You CAN Ace Differential Equations
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Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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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 \(p(0)=p_0\).
\( p' = rp \) |
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\(p(t) = p_0e^{rt}\) |
\(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }\) |
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\(\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }\) |
\(\displaystyle{ \frac{dp}{dt} = r \left( 1-\frac{p}{L} \right)p }\) |
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\(\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} } \) |
\(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }\) |
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\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }\) |
Exponential Growth |
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plot 1 - exponential growth |
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\(p(t) = p_0e^{rt}\) |
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.
autonomous equation |
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A special type of differential equation of the form \(y' = f(y)\) where the independent variable does not explicitly appear in the equation. |
Exponential Growth with Critical Threshold |
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plot 2 - exponential growth with threshold |
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\(\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }\) |
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 \(\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }\). (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 |
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plot 3 - logistic growth |
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\(\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }\) |
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 \(\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }\). 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 |
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plot 4 - logistic growth with threshold \(p(t)\) |
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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
\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }\)
(see the practice problems for a derivation of this equation)
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
Instructions - Solve the following differential equations using the initial conditions, if given.
Solve \( p' = rp \), \(p(0)=p_0\)
Problem Statement |
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Solve \( p' = rp \), \(p(0)=p_0\)
Final Answer |
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\( p(t) = p_0 e^{rt} \) |
Problem Statement |
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Solve \( p' = rp \), \(p(0)=p_0\)
Solution |
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\(\begin{array}{rcl}
p' &=& rp \\
\displaystyle{ \frac{dp}{dt} } &=& rp \\
\displaystyle{ \frac{dp}{p} } &=& r~dt \\
\displaystyle{ \int{ \frac{dp}{p} } } &=& \displaystyle{ \int{ r~dt } } \\
\ln |p| &=& rt+C \\
e^{\ln |p|} &=& e^{rt+C} \\
p &=& ke^{rt} \\
\end{array}\)
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 |
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\( p(t) = p_0 e^{rt} \) |
close solution |
Solve \(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Problem Statement |
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Solve \(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Final Answer |
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\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }\) |
Problem Statement |
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Solve \(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }\), \(p(0)=p_0\)
Solution |
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This differential equation is separable.
\(\displaystyle{ \frac{dp}{(1-p/T)p} = -r~dt }\)
On the left side of the equation, we need to use partial fractions.
\( \displaystyle{ \frac{1}{(1-p/T)p} = \frac{A}{1-p/T} + \frac{B}{p} }\) |
\( 1 = Ap+B(1-p/T) \) |
\( 1 = Ap+B-Bp/T \) |
\( 1 = p(A-B/T) + B \) |
Remember that the variable in the above partial fraction expansion is \(p\). Now we equate coefficients to find the constants A and B.
\(B=1\)
\(A-B/T=0 \to A=B/T \to A=1/T\)
So now our integrals are
\( \displaystyle{ \int{\frac{1/T}{1-p/T} + \frac{1}{p} dp} = \int{-r~dt} } \) |
\( \displaystyle{ \int{\frac{1}{T-p} + \frac{1}{p} dp} } = -rt+C \) |
\( \displaystyle{ \ln \left| \frac{p}{T-p} \right| } = -rt+C \) |
\( -\ln |T-p| + \ln |p| = -rt+C \) |
\( \displaystyle{ \frac{p}{T-p} = e^{-rt+C} }\) |
Before we apply the initial condition to find C, let's solve the last equation for \(p\).
\(\begin{array}{rcl}
\displaystyle{ \frac{p}{T-p} } &=& e^{-rt+C} \\
\displaystyle{ \frac{1}{T/p-1} } &=& \\
\displaystyle{ \frac{T}{p} - 1 } &=& e^{rt-C} \\
\displaystyle{ \frac{T}{p} } &=& 1+e^{rt-C} \\
p &=& \displaystyle{ \frac{T}{1+ke^{rt}} }
\end{array}\)
In the last line, we set \(k=e^{-C}\) just for convenience. Now we will apply the initial condition \(p(0)=p_0\).
\(\begin{array}{rcl}
p_0 &=& \displaystyle{ \frac{T}{1+ke^0} } \\
1+k &=& \displaystyle{ \frac{T}{p_0} } \\
k &=& \displaystyle{ \frac{T}{p_0}-1 } \\
k &=& \displaystyle{ \frac{T-p_0}{p_0} }
\end{array}\)
Our particular solution is
\(\displaystyle{ p(t) = \frac{T}{1+e^{rt}(T-p_0)/p_0} }\)
\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }\)
Final Answer |
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\(\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }\) |
close solution |
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 |
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Solve \(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }\), \(p(0)=p_0\)
Final Answer |
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\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }\) |
Problem Statement |
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Solve \(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }\), \(p(0)=p_0\)
Solution |
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The differential equation is separable.
\(\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }\)
\(\displaystyle{ \frac{dp}{(1-p/T)(1-p/L)p} = -r~dt }\)
Using partial fraction expansion, we have
\(\displaystyle{ \int{ \frac{L}{T(L-T)}\frac{1}{1-p/T} + \frac{T}{L(T-L)}\frac{1}{1-p/L} + \frac{1}{p}dp } = \int{ -r~dt } }\)
\(\displaystyle{ \frac{-L}{L-T}\ln |T-p| - \frac{T}{T-L} \ln|L-p| + \ln|p| = -rt+C }\)
Multiply both sides of the equation by \(L-T\) to get
\(\displaystyle{ -\ln|T-p|^L + \ln|L-p|^T + \ln|p|^{L-T} = (L-T)(-rt+C) }\)
Combining the natural log terms and doing some algebra, we get
\(\displaystyle{ \ln \left|\frac{p^{L-T}(L-p)^T}{(T-p)^L} \right| = (L-T)(-rt+C) }\)
\(\displaystyle{ \frac{p^{L-T}(L-p)^T}{(T-p)^L} = exp( (L-T)(-rt+C) ) }\)
\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = k e^{-(L-T)rt} }\)
Now we apply the initial condition, \(p(0)=p_0\) to get
\(\displaystyle{ k=\frac{(L/p_0-1)^T}{(T/p_0-1)^L} }\)
\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }\)
We can leave this equation in implicit form, i.e. not explicitly solve for \(p(t)\).
Final Answer |
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\(\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }\) |
close solution |