00:08
All right, so we're going to solve the differential equation.
00:14
And this one is a pretty simple, separable differential equation, i would say.
00:25
So we notice that we have two variables.
00:28
So i'm going to start off by multiplying dr and l8, or natural log of h to both sides.
00:39
So we're going to have natural log h, d .h equals r.
00:48
Squared 1 plus r squared t r and we're also going to divide this h squared by both sides so we're going to have l and h over h squared d h equals r square d r all right so now we're ending up we ended up with this and since they both have uh d h or each side has d and their variable that means that this here is a derivative of something.
01:29
So we are going to find the net.
01:32
We're going to integrate both sides.
01:37
So i'm just going to write the same thing, but with an integral.
01:46
All right.
01:48
So i'm going to start off with this side, and then next time i'll do this side.
01:53
So for this side, i'm going to actually integrate by parts.
01:56
So we're going to have u, d, u, and then v, d, v.
02:02
So integration by parts is uv equals uv integral of v d u.
02:23
All righty so now that we have that i'm gonna yeah okay so i'm gonna oh no there we go i'm gonna plug in parts of this here so i'm actually gonna rewrite this as integral l n h times 1 over h squared which is the same thing as this up here so for you i'm actually going to put lnh and then so the derivative of that would be 1 over h something very simple for dv since i need u and db which would be ln h and 1 over h and 1 over h square so i'm going to put 1 over squared for dv and then basically i'm going to find the integral of this here which is going to be negative 1 all right so then we're going to use these here to plug in to this equation here so we're going to have lmh times negative 1 over h minus integral of v which is negative 1 h times du, which is going to be, so then we're going to be left off with negative l and h over h minus integral of negative 1 over h.
04:27
I'll just do negative 1 over h square.
04:32
D .h.
04:33
This would also be d .h.
04:37
All righty.
04:39
And that's just because this is d .h as well.
04:42
All righty.
04:48
So now we have this, and now we're going to have to solve this here.
04:51
This one isn't really too hard.
04:56
I'll go over this one since i really didn't explain too much about how i got this here.
05:02
So i will rewrite this as negative h to the negative 2, since 1 over h is equal to h1.
05:16
This just means that it's reciprocal.
05:22
So we have this here.
05:26
And i'm just going to work on this.
05:28
So we have we have h negative 2.
05:31
And to do the integral, we're going to add 1.
05:36
And then put it over negative 2 plus 1.
05:42
So we have 1 over 3.
05:50
Oh, just kidding...