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
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- 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 equation0075
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 www.educator.com.1685
We will see you next time, bye.1688
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