All language subtitles for Calculus 10_ Modeling with & Solving Differential Equations (Part 1) LM1 Jennifer (Jenny) Wexler-en (1)

JENNIFER WEXLER: Hi.

Welcome to this unit on differential equations.

That's a phrase you may have heard before.

You may even have worked with some differential

equations earlier in a calculus course when you were studying derivatives

or maybe in another discipline.

Differential equations come up in a lot of other disciplines.

Differential equations often provide models for population growth,

also come up in physics, in economics, in engineering,

and so again, this is a phrase you may have heard before.

And so by the end of this unit, you will know, hopefully, a lot more

about differential equations.

So first, we're going to look at differential equations from three

different perspectives-- analytically, graphically, and verbally.

So where we're going to start is to just make sure

that when we say differential equation, we

have some sense of what we mean by that term.

So a differential equation, it has this word "differential" in it,

which you've seen in differential calculus in differentiating a function.

It has to do with derivatives, and that's exactly what

a differential equation is.

It's an equation that has an unknown function, as well as one or more

of that unknown function's derivatives.

And so for example, I could write the differential equation.

I'll just make one up-- y prime plus 5y equals x cubed minus 8.

So we don't know how to solve that.

Actually, I'm not sure I know how to solve that,

but it is a relationship between unknown function y and its first derivative.

And so here's this way that relates them.

There are other notations that we can use.

We can use the Leibniz notation for the derivative.

And that calls out the independent variable x in a slightly different way.

I could use function notation, so this would be f prime of x.

And instead of y, I could write f of x.

And all of these are simply ways to write a differential equation

showing an unknown function and its derivative

in some sort of relationship.

So the first differential equation that we're

going to look a little more closely at is the differential equation y

double prime plus 2y prime minus 3y equals zero.

So we don't have the tools to solve this equation,

but I could propose a solution, and we could see whether or not it works.

So the solution I'm going to propose is the solution y

equals e to the negative 3x.

So notice when I say solution, what I've written here is a function.

My goal in this differential equation, if I'm trying to solve it,

is to find that unknown function.

And so this is my proposed solution.

So I'm going to check it simply by taking derivatives.

I can take the first derivative, e to the negative 3x,

remembering the chain rule.

And I get negative 3e to the negative 3x.

I can take the second derivative, then, again remembering the chain rule.

And now I have enough information to go back to the differential equation

and see whether or not my proposed solution really works.

So my proposed solution would be that there

would be my second derivative plus twice my first derivative, minus 3

times my original function.

So let's see.

I have 9e to the negative 3x minus 6e to the negative 3x,

minus 3e to the negative 3x, and that really is equal to 0,

and so there I have a solution to this differential equation, a function that

satisfies that differential equation.

It happens to be there are others.

Let's see if we can find another one.

I could try y equals e to the x.

That one, since the derivative of the exponential function

is the exponential function, the first derivative

is the second derivative is e to the x, and so when

I go to test it in my original differential equation,

I have the second derivative, e to the x, plus twice the first derivative,

e to the x, minus 3 times the original function.

So 1, 2, 1 plus 2 is 3, minus 3.

That is also 0.

I found another solution.

And so I can verify solutions.

We don't have tools yet.

You will later in the unit have some tools for finding those on your own.

But right now analytically, we can plug in

and use our knowledge of taking derivatives to test a solution.

So there's an analytic approach to differential equations.

So now we're going to look a little bit at a graphical approach,

and so we're going to try another differential equation,

dy dx equals x minus y.

So as I look at that differential equation,

if I start to think graphically, I want to remember

that a derivative is talking about the slope of a tangent line

at some particular point.

And so this differential equation is telling me

about the slope of tangent lines at points x and y, and so if x equals y,

I would expect to see a tangent line with a 0 slope, a horizontal line.

When x was bigger than y, I would expect to see tangent lines

with positive slope.

And when x was less than y, I would expect

to see tangent lines with negative slope.

And so we'll look at a graph to see if there's some evidence that that

could be a solution.

So there I have a graph, and so we'll test this graph to see what I have.

So let's see.

I'll start with the horizontal.

If I look right there, I can see that's a little bit less than 1,

a little bit less than 1.

That really does look like a place where x equals y,

and there's clearly a horizontal tangent there.

I could come out to another point over here is maybe the point

7, 6, and that's definitely a positive slope.

And the scale, that even looks like a slope of 1.

And 7 minus 6 is 1, so that's some pretty good evidence.

I could come over here and find some point over

here and think about, I don't know, about negative 1 and 1/2 minus 7

is negative 8 and 1/2.

That's a very steep negative slope, and so I have some graphical evidence

that what I really do have here is a solution to my original differential

equation.

And so now we're going to look at this graph in a little more detail in a way

that we can actually see the coordinates instead of just estimate them

on a grid.

So here we have a graph that I can move that point p along the curve.

That's the same graph we saw before on the grid.

So if I come over here, maybe I'll stop the point there.

I can see the x and the y-coordinate.

It's definitely a positive slope, and if I look,

if I do a little bit of mental arithmetic, the difference, x minus y,

is a little bit less than 1, and that's what this is telling me.

If I come and move the point over here somewhere,

again, there's the x-coordinate and the y-coordinate,

and a little bit of mental math can get me that slope

and see that the slope of the tangent at that point is also x minus y.

And I'll see if I can get as close to the horizontal tangent.

That's pretty good.

That looks very close to that local minimum point, and those x's and y's

are very, very close together, and so the difference, x minus y,

is very, very close to 0.

So we're pretty close to a horizontal tangent.

So now back at our static graph, if we just

try and incorporate a little bit of information,

here is that point that we tried to trace to where

we got pretty close at about 0.7.

The actual coordinates are 0.693, 0.693.

0.693 some of you may recognize as the natural log of 2.

And the natural log is involved in this curve,

because the equation of the solution that we found

is 2e to the negative x plus x minus 1.

And so if you wanted to at this point, you could pause the video,

and you could use the analytic methods we used in the previous example

to verify that this equation really does satisfy the differential equation,

that this function truly is a solution, because it satisfies our differential

equation up top.

So we've seen an analytic approach, and we've

talked a little bit about a graphical approach,

and so there's one more piece that we're going to look at,

which is a verbal approach.

So here we have a verbal description of a situation.

A population of ants is growing at a rate proportional to the number of ants

present.

So if I look closely at these words, I have

growing at a rate that is talking about a derivative.

Derivatives tell us about rates of change,

and then we have a number of ants.

And so that's our function.

That's our amount or our function.

That's the thing that's actually growing, the thing that's changing.

So I could translate this sentence into the notation

that we saw earlier by saying the derivative of my population of ants,

so p prime, is proportional to the number of ants present.

So here's the derivative or the rate, and that's proportional to the number,

or I could use the Leibniz notation and write dp dt equals k times p.

And so I can see a verbal situation in the notation that we've seen.

And later on in the unit, we'll be able to solve this directly.

Right now we don't yet have the tools, but I could propose a solution

to you of P of t equals Ae to the Kt.

And we could check and see how that solution works

by finding its derivative, which would be Ae to the Kt, and then

the chain rule times K. And I can see that that is my population,

and sure enough, the rate of change, the way that the ants are growing,

is proportional to the number of ants present.

So now you've seen an analytic approach and a graphical approach

and a connection to a verbal situation, and so now you

can play around with all three of those to become a little more comfortable

with these new differential equations.