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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### Separation of Variables

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 0:00
- Separation of Variables 0:28
- Separation of Variables
- Example I: Solve the Following g Initial Value Problem 8:29
- Example II: Solve the Following g Initial Value Problem 13:46
- Example III: Find an Equation of the Curve 18:48

### AP Calculus AB Online Prep Course

### Transcription: Separation of Variables

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

*Today, we are going to talk about a technique for actually solving any particular differential equation that you are faced with.*0004

*The technique is called separation of variables.*0011

*It is a very simple technique.*0013

*It does not apply to every differential equation but of course to the ones that it does, it is very straightforward.*0015

*You just literally separate the x and y variables and you integrate.*0021

*It is that simple, so let us get started.*0025

*Let us look at the following differential equation.*0030

*I will go ahead and work with blue maybe.*0035

*Let us look at the following differential equation.*0042

*Let us look at y' is equal to y² × cos(x).*0051

*Or if you want to put it in terms of dy dx, it is dy dx is equal to y² cos(x).*0059

*For separation of variables, you definitely want to use the dy dx.*0070

*You will see why in just a second.*0072

*The idea is this, if we can rearrange a differential equation*0075

*such that all symbols containing x's are on one side of the equality sign*0091

*and all symbols containing y are on the other, we call this a separable equation or a separable differential equation.*0121

*We can apply this idea of separation of variables.*0153

*Separable equations are very easy to solve.*0158

*They are easy in the sense that all you do is integrate.*0172

*It is as easy as the integration is.*0175

*You just integrate both sides.*0180

*In this particular case, we have dy dx is equal to y² cos x.*0191

*We are going to separate the variables, if we can.*0199

*dx, I’m going to move it over to the right side.*0202

*I get dy = y² cos x dx.*0206

*Now everything that involves of y, I'm going to bring over to this side.*0211

*I’m going to divide everything by y².*0214

*I’m going to end up with dy/ y² = cos x dx.*0217

*Everything involving a y is on the left, everything involving an x is on the right.*0223

*We were able to separate the variables.*0227

*We have separated the variables.*0234

*Now we just integrate both sides.*0248

*Let me write the equation again.*0250

*We have dy/ y² is equal to cos x dx.*0253

*You have an equality.*0259

*Anything you do to the left side of the equality, as long as you do the right side, the equality is maintained.*0260

*We just integrate both sides.*0265

*It is literally that simple.*0269

*What you end up with is, the integral, I’m going to rewrite this y⁻² dy = the integral of cos x dx.*0270

*This one here becomes –y⁻¹ + let us just call it c1, constant, these are indefinite integrals.*0282

*It is equal to sin x + c2.*0291

*C1 and c2 are just constants.*0296

*I’m just going to put them together.*0297

*I’m going to combine c1 and c2 into just a single constant and put it on whichever side I want, in this case, on the right side.*0301

*I’m going to write this as -1/y = sin x + c.*0313

*Now I just solve for y.*0323

*Y is equal to -1/ sin x + c.*0327

*There you go, that is your differential equation.*0342

*This is your function of x.*0344

*In this case, we are actually able to solve explicitly for y = some function of x.*0347

*You would not be always be able to do that.*0352

*You might have to leave it in implicit form because the y and x might be a little too mixed up.*0353

*We will see some examples of that, it is not a problem.*0358

*It is literally that simple.*0360

*C is the constant and it can be anything.*0361

*This is the general solution.*0364

*Remember, a general solution, you have a constant.*0366

*For different values of the constant, you are going to get different curves in the xy plane.*0369

*It is that simple, literally, it is that simple.*0374

*If you can separate the variables, you can integrate.*0378

*Hopefully, you can integrate.*0380

*It is that simple.*0384

*You separate variables, if you can, and two, you integrate.*0391

*You are going to solve the different equation for y, some function y that your looking for.*0401

*That is your unknown.*0405

*Let us take a look at what this looks like graphically, for different values of c.*0411

*Here you have it, here are the different values of c.*0418

*The orange, black, you have a purple, and you have a green.*0420

*When c = 1, you have the black graph, that is this one right here.*0424

*Let me do this in red.*0433

*When c = 1, you get this black curve right here, that is the solution.*0440

*This is one particular number that goes on.*0447

*When c = 2, we have the green graph.*0452

*Now it changes a little bit, now it is this way, that is that one.*0456

*When c = -2, that is the purple graph.*0461

*This time it is above the x axis.*0464

*When c = -1/2, we have the orange graph.*0469

*That is all that is going on here.*0478

*This is the family of solutions for different values of c.*0480

*Once again, in this particular case, this was y =,*0486

*y as a function of x is equal to -1/ sin(x) + c, for different values of c.*0493

*Once you pick a c, it is going to be one particular curve.*0501

*Let us go ahead and do an example here.*0508

*Solve the following initial value problem.*0510

*We remember, an initial value problem is the differential equation and some initial value.*0514

*The x valued does not always have to be 0, it can be any number.*0520

*In this particular case, it happens to be 0.*0523

*They tell me that the curve, when x is equal to 0, it passes through the point 4.*0525

*It passes through the point 0,4.*0532

*I know this much and I know that this is the relationship, that dy dx = x²/ y³.*0536

*Let us see if this is possible to solve.*0543

*Here we have, y’, I'm going to write that as, let me do this in blue.*0546

*I’m going to write that as dy dx is equal to x²/ y³ and we try to separate variables.*0551

*We are going to end up with dy = x² dx/ y³.*0562

*Now I’m going to multiply by y³.*0572

*I end up with y³ dy = x² dx.*0575

*Yes, in this particular case, we were able to separate the variables.*0580

*Now we just integrate both sides.*0583

*Integrate this side, integrate this side, and I end up with y⁴/ 4 is equal to x³/ 3 + c.*0585

*This is my general solution.*0600

*If you want to, you can solve explicitly for y.*0603

*Let me just say, you are welcome to leave it in this form.*0609

*Excuse me, you are welcome to leave it in this implicit form.*0613

*In other words, not y = some function of x.*0628

*Again, sometimes it is not going to be possible, sometimes it is.*0637

*Or if possible, you can solve explicitly for y.*0640

*In this particular case, it ends up being, I multiply by 4.*0658

*I end up with y⁴ = 4/3 x³ + 4c.*0663

*I get y = 4/3 x³ + 4c¹/4.*0672

*This is the explicit, it does not matter, it is a personal choice, however far you want to take it.*0683

*This is the general solution, the one that involves the constant.*0690

*Now let us deal with the initial value problem.*0693

*y(0) = 4 or y(0) = 2, it does not matter.*0696

*I think on my paper I have a different value but it does really matter.*0715

*For the initial value, y(0) is equal to 4.*0724

*Now we have y⁴/ 4, we will write our equation down.*0741

*It is equal to x³/ 3 + c.*0746

*Now we are looking for this particular c.*0752

*We put these in, y is equal to 4.*0756

*It is going to be 4⁴/ 4 is equal to 0³/ 3.*0759

*This is x, this is y, + c.*0769

*C is going to equal 4³.*0778

*4 × 4 is 16, 4 × 16 is 64, c = 64.*0783

*Therefore, our final solution is y⁴/ 4, our particular solution is equal to x³/ 3 + 64.*0788

*That is it, it is that simple.*0802

*This is what the family of solutions looks like for different values of c.*0809

*That is it, like that.*0813

*For a particular value of c, you are going to get a curve that looks like that, one of those curves.*0817

*Let us take a look at another example.*0827

*Solve the following initial value problem.*0830

*This time we have y’ = y sin x/ y³ + 2 with initial value y(0) = -2.*0833

*Let us see if we can separate the variables here.*0842

*We are going to write this as dy dx is equal to y × sin(x)/ y³ + 2.*0845

*When I separate the variables, I'm going to end up with the following.*0862

*I’m going to end up with y³ + 2/ y dy is equal to sin x dx.*0865

*Bring this over here, bring this down here, bring the dx up there.*0883

*I was able to separate the variables.*0887

*Everything with a y on one side, everything with an x on the other side.*0888

*The variables are separated, now we integrate both sides.*0892

*We integrate this side and we integrate that side.*0895

*This integral over here becomes the integral of y² dy, because y³/ y is y².*0898

*I’m going to separate this fraction, + the integral of 2/ y dy is going to equal the integral sin x dx.*0910

*Now I take care of the integrals.*0923

*This one becomes y³/ 3.*0924

*This one becomes 2 × natlog of the absolute value of y.*0929

*This one = - cos(x) + a constant c.*0934

*This is our general solution.*0942

*Now we deal with the initial value y(0) = -2.*0948

*I put these values into here, x is 0, y is 2.*0952

*This is the y, this is my x.*0957

*I put these into here and I solve for c.*0960

*I have y is -2, this is going to be -2³.*0965

*I should do that actually, /3 + 2 × natlog of y absolute value of -2 = -cos(0) + c.*0974

*I end up with -8/3 + 2 ln 2 = -1 + c.*0994

*I move the 1 over, I get c = -5/3 + 2 ln 2.*1007

*This is my value of c.*1014

*Put that back in there and I get my particular solution.*1017

*Our particular solution, in other words, our solution to the initial value problem,*1029

*our particular solution is y³/ 3 + 2 × ln of the absolute value of y = -cos x + 2 × natlog of 2 ln 2 - 5/3.*1036

*There you go, that is it, this is the solution.*1064

*In this particular case, we definitely want to leave it in implicit form.*1067

*In fact, we have no choice because separating the y here and we have a y here as the argument of the logarithm function.*1071

*This is going to be very difficult to separate, if it is possible at all.*1076

*I do not ever think it is possible to separate the y from the x.*1080

*This is a perfectly good, perfectly valid, completely acceptable solution to that particular initial value problem.*1081

*Once again, separate the variables, if possible, integrate.*1093

*And then, put the values in for the initial value problem, if there is an initial value.*1098

*Or else just leave it with the constant c for the general solution.*1102

*Let us see what this one looks like.*1105

*This looks like this.*1107

*This is the particular solution, here is the equation.*1110

*This is what it looks like.*1116

*That is it, for a particular value of x, now you know what the particular value of y is going to be.*1117

*It is that simple.*1123

*Let us move on to the next.*1129

*Find an equation of the curve whose slope at the point xy is x² y, and which passes through the point 3/5, 5.*1132

*We are looking for some y(x), again, some curve.*1147

*This is the verbal description of an initial value problem.*1158

*The initial value problem is the following.*1162

*Find an equation of the curve whose slope of the point xy is x² y.*1165

*What is the slope?*1169

*The slope is the derivative.*1171

*The derivative is dy dx.*1173

*Dy dx, the slope at the point xy is equal to x² y.*1177

*That is our differential equation, and which passes through the point 3/2, 5.*1183

*The initial value is y of 3/2 = 5.*1188

*This is the initial value problem that is represented by this word problem.*1193

*Let us go ahead and see what we can do.*1201

*We have dy dx = x² y.*1204

*We separate variables, if we can.*1206

*We divide by y, move the dx up here.*1208

*We should have something like dy/y is equal to x² dx.*1211

*Now we can integrate both sides.*1222

*We are left with the natlog of the absolute value of y is equal to x³/ 3 + c.*1224

*This is our general solution.*1235

*It gives us a family of curves.*1238

*Now let us go ahead and use the initial value.*1240

*This is our x value, this is our y value.*1244

*The natlog of the absolute value of 5.*1246

*The absolute value 5 is 5, this becomes natlog of 5 is equal to x which is 3/2³/ 3 + c.*1249

*And then, you just solve it from here, not a problem at all.*1262

*This ends up giving the natlog of 5 is equal to 9/8 + c.*1267

*Therefore, c is equal to the natlog of 5 - 9/8.*1275

*Our solution is the natlog of the absolute value of y is equal to x³/ 3 + our value of c which is the natlog of 5 - 9/8.*1286

*This is our particular solution.*1304

*That is it, it is that simple.*1308

*Let us express this another way, it is a logarithm.*1311

*If you want to, you do not have to, this perfectly acceptable.*1315

*If you want to, just to see what something else looks like.*1318

*Let us express this, perhaps your teacher wants you to express it in terms of exponential,*1325

*perhaps the test that you are taking wants you to express it, in terms of exponential, who knows.*1333

*Perhaps you want to.*1336

*Let us express this in another way.*1338

*We have the natlog of the absolute value of y is equal to x³/ 3 + c.*1341

*This is the natlog, just exponentiate both sides.*1352

*What you are left with is the absolute value of y is equal to e ⁺x³/ 3 + c.*1357

*We already found c.*1368

*But if you did not find c, then if you just continued along with this, to put in the exponential form, you can find c.*1370

*We already found c but we could also have come to this point and done,*1380

*We said that y(3/2) = 5, this is our x value, this is our y value.*1408

*Absolute value of 5 is 5.*1416

*We have 5 = e x³/ 3, when I put 3/2 and I cube it and I divide by 3, 9/8 + c.*1418

*I need to solve for c.*1435

*I need to take the logarithm of both sides.*1437

*I get ln of 5 = 9/8 + c which implies that c = ln 5 – 9/2, which is exactly what I got before.*1439

*The exponential version of our solution becomes the absolute value of y is equal to e ⁺x³/ 3 + ln of 5 -9/8.*1457

*This is the other version.*1483

*I like to say something about the absolute value here, because absolute value scares the hell out of people.*1492

*I do not understand why, it used to scare the hell out of me.*1499

*It still does, from time to time.*1502

*Let us just do a quick recollection here.*1504

*Recall the meaning of absolute value.*1505

*The absolute value of y, it means the following.*1518

*It is y, if y happens to be bigger than 0.*1524

*It is –y, if y happens to be less than 0.*1529

*That is the definition of absolute value.*1534

*We have the absolute value of y is equal to e ⁺x³/ 3 + c.*1540

*For y greater than 0, the absolute value of y is just equal to y.*1563

*Therefore, y is equal to just e ⁺x³ + c.*1572

*For y less than 0, we have the absolute value of y is equal to –y.*1589

*Therefore, -y is equal to e ⁺x³/ 3 + c.*1602

*y is equal to -e ⁺x³/ 3 + c.*1612

*That is what the absolute value means.*1623

*Graphically, this is what is going to happen.*1625

*The absolute value sign, the absolute value gives you graphs above the x axis, below the y axis.*1628

*This is the positive, y = e ⁺x³ + c.*1635

*This is negative, e ⁺x³ + c.*1639

*What you have are different values for the constant c.*1643

*In our particular case, which is this one right here.*1649

*The blue, we said the absolute value of y is equal to x³/ 3 + ln 5 - 9/8, that is this blue curve right here.*1653

*Notice how it passes through the point 3/2, 5, and this one here because absolute value.*1665

*It is that simple, separate the variables and integrate.*1677

*And then, solve the initial value problem for the value of the constant.*1680

*If you need to, go ahead and graph it.*1683

*Thank you so much for joining us here at www.educator.com.*1685

*We will see you next time, bye.*1688

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