This page builds on the discussion of exponential growth and decay and specifically applies it to the dynamics of population change, i.e. population growth and decline. In fact, we will touch on basic exponential growth and decay and then extend the discussion to more realistic models. Here are the four topics covered on this page. The initial condition for each equation is \(p(0)=p_0\).
\( p' = rp \) 

\(p(t) = p_0e^{rt}\) 
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\) 

\(\displaystyle{ p(t) = \frac{p_0T}{p_0+(Tp_0)e^{rt}} }\) 
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{L} \right)p }\) 

\(\displaystyle{ p(t) = \frac{p_0L}{p_0+(Lp_0)e^{rt}} } \) 
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\) 

\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = \frac{(L/p_01)^T)}{(T/p_01)^L}e^{(LT)rt)} }\) 
Exponential Growth
plot 1  exponential growth 

\(p(t) = p_0e^{rt}\) 
As discussed on the exponential growth and decay page, we start with the differential equation \( p' = rp \), which solves to \(p(t) = p_0e^{rt}\). In this equation, \(p(0) = p_0\) is the initial population and \(r\) is the growth rate. The graph of \(p(t)\) with \(p_0 = 1\) for various values of \(r > 0\) is shown in plot 1.
There are two problems with this model when we are trying predict population growth.
1. Notice that as \(t\) increases, \(p(t)\) increases without bound. Although this model for population growth may be accurate over short time periods, it is not realistic as space, food supply and other resources would limit the growth in any reasonable situation.
2. Some populations cannot sustain itself if there are too few members. For example, our model shows that even when starting out with only one member, the population increases. Again, this is not realistic.
In order to handle the limitations of resources, we will use another model called logistic growth but first, let's look at how to handle the second issue, starting out with enough members to sustain a population.
Exponential Growth with Critical Threshold
autonomous equation 

A special type of differential equation of the form \(y' = f(y)\) where the independent variable does not explicitly appear in the equation. 
plot 2  exponential growth with threshold 

\(\displaystyle{ p(t) = \frac{p_0T}{p_0+(Tp_0)e^{rt}} }\) 
In order to handle the situation when a population needs more than one member to sustain itself or it will die out, we use another autonomous differential equation
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\) 
where \(r\) is the growth rate and \(T\) is called the critical threshold (or just threshold).
The solution to this differential equation is \(\displaystyle{ p(t) = \frac{p_0T}{p_0+(Tp_0)e^{rt}} }\). (See the practice problems for the derivation of this equation.) Let's discuss what is going on with the equation. We have plotted \(p(t)\) for various values of \(p(0)=p_0\) in plot 2.
When \(p_0=T\), i.e. the initial population is exactly the threshold value \(T\), the coefficient of the exponential term is zero, so \(p(t)=T\) and the population stays exactly at the threshold value for all time \(t>0\). However, if the initial population is either above or below the threshold value, the exponential term \(e^{rt}\) drives the plots away from the threshold. This seems logical since, if the initial population is higher than the threshold value, the population will increase and if the initial population is below the threshold value the population will decrease, eventually dying out. In the second case, there are not enough individuals to sustain the population.
Okay, so far, so good. We have dealt with the second problem of the basic exponential growth model. Now let's address the first problem dealing with limited resources.
Logistic Growth
plot 3  logistic growth 

\(\displaystyle{ p(t) = \frac{p_0L}{p_0+(Lp_0)e^{rt}} }\) 
In the real world, populations are often limited by things like resources and space. To handle this situation, we model the population with the autonomous differential equation
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\) 
where \(r\) is the growth rate and \(L\) is called the saturation level.
If you look carefully at this equation and compare it to the equation for critical threshold above, you will see that they are almost identical except for a minus sign. Even though they are similar, the behaviour of this equation is quite different from the previous one. If you followed the derivation of the threshold equation (in a practice problem), you can derive the solution here as \(\displaystyle{ p(t) = \frac{p_0L}{p_0+(Lp_0)e^{rt}} }\). Notice that the exponential here is \(e^{rt}\).
As we did before, plot 3 shows the solution for various values of \(p(0)=p_0\). Notice that the exponential in this equation drives the solution toward the saturation level (as compared to away from the threshold in the previous section).
Logistic Growth with Critical Threshold
plot 4  logistic growth with threshold \(p(t)\) 

In the last two sections we have addressed both problems with the basic exponential individually. Let's combine the two solutions into one equation. Our differential equation is
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\) 
where \(r\) is the growth rate, \(T\) is the threshold and \(L\) is the saturation level. In this discussion, we will assume that \(0 < T < L\), i.e. the saturation level (limit on resources) is higher than the threshold.
The plot of \(p(t)\) for various initial conditions is shown in plot 4. Notice how the plot tends to diverge away from the threshold but converge towards the saturation limit as we expected. The equation of the solution, in implicit form, is
\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = \frac{(L/p_01)^T)}{(T/p_01)^L}e^{(LT)rt)} }\)
(see the practice problems for a derivation of this equation)
Practice
Solve these differential equations using the initial conditions, if given.
Solve \( p' = rp \), \(p(0)=p_0\)
Problem Statement 

Solve \( p' = rp \), \(p(0)=p_0\)
Final Answer 

\( p(t) = p_0 e^{rt} \)
Problem Statement 

Solve \( p' = rp \), \(p(0)=p_0\)
Solution 

\(\begin{array}{rcl} p' &=& rp \\ \displaystyle{ \frac{dp}{dt} } &=& rp \\ \displaystyle{ \frac{dp}{p} } &=& r~dt \\ \displaystyle{ \int{ \frac{dp}{p} } } &=& \displaystyle{ \int{ r~dt } } \\ \ln p &=& rt+C \\ e^{\ln p} &=& e^{rt+C} \\ p &=& ke^{rt} \\ \end{array}\)
general solution \(p=ke^{rt} \)
apply the initial condition \(p(0)=p_0\)
\(p_0=ke^0 \to k=p_0\)
particular solution \(p(t)=p_0e^{rt}\)
Final Answer 

\( p(t) = p_0 e^{rt} \) 
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Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Problem Statement 

Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Final Answer 

\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(Tp_0)e^{rt} } }\)
Problem Statement 

Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Solution 

This differential equation is separable.
\(\displaystyle{ \frac{dp}{(1p/T)p} = r~dt }\)
On the left side of the equation, we need to use partial fractions.
\( \displaystyle{ \frac{1}{(1p/T)p} = \frac{A}{1p/T} + \frac{B}{p} }\) 
\( 1 = Ap+B(1p/T) \) 
\( 1 = Ap+BBp/T \) 
\( 1 = p(AB/T) + B \) 
Remember that the variable in the above partial fraction expansion is \(p\). Now we equate coefficients to find the constants A and B.
\(B=1\)
\(AB/T=0 \to A=B/T \to A=1/T\)
So now our integrals are
\( \displaystyle{ \int{\frac{1/T}{1p/T} + \frac{1}{p} dp} = \int{r~dt} } \) 
\( \displaystyle{ \int{\frac{1}{Tp} + \frac{1}{p} dp} } = rt+C \) 
\( \displaystyle{ \ln \left \frac{p}{Tp} \right } = rt+C \) 
\( \ln Tp + \ln p = rt+C \) 
\( \displaystyle{ \frac{p}{Tp} = e^{rt+C} }\) 
Before we apply the initial condition to find C, let's solve the last equation for \(p\).
\(\begin{array}{rcl} \displaystyle{ \frac{p}{Tp} } &=& e^{rt+C} \\ \displaystyle{ \frac{1}{T/p1} } &=& \\ \displaystyle{ \frac{T}{p}  1 } &=& e^{rtC} \\ \displaystyle{ \frac{T}{p} } &=& 1+e^{rtC} \\ p &=& \displaystyle{ \frac{T}{1+ke^{rt}} } \end{array}\)
In the last line, we set \(k=e^{C}\) just for convenience. Now we will apply the initial condition \(p(0)=p_0\).
\(\begin{array}{rcl} p_0 &=& \displaystyle{ \frac{T}{1+ke^0} } \\ 1+k &=& \displaystyle{ \frac{T}{p_0} } \\ k &=& \displaystyle{ \frac{T}{p_0}1 } \\ k &=& \displaystyle{ \frac{Tp_0}{p_0} } \end{array}\)
Our particular solution is
\(\displaystyle{ p(t) = \frac{T}{1+e^{rt}(Tp_0)/p_0} }\)
\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(Tp_0)e^{rt} } }\)
Final Answer 

\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(Tp_0)e^{rt} } }\) 
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Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\), \(p(0)=p_0\)
Problem Statement 

Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\), \(p(0)=p_0\)
Final Answer 

\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = \frac{(L/p_01)^T)}{(T/p_01)^L}e^{(LT)rt)} }\)
Problem Statement 

Solve \(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\), \(p(0)=p_0\)
Solution 

The differential equation is separable.
\(\displaystyle{ \frac{dp}{dt} = r \left( 1\frac{p}{T} \right)\left( 1\frac{p}{L} \right)p }\)
\(\displaystyle{ \frac{dp}{(1p/T)(1p/L)p} = r~dt }\)
Using partial fraction expansion, we have
\(\displaystyle{ \int{ \frac{L}{T(LT)}\frac{1}{1p/T} + \frac{T}{L(TL)}\frac{1}{1p/L} + \frac{1}{p}dp } = \int{ r~dt } }\)
\(\displaystyle{ \frac{L}{LT}\ln Tp  \frac{T}{TL} \lnLp + \lnp = rt+C }\)
Multiply both sides of the equation by \(LT\) to get
\(\displaystyle{ \lnTp^L + \lnLp^T + \lnp^{LT} = (LT)(rt+C) }\)
Combining the natural log terms and doing some algebra, we get
\(\displaystyle{ \ln \left\frac{p^{LT}(Lp)^T}{(Tp)^L} \right = (LT)(rt+C) }\)
\(\displaystyle{ \frac{p^{LT}(Lp)^T}{(Tp)^L} = exp( (LT)(rt+C) ) }\)
\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = k e^{(LT)rt} }\)
Now we apply the initial condition, \(p(0)=p_0\) to get
\(\displaystyle{ k=\frac{(L/p_01)^T}{(T/p_01)^L} }\)
\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = \frac{(L/p_01)^T)}{(T/p_01)^L}e^{(LT)rt)} }\)
We can leave this equation in implicit form, i.e. not explicitly solve for \(p(t)\).
Final Answer 

\(\displaystyle{ \frac{(L/p1)^T}{(T/p1)^L} = \frac{(L/p_01)^T)}{(T/p_01)^L}e^{(LT)rt)} }\) 
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You CAN Ace Differential Equations
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

Precalculus 

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Practice Instructions
Solve these differential equations using the initial conditions, if given.