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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.

$9x\left(1-x\right)y''-12y'+4y=0$

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ANSWER

$9x\left(1-x\right)y''-12y'+4y=0\\
 y''-\frac{12}{9x\left(1-x\right)}+\frac{4}{9x\left(1-x\right)}y=0.$\ \
 0 is a regular singular point\\
 $-9x^2y''+4x+9xy''-12y'=0$

 $\ds \sum_{n=0}^\infty
 a_n\{\left[-9\left(\sigma+n\right)\left(\sigma+n-1\right)+4\right]x^{\sigma+n}$

 $+\left[9\left(\sigma+n\right)\left(\sigma+n-1\right)
 -12\left(\sigma+n\right)\right]x^{\sigma+n-1}\}=0$

 Factorizing

 $\ds \sum_{n=0}^\infty
 a_n\{-\left[3\left(\sigma+n\right)-4\right]\left[3\left(\sigma+n\right)+1\right]
 x^{\sigma+n}$

 $+\left[3\left(\sigma+n\right)\right]
 \left[3\left(\sigma+n\right)-7\right]x^{\sigma+n-1}\}=0$

 Reordering

 $\ds \sum_{n=0}^\infty x^{\sigma +n}\{-a_n\left[3\left(\sigma+n\right)-4\right]
 \left[3\left(\sigma+n\right)+1\right]$

 $+\left[3\left(\sigma+n\right)+3\right]
 \left[3\left(\sigma+n\right)-4\right]a_{n+1}\}
 +a_0x^{\sigma-1}3\sigma\left(3\sigma-7\right)=0$

 $\sigma=0$ or $\sigma=\frac{7}{3}$ and
 $a_{n+1}=\frac{3\left(\sigma+n\right)+1}{3\left(\sigma+n\right)+3}\\
 \sigma=0:\ a_{n+1}=\frac{3n+1}{3\left(n+1\right)}a_n\\
 a_n=\frac{3n-2}{3n}a_{n-1}=\frac{\left(3n-2\right)\left(3n-5\right)}
 {3n\cdot3\left(n-1\right)}a_{n-2}\Rightarrow
 a_n=\frac{\left(3n-2\right)\left(3n-5\right)\ldots 1}{3^nn!}a_0\\
 \sigma=\frac{7}{3}:\ a_{n+1}=\frac{3n+8}{3n+10}a_n\\
 a_n=\frac{3n+5}{3n+7}a_{n-1}=\frac{\left(3n+5\right)\left(3n+2\right)\ldots
 8}{\left(3n+7\right)\left(3n+4\right)\ldots 10}a_0$


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