$\%\%-$

$3.y^{\text{'}\text{'}}-x{y}^{\text{'}}-y=0,x_0=1$

Classify Point

$A(r)=10$

$VGR$ are $(6P),x_0=1$ is $(QP)$

assume that $y={\displaystyle \underset{n=0}{\overset{}{}}}{a}_n(k-x_0{)}^n$

$y={\displaystyle \underset{n=0}{\overset{}{}}}{a}_0(x-1{)}^n,y^{\text{'}}={\displaystyle \underset{n=1}{\overset{n}{}}}n{a}_n(x-1{)}^{n-1},y^n={\displaystyle \underset{n=2}{\overset{n}{}}}(n-1)n{a}_n(x-1{)}^{n-2}$

Substitute in $Eq$

${\displaystyle \underset{n=2}{\overset{}{}}}(n-1)n{a}_n(x-1{)}^{n-2}-x{\displaystyle \underset{n=1}{\overset{}{}}}n{a}_n(x-1{)}^{n-1}-{\displaystyle \underset{n=0}{\overset{}{}}}{a}_n(x-1{)}^n=0$

let $x=(x-1)+1$

${\displaystyle \underset{i=1}{\overset{}{}}}(n-1)n{a}_n(x-1{)}^{n-2}-[(x-1)+1]{\displaystyle \underset{n=1}{\overset{}{}}}n{a}_n(x-1{)}^{n-1}-{\displaystyle \underset{n=0}{\overset{}{}}}{a}_n(x-1{)}^n=0$

${\displaystyle \underset{n=2}{\overset{B}{}}}(n-1)n{a}_n(x-1{)}^{n-2}-{\displaystyle \underset{n=1}{\overset{n}{}}}{a}_n(x-1{)}^n-{\displaystyle \underset{n=1}{\overset{n}{}}}n{a}_n(x-1{)}^{n-1}-{\displaystyle \underset{n=1}{\overset{n}{}}}{a}_n(x-1{)}^n=0$

Shiffting indecs

Peel off

$2a_2-a_1-a_0+{\displaystyle \underset{n=1}{\overset{n}{}}}[ubrace((n+1)(n+2)a_{n-1}-n{a}_n-(n+1)a_{n+1}-a_n$ubrace${)}_{\text{v}\text{e}\text{w}\text{r}\text{i}\text{e}\text{n}\text{a}\text{v}\text{e}\text{b}\text{l}\text{i}\text{n}}](x-1{)}^n=0$

to find soulation is

$a_{n+2}=\frac{(n+1)a_n+(n+1)a_{n+1}}{(n+1)(n+2)},n1$

$n=0longrightarrow{a}_2=\frac{a_0+a_1}{2}=\frac{1}{2}{a}_1+\frac{1}{2}*a_1$

$n=1longrightarrow{a}_3=\frac{2a_1+2a_2}{3*2}=\frac{1}{3}{a}_1+\frac{1}{3}{a}_2=\frac{1}{3}{a}_1+\frac{1}{6}{a}_0+\frac{1}{6}{a}_1=\frac{1}{6}{a}_0+\frac{1}{2}{a}_1$

$n=2,a_2=\frac{3a_1+3a_3}{43}=\frac{1}{4}{a}_2+\frac{1}{4}{a}_2=\frac{1}{8}{a}_0+\frac{1}{8}{a}_1+\frac{1}{24}{a}_0+\frac{1}{8}{a}_1=\frac{1}{6}{a}_0+\frac{1}{4}{a}_1$