Given that:

\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)

Let \(y=\sum_{n=0}^\infty a_nx^n\) be a power series expansion of the required solution.

\(y=\sum_{n=0}^\infty a_nx^n\)

\(y'=\sum_{n=0}^\infty a_n(n)x^{n-1}=a_1+2a_2x+3a_3x^2+...\)

\(y''=\sum_{n=0}^\infty a_nn(n-1)x^{n-2}=2a_2+6a_3x+12a_4x^2+20a_5x^3+...\)

Now, find the Taylor series expansions of the given series:

\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)

\(y(0)=2\Rightarrow a_0=2\)

\(y'(0)=6\Rightarrow a_1=6\)

\((x−1)y′′+2y′−4y =0\)

\(\Rightarrow(x-1)(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+2(a_1+2a_2x+3a_3x^2+...)-4(a_0+a_1x+a_2x^2+a_3x^3+...)=0\)

\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(2a_1+4a_2x+6a_3x^2+...)-(4a_0+4a_1x+4a_2x^2+4a_3x^3+...)=0\)

\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(12+4a_2x+6a_3x^2+...)-(8+24x+4a_2x^2+4a_3x^3+...)=0\)

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:

\(2a_2-6a_3+4a_2-24=0\)

\(2a_2-6a_3+4a_2=24\)

\(\Rightarrow2(2)-6a_3+4(2)=24\)

\(\Rightarrow-6a_3=-4(2)+24-2(2)\Rightarrow a_3=2\)

The third degree power series solution is:

\(y(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+...\)

\(y(x)=a_0+a_1x+a_2x^2+a_3x^3+...\)

\(y(x)=2+6x+2x^2+2x^3+...\)

\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)

Let \(y=\sum_{n=0}^\infty a_nx^n\) be a power series expansion of the required solution.

\(y=\sum_{n=0}^\infty a_nx^n\)

\(y'=\sum_{n=0}^\infty a_n(n)x^{n-1}=a_1+2a_2x+3a_3x^2+...\)

\(y''=\sum_{n=0}^\infty a_nn(n-1)x^{n-2}=2a_2+6a_3x+12a_4x^2+20a_5x^3+...\)

Now, find the Taylor series expansions of the given series:

\((x−1)y′′+2y′−4y =0,y(0)=2,y′(0)=6\)

\(y(0)=2\Rightarrow a_0=2\)

\(y'(0)=6\Rightarrow a_1=6\)

\((x−1)y′′+2y′−4y =0\)

\(\Rightarrow(x-1)(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+2(a_1+2a_2x+3a_3x^2+...)-4(a_0+a_1x+a_2x^2+a_3x^3+...)=0\)

\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(2a_1+4a_2x+6a_3x^2+...)-(4a_0+4a_1x+4a_2x^2+4a_3x^3+...)=0\)

\(\Rightarrow(2a_2x+6a_3x^2+12a_4x^3+...)-(2a_2+6a_3x+12a_4x^2+20a_5x^3+...)+(12+4a_2x+6a_3x^2+...)-(8+24x+4a_2x^2+4a_3x^3+...)=0\)

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:

\(2a_2-6a_3+4a_2-24=0\)

\(2a_2-6a_3+4a_2=24\)

\(\Rightarrow2(2)-6a_3+4(2)=24\)

\(\Rightarrow-6a_3=-4(2)+24-2(2)\Rightarrow a_3=2\)

The third degree power series solution is:

\(y(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+...\)

\(y(x)=a_0+a_1x+a_2x^2+a_3x^3+...\)

\(y(x)=2+6x+2x^2+2x^3+...\)