#16 on 1.6 Y prime equal sqrt X + y + 1. If we let Y equal VX, then we know DYDX is v + X DVDX by the product rule. When we substitute here, we realize we can't get the VS on one side and the X is on the other, so we have to try something else. So the next thing we might try is just a straight substitution where we let V equaling X + y + 1. So then DVDX would equal 1 plus dy DX, and the derivative of that one at the end of course is 0. So if we solve for dy DX, we'd get DVDX -1. So now if we go back to the original, we'd get DVDX -1 equaling the square root of V. So once again, we're going to get all the V's on one side and all the X's on the other. So if we divide each side by 1 + sqrt v and we multiply each side by DX, we get DV over 1 + sqrt v equaling DX. And then we're going to integrate each side. So we're going to get the integral of DV over 1 + sqrt v equaling the integral of DX. We don't know how to do the integral of DV over 1 + sqrt v so we're going to do some substitution again. Let's let U equal 1 + sqrt v. So then DU equal 1 / 2 square roots of VDV. So if we multiplied each side by two, we'd get 2 du equaling one over square root of VDV. So when we look at this equation back here, we realize that we don't have the square root of V in the denominator. So we're going to multiply by sqrt v / sqrt v And the reason to do that is we need the DV which is right here and a 1 / sqrt v or the square root of V in the bottom. So that's all equal to 2D U's. Then we need to figure out, because we multiply by root V over root V in order to have the same thing as multiplying by one, we're going to realize that this root V here, if we came back here and solved, we could see that root V is just U -, 1. So right here's this U -, 1, and then finally this one plus root V We started with letting it be plain old U. So now we could factor out the two because a constant could come in and out of the integral. U / U is 1 -, 1 / U du. When we evaluate the integral, we get U minus the natural log of U equaling X + C We're going to substitute that U in for one plus root V. So two 1 + sqrt v minus the natural log of the absolute value of 1 + sqrt v equaling X + C and then the square root of V Remember back here the V is X + y + 1, so square root of V is going to be sqrt X + y + 1. So instead of the square root of E, we get two times the quantity one plus the square root X + y + 1 minus the natural log absolute value 1 + sqrt X + y + 1 equaling X + C. If I divided each side by this two, I'd get this entire thing on the left equaling 1/2 X and then C / 2. Or I could call that a new C. Let's call him C1. I can't easily get an X or Y by itself, so we're going to leave the equation at that step.