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$2 y^{\prime \prime}-\frac{1}{x-1} y^{\prime}+\frac{1}{(x-1)^{2}} y=0$

$(x-1) p(x)=(x-1)\left(\frac{-1}{x - 1}\right)=-1$

$(x-1)^{2} q(x)=(x-1)^{2}\left(\frac{1}{(x-1)^{2}}\right)=1$

$\rightarrow$ Regular singular point

\$x_{0}=1$

$y=\sum_{n=0}^{\infty} c_{n}(x-1)^{n+r}$

$y^{\prime}=\sum_{n=0}^{\infty}(n+r)cn(x-1)^{n+r-1}$

$y^{\prime \prime}=\sum_{n=0}^{\infty}(n+r-1)(n+r)cn(x-1)^{n+r-2}$

$\rightarrow$ Substitution in D.E

$2 \sum_{n=0}^{\infty}(n+r-1)(n+r) c n(x-1)^{n+1-2}-\frac{1}{x-1} \sum_{n=0}^{\infty}(n+r)cn(x-1)^{n+r-1}+\frac{1}{(x-1)^{2}} \sum_{n=0}^{\infty} c_{n}(x-1)^{n+r}=0$

$2 \sum_{n=0}^{\infty}(n+r-1)(n+1)c n(x-1)^{n+r-2}-\sum_{n=0}^{\infty}(n+r) c n(x-1)^{n+r-2}+\sum_{n=0}^{\infty} c n(x-1)^{n+r-2}=0$

$\sum_{n=0}^{\infty}\left[2(n+r)(n+r-1)cn-(n+r) c n+c n](x-1)^{n+r-2}=0 \rightarrow (1)\right.$

When $n=0 \quad \longrightarrow[2 r(r-1)-r+1]=0 \quad \rightarrow$ Indical Equation

$2 r^{2}-3 r+1=0 \rightarrow r_{1}=1 \quad r_{2}=0.5$

From $(1) : cn[2(n+r)(n+r-1)-(n+r)+1]=0 \longrightarrow(1)$

$y=\sum_{n=0}^{\infty} c n(x-1)^{n+r}$

$y_{1}=c_{0}(x-1)$

$y_{2}=c_0(x-1)^{0.5}$

$y=c_{1}(x-1)+c_{2}(x-1)^{0.5}$

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