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Chapter 2, Section 2.2, Additional Question 03Solve the initial value problem5 cos(51) y' = y(0) = 3+Zy and determine where the solution attains its maximum Ta...

Question

Chapter 2, Section 2.2, Additional Question 03Solve the initial value problem5 cos(51) y' = y(0) = 3+Zy and determine where the solution attains its maximum Talue (for 0 < 2 < 0.679 )Enclose arguments of functions in parentheses For example; Sin (2x).y(z)The solution attains maximum at the following value f €_ Enter the exact answer

Chapter 2, Section 2.2, Additional Question 03 Solve the initial value problem 5 cos(51) y' = y(0) = 3+Zy and determine where the solution attains its maximum Talue (for 0 < 2 < 0.679 ) Enclose arguments of functions in parentheses For example; Sin (2x). y(z) The solution attains maximum at the following value f €_ Enter the exact answer



Answers

Solve the initial value problem $$ y^{\prime}=2 \cos 2 x /(3+2 y), \quad y(0)=-1 $$ and determine where the solution attains its maximum value.

Hello and welcome back to another differential equation. Unless we have dy dx is equal to two minus E. To the X. All over three plus two. Y. We're also supposed to find the maximum of this equation as well as its uh simplified exact form. Um So to do the maximum we have to find the critical points and that's when do I. D. X. Is equal to zero or undefined In this one? We only have X. In the numerator. So to South Africa 100 visible two to minus E to the X. And X is equal to l. n. two. That's our critical point. Yeah right there we don't know if it's a minimum or a maximum but you know it's a critical point. Let's continue on with the rest of our problems as other. Just like other separable difference equations. We want the wives on one side and the X is on the other. So I'll multiply it by three plus two. Y. And D. X. Correct. And now I'm going to integrate both sides, luckily these are simple into girls and they don't require too much work in this problem. I'm going to be assuming uh knowledge of basic and girls and uh power rule. If not then how rule is um You take this number here, bring it up there and divide by this number here. No. All right. So we're also going to add C. Uh Now I want to solve for y with completing the square. This is going to have me divide This three x 2 and then square it. We'll add that value to both sides. So we get y plus three halves squared is equal to two X minus E. To the X plus C. Remember the The 9/4, 9/4 that we added is just absorbed into the sea. Next we're going to take the uh huh Squared of this and subtract 3/2 of why is equal to plus or minus the square it of two X minus E. To the X plus C. And we're also going to be subtracting 3/2. The very end like that butt plug in our initial conditions now since we have wild by itself, so we're going to zero is equal to plus or minus. Uhh squared of Well two times 00 when each of the 01 C minus one minus three halves. Okay. And from this we can tell that we will be taking the positive square root. You can get rid of that negative And Saul Saul for seat so we'll get 9/4. This is me adding over the three houses, squaring it. It's equal to see lines one which means see Is equal to 13/4 Great. I'm just gonna plug that into our equation appear so I'd have to be right at remember we're also taking the positive where we get rid of this negative there. That is our uh uh equation for this differential equation up here. Um Now we have to know whether this is a minimum or a maximum. Now looking at this equation inside the square, we can see that this IV the X term is more powerful than the two X term, which means in the long run the value will decrease. So that means this is the maximum because this more powerful term will win in the end. So if we have a graph like this, this Ellen too, we'll be here. We can expect to see uh lower values on the left and right side because these are the X term will win in the long run and because the Y value to approach negative infinity. So that means Xcacel into is a maximum. If you want to find the maximum point, however, we have to plug in Ellen to to the equation. Now this is going to produce a annoying result, so I'll leave that to the viewer. But um, the maximum point is that X equals Ellen, too.

This question covers stopped relating to second order differential equations. Almost equations with constant coefficients. Okay, This question asked you to find a solution and then determine the maximum value of the solution and find the point when solution is cerro. Right? So first we need to find a solution first. Right now the characteristic polynomial. Right? And then you saw for the quality uh quadratic equations And you get to solutions on one equals 1 and R two equal to a half. Okay, so then you can try, the general solution is going to be C1 Exponential of T. Last C two. Its financial 1/2 tea. And then use the initial condition. Then you have to find a distributive of that solution also. And use the initial condition to find C one and C two. So it's gonna be uh C one plus C two echo 22 and see one Plus one. Harrah's is too Echoes to 1/2. Right? And Yourself AC one. You take the first minus the second. So that you have one half of c one equal to 3/2. Right? So that means you want C to equal to three fucking two. The first equation you have C two C one equal to two minor three which is negative one. Right? So your solution is minus each of the T plus three. Each other 1/2 of tea. Right? Right. So uh so in order to find the maximum points of the solutions then you need to take the negative of this. Right? White fine, which is negatively to the T. So it's just gonna be minus uh 3/2 of each of the one half of tea. Right? And then to let quite frame equal to zero. Right? So what frank equal to zero and zero Then what you have is negative 3/2 of of ego to each of the 100 teeth. Right? And this is impossible because each of the T. Is positive. Right? So then there's no solution. Actually this is a negative function. Right? So so why friend is always less than zero whenever to speaker than zero because it goes to zero. So the function is decreasing. Right? So that means whitey is always less than or equal to zero and 10. He is too right? So the maximum is a similar value is to when T Go to zero. Right? And how do you find these solutions For white equal to 0? You need you just need to uh So for this equation three, So sorry, this is three. Each of the one half of tea, right? Equal to zero. Okay, So there are many ways for you to do that. You can just let and far equal to eat to the T one half of T. Right? So unfair square gonna be it to the T. So that you have negative one squared plus three and four equal to zero. Then you see you have two solutions on for you go to zero or on for you go to three. But this is impossible, right? Because each of the T is positive, That means either the one half of T. three, Then T equal to the block of three, divided by 1/2, which is time to write. So it's gonna be a lot of nine. Right? That's it.

The problem is each of this Ustream worry problems have a solution. We're supposed a maximum body on a minimum. While it use not ranch, more players to find it. A stream wallows off the function subject to the given constraint. So that my third of luck arrangement players, we have brilliant off half. It's the Lambda I'm screening of Jean. You're G X Y Z is equal to the square, That's why square us C square Then we have I shall have power tracks is equal to Lambda Hams. Patrick three Posh wax. I shall have partial. Why speak with Lambda Hams? I should be partial. Why? I shall have partial thing is they've got to learn that Ham's impartial thing on DH X squared plus y squared. Plus this choir is he able to nine. What's become two is equal to London times to ACS. Who is he going to? London wants to know why one is to go to London. That has to be X squared. Plus y squared us. This choir is able to nine. Then we have accidents. We go to one over Lambda. Why has he got one over Lambda? See if you go to one over to London. We're plugging Tio. It's the equation with half long day Isaac or two so minus. What up? So when Lambda is we go to a half exercising. Is he going to two? Two? Why? Under the connective half ex ally Thie Is that Goto connective too negative to next year? What therefore has possible is extremely values at this two points. Then we violate value off this two points I've tuned to one. If they could to nine. I've negative too negative to connect. One is they got a negative nine. So, my son, while you're off, is equal to nine. The minimal wily off off too. Next year nine.

So what we're given in this problem is Theis equation or the function rather f of x y z t being equal to X plus y plus Z plus team. So simple enough. And we know that this given function uh, it is subject thio expert plus y squared plus C squared plus t squared. So the grading of F is going to be equal to just 1111 clearly. And then the Grady int of G, which is another function we have and that is X squared plus y squared plus Z squared plus t squared equals one. We know that this is just gonna give us two x two y to z two team. With that, we now set those ingredients equal to each other because we're dealing with LaGrange multipliers. So what we do is we set each of them equal to their respective components, so it's gonna look pretty repetitive. One equals all those things. Um, but then we don't wanna forget adding the lambda. That's what makes thes thes LaGrange multipliers. So we, uh, multiply each component by lambda and then when we solve for Lambda, we'll get one over. Lambda is equal to two x, which is equal to two i, which is equal to two Z, which is equal to two team. So with all that, it's clear that we could just cancel out these twos. And ultimately what we see is that X equals y, which equals E which equals t. So, um, since that's the case, we now use this information, um, to solve a little further. So since X squared plus y squared plus z squared plus t squared equals one, as we already know, we could just say, Hey, let's just let's just set everything equal to X So now we have X squared plus X squared plus X squared plus X squared equals one. So what that means is four x squared equals one or X squared equals 1/4 and X equals plus or minus one half. Now we know that they all have to equal each other. So one possible solution would be f of one half one half, one half, one half. And when we do that, we end up getting two and this is going to be our maximum, because if we take f of negative one half f of negative one half, Uh, negative. One half a negative one half what we end up getting is negative two, and we know that that's going to be the minimum. So these would be the two main values that we want to focus on.


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