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Lecture Comments (24)

1 answer

Last reply by: Dr. William Murray
Mon Apr 4, 2016 11:44 AM

Post by Shih-Kuan Chen on March 31 at 06:05:21 PM

In example 4, why isn't y equal to + or - that whole business?

3 answers

Last reply by: Dr. William Murray
Mon Jan 4, 2016 12:29 PM

Post by Jonathan Snow on December 22, 2015

I have a quick question about the plus or minus K in example 1, why is it plus or minus? If k=e^c, k can't be a negative number right?

1 answer

Last reply by: Dr. William Murray
Fri Oct 30, 2015 4:14 PM

Post by Aisha Alkaff on October 29, 2015

Hello Dr. Murray,
Thank you for your clear explanation. It really helped me a lot.

but i have a question in Example 2:
why wouldn't we just simplify y^2 byt taking the  square root?

1 answer

Last reply by: Dr. William Murray
Sun May 3, 2015 7:38 PM

Post by Tsz Hong Chow on April 29, 2015

Hi, have you taught Bernoulli equations? I badly need this lecture

1 answer

Last reply by: Dr. William Murray
Mon Aug 25, 2014 6:42 PM

Post by Joseph Green on August 24, 2014

will there be any practice problems added to this lecture?

1 answer

Last reply by: Dr. William Murray
Sat Jul 5, 2014 6:07 PM

Post by Josh Winfield on June 30, 2014

Hello,

Why do you not write y' and y as functions of x, but you clearly do for the "coefficients" P(x) and Q(x), it bugs me a little. Is it because you want to highlight that y and y' are the variables your solving for, or just to make the equation a tad more clean. I can't help but leave y and y' as functions of x when i do all problems, is this incorrect?

Regards
Josh

1 answer

Last reply by: Dr. William Murray
Tue Feb 25, 2014 4:49 PM

Post by Ahmed Obbad on February 24, 2014

Hi professor, could you please explain me why the additive constant does not satisfy the differential equation while the multiplicative constant do? what if we keep it as Y=e^(x/2+ c)is this wrong?

1 answer

Last reply by: Dr. William Murray
Tue Jan 14, 2014 11:47 AM

Post by cigdem drahman ozkan on January 8, 2014

Hi professor, Could you kindly explain me; in second example, why we just ignore 2C?  and wrote as new constant K?

1 answer

Last reply by: Dr. William Murray
Fri Jul 5, 2013 10:19 AM

Post by xueping liu on July 2, 2013

Hi Dr.Murray
Your videos are awesome.They helped so much. But I also want to find some exercises for the related topic without buying those books. Do you have any suggestions?

Thanks

3 answers

Last reply by: Dr. William Murray
Tue Apr 16, 2013 8:32 PM

Post by Mohammed Alhumaidi on April 6, 2013

This is an enjoyable experience, at least I do understand something now rather than in class lecturing, why is it the case though Prof, Murray

Separable Equations

Separable Equations (PDF)

Separable 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 0:00
  • Lesson Objectives 0:19
    • Some Equations Are Both Linear and Separable So You Can Use Either Technique to Solve Them
    • Important to Add C When You Do the Integration
  • Example 1 4:28
  • Example 2 10:45
  • Example 3 14:43
  • Example 4 19:21
  • Example 5 27:23

Transcription: Separable Equations

Hi and thanks for joining us here on www.educator.com, my name is Will Murray and I’m doing the differential equations lecture series.0000

We have already had a lecture on linear differential equation, this is the second lecture and we are going to be studying separable equations today.0007

I hope you will join us and let us jump in.0015

The idea of separable differential equations is that you are going to write Y′(x) as (dy)/(dx) and try to separate the variables in the following form.0022

You are going to try get all the y’s on one side of the equation and put (dy) over there and then you try to get all the x’s on the other side of the equation, put the (dx) over there.0032

That will not always work but if you can then it is called a separable differential equation where you can separate it into all the y’s on one side and all the x’s on the other side.0043

If you can do that then what you can do is integrate both sides and you will get some function of y on the left and some function of x on the right.0054

What you can do after that is solve for y(x), generally in to a few steps of algebra to solve for y(x) and you will get that y is equal to some function of x then you will be done.0067

The basic idea of separable differential equation, after we work couple of examples I think it will make a better sense.0083

There is a couple of notes I wanted to mention before we get started, one note is that some equation are both linear and separable.0089

We just had a lecture on linear differential equation and we had a completely different technique for solving that.0098

That was the one linear was the type of equation where you can write it as Y′ + p(y)=q and the we have a different technique for integrating factors for solving linear differential equations.0104

For some differential equations, they are both linear and separable so you can use the technique I just thought you or the linear technique to solve them.0119

If you do have a choice though, you probably want to go with the separation technique.0129

If you do have a choice the separation technique is usually a little bit less work than going through that whole integrating factor business, try to go with separation first, that is your first try.0134

Second very important note is remember that I said you are going to get something in terms of y × (dy)= something in terms of x × dx, and then you are going to integrate both sides.0147

What is important there is that you add the constant when you do the integration and not at any other step, it is very important that you do it exactly at that step of integration.0163

You do not have to add the constant to both sides you can just do it at one side because if you add the constant to both sides then you could just combine the 2 constants on the same side.0179

You only have to add the constant on one side but it is very important that you do it and it is important that you do it when you do the integration.0189

The reason it is so important is because after you do the integration there is usually a couple of steps of algebra for solving for y in terms of x.0197

A lot of times that will end up tangling up the constant into the equation and you have to let that happen.0207

You have to let that constant get tangled up into the solution, this is a real change from what you learned back in calculus 1 and 2.0213

Calculus 1 and 2 which you learned is whenever you do an indefinite integral, you just tack on a (+c) at the end.0226

You can do about 50 problems and then go back at the end and just put (+c) at the end of every single one.0232

Not so in differential equation, you got to be really careful with the constant and be careful when you insert it.0239

The time to add it on there is when you do the integration and every thing you do after that, you have to keep track of the constant as it gets tangled up into the equation.0246

Let us do a few examples and you will see how that constant plays out and how it really becomes part of the solution in a more significant way than it ever did in calculus 1 and 2.0255

For our first example here, we are going to find the general solution to the following differential equation Y′(x)=1/2 y(x).0269

Remember the whole point of separable differential equations is that you write the Y′ as (dy)/(dx) and then you try to separate the x’s on one side and the y’s on the other.0277

Let me write this as (dy)/(dx)=1/2 and that is y(x) that is not y multiply by x.0289

I will just write it as y and let me try to separate the (dy) and y on one side and then I’m going to multiply the (dx) over to the other side.0297

If I pull the y over the left hand side I will get dy/y on the left and move (dx) the other side is equal to ½(dx).0311

I have successfully separated my y’s on one side and my x’s on the other side, I can take the integral of both sides.0323

The integral of (dy)/y is just natural log or technically the absolute value of y.0332

The integral of ½(dx) is just 1/2x, now this is when I’m doing the integration and this is when I’m going to add a (+c).0339

I do not need a (c) on both sides, you do not need a (+c) here because you could just move it over and combine it with the other constant on the other side.0350

It is very important that you have a (+c) on one side or the other and that you incorporate it at this step.0365

Let me emphasize that, must add (c) when you do the integration.0374

Let me really emphasize that you go to do that when you do the integration.0392

The rest of it is just solving for y, in order to undo the natural log I’m going to raise (e) to the both sides, (e)natural log, that is the value of y is equal to,0401

Be careful, this is e (1/2x + c) is not, let me really emphasize this, I will this in red, this is not e 1/2x + c.0413

It is also not equal to e1/2x + ec, it is neither of those things.0429

It is really e(1/2x + c) and the way you resolve that is e1/2x × ec.0438

Ec is a constant so I’m going to call it a new value so I’m going to call it k and on the left hand side we get the absolute value of y is equal to just constant × e1/2x.0448

Y is equal to, technically + or – (ke)1/2x but that is still just a constant, I’m going to call that, what can I use for a new constant?0464

Y is equal to I will use a new constant A(e)1/2x.0483

Let me emphasize here that this is not like calculus 1 or 2 where you can just add on a constant at the end, we would have got in0494

Let me write this in red, we would have got in e1/2x and then we would just tack on a constant at the end.0505

It does not work like that because we really need to keep track of a constant and the fact that it turned into a multiplicative constant and not an additive constant.0514

We are finished with that problem but let me recap the steps here, I wrote Y′ as (dy)/(dx), I wrote y(x) as just y, that is not y multiplied by x.0528

The reason it says y of x here is to remember that y actually is a function of x, I wrote this as ½(y).0537

The whole game for separable differential equations is to get all the y’s on one side and all the x’s on the other side.0547

I moved this y over and I moved this (dx) over this side, so I get dy/y=1/2(dx), integral of dy/y is natural log of the absolute value of y.0555

Integral of 1/2 x is (½ x + c), it is very important that we add the (c) at this step because this is where we are doing the integration.0571

Then we solve for y, so we raise (e) to both sides and you have to be careful with your rules of exponents here e1/2x+c is = e1/2x × ec, not (+) ec.0581

We got the absolute value of y is equal to, ec is another constant and I will call that k.0597

Absolute value of y is = to (ke)1/2x , y just turns into + or – k which I call the new constant A,( Ae)1/2x.0603

I wanted to emphasize here that really the role of the constant, you have to add it as this point and keep track of it.0616

Keeping track of it show you that it is a multiplicative constant and not an additive constant, this would be wrong if you give it as an additive constant.0625

Wrong in the sense that it would not satisfy the differential equation, it would not make the differential equation true anymore.0635

That is why you really have to be careful with that.0641

Let us try another one here, we have to find the general solution to the following initial value problem.0646

We have (yY)′ + x =0, y(0)=3, remember the goal with the separable differential equations is you write your Y′ as dy/dx and we are going to write y × dy/dx + x=0.0655

The idea is to try to separate the variables, get all the x’s on one side and get all the y’s on the other.0677

In this case we get y × dy/dx, I will that x on the other side and get –x and then if I multiply dx by both sides I will get y/dy=-x/dx.0682

Now I have successfully separated my variables x’s on one side, y’s on the other.0700

Take the integral of both of these, the integral of y/dy is = y2/2.0707

The integral of –x/dx is –x2/2, but I just did my integral so I have to put the constant right here.0713

I’m going to try to simplify this a bit, I’m going to multiply both sides by 2 so I get y2 is = -x2 + now 2c is just another constant so I’m going to call it (k).0725

Add the x2 + y2= k and now I’m going to use my initial condition in order to find the value of that constant.0746

I’m going to plug in x=0, y=3, I plug those in y(0)=3, that tells me 02 + 32=k, that tells me k=9.0759

If I plug that back into my general solution there, what I get is x2 + y2=9.0781

That solves my initial value problem, let me remind you of each of the steps that we went through there.0801

We started out with by converting this Y′ into dy/dx and then I was trying to get my x’s on one side and y’s on the other.0806

I got my x on the right, I still have dy/dx on the left so I multiply that dx over to the other side and get x/dx on the right.0816

Now I integrate both sides, integral of y/dy is y2/2, integral of x/dx or –x/dx is –x2/2 + a constant.0827

You do not have to add the constant on both sides but you do have to add it on one side, it is very important that you do it right here when you do the integration.0837

I was going to solve for, try to simplify as best I can, to get rid of those 2 I multiply both sides by 2, get a 2c over here which is 2 × the constant, which is another constant so I called it k.0847

In order to find the value of k, I used my initial condition x=0, y=3, I plug that in right here and I found the value of k is 9.0862

I substitute that back in to the general solution and I get x2 + y2=9.0874

On example 3, I’m given a differential equation Y′(x) + xy=x3, I have to determine if this thing is separable.0885

The way I’m going to do that is I’m going to write my Y′ as dy/dx + xy=x3.0896

I will be wanting to multiply my dx by both side so I’m trying to move and keep my dy/dx on one side and move x3 - xy over the other side.0909

What I would like to the is to factor the right hand side, I want to factor this into something of the form of a function of x × a function of y.0924

I can not factor that into that form and because I can not factor that, this equation is not separable.0948

We can not do this, we can not factor that in that form.0957

This equation is unfortunately is not separable, we can not use our technique of separating the variables to solve this differential equation, it is not separable.0973

That is all really we are asked to find for this example, we are asked to check whether this differential equation is separable.0993

However I will add a note here, in a parenthesis, (However, it is in linear form.)0999

I will remind you what that form is, the linear form for differential equations is Y′ + p(x)/y=q(x).1012

This is in linear form so you could solve this differential equation using the technique we have learned on the previous lecture on linear differential equations.1026

It could be solved using our technique for linear differential equations.1038

That is something that I have a separate lecture here in the differential equation lecture series for linear differential equations.1053

There is another lecture here in the differential equation lecture series for linear differential equations where we learned how to solve this.1073

We used something called the integrating factor to solve the linear differential equation.1081

If you want you can just check that lecture and you can see exactly how to solve this differential equation using that technique even though our separation of variables technique did not work on this one.1086

Let me just remind you what we did here, we started out by writing Y′ as dy/dx, that is a quick first step for all separable differential equation.1099

You put everything over the other side and you try to factor it into something that will easily separate the x’s and y’s.1110

Unfortunately this one, you can not factor it in that form so this differential equation is not separable.1120

However, that is essentially the end of the road as far as separation techniques go but I did notice this original differential equation is in linear form Y′ + p(x)/y=q(x).1126

That was our linear differential equation form so you could use the linear differential equation technique that we learned in the previous lecture to solve this one.1144

You would not give up hope completely on this one, you just have to watch a different lecture to learn how to do it.1154

In our next example, we want to solve the initial value problem 3x – 6y × square root(x2 +1), dy/dx=0 and we also have the initial value of y(0)=4.1161

Again, let us try to separate the variables there, I think I’m going to write 3x= 6y × square root(x2 + 1) dy/dx.1177

I can divide both sides by 3, that will give me 1 here and 2 here, what else I’m going to do?1195

I think I’m going to move the x’s over to the other side, keep the y’s over on the right.1205

I get x/square root(x2 +1) = on the right I’m going to move the dx over to the other side as well, so dx here.1209

On the right I’m going to get 2y/dy, and now I have successfully separated all my x’s to the left and all my y’s to the right, I can integrate both sides.1223

2 separate integrals, one in terms of x and one in terms of y.1237

The y one is definitely going to be very easy that is just y2.1242

The x one looks more difficult, I’m going to do (u) substitution there.1246

I’m going to use u=x2 + 1 and du=2x/dx and if I move that 2 over the other side, I’m going to have 1/2du=x/dx which is what I have in the integral.1251

Let me convert that integral, I see I have 1/2du/square root(u), if I pull that half out to the outside I got ½.1273

The square root of u that is u-1/2du, this is all still equal to y2 and I need to integrate that.1287

Let me remind you the power rule, it says the integral of un/du=un+1/n+1 + the constant.1297

Here my n is -1/2, n + 1 is ½ so I get u1/2/1/2, that is n + 1.1309

I still ½ on the outside so ½ on the outside multiplied by that and I have to add my (+c) right now because now is when I’m doing the integration.1322

This is equal to y2, let us keep going up here, those half cancel each other out and I get u1/2.1337

What was u? u was x2 + 1, so I get the square root of x2 + 1 + a constant is = to y2.1346

Let me just switch my sides here, y2= square root of (x2) + 1 + a constant.1359

I’m ready to use my initial condition to find my constant, now this initial condition is telling me x=0, y=4, I plug those in here.1367

Y2 will be 42=square root of 02 + 1 + (c).1379

I get 16=1 + (c) and so my c =15, I plug that back in here I will get y2=square root of x2 + 1 + 15.1387

If I solve for y, I see that my y is positive here so I’m going to take the +square root, y =square root of all of that.1411

The square root of x2 + 1 + 15 and then the square root of all of that and now I have solved for my y.1421

Let us recap what happened here, I was trying to separate my variables so the first thing was I moved all of this stuff over to the other side.1442

That is how I got 6y root x2 + 1, dy/dx on the other side.1452

I have 3 on one side and 6 on the other, it seems like I should cancel there.1458

If I divide both sides by 3 I get a 2 over on the right, that is where that 2 comes from.1463

I moved the x2 + 1, I divided that over to the denominator and I multiplied that dx over the numerator.1469

That just leaves me with 2y/dy on the right and x2 + 1 on the left.1476

The key point there is I have separated all x’s on the left, all y’s on the right and now I can take the integral of both sides.1484

The integral of 2y is just y2, I’m not worried about the constant there because I know I’m going to add it on the other side as I do my integral for x.1493

That integral turns out to be a little more complicated so I did a substitution u=x + 1, du=2x/dx and so x/dx is ½ du.1502

I know it is that because I have a x/dx there, that turned into ½ du and then I have square root of u in the bottom which is the same as u-1/2.1513

If I integrate that using the power rule back from calculus 1, I raise it by 1, -1/2 + 1 is +1/2.1529

I have to divide by ½ but that cancels out what my ½ out here, that is ½ and that is cancelled out.1540

This is when I do the integration, this is when I add the constant, it is not ok just add the constant at the end in differential equation, you got to add the constant when you do the integral.1549

Add (+c) when you do the integral.1561

The reason you have to do it when you do the integral and not just tack it on the end is because there is several more steps of algebra coming.1575

That (c) gets tangled up in to the equation, in particular it got put under a square root here.1583

You really have to keep track of the (c) as you go through all the algebra afterwards.1590

That is what I was doing here is I was trying to solve for y2 and I was using my initial condition, plugging in y= 4, plugging in x=0 to solve for c.1594

I solved for (c) and I got (c)=15 but then I still have to solve for y.1608

I took the square root of both sides and the (c) goes under the square root, that 15 goes under the square root, I have to keep track of it from now on.1614

I could not just tack it on at the end, I really have to keep track of it and the role it plays in the equation.1623

My final answer there is the square root of x2 + 1 + 15.1630

A little bit complicated there but the basic technique is not too bad.1636

We got one more example here, example 5 is to solve the initial value problem Y′(x) =x2y and y(0)=7.1644

My Y′ there since we are going to treat this as a separable differential equation, I’m going to write that as dy/dx = x2y.1656

I’m not going to worry about that initial condition yet, that will come in at the very end is when you actually use the initial condition to solve for whatever arbitrary constant you find.1667

In the meantime what I’m going to try to do is try to separate all the y’s on one side and all the x’s on the other.1677

That is just a matter of multiplying and dividing, if I divide both sides by y, I will get dy/y on the left, multiply both side by dx I get x2 dx on the right.1684

I have successfully separated the y’s on the left and x’s on the right, I can integrate both sides.1696

The integral of dy/y is the natural log of absolute value of y.1703

The integral of x2 is x3/3.1708

I just did my integration I got to add on a constant right now, that is very important that you add that constant right when you do the integration, not earlier and definitely not later at the end.1713

I’m going to try to solve for y, I’m going to raise e to both sides, e to the natural log of absolute value of y= be very careful here, e to all of the right hand side x3/3 + c.1725

That all gets put in the exponent of e and not just the x3/3 part, I got absolute value of y is equal to1738

My laws of exponents here tell me that this is e to the x3/3 × e to a constant, not added to e to a constant but multiply by e to a constant.1748

Y is equal to (+ or -) e to a constant, think of that as a single constant e to the x3/3.1759

If I think of that as a single constant, I’m going to write that as k × e to the x3/3 because e to a constant is still a constant and (+ or -) a constant is still a constant.1771

I got y = ke to the x3/3.1784

That is the general solution for my differential equation, it is as much as I can derive just using the differential equation.1788

However, I also got the initial condition, I have been given that and so I’m going to use that to go a little bit further and figure out the particular value of that constant.1799

If we have not been given that initial condition we will just stop right now with the general solution but since we have that initial condition y(0)=7.1810

I plug in y=7 and x=0, 7 for y and 0 for x, 03/3, in the 0 that is just 1, that is equal to k.1821

I get 7=k and if I plug that back into my general solution, now that I know what k should be, I get y is equal to 7e to the x3/3.1836

That is my particular solution to the initial value problem, it is a solution to the differential equation and to the initial condition that we are given.1855

Let me emphasize here that it was really a key that I added the (+c) when I did the integration.1866

I could not do it at the end at this step right here at the end like you would in a calculus 1 or 2 problem.1874

In calculus 1, you would just tack on a (+c) at the end and everything is fine.1881

Here you got to add the (+c) right when you do the integration and then what happens is that (c) gets tangled up in the equation after that.1885

Let me go through those steps again just to make sure that they all make sense to you. I wrote Y′ as dy/dx -- that is a very common technique for separable differential equations.1895

That is still equal to x2y, I’m going to cross multiply and cross divide to get all my x’s on the right and my y’s on the left.1907

If I divided that y over to the other side and I multiply dx over the other side, I get this nice separated form with dy/y and x2 × dx.1922

I got all x’s on one side and all y’s on the other I can integrate both sides.1934

Integral of dy/y is natural log of y, the integral of x2 is x3/3 + c.1939

You got to add that (c) when we integrate and then all the steps after that you got to keep track of what is happening to the (c), you can not just tack that (c) on at the end.1947

To undo that natural log I raised e to both sides that gave me just absolute value of y on the left.1958

On the right it gave e to the (x3/3 +c), now you can not write that as e to the x3/3 + c, that is not the same thing, that would be very bad.1968

It is also not equal to e to the x3/3 + ec, that is also bad.1982

We are using the laws of exponents here, xa + b=xa × xb.1988

This is e to the x3/3 × ec, I pull that over to the other side.1998

Because I have an absolute value here, I just made a (+ or -) but that whole thing (+ or -) ec is just on e big constant that I called (k).2005

In order to find the value of K, I used the initial condition, this is telling me right here that x=0, y=7 .2013

I plug those values into this general solution here y=7, x=0 and e0 is just 1 and this just turns in to k =7.2026

I figure out that my k is 7, plug that back into the general solution and I get my specific solution y=7e to the x3/3.2039

That wraps up our lecture on separable differential equations, I hope you have enjoyed it as much as I have.2050

We got a lot of other lectures available on a lot of other topics on differential equations.2058

This is the very first topic, we got all kinds of stuff available on systems of differential equations, on solving differential equations by series.2063

We got stuff on second order differential equations, all the different solution techniques that you are going to learn in any differential equation course.2073

We got some stuff on numerical solutions to differential equations, we got partial differential equations in 4a series.2082

They are all here in the differential equations lecture series on www.educator.com.2089

I hope you will stick around and join me for all those other lectures and enjoy learning differential equations with me.2097

My name is Will Murray and this is the differential equation lecture series here on www.educator.com.2104

Thanks for being with us, bye.2109