Use Taylor series expansion to find the third degree power series solution of the initial...

Given that:
$(x-1)y\prime \prime +2y\prime -4y=0,y(0)=2,y\prime (0)=6$
Let $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ be a power series expansion of the required solution.
$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$
$y^{\prime }=\sum _{n=0}^{\mathrm{\infty }}{a}_n(n)x^{n-1}=a_1+2a_{2}x+3a_3{x}^2+...$
$y^{″}=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}n(n-1)x^{n-2}=2a_2+6a_{3}x+12a_4{x}^2+20a_5{x}^3+...$
Now, find the Taylor series expansions of the given series:
$(x-1)y\prime \prime +2y\prime -4y=0,y(0)=2,y\prime (0)=6$
$y(0)=2\Rightarrow a_0=2$
$y^{\prime }(0)=6\Rightarrow a_1=6$
$(x-1)y\prime \prime +2y\prime -4y=0$

$\Rightarrow (x-1)(2a_2+6a_{3}x+12a_4{x}^2+20a_5{x}^3+...)+2(a_1+2a_{2}x+3a_3{x}^2+...)-4(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...)=0$$\Rightarrow (2a_{2}x+6a_3{x}^2+12a_4{x}^3+...)-(2a_2+6a_{3}x+12a_4{x}^2+20a_5{x}^3+...)+(2a_1+4a_{2}x+6a_3{x}^2+...)-(4a_0+4a_{1}x+4a_2{x}^2+4a_3{x}^3+...)=0$
$\Rightarrow (2a_{2}x+6a_3{x}^2+12a_4{x}^3+...)-(2a_2+6a_{3}x+12a_4{x}^2+20a_5{x}^3+...)+(12+4a_{2}x+6a_3{x}^2+...)-(8+24x+4a_2{x}^2+4a_3{x}^3+...)=0$
Compare the coefficient of $x^0,x^1$ and power:
$(-2a_2+12-8)+(2a_2-6a_3+4a_2-24)x+...$
$-2a_2+12-8=0\Rightarrow -2a_2=-12+8\Rightarrow a_2=2$
Coefficient of x:
$2a_2-6a_3+4a_2-24=0$
$2a_2-6a_3+4a_2=24$