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2.) (1 point) Use reduction of Order to solve this ODE: y" V+yz The solution makes a catenary curve. Find the particular solution for Y(O) = 0 and y(2) = 1...

Question

2.) (1 point) Use reduction of Order to solve this ODE: y" V+yz The solution makes a catenary curve. Find the particular solution for Y(O) = 0 and y(2) = 1

2.) (1 point) Use reduction of Order to solve this ODE: y" V+yz The solution makes a catenary curve. Find the particular solution for Y(O) = 0 and y(2) = 1



Answers

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(t)=\frac{t}{y}, y(1)=2$$

Were given. Why Times y prime, which I'm gonna all d y over dx is equal to two x over two plus y squared squared And then we're told that why of one is equal to minus one. So if we separate our variables, we're going to get why times two plus y squared squared D Y is equal to x dx and then we're gonna integrate these. Okay, this one, we're actually going to buy you substitution. We'll let u equal to plus y squared, which means that do you is to Why do you why? And I have the why and the d y. So when I do this, it's gonna become 1/2 to get correct for this too. And an interval of you squared, Do you? If I integrate that, I get 1/2 times 1/3 you cubed plus C, which will put on the other side that gives you 1/6 of you cubed and you is two plus y squared. So this gives me two plus y squared, um cubed divided by six. The interval of two x is just X squared plus c so that's a little bit easier. So we get this, Okay? And then what we're gonna do is we're going to, um, plugging are one and minus one. So why have one is your a minus one? So I'm gonna kind of go up here so you can, so you little better what we're doing, but we're gonna put in two plus and then why is minus one so minus one squared, all cubed, divided by six is equal to and then X is one. So one squared, which is one plus c. So we get minus one. Squared is one you times. I was ready to be blocked. One, 33 Huge is 27 27. Divided by six is we're gonna end up with. Is he the one plus C 27 divided by six reduces you. Invited by threes. It the nine over where they were going back one. So if you take nine over to extract one, you get seven is equal. So therefore we end up with this two plus y squared cute divided by six is equal to X squared. Plus 7/2. So that's my answer. Now you can expand that two plus y squared if you wish Dio. Um but I don't, so I'm gonna leave it like that

In this question. We are given the differential equation why prime plus two X -1. Why Is equal to 0? And the boundary condition, Why is he going to two annexes? Go to one. Now first we look for our integrating factor μ which is given by E raised to the integral of the coefficient of Y, which is two X -1. D x mm. This is two X squared over the new power to minus X. And ignoring the constant for integration There we have eat of X squared minus six. Is our integrating factor. Not applying everything you know on our differential equation. By the integrating factor we have E to the power of X squared minus six. Why prime plus two X -1. Apply by E to the power of X squared minus X. And why Is equal to zero. And we recognize that the left hand side is equivalent to the derivative of each puff x squared minus X. By why? And this is equal to zero. Now integrating both sides with respect to X. We get e to the power of x squared minus x. Why being equal to see a constant? And writing both sides by E. To the power of x squared minus six. Get see over E to the X squared minus six which can also be read it and see E. Since we want to play the power bi minus one, we get x minus x squared. And this is the general solution. Now plugging in our initial value or boundary value we get since we have to hear and our c E raised to the power one minus one squared is equal to C. E. Races 40 is equal to see, so our C is equal to two. Therefore the final solution is the particular solution will be called to two over. Or since we had, we are now using this definition, so we have two mm to E to the power off X minus X squared. And this is our final particular solution to the given differential equation.

In this question. We are given the differential equation why prime plus two X -1. Why Is equal to 0? And the boundary condition, Why is he going to two annexes? Go to one. Now first we look for our integrating factor μ which is given by E raised to the integral of the coefficient of Y, which is two X -1. D x mm. This is two X squared over the new power to minus X. And ignoring the constant for integration There we have eat of X squared minus six. Is our integrating factor. Not applying everything you know on our differential equation. By the integrating factor we have E to the power of X squared minus six. Why prime plus two X -1. Apply by E to the power of X squared minus X. And why Is equal to zero. And we recognize that the left hand side is equivalent to the derivative of each puff x squared minus X. By why? And this is equal to zero. Now integrating both sides with respect to X. We get e to the power of x squared minus x. Why being equal to see a constant? And writing both sides by E. To the power of x squared minus six. Get see over E to the X squared minus six which can also be read it and see E. Since we want to play the power bi minus one, we get x minus x squared. And this is the general solution. Now plugging in our initial value or boundary value we get since we have to hear and our c E raised to the power one minus one squared is equal to C. E. Races 40 is equal to see, so our C is equal to two. Therefore the final solution is the particular solution will be called to two over. Or since we had, we are now using this definition, so we have two mm to E to the power off X minus X squared. And this is our final particular solution to the given differential equation.

In this question, we are given the differential equation by prime Plus one of our X why Is equal to 0? And we are given the initial condition that 22 right X is going to to and why is it going to to? And we had to find the particular solution to this differential question. So first here we're going to look for the integrating factor, mu is equal to which we can find by raising E to the part of the coefficient of Y, which in this case is one of the X at E to the power of lin X. Which gives us our integrating factor is X. Multiplying everything by X. We have X. Y prime Plus Y is equal to zero. And we can see that this left hand side is equivalent to the derivative wolf, X, Y Is equal to zero and integrating both sides with respect to X. We get our left hand side is X. Y is equal to a constant C. And here our Y will be going to see our X and substituting our point 22. We have two years ago to see over two, giving our see value is for therefore the particular solution To this differential equation will be go to, why is he going to four over X. And this is our final answer


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