P(58−P)=1.16P−0.02P2\displaystyle \frac{dP}{dt}=0.02P\left(58-P\right)=1.16P-0.02P^2

⇒1.16−0.04P=0\Rightarrow1.16-0.04P=0

⇒P=29 \Rightarrow \boxed{P=29 }
Alternatively, recall the fastest growth rate occurs at M2\displaystyle \frac{M}{2}.

P=582=29P=\dfrac{58}{2}=29
The number of a certain type of bacteria grows exponentially from 100 to 200 in 3 hours. In a fermentation process, 50 of these bacteria live. After 9 hours, 150 bacteria die due to feed rate shortage. Approximately how long will it take them to grow from that time to 1410 bacteria?



Practice Question

The growth rate of a certain species of wild cats at time t years is given by dPdt=kP\frac{dP}{dt}=kP. If we know that the initial population in year 2000 was 500, and that there were double that amount after one and a half years, what is the population of wild cats in the year 2009?

Practice Question

A population is modeled by dPdt=2.5P(1−P3500)\frac{dP}{dt}=2.5P\left(1-\frac{P}{3500}\right)
a.) What are the equilibrium solutions?
b.) For what values of P is the population increasing?
c.) For what values of P is the population decreasing?