solve-the-following-initial-value-problems-by-using-integrating-factors-yprimex-y2-x-ex-y02

Question

Solve the following initial-value problems by using integrating factors.
$$
y^{\prime}=x y+2 x e^{x}, y(0)=2
$$

Prathan Jarupoonphol · Accepted Answer

Israel. We are going to solve this differential equation. Why? Prime minus two y equal x times e to the X with initial condition that X equals zero give Why equal minus one. Okay, um so it is stand out. Form P of X is just minus two. Integrating factor would be e to the minus two x We multiply this true left hand side going to become Why times e to the minus two x Oh, this primes equals right hand side will be here will be e to the minus X now And we have to integrate the right hands I So we use it to grow by past again. You is just x Do you is e to minus x d X Now the U is the X and we will be minus E to the minus eggs. Put everything in the formula we get, um to the end. It will be this. So this should be where if you should be very familiar with this technique already, and the term that you see Owen over like x time e to there&#x27;s something should be to it too hard. So practice. Ah, we put this in back in here, and I&#x27;m gonna skip to what? Why? Looks like here. So we shared from E to the minus to the, uh, plus X now, because we we used to have even a mite has two x on the left side. And like we calculate it out, we get this general solution. Now, we put in the condition here. Why is minus one when x is zero? So is zero minus one plus c. So the sea is actually Ciro this term gonna disappear and yeah, why will be just this, um, I thing You can pull this out. Yeah. Um sorry. Let me shake. Yes is correct. Uh, this is Thea Answer. There&#x27;s no constant yet because C is zero. That is the answer. Thank you.

Prathan Jarupoonphol · Answer

Well, we are going to solve this differential equation. Why problem? And that&#x27;s why equals two X square with the initial audition that when x zero, why is zero. So this is in the standout form. We know that p of X is just minus one. So you get in fact, I will be into the in. The girl might just want the ex wishes e to remind us Eggs, right. We multiply this true both side of the equation. The left side gonna become Why? Times integrating factors Prime the right hands. I gonna be two x squared times e to the minus eggs. Now we have to integrate the right hands. I If we can do that, we&#x27;re going to get Why. So let&#x27;s do it first I pool number two out. We will insert it in later. We use our favorite techniques again. You go buy parts so you will be X square. Devi will be C to d minus x d eggs that makes do you, uh, two x, the x and we will be minus he to remind us eight putting them in formula. We get minus x square E to the minus eggs. This will become. Plus, if you grow two eggs either minus x d x and you can see that this is very similar to what we start with. Just chain X choir to eggs, right? We can do the illegal by part again. Yeah, uh, with, like Simula as Simon, we have you asked ex Andy we is the e part after this, you gonna have minus X squared E to the minus eggs, plus Sorry. Minus two x e to the minus. Eggs minus two X into the minus. X plus c So these are four. Turn from this. Now, don&#x27;t forget, we have two times two back in. So I just gonna move to the next page. And here from the original equation, we have that the integrated on the right hand side. Give us this term. So we just Monday pi e to the X into the eggs. Sorry. Okay. True both sides to eliminate them, and we&#x27;re gonna get that. Why is this this term here? Now? We still have initial condition to fire What? Where Lucy is. So when x zero, Why is zero it&#x27;s gonna all be zero left. My last four plus see? And so see will be just number four, right? And so we How did final answer? The particular solution would be minus to a square, minus four x minus full plus full E to the X. And that is the answer. Thank you.

Robert Daugherty · Answer

Section 4.5. Problem to 47 1st order linear ordinary differential equations to solve this equation, we need it in standard form. Why prime plus a function of X times Why is equal to another function of X? Then we will find the integrating factor with the any to the anti derivative of E O. X. So to get this guy in standard form, that is gonna be why prime and then minus one over X plus two one is equal to two plus X over X plus two. And so this is why prime minus one over X plus two. Why is equal to one to find my integrating factor that is going to be e minus one over X plus two d x. So this is e, um to the minus natural log of X plus two, which is e to the natural log X plus two minus one. So this is just one over X plus two. So multiply this entire equation by the integrating factor and you&#x27;re gonna end up with why Divided by X plus two, crime is equal to one over X plus two. Then the key to solving this will be to integrate both sides of this equation. On the left, you end up with Y over X plus two, and on the right you get the natural log of X plus two, plus a constant of integration. And then we have a boundary condition that what y zero is equal to zero. So this means well, first of all, before I do that, let me solve. For why, Why is going to be X plus two Natural log X plus two? Let&#x27;s see X Plus two and you substitute zero. So zero is equal to two after a log of to plus C terms to and so with its gives me to solve this for C ah, you end up at sea is equal to minus the natural log of to. So this final answer is why is equal to X Plus two natural log of X plus two and then minus natural log of to times X plus two. Have you looked at your natural log terms? Um, you could factor out and X plus two here, right this little bit more simply, why is equal to X plus two and then you can have natural log of X plus two over to, and that would be the final answer

Robert Daugherty · Answer

section Ford up five Problem. 245 Here we&#x27;re dealing with the near first order ordinary differential equations and I need to get this equation in standard form. So why prime plus some function of X times Why equal some other function of X? Then I will multiply by integrating factor which is just e to the anti derivative p of x. So to get this in standard form, this is weird to be why prime minus one over X squared plus one. Why is equal to negative one over X squared plus one? So my integrating factor is going to be e and then minus one over x squared plus one x the anti derivative there That&#x27;s just the arc tangent. So this is e to the minus our tension of X. So that is my integrating factor. So this equation reduces down to why times e to the minus arc tangent of X prime is equal to minus one x squared plus one and e to the minus arc tangent of X. Now it just becomes a matter of integrating both sides of this equation trivial. On the left side, it was the whole reason for a designing on this. So this is just going to be why and then e to the minus arc tangent of X. And then when I look at the right side of the equation can make a substitution here, let you, um, equal minus the art tangent of X Then d you going to remind us one over expert plus one DX. So in reality, what you see here with this particular integral, this is just simply the integral of e to the u Do you Ah, in that particular form. So this is going to be when you integrate that you&#x27;ll get e to the U plus a constant of integration. And so that&#x27;s just e to the minus our tension of X plus a constant of integration. So now, in order to solve for why you divide by e to the minus are tended of X, so you will get why is equal to So when I look at this So when I don&#x27;t fight this out, um, uh, you get y is equal to one plus see e to the arc tangent of X. And then I had an initial boundary condition. Wives zero is zero So why zero zero? That means that zero is equal to one plus. See e. The Arc Tangent of zero is just simply zero. So this means that C is equal to negative one. So my final answer is why is equal to one minus e to the arc Tangent of X. That&#x27;s the final answer here.

Prathan Jarupoonphol · Answer

this video. We are going to solve this differential equation. Why? Prom minus y o X equals minus lock eggs over x square with the initial condition that when X is one, Why is one s will? So we have a rented into the standout form already. We know that p of X is minus one over eggs so we can find integrating factors as he to the minor locks. Absolute X wish is one over absolute X. Now, when we multiply this true on both sigh, we can consider the absolute value as just pulling a meal because we have absolute value on both side of the equation. In the case that give miners were Lou off them gonna give my nuts? Why, Lou? So we can cancel the minus one. How? From both sides So we can treated like this. Asked just no mobile e, no meal. Now, when we modified to the left side gonna collapse into this form as usual. What&#x27;s left is to integrate this thing on the right hand side. If we can integrate this, we gonna have that. Why use eggs times that thing that&#x27;d be integrated. So let&#x27;s do it here. I pull minus one out. We were inserted back later on. What we have to do is is obviously this style this home some take the eagle, buy pots used to sew everything. Um, so you will be long off eggs and then we will be one over X Q The ex. So do you will be. Why no eggs the eggs and we will be minus one over two x square. And so we have all the ingredients. We just put them in here. You we will be minus lock eggs over two x square And here we might ask to minus become plus one over two X cube now D X, And this is just pulling a meal so we can integrate it Fly away. It will become minus one over four X square. And don&#x27;t forget the Constanta. Yeah, I would like to write as minors so that when we multiply minus one, true this everything will become Plus, it doesn&#x27;t matter for for the constant is I can be whatever. Now what we have to do, we lock that in and we gonna get Yeah, I skip some steps. That is B I Q X through everything in here. We&#x27;re gonna get long X over two X plus one over four X says, see times, eggs. But it is not over. We have initial condition, Right? So x equal one, Gibbs. Why equal one as well? We have one equal lock off. One year, zero and this one over four. And this is C. So our constant room here will be tree or were four, right? And so we have. So this problem, we just inside this bag, this right here. So the answer to this particular equation is lock X over two eggs, plus one over four eggs, plus tree over four times eggs. And that is the answer. Thank you.

Problem

Solve the following initial-value problems by usi…

Problem 248 Easy Difficulty

Solve the following initial-value problems by using integrating factors.
$$
y^{\prime}=x y+2 x e^{x}, y(0)=2
$$

Answer

$y(x)=4 e^{x^{2} / 2}-2 e^{x}+\sqrt{\frac{\pi e}{2}} e^{1 / 2}\left(\operatorname{erf}\left(\frac{-1}{2}\right)-\operatorname{erf}\left(\frac{x-1}{2}\right)\right) e^{x^{2} / 2}$

Related Courses

Calculus

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Heather Z.

Joseph L.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Gilbert Strang,

Calculus

Grace H.

Catherine R.

Heather Z.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$y(x)=4 e^{x^{2} / 2}-2 e^{x}+\sqrt{\frac{\pi e}{2}} e^{1 / 2}\left(\operatorname{erf}\left(\frac{-1}{2}\right)-\operatorname{erf}\left(\frac{x-1}{2}\right)\right) e^{x^{2} / 2}$

Calculus

Grace H.

Catherine R.

Heather Z.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Gilbert Strang,Calculus

Grace H.

Catherine R.

Heather Z.

Joseph L.

Gilbert Strang,

Calculus