Separable Equations

All right, welcome to another new segment, and in here we are going to focus on the separable equations. Actually, I did start this topic right in the previous segments because I was dealing with cases where we had this g of y, and that made an autonomous equation, right, in the previous segments. Now I'm actually going to relax the rule a little bit, but still not fully. What I will do now is I will be looking at the cases where I can simply go out and separate this in this manner. So this f x comma y can be separated into g of x times h of y, right? So these kind of equations, you can clearly see, if I have this g x of x over here, and if I move a dx over this side, and I just take this h of y and move it down there and take the integral of both sides, I'll be good to go. We'll be able to solve this. All right, so that's what I'm talking over here. It's not very complicated, but let’s let's, you know, in the first module we were discussing the explicit versus implicit solution, and I had a case over there. I was showing an example, if you remember this, you know, very simple. So in here, let's say, you know, you can see what I can do. I can move y here, dx to the side, and take integrals of both sides. You can see how simple this will be. So it's going to be y dy will be equal to minus x dx, right? So take the integral of both sides, right? So this will be y square by 2, right? Will be equal to minus x square by 2 plus the integration constant of C. I can multiply both sides of the equation by 2, so I'm going to get myself y square is equal to minus x square plus D, where this D is 2C, right? And you can see over here, then I get x square plus y square is equal to D. Over there I called this is C, but it doesn't really matter, it's just a constant. So I was able to obtain an implicit solution. Remember over there I said that hey, this is actually interesting because I can go ahead and do this, right? D minus x square, square root. So this is the explicit solution, right? So I can solve these. Okay, so now actually let's go and solve a question, but it's a little bit more demanding, but the fundamental is pretty much this is it. I don't have a lot to talk about in this particular segment except to solve some questions and illustrate how this is accomplished. All right, it's fairly straightforward. So I want to solve a particular question. The reason is I want to make sure that we can work with trigonometry a little bit, and this also particular question that I have will practice you with something called integration by parts. All right, so that's a neat equation that we can take a look at. And you can see right now I escalated real quickly, right? The, you know, the question itself compared to the last question. Well, that is what it is. And this is an initial value problem, and I get myself an initial condition, just a simple y zero is equal to zero, right? The thing is, obviously, I need to first check whether I can, you know, separate these equations to two different sides where one side has only y dependence and the other side only has x dependence. So looking at here, let's see. This e of y can be down here. Let's say this section, this will be the y, this will be the x. Okay, so dx will be there. I can move this cosine x down here. This is only y. Yeah, I can do that. Well, let me go ahead and arrange this to see what is going on. e to the 2y minus y, and then I'll go ahead and divide this by e to the y dy, will be equal to, let's take a look, that will be sine 2x, and I have a cosine of x times dx. Let me take one second to see whether everything is good here. I don't want to solve the wrong equation in front of everybody. Yeah, this is looking good to me. I'm good to go. Okay, so you do see that this is, you know, e to the power of y times e to the power y, right? So I can cancel one of them, so I will get myself e to the y minus y e to the minus y dy. Note that what I did was look at the power. It was in the denominator, so it moves up as a negative sign up here. It will be equal to, now this is something that I would like you to know, okay? 2 sine x cosine x, right? Many of you do know this, but I want you to be on the same page as me, right? So sine 2x is equal to 2 times sine x times cosine x, right? So this will come in real handy. And you can see that, you know, cosines cancel each other out. So let me actually go ahead and write here a little bit of a side point and say that, you know, sine 2x is equal to 2 times sine x times cosine of x. And let me look at cosine of 2x in case I give that to you. Right, that will be cosine square minus sine square. Okay, and also another one that we use fairly often is 1 will be equal to sine square x plus cosine square. Okay? All right, let's continue. Now I'm going to take the integral of both sides. So if I take, basically, let's go ahead and integrate both sides. I don't want to write this over and over again. Integral of e to the power of y will be, as you know, e to the power of y, right? Minus, now integral of, it's kind of a little bit of a tricky over here, dy, will be equal to, on the right-hand side I will get minus minus 2 cosine of x, right? Because the integral of sine x is equal to minus cosine of x, right? So if you're sometimes confused about it, my recommendation is, if you're more comfortable with derivatives as opposed to integration, which many students are, I actually recommend you to take the derivative of this to see whether you end up with that, right? As you know, I believe most of the students give me this feedback. I'm not sure about you particularly, but most of, you know, generalization. So if you take the derivative of this, you will get yourself a minus sine of x, right? So from there you can see minuses cancel and I end up with this particular term. All right, obviously let's not forget the integration constant. Okay, so I'm almost there. Except not really, right? Because this looks like a little bit of a challenging part. And in here, now I'm going to teach you something else. I'm assuming that you don't know this. I'm sure some of you do know this, but that will be just fine. So this type of multiplication, where I don't know, I cannot simply go ahead and take the integral of this, is going to be accomplished by taking a look at a particular way called integration by parts. Let's refresh our minds. The longer version of this, there's a compact version and a longer version. So it's like this: dv dx dx is equal to u times v minus v du dx times dx. Not many students memorize this. It looks complicated. Well, more compactly, I can write this this way and every student memorizes this version, which is perfectly fine. As long as you know what you're doing, it doesn't really matter. v du, okay? So integral of u dv is equal to u v minus v du integral of. So this is something that you need to know. And we have to be careful in noting that, look here, I have a u and I have a dv, so I need to know which one is u, which one is dv. And I'll show you what is the tactic that I use. Okay, let's say that I use u is equal to y, and I use dv is equal to e to the minus y times dy, right? Now you see what is going on over here. When I go to the next step, look at the right-hand side. Look, I have a u, okay, so I don't have to do anything. But I have a du here, so I need to take the derivative of this one. On the other hand, let's look at the other one. Now I selected one term as dv, but on the right-hand side I need to write v, right? So I need to take the integral. So in here, the process is that you're going to take the derivative of this and you're going to take the integral of this, okay? And do you realize that if I take the derivative of this, I will be going down in the power, so I will be getting myself a one, right? So then this term will be able, I will be able to manage. Okay, so my recommendation is, wherever you see y to the power of, or x, doesn't really matter the variable, but you have a power of this, let's say third one, fourth one, fifth one, you need to go down. If this is the fifth one, you need to go to the fourth, third. You see, you may need to do this several times, integration by parts, but that is the tactic that I recommend. Okay, if there's no power over here, then you need to be critical and looking at, okay, can I take the derivative of this? Or, more actually more difficult is, can I take the integral of this? Okay, let's proceed. So this derivative will be du will be equal to one, right, obviously. And I have v, and integral of this will be minus e to the minus y. Okay, let's change colors to be consistent. Then let's look here. e to the, you know, let’s actually write this way. Y e to the minus y dy. So this is this term, right, will be equal to now u times v, okay, u times v, minus integral v. Note that I want to make a correction over here. I said that this is one, right? This needs to be dy, right? My bad. Okay, it's one times dy. I was explaining something else. I get lost my focus there. So it's going to be dy, right? Because if I just simply, you know how I realized I did the error, so you can see the same thing in an exam setting, is when I put one here, it's kind of weird. I don't have a d of something to take the integral of, right? So that's how I realized this. Okay, let's proceed. So from here, then I get myself minus y e to the minus y. So you can see here, minus minus becomes a plus, and the integral of this, plus integral of this becomes a minus e to the minus y, right? Similar logic that I did before, so there's nothing new here. Okay. All right, so I was able to obtain that second term over here, and let me write the whole thing. e to the y minus now this, this whole thing, minus y e to the minus y minus e to the minus y, will be equal to, it was minus 2 cosine of x plus C. I have to verify this. I want to say something to note. Yep, we're good. So let's go ahead and rewrite this one more step. e to the power of y, this whole thing becomes e to the minus y times y plus one. Okay, minus 2 cosine of x plus C. So now I was able to obtain, this is kind of it. I cannot really proceed further than this. I can do some tricks about, you know, e to the y, e to the minus y. We'll show those, but at this point, I'm comfortable with this. Okay, so this will be my final result that I obtained. But one thing I, I am missing over here is the initial value problem. I was given a initial condition, right? So I have an unknown over here, so I can go from this general solution to particular solution. So let's take a look at that one then. I was given that y of zero is equal to zero. So let's do that. e to the power of zero plus e to the minus 0, y 0 plus 1, is equal to... what is cosine? Well, let's write cosine of 0 plus C. What is cosine of 0? It's 1. So I get myself minus 2 over here. How about this? This is 1. What is e? This is 1, right? So this is 1. So 1 plus 1 is equal to minus 2 plus C. So from here you can see that I get myself as C will be equal to 4, right? That's, I think that's what I get. So then I will rewrite this simply and I will kind of get my particular solution. e to the minus y, y plus 1, will be equal to 4 minus 2 cosine of x. This is my final answer for this particular question, and this is a particular solution. So this question is one of those questions where, when I solve it, it will be so easy. But if you're faced with this question in an exam setting, you may be confused. At this point, as I'm covering separable equations, obviously this needs to be solved by the separable equations principles. But in an exam setting, that's not going to be the case. We will have multiple approaches that we'll cover by then, right? And by looking at it, do you see it's separable? Again, once I do it, it will look okay, doable. So I'm just letting you know that things are sometimes not as easy as it looks, okay? So in here, I'm looking at these two terms. So can I take them in parentheses of y? Yeah. I get myself x plus 2, right? Then what is left is, if I have a minus sign like this, I got myself x plus 2, too. You see, this is what I'm talking about. You know, going here to here may be a bit challenging for some. Okay, and I will do the very similar approach down here as well. I picked these two, so I get myself, in y parentheses, x minus 3, right, this time around. Then plus, you can see, x minus 3. So now I can go one more step even. I can take in parentheses of x plus 2 up there, right? x plus 2 times y minus 1, and then I got myself x minus 3, parenthesis over here. Then I got myself y plus 1 this time around, right? Actually, let's go ahead and write dy dx so that it will be easier for me to separate them. Now what I will do is I will go ahead and separate them like this. dy, and I will get myself y plus 1 divided by y minus 1, right? Because as you see, y minus 1 is here. It goes down here. y plus 1 is there. It goes up there, right? Will be equal to, very similarly, I will get x plus 2 divided by x minus 3 times dx. I just simply move dx there, right? So okay, what am I going to do? Again, another tactic that we commonly use is this. I'll show you. y minus 1 plus 2 divided by y minus 1. Do you see where this is going? dy will be equal to x minus 3 plus 5 divided by x minus 3, dx. Okay, so you can see over here I can do this this way. 1 plus 2 divided by y minus 1, dy, will be equal to 1 plus 5 divided by x minus 3, will be equal to, well, times dx, is equal to the other side, right? Well, if I take the integral of both sides, the first term, I mean this term, I get myself y plus, what is the integral of the second term? This, it's going to be 2 times ln of y minus 1, right? Will be equal to, let's do the right-hand side. It will be x, similarly, plus 5 times ln x minus 3. Let's see. Okay, this is better now. x minus 3 plus, let's not forget the C value, okay, integration constant. All right, so then, next step is I'll play around a little bit with this. So it's going to be y plus ln y minus 1 square will be equal to x plus ln x minus 3 to the power of 5 plus a constant of C. Okay, so let's do this way then. Let me have this y minus x. I just move the x to the other side, and I move this to the other side. So I'm going to get ln x minus 3 to the power of 5 minus ln y minus 1 to the power of 2 plus C. Let's do just the right-hand side. So I'll get ln, tell me whether this makes sense, x minus 3 to the power of 5 divided by y minus 1 to the power of 2, right, plus C. Now I'll do the trick that I've been doing 10 times already and we're just in the second week of the classes. So I will do e to the power of both sides. So I will get myself e to the power of y minus x will be equal to e to the power of, that will be x minus 3 to the power of 5 divided by y minus 1 square times, you know, e to the power of C. I'm going to call this D. All right, so basically it's a e to the C. Let's say that this is an exam, this will be, you know, what I expect you to perform. But to me this looks a little bit better. But, you know, that's just an opinion, not a fact. It will be equal to, let's say, this E times e to the y minus x. Okay, so obviously E is one over D, right? So that's what I'm going to box it up. This wraps up the separable equations approaches that we have, okay.