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Here a= 2subject: Linear Algebra and DifferentialEquations...

Question

Here a= 2subject: Linear Algebra and DifferentialEquations

Here a= 2 subject: Linear Algebra and Differential Equations



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This last equation is linear in the (unknown) dependent variable u. Solve the differential equations.
$y^{\prime}-y=-y^{2}$

Okay. What we want to do is we want to step through the process to be able to work with what is termed a newly differential equation of the form dy over dx or Y prime plus some function um Times Y function of X, times Y is equal to another function of X, times Y to the end. And if and it's not equal to 01 then we are going to have to do some kind of a substitution to transform this personally differential equation into a linear differential equation, which then we know how to solve. Um And so we noticed that for this differential equation of X times Y prime plus Y equal to Y to the negative too. We don't um in is not 01 and is actually negative too. So we're gonna have to incorporate this substitution. So we're going to let you whips, we are going to let you equal y raised to the one minus a negative to say plus a two. So this is equal to y cubed so therefore you to the one third is equal to y. We're just gonna keep that noted because we're going to have to do a bunch of substitution then D you over dx is equal to three Y squared dy dx right? Because we're doing implicit differentiation and so Dy over dx is equal to 1/3 Y squared time. See you over dx. But why is you to the one third? So this actually becomes 1/3 you to the two thirds times d'you over dx. So I'm going to substitute that in. So what I'm gonna do is and another thing I'm actually divide everything by X first. So I have dy over dx plus one over X times Y is equal to one over X times wider than negative two. Okay so Dy over dx is one over three U. Two. The two thirds time. See you over dx plus one over X times you to the one third. Because that's what why is is equal to one over X. Times U. To the negative two thirds. Okay. And we need the eu over dx by itself. So we can multiply everything by three. You to the two thirds. So we have t you D X plus three over eggs times you is equal to three over X. Okay. So this is our linear differential equation where this is P. Of X. And this is Q. Of X. Okay so now what we want to do is to go ahead and find a function that we can multiply to everything and remember that function um is equal to E raised to the integral of P. Of X. Which is three over X. Dx. So this is going to be e raised to the three times the natural log of x. Which is actually equal to execute. So we're gonna multiply this by X cubed this by X cubed and this by execute. Okay, so this becomes um X cubed times. See you over the X plus three X squared you is equal to three X squared. Okay. So now when we write this as what did I have to take the derivative of to get this right here and I had to take the derivative of an X cubed times. Are you right Because the derivative of you is see you over D X times the first function plus a derivative of X cubed which is three X squared times you is equal to three X squared. Okay. And now we're going to integrate both sides. So this was just going to be X cubed you is equal to and we're going to integrate three X squared with respect X. So this will become X cubed times U is equal to just execute plus some constant of integration. Okay. Um and now we're gonna divide everything by X cubed. Um and so U. Is equal to one plus C over X cubed. I'm going to a common denominator so this is going to be executed plus C over execute. But remember you is actually why cute? Right. So we have, why cubed is equal to this, execute plus C over X cubed and now why is equal to the Q boot of X cubed plus C. All over an X. And so there is our solution you know

We want to convert the system wide. Double prime plus a Why Prime plus B. Why? Because after tea to ah set of first order equations. So we'll do. Our normal method of my one is equal to why and why To use equal y prime where another differentiation gives us why to prime is the goto Why double time? So we'll have our why to equal why Prime gives us a relationship that why one prime is why too. And, um, we'll also get from this equation our relationship that Ah, if we bring everything to the right, we get that this is equal the f of tea minus b y minus a y prime or as we already established for institutions view I want my A right to. So then our system ends up being following. Why one? Ah, prime is equal to why to and why to prime is echoed A f a t lies a my, uh, B. Why one minus a why to

So what we're gonna do is kind of walk through the process of being able to actually solve a differential equation of the form. Want a glimpse. Y. Prime or dy over dx minus Y. Is equal to X times y squared. Um And this is kind of in what is called a very newly differential equation um where we have dy over dx plus some function of X times Y is equal to another function of X times Y to the end. And for the highest exponent in order for this exponent in for and not equal to zero or one. Then we're going to have to do some kind of substitution to transform this differential equation into a linear differential equation. And so that's what we're going to kind of work with. And we always do this type of substitution where we let U equal why to the one minus in. And so that's what we're gonna do. So I'm gonna let you equal why to the one minus two. Which is why to the negative one. So that implies that why is equal to U to the negative ones then I'm gonna take the derivative. So do you over D. X. Is equal to negative one negative y to the negative too. Dy over D X. Because we're doing implicit differentiation. Um Which means now we're going to solve for dy over dx because that's what why prime is. So we're trying to do some substitution is in here. So that means that dy over dx is going to be actually equal to negative y squared. Do you over dx? And we also know that why is equal to U. To the negative too? So we have negative U to the negative too. G you over dx. And so now I'm going to use these as substitution. So what I mean, Do you yes. Why prime or dy over dx is equal to negative you negative degrees to the negative to do you over dx minus. Usually the negative one. Because that's what Y. Is is equal to X. Times U. To the negative too. And there we have it. And now I recognize um I don't want this um negative. You raised to the negative two in front of to you over dx. I'm gonna divide everything by negative, you raised to the negative two. And so this becomes d'you over D. X. Plus you to you to the first is equal to a negative X. And so now I basically have it as a linear form, right? Um And so what I'm gonna do is rewrite this side um as um kind of a derivative. So D D. X. What do I have to take the derivative of in order to get, nope, that's not what I want to do right now, we're good. Okay, sorry about that. And so now what we want to do is to go ahead and um come up with a function that we're going to multiply everything by and typically we do Vfx is equal to E. Raised to the integral of P. Of X. And remember P. Of X. Is um whatever is in front of you or why? Um in this case it's just a one the X. So that is equal to eat of the X. And so that is what I'm going to multiply to everything inside this equation. Okay. Um And now we're gonna do what we I wanted to do. I now want to rewrite this left hand side as what I would have had to take the derivative of to get it. And so that would have been you either the X, right? Because the derivative, this is a product role. So the derivative of the first term, which is, do you ever D. X times a second function plus the first function times attributed to the second is now equal to negative X. E. To the X. And so now I'm going to integrate both sides. And the integral of the left side is you E to the X. Is equal to the integral of negative X. E. To the X. D. X. Okay. So this is going to now have to be integration by parts, right? So we're gonna do an integration by parts. Um I'm going to let the equal negative X. So the D. V. Is equal to negative dx. I'm gonna let um D W I can't you because I've used it over here. D W is equal to E. To the X. D. X. So W. Is equal to E. To the X. Okay so now what we want to do this actually becomes um negative X. E. X. And then it's minus the integral of W. D. V. So that's going to be plus the integral of E. To the X. Dx. Okay so now this now becomes negative X. Either the X. Plus and eat the X. Plus that constant of integration. And now I can divide everything by each of the X. So I have you is equal to negative X. Plus one plus C. E. To the negative X. Okay. Um that is what you equals but you is one over why and we don't want one of over why we want why? So we're just gonna flip both sides and then I'm just going to kind of make it a little nicer. So um and there we have it. That is our solution.

All right. So we want to do now is convert this third or that disorder or equation into a system of three differential questions that are all first order. So we want to do is make our variables our make. Our derivatives are variables. So we'll start off with why one equals. Why? Why two is equal to y prime. Why three is he quit? A Why Double prime. And since we're not gonna have And since we're not gonna have, um, more than four more than three variables, we're gonna let her avoid atropine. You could have wide three prime. So then this will give us our our preliminary system. So we have, um if we, um, do relationships between each ones we have, we've already done Why three prime is wide, double prime. So then we know that. Why? To why to ah, prime Schickel wide Oh, prime. And why one crime is he coulda y prime. So then why did prime, we already know was just 53 And why prime is just by two. So then our system of equations as a being Why one prime is equal to why to Well, I won, uh, my to prime is equal. Why three? And, um why the re prime? We have been actually evaluated this, but it's just a Whitehall prime. So if we take, um, the other terms and move them over, we end up with what the value of Watcher of Prime is, or in this case, wide three prime. So then, ah will start off with her teeth. We have tea, and then we're gonna write that. And after. So all right back here and then why, uh plus t squared y Primor. He squared minus two squared y two and then e to the t y, which is just eat that. He Why one? So then I wrote it in this sort of way that makes it easy to visualize each room separately. But, um, this is


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