Separation of Variables: How to integrate (x+2y)y'=1 y(0)=2?

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Homework Help Overview

The discussion revolves around the differential equation (x+2y)y'=1 with the initial condition y(0)=2. Participants explore the method of separation of variables and its application to this equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use a substitution method to facilitate separation of variables but struggles to isolate the variables. They question whether the presence of both x and y in the resulting expression disqualifies it from being a separation of variables solution.
  • Another participant suggests that while the equation does not separate in y and x, it can be separated in terms of the substituted variable u and x, which may still satisfy the separation of variables criteria.
  • A different approach using linear differential equations is presented, leading to an alternative solution, prompting further discussion on the validity of different methods.

Discussion Status

The discussion is active with participants exploring various methods to approach the problem. Some guidance has been offered regarding the validity of the substitution method, and an alternative solution has been presented, indicating a productive exchange of ideas.

Contextual Notes

Participants are operating under the constraints of homework rules, which may limit the extent of assistance provided. There is an ongoing examination of the definitions and implications of separation of variables in the context of the problem.

jenettezone
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Homework Statement


Use separation of variables to solve (x+2y)y'=1 y(0)=2

Homework Equations


u=2y+x >>I did not know how to start this, so i looked at the back of the book and it said to use that substitution
y=(u-x)/2, du=2dy+dx, dy=(du-dx)/2

The Attempt at a Solution


so i got the following:

dy/dx=1/(x+2y)
(du-dx)/(2dx)=1/(x+2(u+x)/2)
(du-dx)/(2dx)=1/u

I could not separate the variables from here. Also, according to the back of the book, the answer is supposed to be 2y-2ln|2+x+2y|+4+2ln2=0. But the term -2ln|2+x+2y| has both x and y variables, so aren't the variables not separated? That still qualifies as a solution by Separation of Variables?
 
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dx/dx=1, so you last equation is du/dx-1/2=1/u. Can you separate u and x in that? And, no, the equation doesn't separate in y and x, but it does in u and x and I think that counts as a 'separation of variables' solution after the substitution.
 
I finally solved it, thank you! :smile: :smile: :smile:
 
I got a different answer using linear differential equations.

dy/dx = 1/(x + 2y)
dx/dy = x + 2y

dx/dy - x = 2y

The answer i got was: x = -2y -2 + 6e^(y-2)

Differentiating it again returns me to the original differential
 

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