# second order variable coeff homogeneous ode

• May 6th 2010, 01:31 PM
speedra500
second order variable coeff homogeneous ode
Could someone help me solve the following:

y'' +(1/3x)y' - (1/3x)y = 0

I go through the steps I would normally to solve the DE, but when attempting to find the roots I end up with a function of x which I'm not exactly sure what to do with. It appears that there must be different solutions for different ranges of x. How would I go about solving this?
• May 6th 2010, 01:47 PM
Jester
Quote:

Originally Posted by speedra500
Could someone help me solve the following:

y'' +(1/3x)y' - (1/3x)y = 0

I go through the steps I would normally to solve the DE, but when attempting to find the roots I end up with a function of x which I'm not exactly sure what to do with. It appears that there must be different solutions for different ranges of x. How would I go about solving this?

Are you sure the ODE is $
y'' + \frac{1}{3x} \, y' - \frac{1}{3x} \, y= 0?
$
• May 6th 2010, 02:05 PM
speedra500
yes, confusing... isn't it?
• May 6th 2010, 05:04 PM
speedra500
just so everyone knows, I'm pretty sure this is solved using the Frobenius method
• May 7th 2010, 08:26 AM
AllanCuz
Quote:

Originally Posted by speedra500
just so everyone knows, I'm pretty sure this is solved using the Frobenius method

You are correct, we just need to switch our variables around to obtain the required equation for the Frobenius method: Frobenius method - Wikipedia, the free encyclopedia

$
x^2 y'' + \frac{1}{3} xy' - \frac{1}{3} x y= 0
$

Where,

$p(x) = \frac{1}{3}$

$q(x) = \frac{1}{3} x$

Then continue with the Frobenius method. There is an excellent example on that wikipedia page (first example) that will walk you through all the steps (so I won't repeat them here out of lazyness).