General Solution and Transient Terms

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Homework Help Overview

The discussion revolves around finding the general solution of a differential equation of the form x(dy/dx) + 2y = 3. Participants are tasked with determining the largest interval for the solution and identifying any transient terms present.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the equation by dividing and applying an integrating factor, leading to a constant solution. Some participants question the validity of the approach and suggest separating variables for integration instead.

Discussion Status

Participants are actively exploring different methods to solve the differential equation, with some providing alternative setups for integration. There is no explicit consensus on the correct approach yet, but guidance on separating variables has been offered.

Contextual Notes

Participants are considering the implications of the solution regarding the interval of definition and the nature of transient terms, with some uncertainty about how to express the solution in a useful form.

Geofram
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Homework Statement


Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.


Homework Equations



x(dy/dx) + 2y = 3

The Attempt at a Solution



Divide by x:
dy/dx + (2/x)y = 3/x

The integrating factor would be e2lnx or x2?

Multiplying by IF:

x2(dy/dx) + 2xy = 3x

d(x2y)/dx = 3x

Integrating both sides:

x2y = (3/2)x2

Dividing by x2:

y = 3/2

I've never had a solution like this, which leads me to believe I've done something wrong. If not, would the interval still be x>0? And is 3/2 a transient term since every solution would point to it?
 
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Do you agree that

[tex]\frac{dy}{dx} = \frac{3-2y}{x}[/tex] ?

Then separate the variables and integrate correctly (don't forget about the integration constant).
 
One moment, have to work this out on paper!
 
Alright, using the separation method I got:

x/dx = (3 - 2y )/dy

Integrate both sides:

x2/2 + C = 3y - y2

Now I'm just not sure how to get it in a form that would allow me to figure out the interval and transient terms.
 
Geofram said:
Alright, using the separation method I got:

x/dx = (3 - 2y )/dy

Integrate both sides:

x2/2 + C = 3y - y2

Now I'm just not sure how to get it in a form that would allow me to figure out the interval and transient terms.

This is a little topsy-turvy. You should set this up as
[tex]\frac{dy}{3-2y} = \frac{dx}{x}[/tex]
and then integrate.
 

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