Find the general solution of the first-order linear differential equation. 3 x y^'-y=x^-1

Name: Problem 7, Chapter 9: Calculus for AP
Uploaded: 2020-03-19T06:59:24-08:00
Duration: 5 min 54 s
Channel: Christine Anacker
Description: Find the general solution of the first-order linear differential equation. 3 x y^'-y=x^-1

Find the general solution of the first-order linear...

Key Concepts

Integrating Factor Method

This method involves multiplying both sides of a standard first-order linear differential equation by an integrating factor, which is derived from the function P(x). The integrating factor turns the left-hand side of the equation into the derivative of a product, allowing the equation to be integrated easily to obtain the general solution.

First-Order Linear Differential Equations

These are differential equations that involve the first derivative of an unknown function and can be expressed in the general form y' + P(x)y = Q(x). Their linearity allows us to use standardized techniques that guarantee the existence and uniqueness of solutions under appropriate conditions.

Standard Form Transformation

Transforming a differential equation into its standard form (y' + P(x)y = Q(x)) is an essential step. This process involves algebraic manipulation to isolate the derivative and express the equation so that the coefficient of y' is one, making it straightforward to apply methods like the integrating factor method.

Transcript

00:01 All right, so we're trying to find a general solution for a first order derivative.

00:04 To do that, we're using these three basic equations.

00:07 We have the one we need to get it into, then our general solution, and then how we actually find that other function.

00:15 So let's apply it to a question.

00:18 So we have 3x times y prime minus y equals x to the negative 1.

00:23 So right away we can see that it's not in the format that we actually need.

00:30 It to be in.

00:31 We need y prime by itself in order to figure out what the functions of a and b are.

00:37 So the very first step we're going to do is we're going to divide everything by a 3x.

00:44 And that would leave us with y prime minus y over 3x is equal to, let's write this as 1 over 3x squared.

00:56 So doing that, that gives us our a function.

01:00 That'd be negative 1 over 3x.

01:02 X that gives us our b function which would be one over three x squared and that helps us to find our other a function our little a function because we're going to figure out what the antiderivative of our a x function is so applying that we're looking at the antiderivative of negative one -third x and we know that any time we have an x on the bottom that guy works out to the natural log.

01:38 So i'm going to try to, i'm going to do this another way.

01:41 We have the natural log of x.

01:43 We're going to split it up the one -third in the x, and that would leave us with a negative one -third out front.

01:50 Or we could write that as e -natural log of x to the power of negative one -third.

01:59 And then we know that e -to -the -power of a natural log, those cancel out.

02:03 So we're left with x to the power of negative one third as our little a function.

02:14 So now that we have that, we're going to throw it into our general solution equation.

02:20 So if you forgot what that was, our general solution equation is the inverse of the little a function, and then we're doing the integral of the little a function times the b function plus a constant.

02:37 So filling that out, we would have x the power of negative one -third times negative one.

02:47 Integraled, x the power of negative one -third times, we'll keep this written.

02:56 Well, no, we'll change it a little bit.

02:58 We'll write it as x to the negative 2 over 3, dx plus c.

03:07 So negative 1 3rd times negative 1 leaves us with a 1 third.

03:14 Now we're going to put these xs together...

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