00:01
Can we want to solve this differential equation? i don't know if students can see the ai hints, but i'll tell you that for this one, they're way off.
00:09
Okay, so i'm going to make a substitution.
00:13
Well, first of all, i'm going to multiply through by dy.
00:17
So i get, okay, and then what i'm going to do is i'm going to make a substitution.
00:34
Y is one half times the natural log of u.
00:39
That'll get rid of that exponential.
00:43
The dy there.
00:45
Then dy is one half of du over du over u.
00:57
I'm going to substitute that back in.
01:06
Then e to the two y is just u du over u.
01:28
Then i'm going to multiply through by u and i'm going to get this denominator.
01:42
So i get like that.
01:57
Okay, now i'm going to make another substitution.
02:06
V equals x cubed.
02:09
So dv is three x squared dx.
02:19
If i write this in terms of du dx.
02:22
So let's write it this way.
02:24
So we get minus.
02:29
So you get du dx.
02:32
So we divide by the dx and then rearrange otherwise.
02:37
Minus two over x cubed times x squared plus u times u.
02:50
Okay, so this can be written as minus two over x times u plus or minus two over x squared times u squared or x cubed...