Problem

Solve the differential equations $$(1+x) y^{\prim…

02:38

Problem 5 Easy Difficulty

Solve the differential equations
$$x \frac{d y}{d x}+2 y=1-\frac{1}{x}, \quad x>0$$

Answer

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Calculus 2 / BC

University Calculus: Early Transcendentals

Chapter 16

First-Order Differential Equations

Section 2

First-Order Linear Equations

Discussion

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Watch More Solved Questions in Chapter 16

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Video Transcript

section 16 Tudo five were dealing with first order linear, ordinary differential equations. So I want to get this in standard form. Standard form would be D Y the X. You divide everything by eggs plus two over X. Why is equal to one over X minus one over X squared Now, the integrating factor that I want to multiple by is e to the integral off this expression right here two over x the X that is e to the what to natural log absolute value of X Since excess positive, the absolute values go away and this justice e to the natural log of X squared, which is X squared. So that is my integrating factor. So this equation turns into the derivative with respect to acts of X squared. Why? If you were to use the product rule, you could go from this equation back, so that would be what to acts times D y d x, This is me to acts times why? And, um, And then if you had x squared times, um, do I D x? Okay, So multiply. Then quit our equation by X squared, so this would be equal to X squared over X minus X squared over X squared. And so this is just what is its X minus one? So the equation I'm left with now is the derivative with respect to eggs of X squared. Why is equal to X minus one Now you integrate both sides of the equation X squared. Why is equal to the integral X minus one D x, which is what X squared over two minus X plus a constant of integration. So, therefore, why is equal to 1/2 dividing everything by X squared minus one over x plus C over X squared.

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