Validity of solution to differential equation

[center o bass]

Homework Statement

I solved the differential eqution

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

and I am to discuss the validity of this solution

Homework Equations

Solution

$$y(x) = \frac{c_2\ln|x|}{1-x} + \frac{c_1}{1-x}$$

The Attempt at a Solution

The method i used for solving the equation was to first find one solution
with a powerseries method. This produced the 'geometric series' solution
which only converges |x|<1. The second solution was found by variation of the constant.
And here I got ln|x| in the picture which is only valid for x diffrent from zero.

Would I then be right to conclude that this solution is only valid on the interval 0<x<1 and 0>x>-1?

 

[Gib Z]

Since 0 and 1 are the regular singular points of the differential equation, I would be inclined to think the solution you found only works in $(0,1)$. I would quickly go over the working to make sure there wasn't a detail you glossed over that incorrectly allowed the negative range, and if you can't find any errors, for $-1 < x < 0$ check by differentiating that $$y(x) = \frac{ c_2 \log(-x) + c_1 }{1-x}$$ does indeed satisfy the equation.

 

[center o bass]

Hmm I did check the solution and it did not work for $$\ln |x|$$. I can't really find anything that would kill the solutions in the negative region, but then again I'm not really sure what that would be. What kind of operation would eliminate that region?

EDIT: I rechecked the solution, but for $$\ln(-x)$$ and that did work out.

 

[center o bass]

By the way.. Does this imply that the two regions infact have diffrent solutions in terms of the constants c1 and c2?

 

[HallsofIvy]

What conditions are you given? You should be able to find a solution that is valid for x< 0, another solution that is valid for 0< x< 1, and a third that is valid for x> 1.

 

[center o bass]

No conditions were given and I did check that my solution worked out for ln(-x) as well as ln(x). Doesn't the solutions have to be divided into yet another region according to the radius of convergence of the powerseries for 1/(1-x)

The problem stated that _one_ solution were to be found by frobenius method which should come out to be 1/(1-x), and that the other should be found by variation of the constant.

I then had to discuss the validity of _this_ solution, so I guess I don't have to find solutions in regions where this is not a valid solution.