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

Name: Problem 35, Chapter 9: Thomas Calculus
Uploaded: 2021-06-17T20:37:05-08:00
Duration: 6 min 23 s
Channel: Regina Hays
Description: This last equation is linear in the (unknown) dependent variable u. Solve the differential equations. x y^'+y=y^-2

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

Key Concepts

First Order Linear Differential Equations

A first order linear differential equation is structured in the form u' + P(x)u = Q(x), where u is the dependent variable. This type of equation is central in differential equations as it has techniques—such as the integrating factor method—for finding its general solution systematically.

Integrating Factor

The integrating factor is a crucial function in solving linear differential equations. It is defined as ?(x) = exp(?P(x)dx) and is used to multiply through the differential equation, converting it into an exact derivative. This process greatly simplifies the equation, making it easier to integrate and solve for the unknown function.

Substitution Method in Differential Equations

Sometimes a differential equation that appears nonlinear can be transformed into a linear one through a clever substitution of variables. This method involves defining a new variable to express the equation in a standard linear form, thereby permitting the application of linear solution techniques such as the integrating factor method.

Transcript

00:01 Okay, what we want to do is we want to step through the process to be able to work with what is termed a bernoulli differential equation of the form d .y over dx or y prime plus some function times y, function of x times y, is equal to another function of x times y to the n.

00:21 And if n is not equal to 0 ,1, then we are going to have to do some kind of a substitution.

00:30 To transform this bernoulli differential equation into a linear differential equation, which then we know how to solve.

00:38 And so we notice that for this differential equation of x times y prime plus y equal to y to the negative 2, we don't, n is not 0 .1.

00:52 N is actually negative 2.

00:53 So we're going to have to incorporate this substitution.

00:56 So we're going to let you, we are going to let you equal y raise to the 1 minus a negative 2, say plus a 2, so this is equal to y cubed.

01:14 So therefore u to the 1 3rd is equal to y.

01:17 We're just going to keep that noted because we're going to have to do a bunch of substitutions.

01:22 Then d, u over dx is equal to 3y, squared, d -y, d -x, right? because we're doing implicit differentiation.

01:35 And so d -y over d -x is equal to 1 over 3y squared times d -u over d -x.

01:47 But y is u to the 1 3rd.

01:50 So this actually becomes 1 over 3 u to the 2 thirds times d -u over d -d -s.

02:02 So i'm going to substitute that in.

02:07 So what i'm going to do is, and another thing, i'm actually divide everything by x first.

02:12 So i have dy over dx plus 1 over x times y is equal to 1 over x times y to the negative 2.

02:26 Okay, so dy over dx is 1 over 3 u to the 2 3 3 2 3 to the negative 3.

02:33 Times du over dx plus 1 over x times u to the one third because that's what y is is equal to 1 over x times u to the negative 2 thirds.

02:55 Okay and we need du over dx by itself so we're going to multiply everything by 3 u to the two thirds so we have du dx plus 3 over x times u is equal to 3 over x...

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