General Solution and Transient Terms

[Geofram]

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 e^2lnx 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?

 

[dextercioby]

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).

 

[Geofram]

One moment, have to work this out on paper!

 

[Geofram]

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.

 

[holomorphic]

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.