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(1 point)Given the first order IVP % + 4y52'20 < 2 < 1 21 with initial condition y(D) = 5(1) Find Ihe erplicit solution on Ilie Inlerval 0 < I < 1y...

Question

(1 point)Given the first order IVP % + 4y52'20 < 2 < 1 21 with initial condition y(D) = 5(1) Find Ihe erplicit solution on Ilie Inlerval 0 < I < 1yz)(2) Find the lin ez) -(3) Then Iind Ihe explicit salulian on the inlerval ? > [ sr)

(1 point) Given the first order IVP % + 4y 52'2 0 < 2 < 1 21 with initial condition y(D) = 5 (1) Find Ihe erplicit solution on Ilie Inlerval 0 < I < 1 yz) (2) Find the lin ez) - (3) Then Iind Ihe explicit salulian on the inlerval ? > [ sr)



Answers

In Problems 1 and $2, y=1 /\left(1+c_{1} e^{-x}\right)$ is a one-parameter family of solutions of the first-order DE $y^{\prime}=y-y^{2}$. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. $$y(0)=-\frac{1}{3}$$

Okay we are given a differential equation and a general solution to that differential. We are going to turn the general solution into a particular solution by placing in um a negative one for any excess and that will result in a two for why? So let's go ahead and write the equation. Now. Just notice that the denominator, I just switched the terms around. Um I have that C. E. To the negative acts all over one. Now as we put it to infer are why we're going to put a negative one in for our X. So E. To the negative of a negative one is going to be E. To the first power. And then we have our plus one. So now to solve I'm going to go ahead and multiply up that C. E. Plus one and distribute the two. And as I do that So now I can subtract two from both sides and then divide both sides by A. To E. So we get C equals um a negative one over to E. Show that he's on the bottom there. So now we can go ahead and write our equation and now it's going to be a particular solution because we first see we're going to get that negative one half. Now notice E. Is to the negative one power. So we can write keep the base and add the exponents. So we have a negative one plus a negative X. So we'll just write that as a negative one minus X. We could have also factored out the negative there and did it as a negative parental C. One minus X.

We are given a second order differential And we are also given the general solution to that differential. In the general solution. You'll see a C one and C. Two. We're going to use the initial conditions given to actually solve for that C one and C. Two. So first we are going to be putting in a negative one for X. And a positive five for why? So we can write why equals C. One. Now each of the negative one is like dividing by E. And then E to the negative of a negative one is just eat to the first power. So now to do the derivative E to the X. Is derivative is E. To the X. And then we get an extra negative when we take the derivative of E. To the negative X. Okay, so now we're going to be using our second statement when we put in a negative one for X, we're going to get a negative five out for Y prime. So we'll get negative five equals C one divided by e- C two times e. So notice um we have that C two times E but one is positive and one is negative. So we actually add our two equations together. We see that C two Is actually equal to zero and I should have said there are two of them when I added them together. But either way we're gonna find C1 equals zero. So if C one equals zero, we can solve for C two si two will equal negative five divided by E. And so C two is five divided by because they were both negative, they cross out and then we can rewrite our now our particular solution as Y equals five over E. Um time in E is to the negative X power. So notice that we have to ease here. So we really have an E. To the negative one power and an E. To the negative X. Power. So we could also write this as keep the base and add the exponents. And if they're both negative, we can factor out that negative and write five E. To the negative apprentice E one plus X power. So either of those solutions are correct.

Okay, we are given a differential. Um and then we are also given a solution to that differential. So it's not required for this equation. Um to go ahead and solve the differential instead, We just need to find a particular equation given our wise of two equals 1/3. Um But the actual derivation is shown here, but we'll just be working from the point where we see that why for the differential why equals one over X squared plus C. So now we need to solve for C. Because this is a general solution, but we are finding a particular solution. So notice we're gonna put a two in for X. And then we need to get a one third out for why? So let's go ahead and put that one third in and then one divided by two squared which is four plus C. Now, in order to solve this, notice that um we have a one on the top and we have our fraction. So we can take the reciprocal of both sides. You could also think of this as multiplying both sides by three and also multiplying both sides by four plus C. From here we can subtract four from both sides, and we see that C equals negative one, so now we're just going to go ahead and rewrite this, but then we'll have a X squared minus one there in the denominator.

All right. We want to solve the separation of variables. The equation why prime is equal to two times y minus one. An equation such as this one involving both Y and its derivatives is considered a differential equation. The separation of variables method is one such method for solving these equations, which takes two steps I've already listed below. First. We're gonna isolate the variables on either side of the equation. So we're gonna put terms with Y and left in terms of X on the right. Although there are no X terms immediately. This problem why prime is simply do I. D. X. And so we're gonna put the dx term in the right next. After that, we're gonna integrate both sides of the equation and solve for y. We want our final solution to be in the form Y equals some function of X with the variable isolated on either side. So for y prime equals do I? D. X? Actually the variables as do I over y minus one equals two dx. If we integrate both sides, we obtain on the right. L n y p minus one equals two X plus C. One. The constant immigration taking the natural organisms on both sides is y minus one equals to the to explicitly one. If we simplified to say why equals E to the two X plus one.


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