WEBVTT mathematics/differential-equations/murray
00:00:00.000 --> 00:00:07.000
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.
00:00:07.000 --> 00:00:15.000
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.
00:00:15.000 --> 00:00:22.000
I hope you will join us and let us jump in.
00:00:22.000 --> 00:00:32.000
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.
00:00:32.000 --> 00:00:43.000
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.
00:00:43.000 --> 00:00:54.000
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.
00:00:54.000 --> 00:01:07.000
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.
00:01:07.000 --> 00:01:23.000
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.
00:01:23.000 --> 00:01:29.000
The basic idea of separable differential equation, after we work couple of examples I think it will make a better sense.
00:01:29.000 --> 00:01:38.000
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.
00:01:38.000 --> 00:01:44.000
We just had a lecture on linear differential equation and we had a completely different technique for solving that.
00:01:44.000 --> 00:01:59.000
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.
00:01:59.000 --> 00:02:09.000
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.
00:02:09.000 --> 00:02:14.000
If you do have a choice though, you probably want to go with the separation technique.
00:02:14.000 --> 00:02:27.000
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.
00:02:27.000 --> 00:02:43.000
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.
00:02:43.000 --> 00:02:59.000
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.
00:02:59.000 --> 00:03:09.000
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.
00:03:09.000 --> 00:03:17.000
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.
00:03:17.000 --> 00:03:27.000
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.
00:03:27.000 --> 00:03:33.000
A lot of times that will end up tangling up the constant into the equation and you have to let that happen.
00:03:33.000 --> 00:03:46.000
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.
00:03:46.000 --> 00:03:52.000
Calculus 1 and 2 which you learned is whenever you do an indefinite integral, you just tack on a (+c) at the end.
00:03:52.000 --> 00:03:59.000
You can do about 50 problems and then go back at the end and just put (+c) at the end of every single one.
00:03:59.000 --> 00:04:06.000
Not so in differential equation, you got to be really careful with the constant and be careful when you insert it.
00:04:06.000 --> 00:04:15.000
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.
00:04:15.000 --> 00:04:29.000
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.
00:04:29.000 --> 00:04:37.000
For our first example here, we are going to find the general solution to the following differential equation Y′(x)=1/2 y(x).
00:04:37.000 --> 00:04:49.000
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.
00:04:49.000 --> 00:04:57.000
Let me write this as (dy)/(dx)=1/2 and that is y(x) that is not y multiply by x.
00:04:57.000 --> 00:05:11.000
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.
00:05:11.000 --> 00:05:23.000
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).
00:05:23.000 --> 00:05:32.000
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.
00:05:32.000 --> 00:05:39.000
The integral of (dy)/y is just natural log or technically the absolute value of y.
00:05:39.000 --> 00:05:50.000
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).
00:05:50.000 --> 00:06:05.000
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.
00:06:05.000 --> 00:06:14.000
It is very important that you have a (+c) on one side or the other and that you incorporate it at this step.
00:06:14.000 --> 00:06:32.000
Let me emphasize that, must add (c) when you do the integration.
00:06:32.000 --> 00:06:41.000
Let me really emphasize that you go to do that when you do the integration.
00:06:41.000 --> 00:06:53.000
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,
00:06:53.000 --> 00:07:09.000
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.
00:07:09.000 --> 00:07:18.000
It is also not equal to e^1/2x + e^c, it is neither of those things.
00:07:18.000 --> 00:07:28.000
It is really e^(1/2x + c) and the way you resolve that is e^1/2x × e^c.
00:07:28.000 --> 00:07:44.000
E^c 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 × e^1/2x.
00:07:44.000 --> 00:08:03.000
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?
00:08:03.000 --> 00:08:14.000
Y is equal to I will use a new constant A(e)^1/2x.
00:08:14.000 --> 00:08:25.000
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 in
00:08:25.000 --> 00:08:34.000
Let me write this in red, we would have got in e^1/2x and then we would just tack on a constant at the end.
00:08:34.000 --> 00:08:48.000
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.
00:08:48.000 --> 00:08:57.000
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.
00:08:57.000 --> 00:09:07.000
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).
00:09:07.000 --> 00:09:15.000
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.
00:09:15.000 --> 00:09:31.000
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.
00:09:31.000 --> 00:09:41.000
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.
00:09:41.000 --> 00:09:57.000
Then we solve for y, so we raise (e) to both sides and you have to be careful with your rules of exponents here e^1/2x+c is = e^1/2x × e^c, not (+) e^c.
00:09:57.000 --> 00:10:03.000
We got the absolute value of y is equal to, e^c is another constant and I will call that k.
00:10:03.000 --> 00:10:16.000
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.
00:10:16.000 --> 00:10:25.000
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.
00:10:25.000 --> 00:10:35.000
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.
00:10:35.000 --> 00:10:41.000
Wrong in the sense that it would not satisfy the differential equation, it would not make the differential equation true anymore.
00:10:41.000 --> 00:10:46.000
That is why you really have to be careful with that.
00:10:46.000 --> 00:10:55.000
Let us try another one here, we have to find the general solution to the following initial value problem.
00:10:55.000 --> 00:11:17.000
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.
00:11:17.000 --> 00:11:22.000
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.
00:11:22.000 --> 00:11:40.000
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.
00:11:40.000 --> 00:11:47.000
Now I have successfully separated my variables x’s on one side, y’s on the other.
00:11:47.000 --> 00:11:53.000
Take the integral of both of these, the integral of y/dy is = y²/2.
00:11:53.000 --> 00:12:05.000
The integral of –x/dx is –x²/2, but I just did my integral so I have to put the constant right here.
00:12:05.000 --> 00:12:26.000
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).
00:12:26.000 --> 00:12:39.000
Add the x² + y²= k and now I’m going to use my initial condition in order to find the value of that constant.
00:12:39.000 --> 00:13:01.000
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.
00:13:01.000 --> 00:13:21.000
If I plug that back into my general solution there, what I get is x² + y²=9.
00:13:21.000 --> 00:13:26.000
That solves my initial value problem, let me remind you of each of the steps that we went through there.
00:13:26.000 --> 00:13:36.000
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.
00:13:36.000 --> 00:13:47.000
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.
00:13:47.000 --> 00:13:57.000
Now I integrate both sides, integral of y/dy is y²/2, integral of x/dx or –x/dx is –x²/2 + a constant.
00:13:57.000 --> 00:14:07.000
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.
00:14:07.000 --> 00:14:22.000
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.
00:14:22.000 --> 00:14:34.000
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.
00:14:34.000 --> 00:14:45.000
I substitute that back in to the general solution and I get x² + y²=9.
00:14:45.000 --> 00:14:56.000
On example 3, I’m given a differential equation Y′(x) + xy=x³, I have to determine if this thing is separable.
00:14:56.000 --> 00:15:09.000
The way I’m going to do that is I’m going to write my Y′ as dy/dx + xy=x³.
00:15:09.000 --> 00:15:24.000
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.
00:15:24.000 --> 00:15:48.000
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.
00:15:48.000 --> 00:15:57.000
I can not factor that into that form and because I can not factor that, this equation is not separable.
00:15:57.000 --> 00:16:13.000
We can not do this, we can not factor that in that form.
00:16:13.000 --> 00:16:33.000
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.
00:16:33.000 --> 00:16:39.000
That is all really we are asked to find for this example, we are asked to check whether this differential equation is separable.
00:16:39.000 --> 00:16:52.000
However I will add a note here, in a parenthesis, (However, it is in linear form.)
00:16:52.000 --> 00:17:06.000
I will remind you what that form is, the linear form for differential equations is Y′ + p(x)/y=q(x).
00:17:06.000 --> 00:17:18.000
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.
00:17:18.000 --> 00:17:33.000
It could be solved using our technique for linear differential equations.
00:17:33.000 --> 00:17:53.000
That is something that I have a separate lecture here in the differential equation lecture series for linear differential equations.
00:17:53.000 --> 00:18:01.000
There is another lecture here in the differential equation lecture series for linear differential equations where we learned how to solve this.
00:18:01.000 --> 00:18:06.000
We used something called the integrating factor to solve the linear differential equation.
00:18:06.000 --> 00:18:19.000
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.
00:18:19.000 --> 00:18:30.000
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.
00:18:30.000 --> 00:18:40.000
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.
00:18:40.000 --> 00:18:46.000
Unfortunately this one, you can not factor it in that form so this differential equation is not separable.
00:18:46.000 --> 00:19:04.000
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).
00:19:04.000 --> 00:19:14.000
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.
00:19:14.000 --> 00:19:21.000
You would not give up hope completely on this one, you just have to watch a different lecture to learn how to do it.
00:19:21.000 --> 00:19:37.000
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.
00:19:37.000 --> 00:19:55.000
Again, let us try to separate the variables there, I think I’m going to write 3x= 6y × square root(x² + 1) dy/dx.
00:19:55.000 --> 00:20:05.000
I can divide both sides by 3, that will give me 1 here and 2 here, what else I’m going to do?
00:20:05.000 --> 00:20:09.000
I think I’m going to move the x’s over to the other side, keep the y’s over on the right.
00:20:09.000 --> 00:20:23.000
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.
00:20:23.000 --> 00:20:37.000
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.
00:20:37.000 --> 00:20:42.000
2 separate integrals, one in terms of x and one in terms of y.
00:20:42.000 --> 00:20:46.000
The y one is definitely going to be very easy that is just y².
00:20:46.000 --> 00:20:51.000
The x one looks more difficult, I’m going to do (u) substitution there.
00:20:51.000 --> 00:21:13.000
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.
00:21:13.000 --> 00:21:27.000
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.
00:21:27.000 --> 00:21:37.000
The square root of u that is u^-1/2du, this is all still equal to y² and I need to integrate that.
00:21:37.000 --> 00:21:49.000
Let me remind you the power rule, it says the integral of u^n/du=u^n+1/n+1 + the constant.
00:21:49.000 --> 00:22:02.000
Here my n is -1/2, n + 1 is ½ so I get u^1/2/1/2, that is n + 1.
00:22:02.000 --> 00:22:17.000
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.
00:22:17.000 --> 00:22:26.000
This is equal to y², let us keep going up here, those half cancel each other out and I get u^1/2.
00:22:26.000 --> 00:22:39.000
What was u? u was x² + 1, so I get the square root of x² + 1 + a constant is = to y².
00:22:39.000 --> 00:22:47.000
Let me just switch my sides here, y²= square root of (x²) + 1 + a constant.
00:22:47.000 --> 00:22:59.000
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.
00:22:59.000 --> 00:23:07.000
Y² will be 4²=square root of 0² + 1 + (c).
00:23:07.000 --> 00:23:31.000
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.
00:23:31.000 --> 00:23:41.000
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.
00:23:41.000 --> 00:24:02.000
The square root of x² + 1 + 15 and then the square root of all of that and now I have solved for my y.
00:24:02.000 --> 00:24:12.000
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.
00:24:12.000 --> 00:24:18.000
That is how I got 6y root x² + 1, dy/dx on the other side.
00:24:18.000 --> 00:24:23.000
I have 3 on one side and 6 on the other, it seems like I should cancel there.
00:24:23.000 --> 00:24:29.000
If I divide both sides by 3 I get a 2 over on the right, that is where that 2 comes from.
00:24:29.000 --> 00:24:36.000
I moved the x² + 1, I divided that over to the denominator and I multiplied that dx over the numerator.
00:24:36.000 --> 00:24:44.000
That just leaves me with 2y/dy on the right and x² + 1 on the left.
00:24:44.000 --> 00:24:53.000
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.
00:24:53.000 --> 00:25:02.000
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.
00:25:02.000 --> 00:25:13.000
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.
00:25:13.000 --> 00:25:29.000
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.
00:25:29.000 --> 00:25:40.000
If I integrate that using the power rule back from calculus 1, I raise it by 1, -1/2 + 1 is +1/2.
00:25:40.000 --> 00:25:49.000
I have to divide by ½ but that cancels out what my ½ out here, that is ½ and that is cancelled out.
00:25:49.000 --> 00:26:01.000
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.
00:26:01.000 --> 00:26:15.000
Add (+c) when you do the integral.
00:26:15.000 --> 00:26:23.000
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.
00:26:23.000 --> 00:26:30.000
That (c) gets tangled up in to the equation, in particular it got put under a square root here.
00:26:30.000 --> 00:26:34.000
You really have to keep track of the (c) as you go through all the algebra afterwards.
00:26:34.000 --> 00:26:48.000
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.
00:26:48.000 --> 00:26:54.000
I solved for (c) and I got (c)=15 but then I still have to solve for y.
00:26:54.000 --> 00:27:03.000
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.
00:27:03.000 --> 00:27:10.000
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.
00:27:10.000 --> 00:27:16.000
My final answer there is the square root of x² + 1 + 15.
00:27:16.000 --> 00:27:24.000
A little bit complicated there but the basic technique is not too bad.
00:27:24.000 --> 00:27:36.000
We got one more example here, example 5 is to solve the initial value problem Y′(x) =x²y and y(0)=7.
00:27:36.000 --> 00:27:47.000
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.
00:27:47.000 --> 00:27:57.000
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.
00:27:57.000 --> 00:28:04.000
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.
00:28:04.000 --> 00:28:16.000
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.
00:28:16.000 --> 00:28:23.000
I have successfully separated the y’s on the left and x’s on the right, I can integrate both sides.
00:28:23.000 --> 00:28:28.000
The integral of dy/y is the natural log of absolute value of y.
00:28:28.000 --> 00:28:33.000
The integral of x² is x³/3.
00:28:33.000 --> 00:28:45.000
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.
00:28:45.000 --> 00:28:58.000
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.
00:28:58.000 --> 00:29:08.000
That all gets put in the exponent of e and not just the x³/3 part, I got absolute value of y is equal to
00:29:08.000 --> 00:29:19.000
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.
00:29:19.000 --> 00:29:31.000
Y is equal to (+ or -) e to a constant, think of that as a single constant e to the x³/3.
00:29:31.000 --> 00:29:44.000
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.
00:29:44.000 --> 00:29:48.000
I got y = ke to the x³/3.
00:29:48.000 --> 00:29:59.000
That is the general solution for my differential equation, it is as much as I can derive just using the differential equation.
00:29:59.000 --> 00:30:10.000
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.
00:30:10.000 --> 00:30:21.000
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.
00:30:21.000 --> 00:30:36.000
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.
00:30:36.000 --> 00:30:55.000
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.
00:30:55.000 --> 00:31:06.000
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.
00:31:06.000 --> 00:31:14.000
Let me emphasize here that it was really a key that I added the (+c) when I did the integration.
00:31:14.000 --> 00:31:21.000
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.
00:31:21.000 --> 00:31:25.000
In calculus 1, you would just tack on a (+c) at the end and everything is fine.
00:31:25.000 --> 00:31:35.000
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.
00:31:35.000 --> 00:31:47.000
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.
00:31:47.000 --> 00:32:02.000
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.
00:32:02.000 --> 00:32:14.000
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.
00:32:14.000 --> 00:32:19.000
I got all x’s on one side and all y’s on the other I can integrate both sides.
00:32:19.000 --> 00:32:27.000
Integral of dy/y is natural log of y, the integral of x² is x³/3 + c.
00:32:27.000 --> 00:32:38.000
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.
00:32:38.000 --> 00:32:48.000
To undo that natural log I raised e to both sides that gave me just absolute value of y on the left.
00:32:48.000 --> 00:33:02.000
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.
00:33:02.000 --> 00:33:08.000
It is also not equal to e to the x³/3 + e^c, that is also bad.
00:33:08.000 --> 00:33:18.000
We are using the laws of exponents here, x^a + b=x^a × x^b.
00:33:18.000 --> 00:33:25.000
This is e to the x³/3 × e^c, I pull that over to the other side.
00:33:25.000 --> 00:33:33.000
Because I have an absolute value here, I just made a (+ or -) but that whole thing (+ or -) e^c is just on e big constant that I called (k).
00:33:33.000 --> 00:33:46.000
In order to find the value of K, I used the initial condition, this is telling me right here that x=0, y=7 .
00:33:46.000 --> 00:33:59.000
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.
00:33:59.000 --> 00:34:10.000
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.
00:34:10.000 --> 00:34:18.000
That wraps up our lecture on separable differential equations, I hope you have enjoyed it as much as I have.
00:34:18.000 --> 00:34:23.000
We got a lot of other lectures available on a lot of other topics on differential equations.
00:34:23.000 --> 00:34:33.000
This is the very first topic, we got all kinds of stuff available on systems of differential equations, on solving differential equations by series.
00:34:33.000 --> 00:34:42.000
We got stuff on second order differential equations, all the different solution techniques that you are going to learn in any differential equation course.
00:34:42.000 --> 00:34:49.000
We got some stuff on numerical solutions to differential equations, we got partial differential equations in 4a series.
00:34:49.000 --> 00:34:57.000
They are all here in the differential equations lecture series on www.educator.com.
00:34:57.000 --> 00:35:04.000
I hope you will stick around and join me for all those other lectures and enjoy learning differential equations with me.
00:35:04.000 --> 00:35:09.000
My name is Will Murray and this is the differential equation lecture series here on www.educator.com.
00:35:09.000 --> 00:35:11.000
Thanks for being with us, bye.