00:01
All right, the substitution you make here is u equals y to the 1 minus whatever power this is right here.
00:10
So 1 minus 2, so y to the minus 1.
00:16
So d, u, dx will be minus 1, y to the minus 2, d, d, y to the minus 2, d, y, d, d x.
00:27
So first i'm going to divide by x.
00:30
So i get d, y, dx, minus 1 plus x over x.
00:36
Y equals y squared.
00:41
All right now i'm going to multiply through by negative one over y squared.
00:51
So negative one over y squared, dy, dx minus one plus x over x over x negative.
01:02
Oh, that'll make plus there.
01:03
1 over y equals negative 1.
01:08
All right, so in place of negative 1 over y squared dy, dx, i'm going to put dudx plus 1 plus x over x.
01:22
In place of 1 over y, i'm going to put u equals minus 1.
01:28
Okay, so now it's linear in u.
01:30
We'll solve for you, and then we'll find out what y is.
01:34
Okay, so first of all, i've got to find the integrating factor.
01:38
So it'll be e to the integral, 1 plus x over x.
01:42
So that's e to the integral 1 over x plus 1.
01:48
So that's e to the lnx plus x.
01:53
So that's e to the l and x times e to the x.
01:57
So x, e to the x.
02:02
So i get x, e to the x, d -u -d -x, plus 1 plus x, e to the x, equals negative x, e to the x.
02:16
So this slide should turn into this derivative...
