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6. ,4 Fnd te &5t nanacro 'rur &tr Trk Si @ fW) -...

Question

6. ,4 Fnd te &5t nanacro 'rur &tr Trk Si @ fW) -

6. ,4 Fnd te &5t nanacro 'rur &tr Trk Si @ fW) -



Answers

$$y^{\prime \prime}-5 y^{\prime}+6 y=-6 t e^{2 t}$$

Okay for this problem with friends on the derivative of this s function. So first, we're gonna take a look at this structure and notice that we have this 60 cubed minus five function sitting inside of a 44 X to the 6/5 function. So if I won't find the derivative s prime of tea and first gonna differentiate the outer function so now multiply the outside by 6/5 times 44 of what's inside the function. And then I need to drop the power by one. So go from 6/5. We subtract one or 5/5 and that leaves us with 1/5. And then we need to multiply the derivative of the inside piece. So that's going to give us 18 t squared plus zero. But we don't need to worry about the plus zero now, in order to simplify this, I simply need to take my three numbers right here the 18 44 and 6/5 and we're gonna multiply those all together, and that's gonna come 2 47 52 fifth ce. The numerator is just six times 44 times 18 and we keep the denominator as it is then we have our t squared from on the end of the 18. And we keep the 60 cubed minus five to the 1/5 power. And that right there is the derivative of the S function.

Yeah. Today we're going to trunk. We're going to write, um, quick, System number six as a first order system. What ex prime? Because X was half, Uh, which, in other words, is called It's normal. So this, um So the first step that I want to emphasize is to try and analyze what you have before you actually go about doing anything. So the first observation is that that six is in fact, a second order non homogeneous OD, because it has that t square term there. You'll notice. So it has this t squared term here. Sorry. Um, because it has this t square term, it is non homogeneous. So, um so But our answer should be as a first order system, so we want to get a first order system of this form. So what we d'oh is there's a There's a fairly standard is a standard trick to apply. And the way to go about doing it is to start with, we just replace acts by why, and now we replace X prime by Z, and we re right, um, six. So we rewrite six in this form. So now the main reason I include part see here is just to sort of help me in my, uh in what I do later on. Okay, So, um, s so now Now I use a B and six B, and I write I define a system of equations as such so ins in in 60 here as six C. So you you'll observe that six c is just found directly by applying my replacements from the book. So I just rewrite this, um, using the substitution sze I defined in a B and C Okay, And now the next step, my next step is to while the next step is to write six c in its normal four. Okay, I have a system of equations now. And to do that to write in the normal form what I want to do is what I kind of do. And these personally is the first again analyzed the system I have. So I noticed that it's it's it's two equations. There's too dependent variables y and Z Ah, this means a should be, uh, a matrix should be two by two on and accent half should both be to buy one. Okay, so the next step then is to I guess. Guess and answer. And again, this is, um uh, stand sort of. Ah, a standard trick in in CODIS And this course is that you gas at an answer for a wrist. You make a sort of an educated guess. In this case, my axe matrix are my X factors is just the dependent variables. And my a here eyes, the coordinate is that coefficient matrix. So if you if you look I mean, if your ear a little confused about where hey comes from if you look back at the previous if you look at, um books Sorry. If you look at, um wait a minute. So So if you look at Yeah, if you look at this system here, if the one in the Matrix or the zero in The Matrix comes from, uh, the zero, I guess zero acts here and they're in. Plus, I guess it should be plus plus one z. So the zero comes from here on. The one comes from here. And then, of course, the negative one is here and zero. So I mean, it's fairly. It should be fairly straightforward where these terms air coming from, but just in case, you don't see it. I'm just sort of trying to clarify where those terms are coming from. And that's how I'm getting My A is really just directly from the coefficients and, of course, AF. I guess, after it's fairly straightforward again, it's very similar. Um, it's just the non homogeneous terms in that previous equation and six c and then the next step is to check that check my guess. So I In other words, I look at a x plus af and just by again straightforward matrix multiplication. In addition, what I end up with is, um, the right hand side and the right hand side is exactly, um is exactly right here. Um and you can check that. I mean, it's it really comes to make sure that this is true. Really? What All you do is you have to look back at the given system. Um, the given system here and shock coordinate wise. Really, That that the that this is exactly this the look, I would say that the left hand side here, left hand side is exactly the right hand side. So really, you just look at the system. You look at the system that you're given and and the corn and you can see that coordinate wise. The left hand side is the right hand side. And this is by definition, just ex crime. Okay, Now, we thought in mind this shows that the system that I defined or the system of, um, the system of difference or equations that I defined that, that I should say it this way. That this that this is the correct, Um, this is the cracked normal form for the the system off linear different. You are this the system of linear, different equations I find. But of course, that's not quite the answer. Because I want to make sure that this system is in fact, um six, which was when I was given. And so to check that while all I do is back substitute X for why an ex prime for Z into the second coordinate here, Um, and what I want to get back is this year, and you'll notice that this is exactly six. So that is exactly the the answer. We want it. So, um so in other words, that is the correct answer, and that's what I wanted to do. today. Um, I guess the key here is that this is this is another thing. This is another type of problem where it's a question of just understanding what the steps are and why I'm doing it that way. So it once you understand that, then you should be good in these type of problems. Oh.

In the question the belong differentially question is be square. Why number slash a minus treaty Lyon Nash and duty less. It's why e equals store zeal. No zoom. Why equals two p to the power? Uh, then why Aghbash will be quite odd here to the power R minus one. Why Deborah Dash would be odd And who are minus one B to the power ar minus. Cool. Substituting be value in the question we would get peace. Where? All right, R minus one into the power ar minus two minus three. The on it would be powered on minus one. Last six. State Department articles zero solving This will work back on out of minus one minus three yard. Last six A to the power will still zero. That is odd. Sled minus four. Would I? Plastics would ripple through zero. So now applying the form life, we will get articles 24 Last minus are Nadal Force. Where my ass forging through six. Singh Who won? Upon whom? Everyone here We have applied reform running the Russia. It is our equals two minus be less miners. And that would be square minus four a c upon the way. So I would be grateful for blessed minus, under minus Kate upon you. That is our equals. Toe four plus minus. Who under rule twice I I, uh, upon to that is the routes would be to bless miners on the route. Who I don't know. Therefore, the general form for the given differential equation with revivals to see one peace where boys undergo Do Alan be, Let's see do peace. Where? Sign a notable whom and then he thank you.

This for when we're going to use the black last transform table to evaluate the laugh test transom of the function five minus the two. The two tea plus six t squared. This is done quite easily because the lack last transformers linear. So you only need to compute the last last transfer of one. The hapless transform of each of the two tea and the last blast transform off T square. If you look up your table, you'll find the lap. Last transfer of one is one over s by the lap. Last transformer E to the two t is going to be equal to, um won over s minus two. And the last last transform of t squared is going to be equal to to Over s cute. And if we put this all together, we'll get five the s minus one of the s minus two plus 12 s cute


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