WEBVTT mathematics/differential-equations/murray
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Welcome back to the differential equations lecture here on educator.com.
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My name is will Murray, we are studying partial differential equations and were starting to learn how to solve them we already had a couple lectures sort of warm you up to the idea of partial differential equations.
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We have not actually solved that yet so, today were going to learn how to start to solve them using a technique known as separation of variables so, see what that is all about.
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The separation of variables like as I said , it is a technique for solving some partial differential equations and the idea is that you assume that the function you are looking for remember work in a call that function U of X of T can be written as a product of a function of X only and a function of T only.
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So, we assume that the function were looking for can be written as were to call it X of X of X and T of T by the way were to have lots of equations today where were using X is a little x and T little t .
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Be very careful to keep those straight the little X and little T are the variables X and T are the functions so, let me emphasize that right now, those are functions and the little X and little T are the variables try to be clear when I am writing my own notes which 1 is I am talking about.
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And so, the point of assuming that you have a product of 2 functions like that is it is really easy to take derivatives because of your taking the partial derivative with respect to X that means that all the Ts are constant so, T of T just comes down as a constant we think of that as being a constant .
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So, you just take the derivative of capital X with respect to little X X′ of X and then when you take the second derivative with respect to X your taking capital X″ of X and again the T is just a constant when you are looking at derivatives with respect to T is just the other way around .
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The capital X you think of as being a constant and your taking derivatives of T with respect to little T so, here is the first derivative is T′ and in the second derivative is to″ .
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So, you take these derivatives and you plug them into the partial differential equation and then let us see what happens with that what you trying to do is get all the X is on 1 side and all the Ts on the other.
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It does not always work but it is it works for a lot of these differential equations will see some example lots of examples where does if you can do that then a very interesting thing happens you have 1 function that is only dependent on X 1 function that is only dependent on T and are equal to each other.
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That means if you change T nothing happens on the left because X has not changed so, if you change T nothing happens which means it must be constant so, both of these functions must be constant so, and I call that constant .
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And so, were to end up with a function of X = λ and independently a function of T = λ so, will reorganize these into 2 ordinary different differential equations a function of X = λ and a function of T = λ.
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And hopefully you can solve these ordinary differential equations separately for capital X and T remember X that is a function of little X the variable little X T is a function of little T.
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If you can solve both of those then if you rewrite our original guess remember was new of X and T = X of X and Multiply by T a T so, if you can find X X of X and T of T and you can put them back together and you can compile your solution to the original partial differential equation by multiplying together the solutions to these 2 ordinary differential equations.
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So, that is the idea of separation of variables it is definitely something that will make more sense after we get some practice so, let us try that out on some examples.
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So, in example 1 we are going to separation of variables to convert the following partial differential equation into 2 ordinary differential equations so, remember our guess for all of these for all of these separable partial differential equations is U of X T = capital X of X x T of T .
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It looks like really need to know, some partial derivatives there so, U of X remember the T will be a constant so, that is X′ of X x T of T and U of X X will be X″ of X x T a T and it looks like really need U of T so, that would be now, holding the X is constant so, X of X x T sub' of little T.
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To plug those into my differential equation so, plug-in plugged in to the pit the partial differential equation that we been given so, U of X X is X″ of X T of T + this is a little X now, U of T is X of X and T′ of T = 0 .
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The idea remember is to try to solve this equation in such a way that you get all the X is on 1 side and all the Ts on the other so, I am in a move these terms over to the other side so, I get X″ of little X T of little T = - little X x X of X T′ of T .
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Now, meant across the divide to get all my X on the left and all the T on the right so, capital X″ of X divided by up the - on the left - X X of X = π crust divide T′ of T divided by T of T .
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Notice now, that I got all X is on 1 side all Ts on the other so, got a function of X = to a function of T and so, this must be constant that is the whole idea of separation of variables this must be constant.
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So, then assented = to recall my constant λ and down to split that apart into 2 separate equations so, my - X″ of little X over X X of X X of X must be = to λ on 1 side and T′ of T divided by T of T = λ on the other side.
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Each 1 of those I am not do a little bit of work to make him into a and ordinary differential equation so, X″ of X = a multiply the - over the other side - λ X x X of X and if I just pull the right-hand side over again X″ of X + λ X x X of X = 0 .
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On the other side with the Ts and got T′ of T = λ x T of T and if I move that over T′ of T - λ x T of T = 0 .
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So, I am not actually going to solve these differential equations the important thing is that I have done with the problem asked me which is to separate the partial differential equation into 2 ordinary differential equations 1 for in terms of X and 1 in terms of T so, let me summarize that here .
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We got 2 ordinary differential equations 1 for X and 1 for T of T so, that is all we need to do for that problem me recap how we how we want about it so, we assume that U of XT = X of X x QT that is of an assumption we use for every single problem in this lecture.
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That is always how you start out with separable with trying to separate partial differential equations than the X derivative the all the T constant take the X derivative second derivative is the same way holding T constant the T derivative you hold the X constant and take the derivative of T and you plug them back into the partial differential equation.
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CNX″ of X x T + X x X x T′ = 0 down the work to get all my X on 1 side and all my Ts on the other that is the key part of separating a partial differential equation.
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So, I did some algebra there and got only X is 1 place = to all my Ts but since I have a function of X = to a function of T it must be constant and so, I call that constant λ and then I just separated each of those equations the X = λ the Ts = to λ and each 1 of those I reorganized into a nice ordinary differential equation which you can then solve for X of X for T of T .
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So, let us try out another example in example 2 were going you separation of variables on the following differential equation U of TT + U of XT + U of X = 0 so, let me start out the exact same guess that were going to make every single time for partial differential equations U of XT = X of X x T of T .
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Then let me go ahead and find you what X is can be I Treating the T is constant so, X′ of X x T of T looking at the differential equation I am seeing which derivatives of that have to plug-in.
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U of TT them in a skip a step here and skip right to the second derivative the X will come down is a constant every time and then will have the second derivative of T with respect to T now, U of XT that means you take the T derivative of the X derivative.
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So, the X relative is a peer X′ of X x T of T the T derivative of that the X is now, constitutes X′ of X x T′ of T so, and take each 1 of those 3 derivatives and plug them into the partial differential equation and will see what happens there.
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U TT is X of X T″ of T + U of XT is X′ of X x T′ of T + U of X is X′ of X x T of T = 0 .
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Now, my goal here is to solve this in such a way that I get all the X is on 1 side and all the Ts on the other so, if I look at this is you got X′ here and here them in a factor that out so, X of X T″ of T + X′ of X T′ of T + T of T = 0.
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Remove that term over to the other side so, X of X T″ of T = - X′ of X x that whole stuff in parentheses there in brackets + T of T and now, I am trying to get all my X on 1 side and all my Ts on the other so, I think I am going to cross multiply and divide .
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On the left of me get keep double prime of T divided by that compound term T′ of T + T of T on the right I am getting at - X′ of X divided by X of X and I succeeded in getting all my X alongside only Ts on the other so, this again must be constant must be constant and so, it is got a be = to a constant that I am in a call λ.
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Separate these 2 equations and solve each 1 separately so, my - X′ of X over X of X = λ and on the other side will have T″ of T x T′ of T + T of no divided by T′ of T + T a T is also, = to λ and each 1 of those I am going to sort out into an ordinary differential equation.
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So, here I am going to get X′ of X move the - over the other side - λ X = - λ X of X and if I move that back I get X′ of X + λ X of X = 0 that is a nice ordinary differential equation there is only 1 variable in their which is little X .
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On the right I see I got T″ of T = λ x T′ of T + T of T and if I move those terms over to the left I got T″ of T - λ T′ of T - λ T of T = 0 that is second-order ordinary differential equation in terms of T T of T.
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And so, again I got 2 ordinary differential equations for X of X 1 for X of X and 1 for T of T so, I am done there let me go back over the steps in case you have any lingering questions.
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U of XT were assuming that to be = to X of X x T of T that is the running assumption for all of these differential equations that were going to try to separate it is always separating it into a function of X x a function of T and the point of that is when you take derivatives U of X you just treating the T is a constant CX′ of X and in U of XT down here we take the T derivative of that so, we get X′ of X x T′ of T and U of TT is X of X x T″ of T.
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The reason I took those 3 partial derivatives was because I was looking at the partial differential equation and so, I was planning ahead to see what lives can have to plug-in to the partial differential equation and so, I plugged each 1 of those in and then I am trying to solve it in such a way to get all the X over here all the Ts over there.
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So, I see that I have not X′ of X on 2 terms here so, I factored that out and then if I move that over to the other side then I can cross multiply across divide to get all the Ts on the left and all the X on the right.
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Which tells me that it must be = to some constant which am calling λ so, I separate that out into 2 separate equations = λ and then it is really 2 parallel tracks here 1 ends up with the differential equation in terms of X and 1 ends up with an ordinary differential equation in terms of T.
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In example 3 were to use separation of variables to convert another differential equation into 2 ordinary differential equations so, let me remind you that our guests for every partial differential equation of forgot a use separation of variables is U of XT = X of X x T of T.
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Now, looks like I am going to need the derivatives U X X and U TT so, U X X means were holding T constant and taking the second derivative of X should be to derivatives there x T of T and U TT means were holding X constant so, the X just comes down T″ of T and so, I am in a plugged each 1 of those into my differential equation there. 1930 It looks like I am not the plug-in you by itself as well so, U X X is X″ of X x T of T + U TT is X of X T″ of T + little T x the original U so, that is capital X of X x T of T = 0.
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Now, I want to I I want to try and separate all the X from all the Ts and it looks like I got an X here and it X of X here thing a factor that term out of whenever I cancel got X″ of X x T of T + X of X x the quantity T″ of T + little T x T of T T T = 0.
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Remove that term over to the other side so, I get X″ of X x T of T = - X of X x that compound term T″ of T + T x the original TFT and again I am going to cross multiply and divide so, thing the keep that - sin with the X so, get T1 - X″ of X divided by X so, it really divided both sides way X by - X of X. 2124 On the right-hand side I got T″ of T + little T x T of T divided by T of T and now, I successfully gotten all the X on 1 side all the Ts on the other so, again this thing must be constant and I am a call my constant λ that is kind of the universal and in the when you are using separation of variables.
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And so, I got to a separate into 2 equations - X″ of X over X of X = λ on the 1 side and T″ of T + little T x TT of T over T a T = λ over here on the left and then I multiply my little my X of X out of the denominator and if you bring the - along with the with with the a X of X = - λ x X of X .
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So, if I move that back X″ of X + λ x capital X of X = 0 on the right multiply my T out of the denominator T″ of T + little T x T of T = λ x T of T and if I move that term over the other side the other side of get T″ of T .
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Now, I see about a T in each term here so, right this is T little T - λ just factoring their factoring out a T = 0 and so, what I discovered here is that my original partial differential equation just separated into 2 ordinary differential equations.
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So, 2 ordinary differential equations for X of X and T of T remember were not solving these ordinary differential equations yet will do that in example little later.
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We are just showing how you can start out with the partial differential equation make a good assumption of the beginning and hopefully separated into to ordinary differential equations so, we go back and recap that for you.
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Start out with the original assumption U of XT = X of X x TT a T that is a running assumption we use that in every single example then where you go and that starts to vary on looking over the differential equation I see a many U of XX and U of TT .
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So, for each 1 I hold the other variable constant and just take the second derivative with respect to the corresponding function and then I plugged each 1 of those into the partial differential equation that is what I did here plugged each 1 in.
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And I want to separate the X for the Ts from each other and I noticed I got it X of X and in each of these terms so, I factor that out here and that enables me to pull that hold term that compound turned over to the other side and then we can cross multiply and divide and get all my X is on 1 side and all my T on the other.
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So, since I got a function of X = to a function of T means of them both = constant animate a call that constant λ as usual and so, I separate those into 2 equations in terms of λ and 1 involves X and 1 involves Ts but there is no mixing between the 2 .
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So, each 1 of those I can do a little algebra and rearrange it into a traditional ordinary differential equation 1 in terms of X and 1 in terms of T.
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So, in example 4 were in you separate separation of variables on the heat equation which I said is out 1 of the most important partial differential equations and it is can be the 1 that were to spend a lot of time solving later we still have not completely solve the heat equation so, to take us a couple lectures to catch there.
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So, were going to start solving that right now, are going to apply separation of variables to heat equation and there is a little note here that says will we get the step of using λ we should use - λ instead of λ for the separation constant that really make sense why we can do that at this stage but hopefully in a couple problems down the road from now, you will start to see why we worry is - λ instead of λ.
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So, let us go ahead and try her separation of variables on this 1 starts out just like all the others where you start out with U of XT you assume that it can be written as X of X x T of T and then were to need some derivatives here U so, T X constant so, get X of X x T′ of T and will also, need U X X which means withhold T constant and will get the second derivative of X and then just T of T .
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So, were to take these and plugged them in were to plug them in to our partial differential equation so, what we get here is U of T is capital X of X x T′ of T = α ² on a really care what out is at this point X″ of X x T of T and my goal here is a separate the X on 1 side the Ts on the other.
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It is pretty easy for this 1 just get down X of X divided by X X everyone separate alone across multiplied by the other way around so, get my keys on let T′ of T divided by TT a T = anything to bring my α ² along with my T up at that down there and on the other side of got X″ of X divided by X of X .
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By the way it is not so, obvious at this point y-axis but the α ² with 1 terms or the other there is a lot of different places you put it I am putting it here because I am kind of looking forward to a solution were going to be studying later for the heat equation.
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So, do not worry about it right now, while putting out for ² in this term as opposed to over here with the X just bare with me and you will see will we get to problem later that it is it is a nice convention to have the α ² with the Ts.
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Now, this is a constant because precisely because we got all X is on 1 side we got all Ts on the other and it says were going to use - λ instead of the usual λ th= - λ .
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Remember we can untangle each 1 of those and try to I get some nice differential equations on the X side we get X″ of X divided by X of X = - λ.
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So, if I clear my denominator there I get X″ of X = - λ x X of X and if I move it over to the other side I get X″ of X + λ x X of X = 0.
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On the other side I am going to just keep my T′ of T divided by α ² x T of T this can only that = to - λ I think that is actually going to be the format there is going to be useful to solve later so, just can leave it in that form and what that means is that I have successfully separated my partial differential equation into 2 ordinary differential equations .
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1 in terms of X and 1 in terms of T. I see I left out a′ on my X here; that is X″ of X + λ x X of little X = 0 and here is my differential equation in terms of T. So, it reduced my partial differential equation into two ordinary differential equations .
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X and T now, you want to hang onto these ordinary differential equations because were going use them in the next example let us go ahead and try to solve these ordinary differential equations in the next example so, make sure you understand where these differential equations are coming for the so, the to be ready to understand them and be ready to solve them and use them in the next example.
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So, let me go back and remind you where they came from we start with this generic assumption U of XT is a product of a function of X and a function of T and then we take its derivative with respect to T and with respect to X twice so, each time you take a partial derivative it means hold the other variable constant and you take a derivative with respect whichever variable that is.
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So, U of T hold the X constant and you take T′ of T U of double X that means you hold the T constant and you take the second derivative with respect to X so, we plugged those back into the differential equation and then we try to sort out the X on 1 side and sort out the Ts on the other side.
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So, we do sort those out X it turns out to be fairly easy get them sorted then we can say that the whole thing there both = to the same constant and for reasons that will be useful later it does not really make sense at this point recalling that constant - λ .
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So, this separates out into a function of X = - λ and was we solve that into a regular ordinary differential equation and then a function of T is = λ got a little squished function of T = - λ and just do not need that in that form because that is the form that were getting used to solve later .
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So, the next example you are going solve these 2 ordinary differential equations and were to see what kind of answer we might get to the original partial differential equation.
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So, let us go ahead and start working on that so, were going to solve the 2 ordinary differential equations below that came from the heat equation from the partial differential equation that was the heat equation.
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I gives us a little hint here we want to assume that λ is + to try to find solutions that satisfy the boundary conditions U of 0T = U of LT = 0 so, were to try find solutions that also, satisfy that.
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So, let me go ahead and try to solve this first 1 this is in terms of T this is actually really nicety separable first order differential equation will my α ² over to the other side so, I got T′ of T divided by T of T = - λ α ² .
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Now, I can integrate both sides can integrate both sides here now, the integral because I have T′ of T over T of T the integral that is exactly natural log of T of T and that is both the = to integrate the right-hand side it is were integrating with respect to T so, is just - λ α ² T + a constant .
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I want to solve for T here so, when I raised each of both sides so, get T = E to the - λ α ² T + a constant but remember this is the same as E - λ α ² T x EC and that EC were just think of that as being a K being another constant K and so, my function T actually pretty easy to find here was a constant x E - λ α ² T.
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Let us hang onto that were to be using it later so, before we go on to the next slide let me recap what happened on this slide to try to solve each 1 of these ordinary differential equations that come up in the heat equation and so, far just try to solve the 1 for T try to solve 1 for X on the next slide .
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I am just trying to solve the 1 for T so, I move the α ² over to the other side key observation here is that the derivative of T of T of natural log of T would be 1 over T x its derivative so, the integral of T priority over T of T is very easy it is just natural log of T of T.
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Integrate the right-hand side since were integrating with respect to T I just get - λ α ² x T + a constant remember it is very important to add a constant when you are integrating.
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By the way if it is been a while since you watch the initial lectures here on educator on differential equations might be good to go back and check those out because I am using stuff that we learned back in those early lectures on differential equations .
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So, to get rid of the natural log I raised E of both sides here and now, by laws of exponents remember E of the X + Y = E X x E Y so, since I got a + C here that is been like multiplied by E the C is just another constant so, I called a K and now, if you solve for T we get K x E - λ α ² x T.
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So, we still have not looked at the equation for X at all yet so, that is organized go ahead and do on the next slide so, in were still working on example 5 here we solve the equation for T so, that that is looking good we have to solve the equation for X .
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Now, this is a second order linear ordinary differential equation we had a whole lecture X a 3 lectures on how to solve these earlier on in the differential equations lecture series here on educator.com so, if you do not number how to solve those here is a really quick review.
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You look at the characteristic equation R² + λ = 0 and using of this X of X you thing that is being X to the 0 derivative so, put R the 0s that is why there is no R right here .
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So, were solving trying to solve for λ so, R ²s = to - λ and R = the ² root of - λ the + or - there let me remind you that we were given an assumption here which was that λ that was bigger than 0 .
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For that means that - λ is less than 0 and so, that means the ² root of - λ is complex so, were dealing with complex numbers here it is I x the ² root of + λ.
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So, we can have to deal with complex solutions here now, we did talk about what happens when you get complex solutions the characteristic equation that was back in 1 of earlier lectures here on educator.com back and check it out if you do not remember but the short version is that your answers look like sins and cosine.
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So, our R here is + or - I x root λ and so, my solution X of X is I think in our earlier lectures in C1 and C2 C1 x the cosine + C2 x sin call it A and B for my constants this time so, it is a x well remember if you have a + b I then you go E AT x cosine of B T + C2 η T x sin of BT.
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Sum up all that format simply in a change around what I am calling my constants little bit instead of C1 I am going to use a and B for C1 and C2 this a is just 0 so, you get E 0T which is just 1 so, those terms are both 1 and I get cosine of well my B is the ² root of λ so, I may go constant x quickly go ahead and call that C1 and C2 so, will be too confusing.
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C1 x cosine of the ² root of λ and that is my variable here is X not T so, put it X in there + C2 x sin of the ² root of λ X and let me go ahead and change that C1 and C2is and call them a and B.
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So, a cosine the ² root of λ X + Bsin the ² root of λ X so, that is my X of X and my U my my original function that I was looking for U of XT let me remind you what format that had U of XT was = to X of X this is our assumption.
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We try to separate the variables x T of T and I have already figured out my T of T there is right there and I figured out my X of X so, let me combine those so, that = a cosine the ² of λ X + B x the sin of the ² root of λ X all x K x E - λ α ² T .
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By the way that K drop it in the reason going to drop it is because if I put a K right here it could be absorbed right into the A and B so, just a cancel that K out and put E- λ α ² x T that is my U of XT .
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But I am trying to satisfy my 2 boundary conditions so, let me remind you what those were my first boundary condition was U of XT are sorry U of 0T we label as a boundary condition is probably pin a while since you have seen it is a boundary condition this is given to us as part of the differential equation well as part of the package along with the differential equation was that U of 0T had to be = to 0 .
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So, let us see what we get number that means your plug-in X = 0 let us see what we get if we plug-in X = 0 into our U of XT here so, U of 0T = a x cosine of 0+ B x sin of 0 and all of that gets multiplied by E of the - λ α ² T and that is what is to come out to be 0.
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So, cosine a 0 is just 1 sin of 0 is 0 and so, what we get is a x E - λ α ² T = 0 on the strength of that we can say that a must be = to 0 .
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So, we will not be needing to use that a term now, let us look at how were going to incorporate the other boundary condition and let me remind you what that was so, my U of XT simplifies down a bit when U of XT since our a term is gone away it is just B x the sin of root λ X all of that multiplied by E - λ α ² T .
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And were told by our other boundary condition let me remind you this came as part of the package when we originally got the differential equation so, our other boundary condition was that U of L to had to be 0 remember that represented plug-in X = L which was the right-hand end of the Rod in the heat equation.
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So, let us plug-in X = L we get B x sin of root λ L x E - λ α ² T = 0 now, we do not want it said B = 0 because otherwise our entire solution disappears .
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Also, our E - λ α ² T we do not want that to be 0 because and even function can never be 0s that is never 0 I do not want to set B to be = to 0 because otherwise my whole solution disappears and so, what I am forced to look at is the idea that sin of root λ L = 0 .
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So, I have to figure out what my possibilities there are to make sin of root λ L = 0 so, sin of what = 0 will I know, that sin of any multiple of π in the integer multiple of pies = to 0 so, this will work if I take root λ L = an integer multiple of π where N is an integer .
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And so, the only thing remember L was the length of the Rod in the heat equation so, that is a constant it is established all the way through we can change that so, do not think I can solve for his is λ and so, if I will be L to the other side and ² both sides 4618 I get λ = N² π ² over L ² .
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This will work for any value of N we choose so, we get 1 solution for each N so, I am and I write my X sub and of X remember I had an X for what I am discovering is that for each value of NI get a different solution .
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So, and there is a B there the B could be anything so, do not know, yet what I want B to B but BN sin I have a root λ L there so, remember root λ is N π over L sin of root λ X so, N π X over L and my T = E - now, λ this is rather messy N² π ² over L ² .
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I still have not α ² T in my exponent for E so, let me extend that α ² T such my T of T and so, ultimately get 1 solution for each end my U of an of XT is remember you multiply the X of X and the T of T so, multiply these 2 things together it is B sub N sin of N π X over L E - N² π ² α ² T all divided by L ² .
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Pretty messy that is my solution to the heat equation that is still compatible with both of the boundary conditions that you tend to get package along with the heat equation so, that was quite a mouthful of me go back and recap everything came from.
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We started out with well we had to solve these 2 ordinary differential equations for T of T and X of X so, we solved T of T on the previous slide previous slide so, on the previous side we figured out a form for T of T and we figured out that it was a constant x E - λ α ² T that constant I am X the going to omit that constant and the reason is because down at this step and gets multiplied by X which has Constance attach to it.
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Anyway so, I let that K be absorbed into the Constance for the the the X functions here so, not to worry about that constant what I do have to focus on next is solving this ordinary differential equation for X and I learned how to do that back in our lessons on second-order differential equations we have an earlier lecture on that read here in the differential equations lecture series on educator.com .
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So, if you do not remember how to solve those just check back the earlier lecture and you will see that we did a bunch of second-order differential equations and we had the characteristic equation we had cases where there were real roots where there were complex roots where there were repeated roots so, were through all those different equated different cases in different lectures.
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What happens with this 1 is we get R ² + λ = 0 so, R ² is - λ and so, R turns into the ² root of - λ and were given that λ that was greater than 0 so, - λ is less than 0 so, really trying to here is take the ² of a - number which means we have complex solutions .
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We learned before that will we have complex solutions A+ B I this is the format of our solution except word I using slightly different variables here so, you translate a little bit this C1 and C2 I a relabeled a and B and that is a little confusing as it is not the same as this a and B.
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This a and B will be a was just 0 because it is 0+ or - I root λ and the B this be was root λ so, that is where I got to cosine e of that is the B there and that is the B there also, are variable now, is X instead of T we use T before but now, are using axis and so, and I also, wanted to change the Constance C1 and C2 a want to call them a and B a new a and B .
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So, it is a little confusing with a different variables I apologize for that there is nothing very deeply mathematical going on I am just kind according the solution from the earlier lecture earlier lecture on complex roots on complex roots.
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So, we have now, found our solution for X of X but we want to match the boundary conditions so, we want to compile our solution together with our solution from T of T so, remember this is our guess with separation of variables.
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U is the product of X of X and T of T so, here on multiplying them together and are boundary condition was the U 0T = 0 which means you plug-in X = 0 so, I plug-in X = 0 here and that when the plug-in X = 0 you get the cosine a 0 is just 1 the sin of 0 is just 0 and so, you get a x E the something = 0.
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What you do something this term is never = to 0 so, we must have a = 0 which means or hold term here drops out of our solution so, were just down to the simpler solution Bx the sin term x that exponential term .
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Now, we bring in our second boundary condition which says that when you plug-in X = L then you must be getting 0 so, if we plug-in X = L here then we get complicated expression = 0 now, the E term again can not be 0 that that can not be 0.
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I do not want B term to be 0 because it B = 0 in my entire solution disappears and I just get 0 so, I do not want to be the 0 which means I must have sin = to 0 .
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So, if sin = 0 then I have to think about what angles have sin 0 and I remember that sin of any multiple or any integer multiple of π = 0.
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That means root λ L must be N π for some integer and an exit is works no matter what and is as long as it is an integer sorted get a whole family of solutions with different integers if I saw for λ I get λ is N² π ² over L ² and I am just going to plug those back into my solution here back in for root λ here and back in for λ here.
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So, the X part gives us n root λ is N π X over L the E part that the T part gives us E - λ which is - N ² π ² over L ² and we what we multiply those together and putting subscript and 0 because remember getting different solutions for each value of that so, were to have a whole bunch of solutions here .
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So, we have BNx the X part x the T part so, that is our solution to the partial differential equation that also, satisfies both boundary conditions that is actually officially the end of this lecture here but let me go ahead and give you little Teaser for the next lecture.
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We write down what we just figured out U of an of XT = BN x sin of N π X over L just figure this out E - N ² π ² α ² T over L ².
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Now, let me remind you what we did there were trying to solve the heat equation and the original heat equation me make sure I write it down correctly was U of T = α ² U of X X what we have done now, as we solve that using separation of variables.
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We also, had 2 boundary conditions U of 0T = 0 and U of LT = 0 and what were doing in this example was confirming a solution that satisfied both of those boundary conditions.
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Now, there was 1 more condition which was the initial condition, U of X 0 = F of X and what we have not done yet is satisfied that initial condition so, that take quite a bit of work that is going to be the content of the next couple of lectures here on educator.com .
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Were to finish off the solution of the heat equation , we are to try to make it match that initial condition so, hope you stick around and watch those lecture so, we can finish solving the equation together.
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In the meantime, this is the end of our lecture on separation of variables, and I will just remind you that this is part of the differential equations lecture series here on educator.com. My name is Will Murray, thanks very much for joining us, bye bye.