AP Calculus AB Introduction to Differential Equations

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to start our discussion of differential equations.0004

Let us jump right on in.0008

Let us going to start with what a differential equation is, it is very simple.0013

A big deal was made about differential equations, I’m not exactly sure why.0017

It is just one of those things that has become sort of mythical, like they are difficult crazy, insane, esoteric things.0021

They are not, it is just an equation like any other equation.0028

Except instead of solving an algebraic equation where you are searching for a number x = 5,0031

here the unknown is actually just a function.0037

We are trying to recover a function that we do not know what it is, that is all.0040

I will stick with black here.0046

A differential equation which we will often just call a de, is an equation that involves one or more derivatives of an unknown function y(x).0048

That is it, that is all a different equation is.0101

It is just an equation that involves one or more derivatives of an unknown function x and it can look like anything.0103

It can look like anything.0113

Sometimes it involves only the derivatives of y(x).0123

Let me go on to blue, actually.0128

Sometimes it involves only the derivatives, and of course, other functions of x, other constant thing like that.0131

Only the derivatives of y(x).0147

The examples would be something like dy dx - 4x = 3x, that is a differential equation.0155

It involves the derivative of an unknown function.0167

y is the functional that we are trying to recover, y(x) = sin(3x)² + 9, something like that, whatever it is.0170

That is what we are trying to elucidate.0178

We might have d² y dx², it might involve a second derivative, + 6x = dy dx.0181

Now the second derivative and the first derivative are involved in the equation.0192

How do we solve this to recover the function y(x).0195

These are two differential equations that involve only the derivatives of y(x).0199

Sometimes the de involves derivatives of y(x), as well as the function y itself which we do not know.0207

We just call it y.0231

As well as the function y(x) itself.0234

Some examples of that are dy dx = x × y(x).0244

Or we might have dy dx² – dy dx = y(x) × sin of y(x).0256

Here we have the derivative, as well as the unknown function itself.0276

Here we have the first derivative² - the first derivative.0283

It is telling you that when I do this, when I take the first derivative² of whatever function it is and0288

if I subtract the first derivative from that, it actually equal the original function × the sin of the original function.0295

That is it, that is all this says.0303

It is just an equation that involves the derivatives of an unknown function.0304

Sometimes the function itself, sometimes not.0308

It is an equation, we are solving for y.0311

We are trying to find the function y(x) = something, that is all we are doing.0314

Once again, our job is to find the unknown function y(x).0325

We are just trying to find a function, y = x², y = sin x, y = e ⁺x, whatever it is.0343

Let us see what we can do.0358

Do I want to write this thing?0362

I might as well, I have it here.0367

Again, to reiterate, for years, we have been solving algebraic and trigonometric functions for an unknown number.0369

Every time, all these years in math, we have been trying to solve for x or t, or z, whatever it is, for unknown variable, for an unknown number.0398

In other words, the variable x is going to be some unknown number.0412

We have something like x² + 4x + 5 = 0.0415

x is equal to this, this, this, whatever it is.0421

Or we have 2 sin x + cos x is equal to 3.0424

Now our unknown is not a number, it is an actual function.0431

Now our unknown is a function y(x).0441

When writing a differential equation, we often use the prime notation instead of a dy dx and dy dt notation.0456

We often you the prime notation of differentiation.0472

There is a prime notation and we would leave off the independent variable, usually x.0481

You leave off the independent variable from the notation y(x).0500

We do not really write y(x), we will just write y for the function.0509

Our sample equations, the four equations that we mentioned on the first page, would look like this.0520

Our sample equations would look like y’ – 4x² = 3x, y” + 6x = y’, y’ = xy, y’² – y’ = y sin y.0532

It is a lot cleaner to use this prime notation.0571

Now take note, y’² is not the same as y”.0575

y” is the second derivative.0591

y’² is the first derivative², be very careful with that.0594

Always remember, I will repeat this several times.0601

y is a function not a number.0606

We are looking for a function of x.0617

The question is, what are these differential equation is saying, what do they mean?0626

What are these de is saying?0634

If I have something, let us take the first equation y’ - 4x² = 3x.0647

What does this equation mean?0657

This means that y(x) is a function such that, when I take the first derivative then subtract 4x², I am left with 3x.0662

That is all this is saying.0718

A differential equation establishes a relationship between the derivative and other things.0720

Those other things might be functions of x.0726

They might be the original function y(x) itself, that is all it is.0728

That is all any equation is.0732

It is a relationship between the different parts.0733

In this particular case, let us say some scientist collected some data.0736

When they analyze that data, they found a relationship between the rate of change of y.0741

That is what a derivative is, it is a rate of change.0746

The rate of change of y - 4x² actually ends up being equal to 3x.0750

We want to find out what y is, what function satisfies this differential equation?0756

What function, when I take the first derivative, subtract 4x² from it will actually give me 3x?0762

That is what we are doing, that all we are doing.0768

We are trying to find a function that actually does this.0771

Let us go to, I’m going to try a function, I’m going to ask you to verify that it actually solve the differential equation.0777

Verify that y(x) = 4/3 x³ + 3/2 x² + 19, is a solution of the de that we are just working with, which is y’ – 4x² = 3x.0790

Verify that this function that I gave you is a solution of the de.0825

How do you verify that a particular function is a solution of a different an equation?0832

You take the derivatives, you plug it into the equation, and you will see if the left side actually equals the right side.0837

This equality, verify it just like we did in trig identities.0845

You are actually verifying that the left side = the right side.0848

That is how you do it.0853

You take derivatives, however many you need, whatever the differential equations says.0855

Take derivatives and substitute into the de to check that equality holds.0871

In this particular case, this is the function, so I find y’.0896

y’ is going to equal 4x² + 3x.0901

Now I substitute this into the differential equation.0914

We have y’ which is equal to 4x² + 3x - 4x².0925

The questions is, does is equal 3x?0936

4x² - 4x², you get 3x = 3x.0939

Yes, this confirms that this is a solution of that differential equation.0944

Let us take a look at the second differential equation, that was y’’ + 6x = y’.0956

What is this one saying?0969

This says y(x) is such a function that differentiating twice then adding a 6x,0977

actually gives you the first derivative of the function.1011

This is saying find the function y(x), that when you take the second derivative of it and then you add 6x to it,1022

you actually end up getting the first derivative of the function.1029

What function satisfies that equality?1033

Here is the verification.1037

I put it to you, verify that the function y(x) which is equal to 4e ⁺x + 3x² + 6x + e ⁺π is a solution of this differential equation.1041

Verify that.1073

How do you verify it?1074

You take derivatives, you plug it back in, and if you see if the left side = the right side.1075

Let us go ahead and do that.1081

This is our y, let us go ahead and take the first derivative.1082

y’ is equal to 4e ⁺x + 6x + 6 y”.1086

Because it involves double prime, it is equal to 4e ⁺x + 6.1096

Now I substitute.1105

Now put these into the differential equation to see if the equality holds.1116

y” that is going to be 4e ⁺x + 6, that is the y” part, and then, + 6x.1129

The question is, does it equal the first derivative which is 4e ⁺x + 6x + 6.1145

Yes, this is a solution of that differential equation.1157

I have verified it by taking derivatives, plugging it in and showing that the left side = the right side.1168

Let us do one more of these.1180

Let us try the third differential equation.1183

The third differential equation was y’ = xy.1188

This says, when I differentiate y one time, when I differentiate y(x), I get x × the original function y.1193

Let us do a verification.1227

Verify that the function y(x) = 5e ⁺x²/ 2 is a solution of this differential equation.1230

y’ is equal to 5e ⁺x²/ 2 × the derivative of that which is going to be 2x²/ 2 which is x.1256

y’ is equal to 5x e ⁺x²/ 2.1272

Now I substitute into the differential equation.1278

Put this into the differential equation and check to see.1290

y’ that is equal to 5x e ⁺x²/ 2, the question is does that equal x × y?1293

y was 5e ⁺x²/ 2, yes it does, that is a solution.1305

Let me go to blue, I think blue is my favorite color for these.1321

Instead of y(x) = 5 × e ⁺x²/ 2, I could have given you the following.1325

I could have given you y(x) = c × e ⁺x²/ 2, where c is any constant or c is any constant.1339

You can verify that this where c is any constant, it does not have to be the 5 will work.1365

Since c can be any number, this y(x) = c e ⁺x²/ 2, it represents an infinite family of functions, an infinite family of solutions.1374

We call it the general solution.1418

Once you have verified that, if you just use c instead of 5, it still satisfies the differential equation.1438

You are just going to get cx e ⁺x²/ 2 for our y’.1446

It satisfies it.1450

Basically, c could be any number.1452

Whatever c is, y could be 1 e ⁺x²/ 2, 5e ⁺x²/ 2, -15e ⁺x²/ 2, π e ⁺x²/ 2, it does not matter.1454

What it gives you is an infinite family of solutions.1468

We call that a general solution.1471

Anything that involves a constant, we call that the general solution.1472

This y(x) = 5e ⁺x²/ 2 is called a particular solution.1479

In differential equations, we speak of general solutions and we speak of particular solutions.1494

Notice that the thing that I want to show you in just a moment, that y (x) = c × e ⁺x²/ 2 is a family of curves.1503

Because this function y = is just a function of x, y = a function of x.1520

We know that y as a function of x is just a graph.1526

It is just a curve in the xy plane, that is it.1527

It is already a function, we can graph it.1530

It is a family of curves.1532

There you go, that is it.1543

A solution to a differential equation is also a curve in the xy plane, in the Cartesian coordinate system,1548

because it is just a function of x.1556

You are looking for some unknown function of x.1558

A function of x is a curve, that is it, it is all it is.1561

In this particular case, this curve right here, this is when c is equal to 2.1563

This curve represents when c is equal to 3.1572

This curve represents when c is equal to 4, and so on.1576

This curve right here, this was our y = 5e ⁺x²/ 2, that is it.1582

When a differential equation is given alone, it is the general solution that we are finding.1607

In other words, the one with constants.1631

In this particular case, y(x) = ce ⁺x²/ 2.1642

That is what this is, it is a family of solutions.1649

When a differential equation is given with a set of initial conditions, when a de is given with a set of initial conditions,1658

I will describe what those are in just a minute.1677

In other words, a certain function of y for a certain function of x that I know,1684

conditions such as y’ = xy, that is the differential equation and y(1.5) is equal to 7.2.1689

Now I do not know the function but I know an initial condition.1704

I know that at 1.5, when x = 1.5, I know the value of y is 7.2.1710

This is called an initial condition.1715

When a de is given with a set of initial conditions, in this case, just one initial condition, this is called an initial value problem.1720

In other words, you will also see it as just plain old ivp.1743

After we find the general solution, we use the initial condition to find a particular solution.1748

We use the initial condition to find, find a particular solution, to find a specific value for c.1784

Once we find the general solution, we use the initial condition next in the general solution, to find the specific value for c.1795

For y’ = xy, we found that y(x) is equal to some constant × e ⁺x²/ 2.1816

They tell us that, another thing that we know is that y(1.5) = 7.2.1833

Let us put 1.5 in for x.1838

y(1.5) which is equal to c × e¹.5²/ 2, they are telling me that it actually = 7.2.1841

I solve this equation for c.1854

c = 2.337, my particular solution in this case is y(x) is equal to 2.337 e ⁺x²/ 2.1863

This is my particular solution.1878

Find the general solution, use the initial conditions to find c.1883

This is your particular solution for your particular task at hand.1888

That is this curve right here.1894

This curve is y = 2.337 e ⁺x²/ 2.1896

In this particular case, 1.5, 7.2, that is that point right there.1907

An initial condition is a point through which the particular solution passes, when you are looking at an actual curve of the solution.1919

From a family of solutions, we have reduced it to one, by use of an initial condition.1931

Let us see, let us move on to do some examples.1939

Again, in this lesson, we are only concerned with introducing you to differential equations,1945

having you do some basic verification, a little bit of manipulation.1949

In the following lessons, we will actually start working on how to solve these differential equations.1953

Verify that the family of functions y(x) = c/ x³ + 5x/ 4, is a solution of the differential equation xy’ + 3y = 5x.1959

How do we do a verification?1970

We find the derivatives, however many we need, first, second, third, whatever.1972

We put them in and we verify that left side is actually equal to the right side.1976

y(x) = that, let me work in blue.1984

I have got y(x) is equal to, I’m going to write it as cx⁻³ + 5/ 4x.1987

y’ is equal to -3x⁴ + 5/4.1998

Now I substitute into the de, substitute into the differential equation.2010

I have got x × y’ which I just found which is -3x⁻⁴ + 5/4 + 3y + 3 × y which is c/ x³ + 5x/ 4.2024

The question is does it equal 5x?2055

This is x × -3/ x⁴ + 5/4 + 3c/ x³ + 15x/ 42064

which = -3c/ x³ + 5x/ 4 + 3c/ x³ + 15x/ 4, 20x/ 4 5x.2084

Yes, this is a solution of the differential equation.2112

Notice this has a constant in it.2118

This is a family of solutions for this particular differential equations, with different values of c.2125

I have the different values of c over here.2131

In one of the cases, one c is equal to 1, I have the black curve.2134

That is this thing, that is this one.2138

That is a solution of the differential equation.2144

It is a graphical solution of the algebraic equation.2147

When y = 10, I have the orange curve.2150

Your orange curve is right there, and so on.2153

When c = 20, when c = -1, -10, -20, this is a family of curves.2159

For what values of k does the function y = k ⁺x satisfy the differential equation 3y” + 6y’ - 9 = 0.2170

For what values of k, interesting.2181

We have a function, we have the differential equation.2190

Let us just differentiate twice, plug it in and see what happens.2194

y is equal to e ⁺kx, y’ is equal to ke ⁺kx, and y” is equal to k² e ⁺kx.2202

Let us put these into the differential equation.2219

I have got 3 × k² e ⁺kx + 6 × y’ which is ke ⁺kx - 9 is equal to 0.2222

There is a little bit of mistake here, sorry about that.2254

Let me go to black, I forgot my y, I apologize for that.2257

This is supposed to be -9y is equal to 0.2265

Now let me go back to blue and finish this off.2274

-9 × y which is e ⁺kx = 0.2278

I have the function, I have the differential equations for what values of k I have differentiated.2288

I have this equation k.2293

Let me factor out the e ⁺kx.2295

I’m left with 3k² + 6k - 9 is equal to 0.2298

e ⁺kx is greater than 0 for all x.2309

Therefore, this is equal to 0.2314

3k² + 6k - 9 is equal to 0.2323

Let us go ahead and divide by 3 and make it a little easier on myself.2327

k² + 2x - 3 = 0, this one happens to factor, if not, no big deal.2330

We will just a graphing device or quadratic equation.2338

We have got k + 3 – 1.2345

We have k = -3, we have k = 1.2353

You have y is equal to e ⁻3x.2363

y is equal to e ⁺x.2370

When k is a -3 or 1, those two values of k satisfy this particular differential equation.2372

That is it, nice and straightforward.2383

Verify that the family of functions y(x) = c × e¹/ x + 7/ x + 7 is a solution to the differential equation x³ y’ + xy = 7.2388

Then, solve the initial value problem differential equation + y of 5 = 5.2400

Let us see what we can do.2409

We have the y = this thing.2410

Let us go ahead and find y’.2415

y’ = the derivative of this.2417

It is going to be c × e¹/ x × -1/ x² - 7/ x² + 0.2421

Now we go ahead and we substitute this into here.2438

We are going to write x³ × -c/ x² e¹/ x - 7/ x² + x × y + x × y2443

which is c × e¹/ x + 7/ x + 7.2465

The question is does that equal 7?2475

Here we are going to get –cx e¹/ x - 7/ x + cx e¹/ x.2480

What is going on, I’m losing my way here.2484

x × c ⁺7x/ 7.2516

Wait a minute, I have got all these symbols floating around.2527

We took the derivative, - c/ x² that is correct, - c/ x² e¹/ x.2532

This is -7/ x², that was our y’.2542

x³ × that, this is not 7/x, this is 7x x³.2547

There you go, + x × that, perfect.2562

+ 7 + 7x, sorry about that, it happens a lot.2572

Here we have – cx e¹/ x + cx e¹/ x - 7x + 7x.2579

Yes, we are left with 7.2587

It definitely equals that, great.2589

Now let us go ahead and solve the initial value problem.2593

y(5) = 5, we have y(x) = ce¹/ x.2598

We just verified that this is a solution, + 7/ x + 7.2612

They tell me that y(5) which is equal to ce¹/5 + 7/ 5 + 7, they are telling me that actually = 5.2620

When I solve for c, I get 2.78.2636

Therefore, our particular solution is y(x) is equal to - 2.78 e¹/ x + 7/ x + 7.2641

Now I have a function, the unknown function that I actually solve for a particular situation.2658

Now no matter what value of x I have, I can tell you what y is going to be.2665

This represents the family of solutions.2671

This is the family y(x) = ce¹/ x + 7/ x + 7.2676

That is all of these, the black.2697

When c is equal to 1, we have the black curve right here.2699

When c is equal to -1, we have the blue curve.2706

It looks a little different, that is here.2711

It comes up like that.2713

When y is equal to 10, we have the purple curve.2716

This one right here, that right there.2721

This is the family of curves.2724

Of course, the particular solution that we found.2730

This is y = - 2.78 e¹/ x + 7/ x + 7.2734

This is the particular solution to our initial value problem which was x³ y’ + xy = 7, y(5) = 5.2753

5,5, here is the point, 5,5 that is what this tells me.2770

The curve passes through that point.2776

I have an initial value.2780

The general solution, I have an initial value.2781

I can actually find a specific curve that satisfies this differential equation.2784

That is it for our introduction to differential equations, I hope that made sense.2791

Thank you so much for joining us here at www.educator.com.2794

We will see you next time, bye.2796