For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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## Table of Contents

## Transcription

### Introduction to Differential Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction to Differential Equations
- Overview
- Differential Equations Involving Derivatives of y(x)
- Differential Equations Involving Derivatives of y(x) and Function of y(x)
- Equations for an Unknown Number
- What are These Differential Equations Saying?
- Verifying that a Function is a Solution of the Differential Equation
- Verifying that a Function is a Solution of the Differential Equation
- Verify that y(x) = 4e^x + 3x² + 6x + e^π is a Solution of this Differential Equation
- General Solution
- Particular Solution
- Initial Value Problem
- Example I: Verify that a Family of Functions is a Solution of the Differential Equation
- Example II: For What Values of K Does the Function Satisfy the Differential Equation
- Example III: Verify the Solution and Solve the Initial Value Problem

- Intro 0:00
- Introduction to Differential Equations 0:09
- Overview
- Differential Equations Involving Derivatives of y(x)
- Differential Equations Involving Derivatives of y(x) and Function of y(x)
- Equations for an Unknown Number
- What are These Differential Equations Saying?
- Verifying that a Function is a Solution of the Differential Equation 13:00
- Verifying that a Function is a Solution of the Differential Equation
- Verify that y(x) = 4e^x + 3x² + 6x + e^π is a Solution of this Differential Equation
- General Solution
- Particular Solution
- Initial Value Problem
- Example I: Verify that a Family of Functions is a Solution of the Differential Equation 32:24
- Example II: For What Values of K Does the Function Satisfy the Differential Equation 36:07
- Example III: Verify the Solution and Solve the Initial Value Problem 39:47

### AP Calculus AB Online Prep Course

### Transcription: 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 and*0288

*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/ 4*2064

*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 × y*2443

*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

3 answers

Last reply by: Professor Hovasapian

Mon Jan 29, 2018 11:15 PM

Post by Magic Fu on January 23 at 12:23:09 PM

can you go over Euler's method?

1 answer

Last reply by: Professor Hovasapian

Wed Oct 26, 2016 7:42 PM

Post by Tiffany Warner on September 28, 2016

Hello Professor Hovasapian,

With Example 1, I am really confused. In the last line of solving, a C reappears in the problem. x((-3/x^4)+(5/4)) becomes (-3C/x^3) + (5x/4). How come the C got thrown in? I'm probably missing something obvious but I keep looking it over and I'm not seeing it.

Thank you.

3 answers

Last reply by: Professor Hovasapian

Mon Jul 25, 2016 7:07 PM

Post by Peter Ke on July 23, 2016

At 40:40, how is this ----> http://prnt.sc/bwjrof the derivative of y?

I thought it was:

http://prnt.sc/bwjrof

Please explain.