00:01
Okay, we wanna solve this differential equation given this initial condition, okay? and i'm gonna rearrange it so it looks like that.
00:25
And i'm gonna notice that this is a homogenous equation with the sense that every term has the same power.
00:34
And i'm gonna make a substitution, x is equal to b times y.
00:41
So dx is ydv plus vdy.
00:50
I wanna point out that if i did it the other way, y is equal to v times x, at our initial point, x equals zero and y is equal to one, we wouldn't be able to solve for v.
01:05
But going this way, we can.
01:08
So it does make sense.
01:12
So let's substitute in.
01:14
So we get v squared y squared plus y squared times dy is equal to two v y squared dv plus v.
01:40
Then divide through by y squared, we get x squared plus one times dy.
01:47
Then i'm gonna subtract this term minus two v.
01:54
I'm dividing by y squared, ydx plus vdy.
02:02
And that equals zero, okay? then collect my terms.
02:08
So i got a minus two v squared dy over here.
02:21
Oh, this is a v, not an x...