OK, on 52 of 1.6 we've got y ^3 * y double prime equal 1. The fact that we have AY here means that we have to use the chain rule concept. So we have our Y prime equaling DYDX or P. So our second derivative is going to be DP in terms of DX by chain rule. That's DPD y * d YDX. And remember this DYDX is really just P, so we can think of that as DPDY times the P. So this y ^3 * d PDY P = 1. So if we get the DP and the P on the side and the 1 / y ^3 d Y on the side, and we integrate each side, we can see this is 1/2 P squared equaling negative one half y ^2 because that's really wide in the -3. So we add 1 to the exponent, we get -2, and then we divide by the new exponent plus AC. So then we can think about having this become, let's multiply everything through by a two to get rid of the twos. First of all, and I'm going to put this over now, I'm not, I'm going to leave it as 2C. OK, so now what we're going to do is we're going to square root. So we get P equaling square root of two Cy squared -1 all over y ^2, technically positive or negative. So then that P is really our DYDX and that's going to equal this. So then we're going to get square root y ^2 / 2 Cy squared -1 DY equaling DX. So the square root of the Y ^2 is going to give us Y. We're going to get square root. We could call that two CA new C if we wanted C one y ^2 -, 1 DY equaling DX. So if I let U equal my C one y ^2 - 1, then DU would equal to C1YD Y or 1 / 2 C one DU equals that YDY which is right there. So we get an integral of 1 / 2 C one 1 / sqrt U DU equaling the integral of DX. So 1 / 2 C one we could call some new constant. Once again, this is going to give us square root of U. Actually let's do that in two steps, 1 / 2 C one. So we're going to get sqrt U * 2. And then we can take all of this together and make it a new constant, call it C ^2 U equal X + C What number are we up to? C2. So then C square root C one y ^2 - 1 equal X + C two if we wanted to then divide by C square root C one y ^2 - 1 equal X / C + a new sum C square each side C one y ^2 - 1 X over C + C three squared C1Y squared equal. Now it's just a bunch of busy algebra kinds of stuff. We could actually foil this out even if we wanted to and have all sorts of cool things going on. Y ^2 would equal one over C 1 * X / C + C ^3 ^2 + 1. So if we really foiled all this out, we get one over C1. We'd have X ^2 / C ^2 + 2, C three over CX plus C 3 ^2 + 1 / C one. So then we could pull together a bunch of these constants, this 1 / C ^2, 1 / C, one times X ^2 / C ^2. All of these CS could come together column AX squared when we multiply this one over C1 and the two C ^3 / C which would give me abx and then all the rest I could combine together to get AC.