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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

Transcription: Separation of Variables

Hello, welcome to, 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 sign0091

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

We will see you next time, bye.1688