Few questions about series solution of ODEs

[gikiian]

Consider the ODE x(x-1)y''-xy'+y=0.

I need help in identifying the method of solution (power series or frobenius) for this ODE.

Using the formulae \stackrel{limit}{_{x→x_{o}}}\frac{q(x)+r(x)}{p(x)} and \stackrel{limit}{_{x→x_{o}}}\frac{(x-x_{o})q(x)+(x-x_{o})^{2}r(x)}{p(x)} , where p(x)=x(x-1), q(x)=-x, and r(x)=1, I have worked out the following:

(Eq. 1) \stackrel{limit}{_{x→0}}\frac{-x+1}{x(x-1)}=\stackrel{limit}{_{x→0}}\frac{-1}{x}=\frac{-1}{0}=∞ i.e. diverges

(Eq. 2) \stackrel{limit}{_{x→0}}\frac{(x-0)(-x)+(x-0)^{2}(1)}{x(x-1)}=\stackrel{limit}{_{x→0}}\frac{-x^{2}+x^{2}}{x(x-1)}=0 i.e. converges

Hence x_{o}=0 is a regular singular point.

(Eq. 3) \stackrel{limit}{_{x→1}}\frac{-x+1}{x(x-1)}=\stackrel{limit}{_{x→1}}\frac{-1}{x}=\frac{-1}{1}=-1 i.e. converges

Hence x_{o}=1 is a singular point.

My question is: How do I use this data to find out which solution method to use?

I am guessing that since the problem equation has a regular singular point besides having a singular point, we will drop the power series method and use the Frobenius method of solution. Am I right?

Another question: Which regular singular point do we choose if there are more than one regular singular points for an ODE?

 

[JJacquelin]

Hi !
An obvious solution of the ODE is y=x.
So, in order to find a second solution, let y=x*f(x) where f(x) is an unknown function. Bring it back into the ODE
This leads to a first order ODE easy to solve, which gives f '(x). The integration leads to f(x)=ln(x)+1/x, then y(x)= 1+x*ln(x)
The general solution of the ODE is y=a*x+b*(1+x*ln(x)) where a and b are constant.