Here's #14 on 1.6. Remember we're going to start with Y equal VX, so DYDX equal. Doing the product rule. The derivative of X is 1, so just v + X DVDX. So instead of the Y, we put in VX, instead of the Y prime v + X DVD X + X equal the square root X ^2 + v ^2 X ^2. So doing some manipulation here, we're going to get v ^2 X plus VX squared DVDX plus X equal. If we factor out an X here, we get an X sqrt 1 + v ^2. So now we're going to let's go ahead and divide everything through by an X so we get v ^2. Now remember, this means it's an equivalent. It's not necessarily equal, but we're going to do it to see if we can get anything a little more simplistic. So if we divide everything through by an X, we get something that looks like this. If I take everything without the DVDX to one side, I'm going to get square root 1 + v ^2 -, 1 -, v ^2. So then if I think about needing to take all of the V's to one side and everything else to the other, I'm going to V divided by square root 1 + v ^2 -, 1 -, v ^2 equaling DX divided by X. So now we could think about pulling out a negative here and realize that the right side when I integrate, oops, I lost my DV, sorry, there's DV. This right side, when I integrate it, is just going to end up with the LNX plus C. So let's think about doing AU sub here. If we have AU equal 1 + v ^2, we get DU equaling 2 VDV or 1/2 DU equal VDV. So we'd have our 1/2. I'm going to put it out in front DU over. Here's sqrt U -, U equaling the natural log absolute value of X plus some C Well, we don't really know what to do with this either, so let's factor out a sqrt U. And if we did that, we get 1 -, sqrt U left. And let's think about doing another substitution here. So if we have this, we could think of V equaling 1 -, sqrt U. So DV would then equal -1 / 2 square roots of U du. So we would get -2 DV. Oh no, let's leave that there. So we have -1 half square, DUDU. So when I look at this, here's my 1/2, there's my du, here's my square of U. So all I really need is a negative. So over here we're going to get negative DV over V and then we'll take the integral of that. So this is going to be negative. The natural log of V equal the natural log of X. We could call that C the natural log of some other constant. So now we have the negative natural log of 1 -, sqrt U equal the natural log of X plus the natural log of some C1, let's say. So from there we can see that this is just really going to be. If I take that negative up on top, we could then use the natural log rules and we could get 1 / 1 -, sqrt U equaling X * C one, and then the square root of U Back here, if we thought about square root of U, that'd be sqrt 1 + v ^2. So we're going to look at 1 / 1 -, sqrt 1 + v ^2 equaling XC1. And then we got to get rid of the V, so we get 1 / 1 -, sqrt 1 + y ^2 / X ^2 equaling XC1. Common denominator here we're going to get one minus the square root X ^2 + y ^2 / X ^2 equaling XC1. So then we can pull out an X out. So we get one over. This is going to be X minus the square root X ^2 + y ^2 all over X. Just getting a this one to turn into X / X and that square root coming out equals XC one. So then we're going to 1 divided by means we take the reciprocal and then we could divide each side by plain old X and we could cross multiply. So we would get X minus the square root X ^2 + y ^2 times the XC1 equaling X. So just cross multiply to start and then cancel the X's. So we'd get X -, sqrt X ^2 + y ^2 * C one equaling one if we divide by C1. All I'm doing here is doing some simplification. I could get 1 / C one, which we could call some new CC two. So if I subtract this constant over and add sqrt X ^2 + y ^2, then we could actually square each side. So we could have X + C two quantity squared minus X ^2 equaling y ^2 by squaring each side. And then actually we could foil this out and the X ^2 would cancel. We get X ^2 + 2 XC 2 plus C 2 ^2 -, X ^2. So our y ^2 would eventually end up being 2X. We could call. We'll just leave it as C2 plus C2 squared. So that's one possible answer.