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Obtain the first nonzero terms in a power series solution about $x=1$ for the D.E.
$\frac{d^{2} y}{d x^{2}}+(x-1) y=1, y(1)=1, y^{\prime}(1)=2$

Let $x-1=t \quad \rightarrow \quad x=t+1 \quad$ then

$\frac{d^{2} y}{d x^{2}}+t y=1$

$\sum_{n=0}^{\infty} a_{n}(n)(n-1) t^{n-2}+\sum_{n=0}^{\infty} a n t^{n+1}=1$

$n \rightarrow n+2$

$n \rightarrow n-1$

$\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1) t^{n}+\sum_{n=1}^{\infty} a_{n-1} t^{n}=1$

$a_{2}(2)(1) t^{0}+\sum_{n=1}^{\infty} a_{n+2}(n+2)(n+1) t^{n}+\sum_{n=1}^{\infty} a_{n-1} t^{n}=1$

$2 a_{2}+\sum_{n=1}^{\infty}\left[a_{n+2}(n+2)(n+1)+a_{n-1}\right] t^{n}=1$

$2 a_{2}=1 \rightarrow a_{2}=1 / 2$

$a_{n+2}(n+2)(n+1)+a_{n-1}=0$

$a_{n+2}=\frac{-1}{(n+2)(n+1)} a n-1 \quad n \geq 1$

$n=1 \rightarrow a_{3}=\frac{-1}{3 \cdot 2} a_{0}=\frac{-1}{6} a_{0}$

$n=2 \rightarrow a_{4}=\frac{-1}{4 \cdot 3} a_{1}=\frac{-1}{12} a_{1}$

$y=\sum_{n=0}^{\infty} a_{n} t^{n}=a_{0}+a_{1} t+a_{2} t^{2}+a_{3} t^{3}+a_{n} t^{4}+\cdots$

$y=a_{0}+a_{1} t+\frac{1}{2} t^{2}-\frac{1}{6} a_{0} t^{3}-\frac{1}{12} a_{1} t^{4}+\cdots$

$t=x-1$

$y=a_{0}+a_{1}(x-1)+\frac{1}{2}(x-1)^{2}-\frac{1}{6} a_{0}(x-1)^{3}-\frac{1}{12} a_{1}(x-1)^{4}+\dots$

$y(1)=1 \rightarrow a_{0}=1$
$y^\prime(1)=2 \rightarrow a_{1}=2$

$y=1+2(x-1)+\frac{1}{2}(x-1)^{2}-\frac{1}{6}(x-1)^{3}-\frac{1}{12} \cdot 2(x-1)^{4}+\cdots$