And 1.654, we can see that we have YS but no XS. So we're going to have our Y prime equal DYDX, which equals P Our Y double prime is going to be DPDX, and by the chain roll, that's DPD y * d YDX. But remember this DYDX is really just P so we're going to take out that DYDX and put in P. So now we're just going to do straight substitution so we get y * P DPDY equaling 3P squared. So taken all the PS to one side, we're going to get one over PDP equaling 3 / y DY. If we integrate each side, we can see that that's going to turn into the natural log of P equaling 3, the natural log of Y plus some constant. And then from there, let's actually call this constant just some natural log of a constant. So then we can see that P is going to equal y ^3 times RC or Cy cubed. So then that P is really just our DYDX and that's equaling our Cy cubed. So getting all of our YS on one side, we get D y / y ^3 equaling CDX and a great we get -1 / 2 Y squared equaling CX plus C1. So then we would get -2 Y squared equaling one over CX plus C1, divide by two -2 we get y ^2 equaling -1 over this 2 could get absorbed into some new variables, so we could get y ^2 equaling, and the negative actually could also. So we get one over. Let's call it AX plus B. If we wanted to solve for Y, we just take the square root so we get positive -1 over AX plus B.