Applications of First Order Differential Equations – Exponential Growth: Part 1


– [Narrator] Welcome to
the first of several videos on the applications of first
order differential equations. This video will focus
on exponential growth. The differential equation
dP/dt equals k times P or P prime of t equals k times P of t models exponential growth, which means the rate of growth is proportional to the
population or amount. So looking at these
again, this is telling us the change of P with respects to t is equal to some constant k times P, or the derivative is equal to a constant times the original function P of t. And this can also be used to
model continuous interest. While we will solve this
differential equation, find P of t, the only function that satisfies
this differential equation is P of t equals P sub zero times e raised to the power of k t. We should recognize this
exponential function from our algebra class where P sub zero is the initial amount, k is the growth rate and t is the time. But let’s start by looking at an example. A population of a small town grows proportional to its current population. The initial population is
5000 and grows 4% per year. This can be modeled by
the differential equation dP/dt equals 0.04 times P, and P of 0 is equal to 5000. We want to find an equation
to model the population, determine the population
after three years, then determine how long it will take for the population to double. So to find the equation
to model the population, we’re going to solve this
differential equation. So if we write this in differential form, we would have dP is equal to 0.04 P dt. We want the P on the
left side of the equation so we’re going to multiply
both sides by one over P. So we’d have one over P dP is equal to 0.04 dt. Now we can integrate both
sides of this equation. The antiderivative of one over P would be the natural
log absolute value of P. But we know P has to be positive because it’s the population. So we have the natural log of P. We’ll put the constant of
integration on the right side. So this is going to be equal to the antiderivative of
0.04 with respects to t would be 0.04 t plus the constant of integration. Now once we have this
equation, if this is true, then e raised to the
power of natural log P must also equal e raised
to the power of this sum. By doing this, the left side
simplifies very nicely to P, which we know is going to be a function of t. So we’ll write P of t equals… Now because we have a sum here, this would be e to the 0.04 t times e to the C. But e to the C is just a constant. So let’s let e to the power
of C equal P sub zero. So now we have an exponential function that would satisfy this
differential equation. It’s going to be P of t equals P sub zero times e raised to the power of 0.04 t, which does fit the form of
the exponential function that we discussed on the
previous slide, also given here. But now let’s go ahead and verify that P sub zero would actually be 5000. So P of zero would be equal to times e to the zero power, which again we know must equal 5000. Well e to the zero is equal to one, so we do have P sub zero equals 5000. Giving us our particular solution to the differential equation, we have P of t is equal to 5000 times e raised to the power of 0.04 t. Again which matches the
exponential function that we knew would be
our solution given here. Now let’s go ahead and use this function to answer a couple questions. We want the population after three years. So we’re just going to
substitute three for t. You need a calculator for this part. So 5000 times e raised to the power of 0.04 times three. So the population will be approximately, we’ll round to the nearest person, so we’re going to round down: 5,637 people. The last question asks us to determine how long it will take for
the population to double. So again, starting with
our population function, the initial population is 5000. 5000 times two is 10,000. So we’re going to set P of t equal
to 10,000 and solve for t. Divide both sides by 5000. This would be one, 10,000
divided by 5000 is two. This will always be two when trying to find the doubling time. And to solve for t, we’ll have to use the properties of logarithms. We’ll take the natural log of
both sides of the equation. So on the left side we have
natural log two equals- Here we can apply the power
property of logarithms. We can take this exponent
here and move it to the front. So we have 0.04 t times natural log e. But natural log e is equal to
one, so that simplifies out. So now we just divide by 0.04 and go to our calculator to find the approximate doubling time. So we have natural log
two divided by 0.04, which happens to be our growth rate. So the doubling time is
approximately, let’s say 17.3 years. I do want to point out there’s a shortcut for determining doubling time. Let’s call it capital T. The doubling time is
always going to be equal to natural log two divided by
the growth rate, we’ll call K. Okay, I think we’re short
on time, so I’ll go ahead and take a look at the
second example in part two. I hope you found this helpful.

4 thoughts on “Applications of First Order Differential Equations – Exponential Growth: Part 1

  1. Hi, i tried using a calculator technique and got an answer of 5624.32 . What should be the value at 1st year to get 5634 ??

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