00:01
All right, so we've got to find the implicit general solution.
00:03
I think this is the problem.
00:06
So, d .y over dx is going to equal x minus 9 times y to the 8th, all of that over x squared times 2y to the 6th minus y.
00:22
I think that's what we want.
00:24
All right.
00:25
So what we've got to do is kind of almost like cross -multipline.
00:31
All of these ys all of this i'm going to multiply by the reciprocal 2 y to the 6 minus y over y to the 8th i'm going to multiply by this on both sides and then i'm going to multiply by dx on both sides okay i'm going to multiply by the dx so that the dx reduces out here and all of this will reduce out on the right so dx and d .y will be on the same side.
01:06
So we'll have 2y to the 6th minus y over y to the 8th on the right times dy equals x minus 9 over x minus 9 over x squared dx.
01:25
All right.
01:26
So now we just have to integrate both sides.
01:30
I'm going to simplify this down a little bit.
01:32
I'm going to make this 2y to the 5th minus 1 over y to the 7th dy and then i can't do much on the right.
01:43
But to integrate to these, i'll split them up.
01:48
So i'm going to integrate 2y to the 5th over y to the 7th minus 1 over y to the 7th, and i'm going to integrate x over x squared minus 9 over x squared and split them up because now i could split up the integral so so really i just have to integrate 2 over y squared d y and integrate 1 over y to the 7th d y and integrate 1 over x d x and integrate 1 over x d x and minus 9 over x integrate 9 over x squared d x okay um so this first one will just be uh you know if you write it as 2 times y to the negative second it'll be uh 2 y to the negative first over negative 1 because that's the new exponent so really it's going to be 2 over y to the first but with the negative so it'll be negative 2 of 2 of the 1 over y this one here will be turn into a plus um that'll be one over y to the sixth but it'll be a six on top i'm messed up hold on i want to make sure i'd so if we write this as y to the negative seventh it'll become why to the negative six over negative six yeah so it'll be one over seven 6 y to the 6th, 1 over 6 y to the 6th, okay, equals the antiderivative of 1 over x is natural log of x minus, and then it'll be negative 9 over x, so make that a plus 9 over x.
04:06
And then there's this c value we have here.
04:09
Okay, so now it's like, oh my goodness, what the heck do i do? with this.
04:14
It looked like it wanted all the cs and stuff on the same side...