This last equation is linear in the (unknown) dependent variable u. Solve the differential equat

Name: Problem 34, Chapter 9: Thomas Calculus
Uploaded: 2021-06-17T20:18:03-08:00
Duration: 7 min 35 s
Channel: Regina Hays
Description: This last equation is linear in the (unknown) dependent variable u. Solve the differential equations. y^'-y=x y^2

This last equation is linear in the (unknown) dependent...

Key Concepts

Integrating Factor Method

The integrating factor method is a technique used to solve linear first-order differential equations of the form u' + P(x)u = Q(x). By multiplying the entire equation by a carefully chosen function called the integrating factor, the left-hand side becomes the derivative of a product, allowing the equation to be integrated directly to find the solution.

Linear First-Order Differential Equations

These are equations that involve the first derivative of an unknown function and can be written in the standard form u' + P(x)u = Q(x). Such equations are significant because they exhibit a structure that allows for systematic solution methods, such as the integrating factor technique, which turns the ODE into an exact differential equation.

Substitution Method in Differential Equations

The substitution method involves replacing the original dependent variable with a new variable to simplify the structure of a differential equation. In the context of Bernoulli equations, substituting u = y^(1-n) transforms the original nonlinear ODE into a linear ODE in terms of u, which is generally easier to solve.

Bernoulli Differential Equation

A Bernoulli differential equation is a special form of a nonlinear first-order ODE expressible as y' + P(x)y = Q(x)y^n, where n ? 0 or 1. The significance of this form lies in its transformability into a linear differential equation by an appropriate substitution, typically u = y^(1-n), simplifying the process of obtaining a solution.

Transcript

00:01 So what we're going to do is kind of walk through the process of being able to actually solve a differential equation of the form, what, whoops, y prime, or d y over d x minus y is equal to x times y squared.

00:19 And this is kind of in what is called a renewly differential equation where we have d y or dx plus some function of x times y.

00:30 Is equal to another function of x times y to the n.

00:33 And for the highest exponent in, or for this exponent in, for n not equal to zero or one, then we're going to have to do some kind of substitution to transform this differential equation into a linear differential equation.

00:50 And so that's what we're going to kind of work with.

00:53 And we always do this type of substitution, where we let you equal y to the one, minus n.

01:01 And so that's what we're going to do.

01:03 So i'm going to let you equal y to the one minus two, which is y to the negative one.

01:12 So that implies that y is equal to you to the negative one.

01:18 Then i'm going to take the derivative.

01:20 So du over d x is equal to negative one, negative y to the negative 2 dy over dx because we're doing implicit differentiation which means now we're going to solve for dy over dx because that's what y prime is so we're trying to do some substitutions in here so that means that d y over dx is going to be actually equal to negative y squared du over dx and we also know that y is equal to u to the negative two so we have negative u to the negative two d u over d x and so now i'm going to use these as substitutions so what i'm going to do is y prime or d y or d x is equal to negative u negative raise to the negative two d u over d x minus u to the negative one because that's what y is, is equal to x times u to the negative two.

02:37 And there we have it.

02:39 And now i recognize, i don't want this negative u raised to the negative 2 in front of to u over dx.

02:47 So i'm going to divide everything by negative u raised to the negative 2.

02:51 And so this becomes du over dx plus u to the first.

02:58 Is equal to a negative x.

03:03 And so now i basically have it as a linear form, right? and so what i'm going to do is rewrite this side as kind of a derivative.

03:19 So, d, d, x, what do i have to take the derivative of in order to get, nope, that's not what i want to do right now.

03:33 We're good.

03:34 Okay, sorry about that.

03:37 And so now what we want to do is to go ahead and come up with a function that we're going to multiply everything by.

03:53 And typically we do v of x is equal to e raised to the integral of x.

04:00 And remember p of x is whatever is in front of you or y.

04:06 In this case, it's just a 1, the x.

04:10 So that is equal to e to the x.

04:17 And so that is what i'm going to multiply to everything inside this equation.

04:29 Okay...

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