[boş_ses] Hello. In this section we have reached the end of our lesson. Our subject differential equations. Now those of you who have never found differential equations, who does not know very well, but can also be seen that; but this term deterrent, repellent, whether or not a subject we will try our very unfamiliar feeling. Because the use of this matrix, the equation, we the unknowns, we can return to the simple constant coefficient differential equations. Our job easy for him. Differential equations, for they have never seen before; easily a subject, and it works well everywhere can learn. Already here in the first four sub-headings, as part of a preliminary, subject to introduce to you, if the differential If y'all ever seen in this equation to explain briefly, because the issue shortly. It's also no longer is. The other two issues; one of them, Why you need a set of differential equations? Introducing it. This last, in relation to each other actually three and four sub-headings; when given a diferransiyel equation; a top-order differential equation, n'yinc the order, it is the matrix, rotating the first order equations. From here also directly given to the first order differential equations, but the unknowns of an equation. So that the difference between these two; here's one derivative, A maximum of the first derivative; but many There are unknown; If there are unknown but high-order one here. Between the two, one, in some cases, the other can be useful in other cases, we see that she respects. After creating this infrastructure we have seen in the two sections, Using powerful tools, will solve the differential equations. These two chapters we have seen before; 12. In part, even in page 195, given a complete overview of a page; diagonalizing here we go into a matrix solution, We will use various means and here diagonalization.Lines We also reinforced by examples. Again we saw in the previous section; 13. we saw in section 249 concretely. we provide a summary of the page; We will take advantage of matrix functions. Using these solutions we will get. Both methods will develop as well as we will see concrete examples. So along with being called differential equations, matrices our main topic again. Differential part of the overall structure able to fold into a short time, it is easier to maintain our business. Now before the first order, Let's start with the differential equations with constant coefficients. Once differential equations what does that mean? Differential equations of an unknown function y, and the availability of various derivative itself. Highest derivatives, whatever we say it's from that order. We call ours here because there is only the first derivative of the first order, There are also functions itself. With constant coefficients, unknown factor because the numbers are fixed; y base multiplier number, one; y is a number of a number of multipliers. These variables also be numbered, It was possible to be variable functions; but that does not interest in the species matrix. Already not able to benefit from the matrix, in the way you see it. This b is a given function. This task shows external influences. There's a system that works by itself, but there is also external influences, It is given this structure. x instead; if you think that in times like this time, It will show an evolution; but one thing's evolution You also need to know where to start to watch. We show it here, the value to 0 instantly; it also We call the initial condition, call the initial value; in it the evolution of the event is fully traceable situation. So here are many constants; 1, a constant, y0 constant, a single variable function b (x). Quite simply a kind. As we approach solving it. First, the equation will be easy to wonder if it was not for me? We ask. It would be really easy. Because due to a constant; he says, when there is no b, It is a function that when you take the derivative itself to get a solid. We know that such a function is an exponential function of this kind. Here because it is a; YHA, the simple equation is the solution of the name; also it called homogeneous, Turkish as well; e to the ax of the time we have to take the derivative again over ax, but also a number of times before falling here. Yh derivative As you can see, a time YHA. Because e to the ax was yh'y. Therefore, now the equation, we can find the usual solution is not b. It has meaning in terms of a linear processors, The presence of this null space this means differential operator. Now, why we do it, We have two kinds of reason; One important on its own, it; but also helpful in finding solutions when the equation is b. As we embark on a path optimism; When we say that b is that, çöüm of the full equation, assume that YH times. Why we are optimistic, we are optimistic due to the following; YHA, an information carrying quite a few, including a base year and y, It carries information; so when you use the information, unknown to the rest come from more than a simple equation; and already experiences but we can hope the show; This boldly optimistic approach. So y is the time to call them ax. Because homogeneous, simple solution to ax them. We do not know where we will look forward to. Next, by that time u e to ax; u v times mean the product; It means taking the base year, It crossed the first derivative of the first derivative of the second plus the second time here. So we take derivatives. u derivative; See here multiplied by two; In order to get such elegant is it she did; the second term, once u e to ax derivative. It stands, a time derivative of e to ax the ax to them. Let's place this in our equation. Our equation where y + b = moon base. y base of the left side; We calculate here, we put here to bring the base year. It consists of two terms, u üssül terms and u'l term; It is equal to the right side; Once a year in the right-hand side. But we now know y structure, We know partially, if at all; once u e to the ax, we are writing this, plus where b. We see here the reward of our optimism, with the first term on the left side, The first term on the right side as each other; thereby taking each other and leaving only the term kalııy u'l; u base is equal to b and e to the ax. This is a problem because a lot easier integration He's reached to the problem; We know the base. e to the right side of X. If a split; Of course The share of e to the minus future as ax; Once b. So the issue has become a integral problems. If we account for this integration; definite integrals, indefinite integrals, various There are variations, but they all come in the same way, we calculate the integral. There is a sense of putting the base x x base is changing because, not having a job after graduating variable; x x base are putting in place to bring in the end, therefore. If you do not put a bad habit but will also results can be obtained in the same manner. When we insert equation that equals y had a homogeneous solution, plus new solution we get it. This solution can be interpreted as follows; here is a simple the first term solution, call a special solution to a second term, because this is a solution, here we come to definite integrals, We want to know why a particular right here in the initial conditions. Y is zero initial condition y equal to zero. If we put y e to zero as you can see here ax because there will be an exponential zero, we will have a car here. This particular solution we can arrange for it to be zero. We organize as follows: leaving the bottom of uncertain location, the lower limit, x base we started from scratch. Indeed, when we put zero-zero-zero instead of x it is going integration, an integral value to the lower and upper limits for the negative result is the same Two of the following value would be zero for the same. This structure is slightly more compact bulk structure, but no harm in leaving a very uncertain integral. This simple solution in a secure çarpılmış of the solution. Next to it, we obtain the general solution of attaching this particular solution. So overall this solution; The most common solution that provides this equation. e to a solution, but it secures an ax çarpsa you again when you take the derivative itself a'yl to çarpılmış the future, We are bringing a car to find it here in the most general sense. We have already provide the initial condition, To ensure the initial condition uncertainty where it should be Keep flexibility to ensure we have the initial condition. In this way they are found all the overall solution. This solution is the only solution to the differential equation initially. Another solution can be proved that this is not, we will enter it here. It requires a bit more work to find a private solution, simple solution supremely simple. Because that's what you are looking for a term that you get a solid derivative itself, it is also exponential. So it's simple solution immediately. There is a second method to find specific solutions, but this second The first method is not broad enough, there's no restriction on b. b or whatever you bring, you calculate this integral. Do Hesaplayamasa by at least such an integral result of it is certain that no specific solution. If you recall again that the right term in the equation, those on the right side b (x) of a given term. Given the strength of the term if x, exponential function, cosine, sine, hyperbolic cosine hyperbolic sine function types, or multiplied by a certain number of them are multiplied with each other, collected compositions In particular solution would remove one of these functions. This function is called exponential functions on the type of family. So the right side of exponential functions in earnings rather than specific solutions that the exponential function. But right in front of some given those numbers are, Although such functions in a special solution, i.e. structure Although we do not know what will happen with the coefficients in front initially, This method is called the method of undetermined coefficients for her. Soon we will start just one example, in doing so, it We will calculate a special solution in two different method and differences, similarities, advantages to be seen. Although this latter method is usually more restricted, but because of this b right side of the function, the exponential functions a method that works regardless of the family, it is limited to maintenance. But an easier method, because the first method, an integral You have to figure, well, that is quite a job to calculate integrals. Here integral take, there is no need to calculate. Now let's just our example. Our example is a simple example but do not challenge these kinds of problems already, just simple numbers selected. y is equal to the base year. If you get this part just because of this simple solution, homogeneous solution. B is also a plus e to the x, consists of two. Given this function, it is unknown natural derivative of y in the y unknown. We say we need to get a value of zero initial condition of this function. We solve this problem in four steps, the first step to find simple solutions. Lean solutions that without them this Part B base year equals y derivative itself equal function to the above, When we take the derivative of e to the x is equal to itself, it makes this place so. Or, if you move to the previous solution Another approach equals a in our present problem. Y here in a number in front of this one. In any event e to the x, we find it çarpsak but also of any number again solution, because when you take the derivative of it once again c e to the x is to the left c times e to the x-side, right in the c e to the x times. So the simple solution that is not much of a challenge. Special solutions we find in two ways, one with the integral formula. We use the formula to bring the integration, there were simple solutions, I had the opposite simple solution inside, e to the x it comes from molasses, and B function. We calculate this integral. Let's open it here in terms of calculating this integral; to one hit the minus X. e to the minus x is coming, where e to the minus x x plus two over here to there, so hit the e via x occurs. This integration is supremely easy to get. E to the x minus the integral of the first term but got a minus because derivative when you get to the minus minus needed here to give you x. e to the x integral of e to the x again. From here we get the product, You have to look at e to the minus x where x The first term comes from a minus for them to be given a multiplication. In the second term over here to the second term to X. e to the x when multiplied over two x turns out, that is to say we have found our custom solution. I showed two base to find solutions so special with the second method, namely the uncertain coefficients, just to make this comparison. We say; the right type of exponential functions. The contribution will be a fixed one specific solution, but we do not know the value of fixed call it a B for his already uncertain the coefficients to be with him, but we know we do not know the structure initially. In order to balance the constant on the right y must be a constant. E via e over again, but we need to balance two x two x The number at the beginning we do not know yet. These coefficients, these coefficients are uncertain, The name of the method is already coming from there now to find this structure after we put in the equation y, we have set. Next year we need to base that you're here on the fixed base year as derivative falls to zero. e to two times e to the x is going from the two x two derivatives, B of course, also here there are two, it means that base year only consists of these terms. We know that the y term here. Let's put it after the left side of the base year, we put the base year left. There are constant and exponential functions, they are here. There's a special solution. This after we put in a special solution. [BOŞ_SES] when we place them We can easily see that only a B no on the left side, it's on the right. To balance one on the right, there must be a minus B1. Two B2 on the left side, the left side has a B2. If we take it to the left, staying here B2. On the right side there is also e to the 2x. When you put in it, e to the 2x coefficient should also understand that a. If the car once we found our solution because simple solutions plus special solutions. c times e to the simple solution of x, custom solutions as well B1 negative A B2 is a finding, we have achieved here and the same solution as well We see here that we have achieved, it needs to be said. We're one more step here. Third step. This need to find out what the c. We use the initial conditions to find out what the c. x equals zero, y is one. Where x is zero we put in place, for having an above zero, only the first term remains c. Not already a minus x to a second term. e to the 2x de e to zero in a more revenues. As we see here, they brought the right and the left one with each other by, c must also be one. Therefore, as you can see here the one and only solution to the equation, e to the x minus 2x plus to emerge as above. Now we see something very deep here in terms of differential equations, but only unknowns in terms of differential equations with constant coefficients, also solve the equations we see that the right side is not very difficult. Now many in this application much a part of the equation but There are factors in the case, it has the contribution. They also interact with one another. Y1, Y2, wool, it shows each event, If a show case, in the first case you can interact with it, You can interact with others, it can interact with the outside world. When we get here is a typical one, i'yinc we see it. Have interaction with it, to have interaction with the other component, There are interactions with other contributions, have interaction with the outside world. This y interaction with the outside world, regardless of year. Because now we have only one variable to the equation first, It had an initial condition. Here all state vectors, We need to give initial conditions for state components. They are also the number one zero Y1, Y2 zero, yn given as zero. In fact if you look at the left side of this equation it is a derivative of a vector. On the right side we see in it more equations. This structure is a matrix multiplication of the vector y. There is also a plus that he's. The one in the initial condition here, the vector y is the value of the zero instantly. As Gördüğüz, this one can see the matrix equation into the equation. Without the base year would be algebraic equations. To find the base year, it's a differential equation of the team. If we visualize it a bit more, Once the base is equal to y h y work consists of: Sol y base vector forming components, have the right matrix. This is multiplied by the vector y. Plus they vector indicating that external influences. So, we are faced with an equation of this nature. Keep them out of chemical reactions, biological systems to interact with each other as parts of the machine elements, The interactions of the electrical circuit, a team faced in a planning event or initially it may have a differential equation, When it can be an existing structure discretization. So they encountered equations. A second path may be a higher order equation. Here we will start from the most simple. Get second order equations. The two unknowns can call the resort to download the first order equations. This one can see the benefits. To do so, y y1 to say, the first component of the vector. The second component of the vector is to say y derivatives. Wherein the second derivative is Y2 derivative. Let's put them in the equation is that after startup. Y2 y2 base derivative, Y2 derivative, a1 times the base year, see here Y2 y base. Y2 can put it here. and y1. As you can see, we are creating two parts of this differential equation. One of them is here. Y1 Y2 is equal to the base. This equation. Secondly, we have obtained the equation here. To make it a little more structure to our convenient, base left in the early years and let you get a multiplier. Let's also right Öbürkü minus sign will appear. y to bring a side of the base, a2 divides them and we get on the right side of this equation. If we write these two matrix equation, see the first line of the matrix is zero and no We have to, because we want to be Y1 Y2 of the base. Means to multiply the vector of this year, we invested take this vector, carpacagız y1i zero, one with its carpacagız Y2. Indeed, once staying on the right side Y2. That equation. In this second equation, the coefficient in front of our Y1A will be zero minus a2. this in front of Y2. Again this year took the deposit hit the vector indeed these terms We are creating positive terms on the right side. There are zero on the right side in the first equation. None, so here comes zero. In the second equation, h is a2 division. As you can see, this is the kind of brew a matrix equal to the base year plus a term that indicates a vector cross placed outside influences. What we earn here? Initially there was a higher order differential equations, second-order equations. We have them, but we have one in our hands. Instead, an equation with two unknowns We came to the first order, but the order of the equation. This matrix can be won with great ease working in many cases With this equation. He respects that, given the equation of the first order, low, making the low-order matrix equation It can be useful in many cases to make differential equations. This generalization would be as follows: n'yinc a variable equations of order. But high order. Here again, as we have done before, y of tanımlasak as y1, y2 tanımlasak the derivative of y. O zaman ki madem and y1'di, Y1 is equal to the base we obtain a differential equation such as Y 2. If we take up this n'yinci, just in the last second equation As in the order, we find the term of this year üssül it a little fast because no harm comes to pass in the same building; off such a matrix, See it left only in the base year, where only the base year, the base year, y base, only consists of base year. A matrix multiplier times a year by the sheer equals. As you can see from this matrix it is equal to the initial base Y1 Y2. Just to bring Y2, where there must be a. Everywhere also have to be zero. In this way, an on the diagonal, Coming to a matrix structure, which is filled with the last line. And h is zero, and the last non-zero function. If we visualize than that, we get the following. where n is a vector y base of the derivative component. Matrix vector multiply y plus hi vector. Reset but so important structure here and solution methods will be the same. This convenient technique that can benefit from improved zero etc. but y is a matrix multiply essence. Wherein the vector is equal to a given vector derivative. I want to stay here for now. Let's think a little. If we do we summarize, we have created an infrastructure, We enter into more solution methods. We remembered the first order differential equations with constant coefficients or We provide an environment to ensure no re-learning the unfamiliar. We talked about how the equations that arise. A variety of naturally occurring element itself, It represents an event indicating the interaction of factors The equation can be a higher order or, to the extent n'yinc with only one first-order equation, but the one with the unknown, So you win, you increase the number of unknowns from a place somewhere. But I think that this method of use matrix of the first order only We will see that it has become solves the equation.
