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Solveay e* +y + 3xy?(y2 + xy)dx Zxy _ z)dy= 0,y(1) =2(2x + y)dx + ( = 1)dy = 0, y(0) = 2...

Question

Solveay e* +y + 3xy?(y2 + xy)dx Zxy _ z)dy= 0,y(1) =2(2x + y)dx + ( = 1)dy = 0, y(0) = 2

Solve ay e* +y + 3xy? (y2 + xy)dx Zxy _ z)dy= 0,y(1) =2 (2x + y)dx + ( = 1)dy = 0, y(0) = 2



Answers

Solve the differential equations $$2 y^{\prime}=e^{x / 2}+y$$

Hello. We have to solve that given defensive equation that is three A viable less. There's two ideas man as well. He calls to zero so we can regard this equation using the notation using that Dean addition for the derivatives. So this will be three days car by There's two white man is close to zero now we will write the auxiliary equation. Yeah that is three M squared plus two M Man is very close to zero so that we can at three AM mr plus three a.m. minus M. Manager's managed to close to zero. Now we will take common three M. So this will be M plus one minus of one. So that is M plus one equals to zero. So three M minus one multiplied with M plus one equals to zero Children. So from here by sorting it m maybe even buy three. I am too is -1. So sure listen can bitterness By cost to seven into the power and Monex because you do you do the power of MPX. So solution will be by close to seven into the power of X. Y. Three. Yes plus you to U to the power of minus X. Mhm. So this is the answer. I hope you understood. Thank you.

In order to solve this equation, we want to get into the form y prime plus y p of X is Q backs as listed in the textbook and order. To this, we must divide both sides by X to get wide prime minus two. Why over acts is acts. Now we know we must multiply each of these terms by the constant of integration. So each the integral is night of negative to over Axe de Axe gives us eat the negative to natural law of acts which gives us eat the natural of X to the minus two, which is acts to the minus two because again, each the natural log simply gives us once This can literally just be crossed off. Okay, Now that we have this, we know we're gonna be multiplying the integrating factor by each of our subsequent terms. Because then were we able to integrate the right side So you can see we're just multiplying each other charms by the integrating factor. Okay, cool. Now that we've got this, we know the left hand side remains. Why axe to the negative too? However, the right hand side we now need to integrate DEA vacs over acts which is the same thing is integrating one over ox, which is natural log of X plus C. Now, lastly, for a last step, we know we must divide both sides in order to get why by itself. So we're dividing each of these terms by X to the negative too. So we end up with why is X squared times natural log of acts plus see are integrating factor.

So in this equation, we're going to have i m Which will be to why square times eating the two X Well, the race to the two X plus three times X squared. And for the end, we're gonna have to times why times e race, the two X and so from the derogative of each of them, we can already determined that they're both exact. So now, going into the they're integral when we take the integral of em and we take it with respect to D X and we're taking it with respect to DX because the M is being multiplied by D X, Um, we should get something along the lines of what? Why squared three times e to the e rates of the two X. And it's going to just be this because when we had, when we're taking an integral of e raise to some constant times, the variable, we're just going to sort of divide it by the concept. So in this case, it was the two. So we're just going to be left with one. So we're just gonna be have this right here, and then it's very really simple. Enter world. We're going to just have the integral for the integral of three times X squares. Just going to be X cubed. No, that's a bad three. Let me write that. There you go. Okay. And so now for the integral off en with respect to D X, what we're going to get is that we're going to B O A. Oops, not the X. And that, um, de y What we will get is that we will have to times why square times e to the to thanks. And so when looking at this hold on a second, I think I mentally Oh, no, my ah, I was looking at this right here for a little bit. That's actually not too there. Since the tour is right here and we're dividing it by two, it's gonna just be one again. So anyways, when we have what we have are constant right here. We're going. Teoh, bring this. We're going to start with the ammo we're going to have why Squared multiplied by E race the two X and we're going to cross this part off. And since this is also exactly the same, we're gonna cross this off, and so we're gonna add this to X cubed. And so that's going to be our final and

16 meant to problem number eight were asked to differentiate. Should need to solve this difference or equation. So it is a leading your difference equation in his first order of since only have one derivative. So what you gonna recognize? You can put it in standard form Or you could recognize that e to the two x times y How would you differentiate this with respect to X? If you differentiate this with respect to acts, that would just be white way you use the product rule. So that's two e two the two x times y plus e to the two x do y d x. And so what you see there, this is the exact thing you see on the left side of this equation. OK, so this equation is the derivative with respect toe acts of whyy to the two X is equal to two X. Now I can integrate both sides of this equation. So when I integrate both sides of this equation, I get why e to the two X is equal to, and I've got to integrate what two x the X, which is just X squared plus constant of integration. So why E to the two X is equal to X squared plus C. Therefore, why is equal to X squared e to the minus two x plus c e to the minus two x.


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