All language subtitles for Calculus 10_ Modeling with & Solving Differential Equations (Part 1) LM2 Jennifer (Jenny) Wexler-en

JENNIFER WEXLER: So in the last part, we looked

at differential equations in an introduction in three different ways;

analytically, graphically, and verbally.

In this part, we're going to focus a lot more on the analytic piece,

although you will see at least a little bit of a connection

to a graph along the way.

So in our differential equations, when we defined them, we had solutions.

And as of the last segment, we could verify solutions

and look for graphical evidence but we didn't

have a way to actually find any solutions on our own.

And so in this part, we're going to look a little more carefully

at some particular types of differential equations

for which we can find the solutions.

To start, we'll look at the differential equation dy dx equals x plus 1.

So this equation says that the derivative of some function that we

don't know is equal to x plus 1.

But if we think back to what we know about anti-differentiation,

we do know that function.

The function whose derivative is x plus 1 is 1/2 x squared plus x.

And then we add a constant of anti-differentiation

because if I shift this function vertically, up or down,

its slope, its derivative, is still x plus 1.

And so what I have here is something called a general solution

to this differential equation.

Because it encompasses many, many, many different solutions.

And we can visualize those solutions in their graphical form.

And so I can see all of these parabolas described by this solution

to our differential equation.

You can see in the graph, some of them are highlighted.

Here is a curve that goes through the point negative 2, negative 3.

And so within this family of solution curves,

I can zero in on one particular solution curve.

That would be the parabola, in this case it's a parabola, that

contains negative 2, negative 3.

And because I have an equation, a general solution

for this family of solution curves, I can find that particular solution curve

by substituting in negative 3 for the y and negative 2 for the x

and find what constant determines that particular curve.

So negative 3 equals 2 plus negative 2 plus c, my constant

is c for this particular curve right here,

and so the particular solution curve that is highlighted in red

is y equals 1/2 x squared plus x minus 3.

And so we have a general solution that is all of these curves and more.

The screen is only showing maybe 8 of them.

And then we have particular solutions where we know a point

and we can zero in on one particular curve.

So we found one.

We didn't have this ahead of time, we didn't verify it,

we were just given the differential equation,

and we knew enough about anti-differentiation

to find the solution.

So let's look at a couple more like that.

Let's see, the one we just had was dy dx equals x plus 1.

And it gave us the solution y equals 1/2 x squared plus x plus c.

I'm going to look just at the general solutions here.

So let's try another, dy dx equals cosine of 2x.

So if you think about anti-differentiation,

what function will give this out as its derivative?

That function is sine, and we need a 1/2 because there's a chain rule piece.

So the function 1/2 sine 2x will give out this derivative.

I can shift that sine curve up and down, and I will still have solutions

to this differential equation.

Try one more.

dy dx equals kx squared.

Notice this differential equation has a constant in the differential equation,

but we still know anti-differentiation techniques

that allow us to find a function for which this is its derivative.

This is quadratic, so our solution function is cubic, 1/3 kx cubed.

And, again, the general solution, these curves, these cubics, all

are solutions to this differential equation.

So we've seen a set of differential equations

that we can solve using the anti-differentiation techniques

that we already have.

We're going to look at a different kind now.

So our next differential equation looks a little bit different.

dy dx equals y.

Let's pause for a minute to think about the difference.

All of these that we solved, the derivative

involves only the independent variable.

But this new one, the derivative is actually

involving the dependent variable, the function itself.

So that's a little bit different.

But if I think about what this equation really says,

it says I'm looking for a function where the derivative is the function itself.

And we actually have a function where the derivative is the function itself.

y equals e to the x.

And so right now, just by thinking about derivative equals function,

I have a particular solution to this differential equation.

I don't have a general solution yet, so I'm

going to try to find one that shows all of the solution curves.

And I'll try the same thing I tried before.

Perhaps the general solution is y equals e to the x plus c.

I can vertically shift it.

If I go back to what we've learned before, though,

when I try to verify this solution and I take

the derivative of this proposed general solution,

the derivative is not equal to the original function.

And so this isn't the general solution to this particular differential

equation.

If instead I use a multiplying constant then, now, when I try to verify

and I take the derivative, I can see that the derivative of a times

e to the x is a times e to the x.

So the derivative is the original function,

and I found my general solution.

Notice, by the way, that a could be positive or negative or 0.

And all of those particular solutions would

satisfy this differential equation.

So we're going to look at those graphically

a little bit to see how the slope is equal to the y-coordinate

as we look at these curves.

So now we have a graph of y equals a e to the x and its starting point.

I'm going to drag the parameters around a little bit in a moment,

but at the starting point, the a is 1.

So we just have the curve y equals e to the x.

And we have the point 0,1, and we can see that the slope at that point is 1.

If I drag the point a little ways away, now the point, the y-coordinate,

is 7.389, the slope is 7.389 to three decimal place accuracy.

And anywhere I move that point, the y-coordinate and the slope agree,

just like the differential equation said they would.

If I change the constant to make the graph steeper

or to vertically reflect it so that I have a negative a value,

notice the y-coordinate and the slope still agree.

And I can even make the a value 0, pretty close to 0.

If I make that multiplying constant 0, then I have the horizontal line y

equals 0.

And sure enough, the y-coordinate and the slope still agree.

This isn't a particularly interesting solution to the differential equation,

but it is a member of the family of solution curves.

So we've seen a number of analytic examples

to solving differential equations that we can solve on our own.

When the differential equation is a function

only of the independent variable, then we

can use our anti-differentiation techniques.

And we've seen one example, the exponential function,

that we know a little bit about and that we can solve by hand as well.

So now before going on to the next part, you'll

have a chance to try the techniques that we've just

seen, including finding solutions on your own

using anti-differentiation techniques and things you already

know about derivatives of functions.

And you'll also have a chance to look at particular solutions

and general solutions together finding constants of anti-differentiation

when you're given a point on a particular solution curve.