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\def\So{\Sigma_{n = 0}^\infty}
\def\S{\Sigma_{n = 1}^\infty}
\def\St{\Sigma_{n = 2}^\infty}

\def\so{\Sigma_{n = 0}^k}
\def\s{\Sigma_{n = 1}^k}

\def\hr{\vskip 5pt \hrule \vskip -0pt}
\def\hb{\hfil \break}

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\def\u{\vskip -5pt}

{\bf 5.5 Series Solutions Near a Regular Singular Point, Part I }

Theorem 5.3.1:  If $p(x)$ and$q(x)$ are analytic at $x_0$  (i.e., $x_0$ is an ordinary point of the ODE \hb
$y'' + p(x)y' + q(x) y = 0$), then the 
general solution to this ODE is 
\vskip -6pt
\centerline{$y  = \So a_n(x - x_0)^n = a_0\phi_0(x)  + a_1\phi_1(x)$}
\vskip -7pt
where $\phi_i$ are power series solutions that are analytic at $x_0$.  The solutions $\phi_0, \phi_1$ 
form a fundamental set of solutions.  The radius of convergence for each of these series solutions is 
at least as large as the minimum radii of convergence of the series for ${Q \over P}$ and ${R \over 
P}$. 

If you prefer a power series expansion about 0, use $u$-substitution:  let $u = x - x_0$.  Then $p(u + x_0)$ and $q(u + x_0)$  are analytic at $0$




(Semi-failed) attempt to transform 5.5 problem into 5.4 problem:

5.5:  $y'' + p(x) y' + q(x)y = 0$

 $x^2y'' + x^2p(x) y' + x^2q(x)y = 0$

 $x^2y'' + x[xp(x)] y' + [x^2q(x)]y = 0$ where $xp(x)$ and $x^2q(x)$ are functions of $x$.

5.4:   $x^2 y'' + \alpha x y' + \beta y = 0$ where $\alpha, \beta$ are constants.


Combine 5.3/5.4 methods.


Defn:  $x_0$ is a regular singular value if $x_0$ is a singular value and $xp(x)$ and $x^2q(x)$ 
are analytic at $x_0$.  A singular value which is not regular is called irregular.


Examples:

$y'' + {y' \over x} + {y \over x^2} = 0$,   regular singular value:  $x = 0$.

$y'' + {y' \over x^2} + {y \over x} = 0$,   irregular singular value:  $x = 0$.

$y'' + {y'} + {y \over x^3} = 0$,  irregular singular value:  $x = 0$.


If $p(x)$ and $q(x)$ are rational functions, then $xp(x)$ and $x^2q(x)$ are analytic iff $lim_{x \rightarrow 0} xp(x)$ and iff $lim_{x \rightarrow 0} x^2q(x)$ are finite.  (i.e., after reducing fractions, $x$ is not in the denominator.

Ex:   $p(x) = {1 \over x}$ implies $xp(x) = {x \over x} = 1$

Ex:   $p(x) = {1 \over x^2}$ implies $xp(x) = {x \over x^2} = {1 \over x}$


If $x_0 = 0$ is a regular singular value of the linear homogeneous DE,  $x^2y'' + x[xp(x)] y' + x^2 q(x)y = 0$ (*)

Suppose $y = x^r  \So a_nx^n = \So a_n x^{n + r}$  is a solution to (*).

$ y' =   \S (n+r)a_n x^{n + r-1}$   and $ y'' =  = \St (n+r)(n+r - 1) a_n x^{n + r-2}$   

 $ x^2\St (n+r)(n+r - 1) a_n x^{n + r-2} + x[xp(x)]  \S (n+r)a_n x^{n + r-1} + [x^2q(x)]y \So a_n x^{n + r}$   = 0$ 

$ \St (n+r)(n+r - 1) a_n x^{n + r} + [xp(x)]  \S (n+r)a_n x^{n + r} + [x^2q(x)]y \So a_n x^{n + r}$   = 0$ 


$ \St (n+r)(n+r - 1) a_n x^{n + r} +   xp(x)(1 + r)a_1x^{1 + r} +  [xp(x)]  \St (n+r)a_n x^{n + r} + 
[x^2q(x)]a_0x^r   + [x^2q(x)]a_1x^{r+1} +   [x^2q(x)] \St a_n x^{n + r}   = 0$ 


$     [x^2q(x)]a_0x^r   +(xp(x)(1 + r)a_1 +  [x^2q(x)]a_1)x^{r+1} +   
\St [(n+r)(n+r - 1) a_n x^{n + r} + [xp(x)]  (n+r)a_n x^{n + r} + 
[x^2q(x)]  a_n x^{n + r}   = 0$ 




\end


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