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### Course: AP®︎/College Calculus BC > Unit 7

Lesson 8: Exponential models with differential equations- Exponential models & differential equations (Part 1)
- Exponential models & differential equations (Part 2)
- Worked example: exponential solution to differential equation
- Differential equations: exponential model equations
- Differential equations: exponential model word problems

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# Exponential models & differential equations (Part 1)

Assuming a quantity grows proportionally to its size results in the general equation dy/dx=ky. Solving it with separation of variables results in the general exponential function y=Ceᵏˣ.

## Want to join the conversation?

- Even though we can never have a negative population in practice, do we really need to assume that the population is positive in order to get rid of the absolute value sign, mathematically speaking? Since our new constant C could be either positive or negative, doesn't that simply account for all of the values our population could be in and of itself, making the absolute value sign redundant?(13 votes)
- Well, the mathematical reason would be this:

The sign of P only depends on the sign of constant C, so you can notice that your initial population, P(0), is equal to C (just make the substitution).

Now, the initial population is always positive, therefore P is always positive.(21 votes)

- I don't understand why C only appears on the right side of the equation. Can somebody please elaborate?(8 votes)
- As Hengru said, you can combine the C's to make another C. Since C1 and C2 are arbitrary constants anyways, adding/subtracting them shouldn't do anything since something arbitrary + something arbitrary is still something arbitrary.(4 votes)

- In the final form of the equation, P = Ce^kt, shouldn't the caveat C > 0 be included, since C is really just e^C, which is always positive and nonzero?(5 votes)
- Mathematically that is not really a restriction; if for example
`C₁ = 𝒾π`

, then`C = e^(𝒾π) = -1`

. In the context of the problem we know that`C₁`

is never going to take imaginary values, but as a general solution to a differential equation, the restriction is not valid.(9 votes)

- At4:58why is it Kt and not K^2/2?(5 votes)
- It's because K is a constant and we're integrating *with respect to t" - the dt bit in

∫ K dt

If we were integrating with respect to (a variable) K,*then*we'd have

∫ K dK

and you'd be right, the integral would be K²/2(8 votes)

- What is K and why do we need to include it?(7 votes)
- I think there are some videos in Pre-Algebra that discuss K. Just search up the Constant of Proportionality, and maybe also look at the video on Direct Vs. Inverse Variation.(2 votes)

- In the first equation : dP/dt = kP. P is a function or a real number ?

dP/dt is a slope , kP is a number , why are they equal ?(2 votes)- P is a real number. Multiplying by k allows you to go from a number to a slope.(8 votes)

- The mass of an isotope decreases at a rate that is proportional to the mass at that time.

The mass of the isotope was 40 grams initially, and it was 10 grams after 16 days.

What was the mass of the isotope after 20 days?

Should my model be look like this: dM/dt = - kM

Is my model correct?(2 votes)- Yes that is correct: Directly proportional means that the equation takes the form y = kx, where y is said to be proportional to x. In this case, it says the rate that the mass decreases is proportional to the mass. The rate is the derivative which must be equal to -kM where k is just a constant. Note that it is -kM since the mass is decreasing.

Thus, the model becomes dM/dt = -kM(2 votes)

- When integrating 1/p, is it necessary to write ln|x|? or could you say the integral is ln(x) since population can't be negative?(1 vote)
- The antiderivative of 1/x is ln|x|. Here is a video that might help explain the intuition behind why this is the case:

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1(2 votes)

- Would another solution to the differential equation in the video be P=C/1-kt?

at2:23I tried doing the following:

dP/dt = kP \\-kP

-kP +dP/dt = 0 \\Integrate both sides with respect to t

-kPt+P = C

P(1-kt) = C \\divide by 1-kt

P = C/(1-kt)

However if we differentiate the following equation with respect to t we get:

dP/dt = (0*(1-kt)-(-k)*C) / (1-kt)^2

dP/dt = kC/(1-kt)^2 \\ substitute for P = c \(1-kt)

dP/dt = kP/(1-kt) not equaling to dP/dt = kP which was my initial condition.

I`d be happy if someone can help me understand where I made a mistake :)(1 vote)- Integrate -kP with respect to t won't result in -kPt, because P is a function of t.(2 votes)

- I don't really get where dP/dt=kP (as in the kP part; I don't understand why we use a linear change instead of any other sort of change).(1 vote)
- Because linear change is the simplest type of change, so this is a more appropriate example for an introduction to differential equations.

Also, these types of relationships tend to show up in nature a lot, e.g. with Newton's law of cooling.(2 votes)

## Video transcript

- What I'd like to do in
this video is start exploring how we can model things with
the differential equations. And in this video in particular, we will explore modeling population. Modeling population. We're actually going to
go into some depth on this eventually, but here we're going to start with simpler models. And we'll see, we will stumble on using the logic of differential equations. Things that you might have seen in your algebra or your precalculus class. So, in some level, what
we're going to do here is going to be review, but
we're going to get there using the power of modeling
with differential equations. So let's just define some variables. Let's say that P is
equal to our population. And let's say that t
is, let's say that t is equal to the time that has passed in days. In days, it could have
been years or months. But let's say we're
doing the population of insects that reproduce quite quickly. So days seem like a nice
time span to care about. Now, what would be a reasonable model? Well, we could say that
the rate of change, the rate of change of our
population, with respect to time, with respect to time is, well,
a reasonable thing to say is that it's going to be proportional to the actual population,
the actual population. Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing
is going to be more, or a thousand insects is
going to be more insects per second, per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population, with respect to time, is going to be proportional
to your population. And so, you know, sometimes you think of differential equations as
these daunting complex things, but notice we've just been able to express a reasonably not so complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed
that, we can actually try to solve this differential equation. Find a general solution, and
then we can try to put some initial conditions on there or
some states of the population that we know to actually
solve for the constants to find a particular solution. So how do we do that? And I encourage you to
pause the video at any time and see if you can solve
this differential equation. So assuming you, at least,
maybe have had an attempt at it, and you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and
all the differentials involving that variable on one
side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the
derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change
in P over change in time. This is our instantaneous
change, but for the sake of separable differential equations or differential equations
in general, you can treat, you can treat these, this
derivative in Leibniz notations like fractions, and you can
treat these differentials like quantities because we
will eventually integrate them. So let's do that. So, we want to put all
the Ps and dPs on one side and all the, all the things that involve t or that I guess don't
involve P on the other side. So, we could divide both sides by P. We could divide both sides
by P, and so we'll have one over P, and you have one over P here and then those will cancel, and then you can multiply
both sides times dt. We could multiply both sides times dt. Once again, treating the
differential like a quantity which isn't, it really isn't a quantity. You really have to view this as a limit of change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But for, once again, for the sake of this we can do this, and when we do that we would be left with one over P dP is equal to, is equal to k dt, is equal to k dt. Now we can integrate,
integrate both sides. Because this was a separable
differential equation, we were able to completely
separate the Ps and dPs from the things involving ts or, I guess, the things that aren't
involving Ps, and then if we integrate this side,
we would get the natural log, the natural log of the absolute
value of our population, and we could say plus
some constant if we want but we're going to get a
constant on this side as well so we could just say that's
going to be equal to, that's going to be equal to
k, it's going to be equal to k times t, k times t plus some constant. Plus some constant. I'll just call that C one, and once again I could've put a plus C
two here, but I could've then subtracted the
constant from both sides and I would just get the
constant on the right hand side. Now, how can I solve for P? Well, the natural log of
the absolute value of P is equal to this thing right over here. That means that's the same thing. That means that the absolute value of P, that means that the absolute value of P is equal to e to all this business, e to the, e to the, let
me do in the same colors, kt, kt plus, plus C one, plus C one. Now this right over
here is the same thing. Just using our exponent
properties, this is the same thing as e to the kt, e to the k times t times e, times e to the C one, e to the C one. Now this is just e to some constant, so we could just call this, let's just call that the constant C. So this is all simplified to C e, C e to the kt, to the kt. And if we assume our
population at any given time is positive then we could get rid of this absolute value sign, and
we have a general solution to this, frankly, fairly
general differential equation. We just said proportional. We haven't given what the
proportionality constant is, but we could say, if we
assume, positive population that the population is
going to be equal to some constant C times e to the kt power. To the kt power, and the
reason why I said that you've seen this before is this is just an exponential function,
and it's very likely that in algebra or in precalculus class you have modeled things
with exponential functions, and my guess is that
you've modeled things, modeled things like population. The reason why this is interesting is you now see where this is coming from. The underlying logic that's just driven by the actual differential equation. The rate of change, with respect
to time, of the population. Well, maybe it's just
proportional to population. So I'll leave you there,
and in the next video we'll do what you probably
did in the 10th or 11th grade or maybe later in your
life, it doesn't matter when you did it, where we actually look at some initial conditions to find
a particular solution.