Welcome to Calculus. I'm Professor Greist,
we're about to begin Lecture 19 on More O.D.E.S. In this lesson
we'll consider some new families of ordinary differential equations, their
solutions, and what they're good for. Along the way we'll learn some new
methods and some new applications. Among the many differential
equations that you may encounter, a few families are worthy of note. Some ODEs are autonomous,
meaning that the right-hand side depends on x alone and
does not depend on t. Those equations that do depend on t
are called non autonomous equations. They can be very difficult to solve. Some, however, are solvable by
the following separation method. If dx/dt is in the form of a function
of x alone times a function of t alone, then following the pattern of what we did
for the linear differential equation, we can separate out all of
the x terms on the left. All of the t terms on the right
then it makes sense to integrate both sides if these
integrals are computable and if you can solve for
x then you're in good shape. Let's look at an example. If you consider a body that is
falling under the action of gravity, you'll notice different behavior
depending on its shape. This is due to drag force. If we assume a drag force model
where that force is proportional to the velocity of the body then,
what do we have? Well, our body is being acted on by
a gravitational force of magnitude m, the mass, times g,
gravitational constant pointed down. Pointed upwards is the drag force that is
assumed proportional to the velocity, v. By some constant of proportionality,
kappa the drag coefficient. Newton's law says that the mass
times the acceleration, dvdt, is equal to mg minus kappa times v. Here we have an autonomous and separable,
ordinary differential equation. By doing a little bit of factoring and algebra, we can move
to the left hand side, the quantity v minus mg over kappa,
all in the denominator. On the right hand side,
what is left over is negative kappa divided by m dt integrating both sides. What do we do? On the left, you have to think
a moment to see that the entire derivative of dv over v
minus constant is in fact, the logged of quantity
v minus that constant. On the right-hand side
the integral is very easy. It is negative kappa over m times t. We'll put our constant of
the integration on the right side. Now, to solve for v, what do we do? We exponentiate both sides. On the left we have v- mg over kappa. On the right we have E to
the negative kappa over M times T, that's times E to the constant
of integration we're just going to call that a new
constant with the same name C. Now solving for V what do we have V is
constant E to the negative kappa over MT. Plus mg over kappa. Let's think about this for a moment. If we take a limit as t goes to infinity,
what happens? Well the exponent in
the exponential is negative. That means that term goes to zero. And goes to zero rapidly. What remains is a constant velocity,
equal to m times g over kapa. This is called the terminal velocity. Depending on the value of kappa. It may be very fast or rather slow. Separable ODEs also arise in analysis
of a fluid leaking from a tank. This is regulated by
Torricelli's Law which relates h, the height of the liquid with A,
the cross sectional area at the top. The law states that the rate of change
of h with respect to t is proportional to the square root of h, and
inversely proportional to A. If we consider say a cone, or anything where the cross sectional area
is quadratic in h by some constant. Maybe pi, maybe not. Then we obtain h prime is proportional
to h to the negative three halves. This is separable. And moving the h to three
halves over to the left side. And then integrating both sides,
yields on the left, h to the five-halves times two-fifths. And on the right, a constant times
t plus an integration constant. Solving for h, gives a linear function of
t raised to the two-fifths power where the constants will depend on
the physics in your initial condition. What this implies is that h approaches
zero very rapidly at the end. How rapidly? It is in O of the final
time minus t raised to the two-fifths power,
that comes into zero very very sharply, when you're looking at something
to a power of two-fifths. Let's move on to a different
class of nonautonomous ODEs. These are the linear nonautonomous ODEs. There of the form dx dt
equals A of t times x times B of t where A and
B are functions of t. For example, dx dt equals tx plus
one half t squared is of this form. We say that it's a linear equation
because if you suppress the t dependence you get something of
the form dxdt equals ax plus b. That right hand side should
remind you of the equation. The line, well,
how do we solve such a problem? The method that we will introduce
is that of the integrating factor. It's a little tricky so hang on. Let us denote that
integrating factor by I. We are going to multiply both sides
of this equation by an as yet unknown function I. Rearranging terms a little bit and
multiplying through gives us I times dx dt minus AI times x. And on the right I times B. Now what is this good for? Well if you look at that left hand
side and keep in mind the product rule then you see something that should remind
you a little bit of the product rule if, if we had something that looked
like I times dx dt + dI dt times x. Well then, we would recognize that
that's just the derivative of I times x. And that's something that's
gonna be helpful to us. So what would we need to be
true in order to have this? We would need to have dI dT = -AI. Now, focus on that because
what we're doing is deriving a differential equation for
this unknown integrating factor I. That is separable. That is solvable by the standard
techniques that we've been using. When we solve that equation,
what are we going to get? I is e to the integral of negative A dt, which we could write in shorthand. As E to the negative integral A. If we choose such a function I and
multiply both sides of our original equation by this then what we obtain is
on the left the derivative of I times X. On the right I times B. Now we can integrate both sides,
because differentiation and integration undo each other. On the left we have I times X. On the right, the integral of I times B. We can solve for x and obtain,
after substituting in our formula for I. A scary looking but very useful equation. The solution is x is e to
the integral of a times the integral of b times e to
the negative integral of a. Let's demonstrate this formula in
the context of a mixing formula. Assume you have a tank that holds
a 1000 gallons and is 90% full. An additive is injected at
a rate of 10 gallons per minute. The solution is mixed well and then drained out at a rate
of 5 gallons per minute. The question is what is
the concentration of the additive at that moment when the tank is full? There are a couple of
features to this problem. We're gonna have to analyze. First of all, we're talking
about when the tank is full so we've gotta be concerned with volume. Also, we are told the rate
of change of the volume, that coming in and that going out. Lastly, what we're asked about is the
concentration of the additive in the tank. Let's set a few variables. First of all,
V is a function of t is our volume. Next comes Q(t), that's the quantity of
the additive that is put into the tank. So that when we talk about
the concentration, C(t), that is simply Q(t) over V(t). Let's derive the differential equation and
then solve it. By considering flow rates coming in and
out, its easy to see that the volume is
the initial volume, 900 plus 5T, meaning that the tank will
be full at precisely T = 20. If we consider the time rate of
change of the amount of additive, Q, well that additive comes in at
a rate of 10 gallons per minute, but the fluid is drained out at
a rate of 5 gallons per minute. That drained out fluid
is not pure additive, it depends on a concentration, C. Thus we derive a differential
equation on Q, as follows, dQ dt + Q /180 + t = 10. This is a linear, nonautonomous ODE, where the function a is -1 /180 + t. That means our integrating factor is e to the integral of dt/ 180 + t Of course that integral is simply the log of 180 + t, and when we exponentiate that
we obtain simply 180 + t. If we therefore multiply
the differential equation by this integrating factor
we obtain Very simply 180 plus t times dQ/dt plus Q
equals 10 times quantity. 180 plus t, the left hand side is the
derivative of Q times quantity 180 plus t. Integrating both sides gives us
on the left 180 plus t plus q. On the right 1,800 t plus five t squared,
plus a constant. You can show based on
the initial condition that there's no additive at the beginning,
that that integration constant is zero. Now we almost have
the solution to our problem. What do we need to do? We solve for Q as 1800 t plus
5 t squared over 180 plus t. But we're asked about the concentration
at the time when the tank is full. At T equals 20 we get the amount
of additive to be 190. Therefore the concentration
at that time is 190. Divided by the volume of the tank,
that means that the additive is present at
19% when the tank is full. There are many similar types of problems
for which integrating factors are helpful. Some of them are like this
involving mixing and tanks. Others come from entirely
different applications. Indeed, this is just the beginning of the wonderful subject
of differential equations. Where there many more families of
differential equations that require specialized solution. A few of these are doable with
the tools that you now have in hand. But many of these differential equations
require multi variable calculus. Some can be very difficult
to solve indeed. We now have in hand
several types of ODE's, their solutions, and their applications. That's the good news. The bad news is most differential
equations are hard to solve. Some seem impossible. In our next lesson, we'll consider what to do when
the solution seems to slip from our grasp.