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QUESTION

Obtain at least one solution of the form $$
y=x^{\sigma}\sum_{n=0}^{\infty}a_nx^n $$ for each of the following
differential equations. Where possible, obtain a second
independent solution of the same form, or comment on why it is not
possible to do so.

$2x\left(1-x\right)y''+\left(1-x\right)y'+3y=0$

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ANSWER

$2x\left(1-x\right)y''+\left(1-x\right)y'+3y=0\\
 y''+\frac{1}{2x}y'+\frac{3}{2x\left(1-x\right)}y=0$ so 0 is a regular
 singular point. We group terms together that will give the same
 powers of $x$:

 $2xy''+y'-2x^2y''-xy'+3y=0$

 $\ds \sum_{n=0}^\infty a_n\left\{\left[2\left(\sigma +n\right)\left(\sigma+n-1\right)\ds +\left(\sigma
 +n\right)\right]x^{\sigma
 +n-1}\right.$

 $\left. +\left[-2\left(\sigma+n\right)\left(\sigma+n-1\right)
 -\left(\sigma+n\right)+3\right]x^{\sigma+n}\right\}=0$

 Factorising  gives

 $\ds \sum_{n=0}^\infty a_n\{\left(\sigma +n\right)\left(2\sigma+2
 n-1\right)x^{\sigma+n-1}$

 $-\left[\left(2\sigma +2n-3\right)\left(\sigma
 +n+1\right)x^{\sigma+n}\right]=0$

 Reordering by powers of $x$ gives

$a_0\sigma\left(2\sigma-1\right)x^{\sigma-1}$

$+\ds \sum_{n=0}^\infty x^{\sigma+n}\left\{a_{n+1}\left(\sigma +n
+1\right)\left(2\sigma +2n +1\right)\right.$

$\left.-\left(2\sigma
+2n-3\right)\left(\sigma+n+1\right)a_n\right\}=0$

$\sigma\left(2\sigma-1\right)=0 \Rightarrow \sigma =0$ or
$\sigma=\frac{1}{2}\\ a_{n+1}=\frac{2\sigma +2n-3}{2\sigma+2n+1}\\
\sigma=0:\ a_{n+1}=\frac{2n-3}{2n+1}a_n\\
a_n=\frac{2n-5}{2n-1}a_{n-1}=\frac{2n-5}{2n-1}\frac{2n-7}{2n-3}\frac{2n-9}{2n-5}\ldots\frac{-1}{3}\frac{-3}{1}a_0\\
\Rightarrow a_n=\frac{3}{\left(2n-1\right)\left(2n-3\right)}a_0$
after all cancellations.

$\sigma=\frac{1}{2}:\
a_{n+1}=\frac{2n-2}{2n+2}a_n=\frac{n-1}{n+1}a_n\\ a_1=-a_0,\
a_2=0,\ a_3=0,\ldots$

General solution is $$y=A\sum_{n=0}^\infty
\frac{1}{\left(2n-1\right)\left(2n-3\right)}x^n+B\left(x^\frac{1}{2}-x^\frac{3}{2}\right)$$

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