Few questions about series solution of ODEs

vendredi 29 novembre 2013

Consider the ODE [itex]x(x-1)y''-xy'+y=0[/itex].

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

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

Quote:

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

(Eq. 2) [itex]\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[/itex] i.e. converges

Hence [itex]x_{o}=0[/itex] is a regular singular point.

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

Hence [itex]x_{o}=1[/itex] 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?