Problem

Solve the given differential equation. $$\frac{d …

07:56

Problem 9 Medium Difficulty

Solve the given differential equation.
$$\frac{d y}{d x}=\frac{x\left(y^{2}-1\right)}{2(x-2)(x-1)}$$

Answer

$\frac{(x-1)+(x-2)^{2} c}{(x-1)-(x+2)^{2} c}$

Related Courses

Calculus 2 / BC

Differential Equations and Linear Algebra

Chapter 1

First-Order Differential Equations

Section 4

Separable Differential Equations

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Video Transcript

So here we are, considering the differential equation. Do I d. X is equal to x times the quantity Why squared minus one all divided by two times the quantity X minus two times the quantity X minus one. Okay, so what? We can rewrite this differential equation as well. One over. Why squared minus one? Um, de Y is equal to that. See? 1/2 times. Well, times acts over X minus two times X minus one. Um, de X. Okay, so now we have a separate vote differential equation, and we can then integrate both sides. So we're just gonna integrate, dropping an integral We're gonna integrate the left side, and I'm gonna integrate threats. Okay, Well, to do this integral here, we're gonna want to find the, um, partial fraction decomposition to the partial fraction of X over X minus two times X minus one. Right now. To do this, we're gonna set X over X minus two times X minus one equal to or some constant A over. Um, X minus two over the first factor. Eso over X minus two and then plus, um, be over the other factor of X minus one. Okay, so to do this. Um, Well, multi. That's how we get acts is equal to a times X minus one plus B. Um, Times X minus two. Okay, we distribute this out, um, and then we factor to get X is equal to a plus B X minus. Hey, minus to B. Okay, we combine our like terms. And what do we get? We get a plus. B is equal to one. Um, and we get negative. A minus to B is equal to zero. Okay. And this implies that Well, a right is equal shoe. Um, negative to be right over a So a is he go to negative to be okay. There. We go ahead and substitute this in our equation. Um, up here. Right. If you speak with one to get negative B plus beak was one or negative. I'm starting in native to B plus B, um is equal to one. Okay, that means that negative b looks negative. Negative beat negative. Be, um, is equal to one. Which means that B writes implies that B is equal to negative one. Okay, so now we can go ahead and substitute b equals negative one into a plus. B equals one. So we get a, um Well, a minus one equals one. So therefore a is equal to two. Okay, so we have our partial fraction decomposition of X over X minus two times X minus one is equal to two over X minus two. Um, minus one over X minus one. Okay, so therefore, we can then do our inner grow. Um, and we get the integral of one over Why squared? Well, minus one squared D y. Well, that is equal to well, the integral of well, 1/2 of the 1/2 Qin. Well, so we get with the off off 1/2 times X over X minus two times X minus one Um, the axe. Right. But then well, this becomes we have 1/2 1/2 times the integral off. Whoa of, um um one over. Why? Minus one minus one over y plus one de y is equal toe 1/2 times the integral off whips of two over X minus two minus one over X minus one DX. Okay, so well, we get I mean, we can break this up as the inner girls of the inter grow off of one over. Why, Sorry? This just froze for a minute. Here. Get my well. Okay, let's try it sometime. All right. So you have the integral of one over. Well, why? Minus one B y, um, minus the integral of one over of one. Over. Why, plus one. Do y. All right. This breaking that up that was equal to the inner grow off. Two over. X minus two. The axe minus the integral of one over X minus one DX. Yeah. Um, so right, we evaluate this and what we get, we get the, um the natural log of the absolute value of why minus one minus the natural log of the absolute value of why plus one is equal to two times the natural log of the absolute value of X minus two, um, minus the natural. Log off the absolute value of X minus one. Um, plus the natural log of, well, the absolute value of our constancy. Okay, so then this We can then see that. See? Put this as just a natural log. So the natural log off the absolute value off. Why? Minus one over. Why? Plus one is equal to while the natural log of see times. Ah, the quantity X minus two squared over X minus one you got? And then Well, to get rid of the natural log, you can exponentially eight both sides to take e raised to financial log of all this stuff just exponentially. Eight. Both sides Well, e to the power of natural Laugesen. This cancel out and we just end up with why minus one over why? Plus one is equal to see times X minus two squared over X minus one. Okay, so we get why minus one, right? Mom plans to buy one plus one. We get why minus one is equal to well, see times X minus two squared over X minus one and then times while plus one the multi. Both both sides through my plus one. Okay, so, um, we get why, minus one Hopes we get why minus one. So I minus one, um, is equal to see times X minus two. Squared over X minus one times. Why? Plus c times X minus two. Squared over X minus one. Okay, um, and then well, let's see. We can do that because we have what we have. Why? Minus, um, see, Times X minus two Squared over X minus One times Why equal? See Times X minus two, squared over F minus one plus one and then fashion. Or why we get to get why. Times. Let's see one minus see times X minus two square Ah, over X minus one is equal to see Time's X minus two squared over X minus one. This one okay. And then continuing down. Um, uh huh. Let's see. Let's see, we got well, we end up with why is equal to see Times X minus two squared plus X minus one all divided by X minus one times X minus one Um, divided by X minus one minus C times the quantity X minus two square. Okay, so if you just, um, clean this up, we grow a little bit. Here we got why is equal too well, see times X minus two square, plus the quantity X minus one all divided by X minus one minus C times X minus two square. Okay, so, um, the general solution, right? Hence the general solution of this given differential equation. Once we start for why, um, we can clean this up. Many kind of put the terms in a different order. Here we get is equal to we have why is equal to X minus one plus C times X minus two Squared all over X minus one minus C. Times X minus two squared. OK, so there would be our general solution to the differential equation.

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