91Ó°ÊÓ

Recall that a point x=a is an ordinary point if the coefficients P(x) and Q(x) of the differential equation of the form \(y'' + P(x)y' + Q(x)y = 0\) are analytic at x=a. If either P(x) or Q(x) have a singularity but still satisfy the condition that (x-a)P(x) and \((x-a)^2Q(x)\) are analytic functions, then x=a is considered a regular singular point. In all other cases it is an irregular singular point. For the given differential equation, we can write it in the form: $$y''+\frac{1}{x}y'+\left(x-\frac{1}{4x}\right)y=0.$$ The coefficients are: $$P(x) = \frac{1}{x}$$ and $$Q(x) = x- \frac{1}{4x}$$ To check the behavior around x = 0, we need to see whether these coefficients are analytic (i.e., have convergent power series expansions) at x = 0. Clearly, P(x) is not analytic at x = 0. However, we can check if the conditions for regular singular point are met: $$(x-0)P(x)=x\frac{1}{x}=1,\ \ (x-0)^2 Q(x) = x^2\left(x-\frac{1}{4x}\right)= x^3-\frac{1}{4}$$ Both of these expressions are analytic at x=0, therefore x=0 is a regular singular point.

Since x=0 is regular singular point, we will use the method of Frobenius to find linearly independent solutions. In this method, we assume the solution has the form: $$y(x) = x^r \sum_{n=0}^\infty a_n x^n$$ Plug this assumption into the differential equation and equate terms with the same power of x: $$(x^2 \sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2})$$ $$+(x \sum_{n=0}^\infty (n+r)a_n x^{n+r-1})$$ $$+(\left(x^{2}-\frac{1}{4}\right) \sum_{n=0}^\infty a_n x^{n+r})$$ $$= 0$$ This results in the following indicial equation for r: $$(r-1)r = 0$$ This gives us two values for r: $$r_1=0, r_2=1$$ Now we can construct two linearly independent solutions, each corresponding to a different value of r: $$y_1(x) = \sum_{n=0}^\infty (a_n x^n)$$ $$y_2(x) = x\sum_{n=0}^\infty (a_n x^n)$$ Substitute these two solutions back into the differential equation, and two recurrence relations can be established to find the coefficients. Solving the recurrence relations and finding the first few coefficients will give us: $$y_1(x)=1-\frac{1}{8}x^2+\frac{1}{384}x^4-\dots$$ $$y_2(x)=x-\frac{1}{8}x^3+\frac{13}{384}x^5-\dots$$ These two series solutions are linearly independent and form a fundamental set of solutions to the given differential equation.

The radius of convergence for the Frobenius series solutions is determined by the nearest singularity to the regular singular point x=0 in the P(x) and Q(x) coefficients. Since P(x) and Q(x) do not have singularities except at x=0, the Frobenius series solutions converge for all values of x. Thus, the maximum interval of validity for the obtained solutions is: $$(-\infty, \infty)$$