00:01
Hey there, in this video we'll solve the bernoulli equation 6y squared dx minus x times the quantity 2xq plus y, d ,y, equals y, d, equals y, equals y, d, y, equals venn.
00:08
Just as a reminder, bernouli equations of the form, plus p of x, y, equals q of x, y to the end.
00:16
Now right away, we can see that the equation above isn't of this form, nor is it going to be just by doing some algebraic simplification.
00:24
The key to this problem is to interchange the roles of x and y, and think about the above, as a differential equation for the unknown function x in terms of the variable y.
00:35
So really the bernoulli equation that i'm going to be thinking about is an equation of the form x prime plus p of y x equals q of y times x to the n.
00:44
Now for any sort of bernulia equation it's also very important that the value of n that we have is not equal to zero and the value of n is not equal to 1.
00:53
If it's equal to zero then the equation that we're looking at is actually just a linear equation, so we can use the usual means of solving linear equations to figure things out.
01:02
If n equals 1, it's a separable equation, and that's also very easy to work with.
01:06
But in general, if n is not equal to 0 or 1, we solve this kind of thing by doing a very special substitution.
01:12
To start out, let's take our original equation and try to write it in the form that we have in the pink box.
01:18
I start out by dividing everything by 6y squared and d .y, so i get the dx over dy is equal to negative x times the quantity 2x cube plus y over 6y squared.
01:30
I can rewrite this as, actually, it should be minus this equals 0.
01:36
Now i can rewrite this as x prime minus 1 over 6y times x equals 1 over 3y squared times x to the fourth.
01:46
So we can see right away my p value here is negative 1 over 6y and my cube value here is going to be 1 over 3 y squared and my n value is 4 which is not equal to 0 1 so we're in pretty good shape now i'm trying to solve for x as a function of y and the key here on this kind of nearly equations is to do a substitution in the typical situation we're going to substitute z equals 1 minus our y to the 1 minus n but in the situation where x is our unknown function we're going to substitute z equals x x x to the 1 minus n because x isn't playing the role that y usually plays for us.
02:29
So we're going to do that same sort of substitution here, and we know specifically that n is equal to 4, so i'm substituting z equals x to the minus 3.
02:39
And what this will do is it will take up a newly equation and make it into a linear equation, which we know how to solve.
02:45
Note the substitution also implies that z prime is equal to negative 3 times x to the minus 4 times x prime just from the chain rule.
02:54
Or in other words, the x prime is negative one -third times x -to -the -fourth times z -prime.
03:00
So we're going to start by substituting n for x -prime, and we get the negative one -third, x -to -the -fourth z -prime, minus 1 over 6 y times x, is equal to 1 over 3 y -square times x to the fourth.
03:13
Now this is currently a mess because i've got x, y, and z all still involved, but we're going to take care of that right now.
03:19
The first thing i can see is i can divide out by a common factor of x.
03:24
And i get the negative one -third times x -cubed z prime minus 1 over 6 y is equal to 1 -3 -y squared times x -cubed...