
Title: Separation of Variables

Series: Ordinary Differential Equations

YouTubeTitle: Ordinary Differential Equations  Part 6  Separation of Variables

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Subtitle in English
1 00:00:00,429 –> 00:00:04,477 Hello and welcome back to ODEs.
2 00:00:04,677 –> 00:00:09,930 The video series where we talk about the theory of these equations and how to solve them.
3 00:00:10,571 –> 00:00:17,658 Indeed in today’s part 6 we will talk about the very important solving method called the separation of variables.
4 00:00:17,986 –> 00:00:25,249 It tells us that we can solve a nonautonomous ODE, if we can separate the two variables
5 00:00:25,449 –> 00:00:31,238 and it turns out that the whole procedure is very similar to the method of the last video.
6 00:00:31,438 –> 00:00:33,214 However before we start
7 00:00:33,314 –> 00:00:39,696 I really want to thank all the nice people who support this channel on Steady, here on YouTube, via PayPal or by other means
8 00:00:39,896 –> 00:00:46,342 and please don’t forget to download the PDF version and the quiz for this video with the link in the description.
9 00:00:46,700 –> 00:00:52,932 Ok, then I would say let’s immediately start by considering a nonautonomous ODE.
10 00:00:53,132 –> 00:00:56,929 So you already know we consider explicit forms.
11 00:00:56,971 –> 00:01:05,635 So we can write this ODE as x dot is equal to a function w, that has two variables as its input.
12 00:01:06,414 –> 00:01:11,343 The one is the time variable t and the other one our function x
13 00:01:11,471 –> 00:01:19,343 and now you know, if t is actually involved in this function w, then we speak of a nonautonomous ODE
14 00:01:19,643 –> 00:01:26,654 and now the question for this method of today is: “can we separate both variables on the righthand side?”
15 00:01:27,186 –> 00:01:32,543 Now, what this question actually means I can immediately show you with an example.
16 00:01:32,829 –> 00:01:40,257 Let’s say we have x dot is equal to t^3 times the function x^2
17 00:01:40,874 –> 00:01:45,100 Then in this case we have the separation of variables,
18 00:01:45,186 –> 00:01:52,023 because we have a product where in one factor there are only t’s and in the other factor there are only x’s.
19 00:01:52,223 –> 00:01:59,372 Of course this is not always possible, but if it is, we can solve this onedimensional ODE
20 00:01:59,929 –> 00:02:04,684 and the method we have for that we simply call separation of variables
21 00:02:04,943 –> 00:02:15,106 and now you already know it’s applicable if the ODE can be written as x dot is equal to g(t) times a function h(x)
22 00:02:15,486 –> 00:02:22,361 and as you might recall from the last video we can state an ODE as an initial value problem.
23 00:02:22,561 –> 00:02:26,726 This means that we search for solutions that satisfy a given value.
24 00:02:27,057 –> 00:02:33,932 For example we could say at the point t_0 the solution should have the value x_0.
25 00:02:34,132 –> 00:02:41,500 So now we actually choose a point in time t_0 and then we say it should have a given value at this point
26 00:02:42,700 –> 00:02:48,732 and now please recall from the last video. This is exactly what we call an initial value problem.
27 00:02:48,932 –> 00:02:57,457 Last time we have discussed that for an autonomous ODE and now we take a nonautonomous ODE of exactly this form.
28 00:02:58,029 –> 00:03:02,945 So two functions g and h are involved and both should be continuous
29 00:03:03,557 –> 00:03:10,511 and now I can tell you we will do exactly the same steps as in the last video, just with one small addition.
30 00:03:10,711 –> 00:03:17,553 First let’s immediately go to the interesting solutions. Where h(x_0) is not 0.
31 00:03:17,753 –> 00:03:25,952 Otherwise you should see, there is a 0 here on the righthand side and a constant function will be a solution immediately.
32 00:03:26,152 –> 00:03:31,065 In other words this is actually the fun case, where we have something to do
33 00:03:31,386 –> 00:03:36,855 and indeed the first thing we can do is to bring h(x) to the other side.
34 00:03:37,055 –> 00:03:42,504 We know in a neighbourhood of x_0 we can divide by h(x).
35 00:03:43,300 –> 00:03:49,042 Ok, then let’s see what happens if we have a solution of this ODE and let’s call it alpha
36 00:03:49,242 –> 00:03:53,037 and you know t_0 should lie in this domain of definition.
37 00:03:53,437 –> 00:03:58,999 This is a requirement of the solution t_0 should be inside this interval.
38 00:03:59,000 –> 00:04:06,483 Moreover alpha should solve the initial value condition. So alpha at t_0 should be equal to x_0.
39 00:04:07,283 –> 00:04:12,339 In other words, here we can just write alpha instead of x in the ODE.
40 00:04:12,739 –> 00:04:20,050 However instead of t I now want to write s. Simply because we want to do an integration again.
41 00:04:20,450 –> 00:04:25,229 In other words the fundamental theorem of calculus comes into the game again.
42 00:04:25,929 –> 00:04:29,730 This means here we just integrate on both sides
43 00:04:30,057 –> 00:04:34,696 and indeed now we will integrate from t_0 to t.
44 00:04:35,114 –> 00:04:41,640 This makes sense because t_0 was our fixed point in time and t is just an arbitrary point
45 00:04:42,129 –> 00:04:47,849 and now you already know on the lefthand side here we can do a nice substitution.
46 00:04:48,429 –> 00:04:54,712 Then the whole thing looks much easier, because we just have to integrate the function 1/h
47 00:04:55,214 –> 00:05:00,064 and please don’t forget the boundaries are now x_0 and alpha(t)
48 00:05:00,329 –> 00:05:08,765 and then on the righthand side of course nothing changes, but we see on both sides we need antiderivatives to solve the thing.
49 00:05:09,257 –> 00:05:19,638 So let’s use capital F for the antiderivative of 1/h. So for the one on the lefthand side and capital G for the one righthand side
50 00:05:20,014 –> 00:05:25,557 and at this point let’s say it again. There the fundamental theorem of calculus is at work again
51 00:05:26,157 –> 00:05:34,014 and indeed for this theorem it does not matter which antiderivative we choose, because they only differ by an additive constant
52 00:05:34,743 –> 00:05:40,979 and in fact this helps us, because we can put these 2 constants here and there into a single one.
53 00:05:41,179 –> 00:05:48,091 In other words we simply write F is equal to G + a constant c.
54 00:05:48,429 –> 00:05:51,552 So in this equation here c could be any real number
55 00:05:52,229 –> 00:05:58,196 and with that you see we have solved the problem. We simply have to take the inverse of F now
56 00:05:59,543 –> 00:06:02,390 because then we have alpha of t on the lefthand side.
57 00:06:02,929 –> 00:06:05,443 So maybe it looks a little bit complicated,
58 00:06:05,514 –> 00:06:11,857 but what you should recognize is that the whole procedure here gives us the solution alpha(t)
59 00:06:12,008 –> 00:06:15,913 and that’s all. This is the method of separating variables.
60 00:06:16,214 –> 00:06:20,443 However as I have told you in the last video, you don’t have to memorize this formula,
61 00:06:20,554 –> 00:06:24,142 because you just have to know what to do in an example
62 00:06:24,457 –> 00:06:28,610 and that’s exactly what I want to show you now.
63 00:06:29,000 –> 00:06:33,089 So let’s consider a very standard ODE. x dot is equal to x.
64 00:06:33,686 –> 00:06:38,338 However now the factor in front of x should depend on t
65 00:06:38,538 –> 00:06:44,172 and maybe let’s look in the case that we have 1/3 times t^3
66 00:06:44,372 –> 00:06:50,365 and moreover let’s also write our initial value condition as x(0) is equal to x_0.
67 00:06:51,065 –> 00:06:58,138 Ok and now you already know what you should do here is to rewrite x_0 as dx/dt.
68 00:06:58,538 –> 00:07:02,213 We do that, because it helps separating the variables,
69 00:07:02,613 –> 00:07:08,447 because now we can just informally multiply dt to the righthand side
70 00:07:08,947 –> 00:07:13,894 and moreover we can also bring x to the lefthand side by dividing by x.
71 00:07:13,895 –> 00:07:19,203 So then it looks strange like that, but then we can just integrate on both sides
72 00:07:19,514 –> 00:07:25,438 and now you know this is just a short notation for the antiderivatives we have discussed before.
73 00:07:25,871 –> 00:07:33,675 So indeed everything is correct here and we can just use an antiderivative on the lefthand side and one on the righthand side
74 00:07:34,075 –> 00:07:39,248 and there you should know, we have the natural logarithm of the absolute value of x on the lefthand side
75 00:07:39,971 –> 00:07:48,889 and 1/12 times t^4 on the righthand side and please NEVER forget to add the constant in this step here
76 00:07:49,089 –> 00:07:56,581 and moreover at this point you should see this is definitely correct, because it’s exactly this formula here in a general case.
77 00:07:57,171 –> 00:08:02,386 However what I want to tell you here is that you don’t need to remember this formula at all,
78 00:08:02,500 –> 00:08:06,008 because you just need to remember these calculation steps.
79 00:08:06,308 –> 00:08:11,507 I think that’s much easier, because it’s a natural calculation anyway.
80 00:08:12,043 –> 00:08:18,508 Ok, then in the next step let’s use the inverse function of the natural logarithm. Which is the exponential function
81 00:08:18,586 –> 00:08:24,425 and moreover let’s also write alpha(t) instead of x to denote our solution
82 00:08:25,100 –> 00:08:30,436 and then we see this exponential function here is the absolute value of our solution.
83 00:08:31,036 –> 00:08:36,245 Hence in the last step here we just have to satisfy the initial value condition.
84 00:08:36,645 –> 00:08:41,030 In other words we have to find the correct value of our constant c.
85 00:08:41,530 –> 00:08:48,042 So we do that and then it turns out that we have the exponential function with the factor x_0 in front of it
86 00:08:48,943 –> 00:08:53,409 and there we have it. This is the solution of the initial value problem given here
87 00:08:54,157 –> 00:08:59,745 and you see we have done the separation of variables exactly in this step here.
88 00:09:00,271 –> 00:09:04,739 Ok now in order to close this video, let’s look at another example.
89 00:09:05,357 –> 00:09:11,426 Now, let’s take x dot is equal to sin(t) times e^x.
90 00:09:11,826 –> 00:09:18,626 So in fact it’s more complicated than before, but we also recognize that the 2 variables are separated.
91 00:09:19,157 –> 00:09:23,813 In other words we should be able to do our informal step here again.
92 00:09:24,113 –> 00:09:30,949 So we have dx divided by e^x is equal to sin(t) dt
93 00:09:31,249 –> 00:09:36,318 and then you already know we write the integration symbols and go to the antiderivatives.
94 00:09:37,019 –> 00:09:40,915 So on the lefthand side we have e to the power x
95 00:09:41,371 –> 00:09:44,972 and on the righthand side we have cos(t)
96 00:09:45,872 –> 00:09:49,767 and again please don’t forget to add our constant c here
97 00:09:50,157 –> 00:09:56,283 and then in the next step we just have to find the inverses again to find our solution x
98 00:09:56,680 –> 00:10:01,237 or to say it in better terms: to find our solution alpha(t)
99 00:10:01,537 –> 00:10:07,978 and for that we get  the natural logarithm of cos(t)  our constant c
100 00:10:08,778 –> 00:10:13,599 and there I can tell you. Sometimes it’s helpful to change the constant by a factor.
101 00:10:14,043 –> 00:10:17,853 So in other words you simply introduce a new constant.
102 00:10:18,553 –> 00:10:22,217 So we just change the name and then we don’t have this annoying  sign there.
103 00:10:23,070 –> 00:10:30,189 This is helpful, because now we want to find the correct constant in order to fulfill our initial value condition.
104 00:10:31,089 –> 00:10:35,846 So we put in 0 in the lefthand side and this should be equal to x_0
105 00:10:35,847 –> 00:10:40,070 and now we simply solve this equation for c tilde.
106 00:10:40,670 –> 00:10:45,515 Now, since the cos(0) is 1 we immediately have the solution here.
107 00:10:46,043 –> 00:10:49,630 Namely, c tilde is given as an exponential function.
108 00:10:50,230 –> 00:10:57,730 More precisely it’s e to the power x_0 and then we have to subtract the cos(0).
109 00:10:58,929 –> 00:11:04,799 Ok and then you see, we put that in and then the whole initial value problem is solved
110 00:11:05,199 –> 00:11:10,799 and with that you now know the method of separating variables to solve an ODE
111 00:11:10,999 –> 00:11:16,699 and in the next video I will show you how you can solve so called linear ODEs.
112 00:11:17,099 –> 00:11:22,700 These are also very common and therefore it’s very important to know how to solve them
113 00:11:23,099 –> 00:11:28,200 and with that I really hope that I see you in the next video. Have a nice day and bye bye!