Solve the given differential equation. y^'+2 x^-1 y=6 y^2...

Key Concepts

Integrating Factor Method

Once the differential equation has been transformed into a linear form, the integrating factor method is used to solve it. This technique entails multiplying the entire differential equation by an integrating factor—typically an exponential function of the integral of the coefficient function—to transform the left-hand side into the derivative of a product, which can then be integrated directly.

First Order Linear Differential Equations

First order linear differential equations are those that can be written in the standard form y' + p(x)y = q(x). These equations are important because they can be solved systematically using the integrating factor method, a process that reduces the problem to finding an antiderivative after transforming the equation into an exact derivative.

Bernoulli Differential Equation

A Bernoulli differential equation is a first-order differential equation of the form y' + p(x)y = q(x)y^n, where n is any real number. Although nonlinear for n ? 0 or 1, these equations can be transformed into linear equations through an appropriate substitution, which simplifies the process of finding a solution.

Substitution Method for Nonlinear Equations

This method involves introducing a new variable to replace a function of the original variable, thereby reducing a nonlinear differential equation into a linear one. In the context of Bernoulli equations, the substitution u = y^(1-n) converts the equation into a linear form, making it more tractable for further solution techniques.

Transcript

00:01 Hello, so today we're given a problem and we're asked to solve it.

00:04 So looking at it, we see that we have a y prime, so a d .y over dx, and that should make us think that perhaps we're going to use separation of variables because we have an x component and a y component.

00:17 Okay, so let's start working away at this problem.

00:21 So first, what i just like to do right off the bat is take that x to the negative one and just put it in the denominator.

00:32 That just makes it a little nicer to work with in this scenario.

00:40 So let's just remind ourselves that y equals v of x times x, and that v equals y over x.

00:52 So let's put this all in the expression of y prime, which involves us subtracting that expression over to the right -hand side.

01:02 So they get y prime or dy over dx equals 6y squared x to the fourth minus two times y over x.

01:20 And then we can plug in our v's and our y.

01:26 We can use our expressions there to plug in.

01:31 So d over dx plug in for y there, we get v of x times x.

01:39 Equals 6 times vx squared or v squared x squared x squared times x to the fourth minus 2b now we've done this a few times so i'm just going to give the left side derivative just straight away so we get and we'll just erase that little accidental mark so we get x times dv over d f x x plus v and that equals 6 v squared x to the 6 because x squared times x of the 4th is 4 plus 2 which is 6 and then minus 2v now we can subtract the v from both sides and making a break line this gives us x v v over d x x to the 6 minus 3b and then we can take that x from the left hand side and well we'll just divide the entire expression by one over x to you transfer the x from the left hand side to the right -hand side expression and then that gives us dv over dx equals 6b squared x to the 5th minus 3v over x so you might be looking at this and you're thinking, well, how are we supposed to separate the variables? i don't really see how that can be done.

03:30 And, well, it's going to be really tricky unless you recognize that this is actually a bernoulli's equation.

03:45 And if you want to like learn more about bernoulli's equation, they use that a lot in fluid dynamics and then also thermodynamics.

04:03 But for our just strict mathematical purposes, i'm just going to show you how you can solve this type of equation.

04:12 So first, let's put it into our generalized format, and our generalized form is going to look something like y prime plus some function p times y equals some function q times y to some power.

04:36 Let's see if our expression looks like that.

04:41 So we have dv over dx, which is our v prime.

04:47 And then if we add this 3v over x to the left, this gives us plus 3 over x, v, v, equals 6, x to the 5th, v squared.

05:13 So if you look at it, we have our, we have our v component, or our derivative, or v prime, we have our v and our v to some power n.

05:31 And then if we isolate our function px and our function qx and our function qx and our n, we can pull out 3 over x, which is our p of s, x our q of x and then our n well what was the purpose of doing that well the purpose of doing that well first let's write these out so p of x equals 3 over x q of x equals 6x to the 5th and n equals 2 and then this might get a little confusing but we then have a little substitution rule, v equals y to the 1 minus n.

06:51 And so in our case, our original v and our problem set is the y in the solution problem set, and that's being set to a new v.

07:05 So i'll just put the slash just for future reference that it's different from our original v.

07:13 Anyways, solving this, we want our v or the quote unquote y to be substituted in for and to do that.

07:30 Our v is then going to be equal to 1 over 1 minus n, which allows us to create this substitution formula, which is really a part that we're interested in 1 over 1 minus n b prime plus p of x v equals our q and then when we substitute in we get 1 over 1 minus 2 which is just negative 1 so we get negative b prime plus 3b over over x equals 6x to the fifth.

08:31 And the interesting part of this is that once we're down to this form, we can take this expression right here, the 3bx, in particular the 3x, the p of x part, and we have something that we're going to do.

08:56 We're going to define a new function, mu of x, equals e to, the integral of p of x and i should note that to get this formula correct we need to multiply the whole thing by negative one so that we don't have that negative one attached to the v prime term so we have v prime now minus three x three over x equals negative six x so mu of x equals e to the integration of our p of x d x.

09:40 So when we plug on our p of x, our 3 over x in this case, we get e to the integral of negative 3 divided by x dx...

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