Find at least the first four nonzero terms in a power series expansion about ${\mathbf{x}}_{\mathbf{0}}$for a general solution to the given differential equation with the given value for ${\mathbf{x}}_{\mathbf{0}}$,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}{\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{1}$

The given differential equation is,

$\mathrm{y}\text{'}\text{'}+(3\mathrm{x}-1)\mathrm{y}\text{'}-\mathrm{y}=0$

From the above equation, p(x) = 3x-1 and q(x) = -1; thus, the equation does not have any singularities, and the point${\mathrm{x}}_0=-1$is an ordinary point.

$\begin{array}{l}\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}}\\ \mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}-1}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right).{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}-2}\\ \mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}-1}\\ \sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right).{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}-2}+(3\mathrm{x}-1)\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}}=0\end{array}$

Let $\mathrm{x}+1=\mathrm{t}\Rightarrow \mathrm{x}=\mathrm{t}-1$

Substituting,

$\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right).{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+(3\mathrm{t}-4)\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$

Separating each term of the above expression’

$\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right).{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+3\mathrm{t}\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}-4\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$

Expanding the individual series contained by each tern of the above expression,

$\begin{array}{l}(2{\mathrm{a}}_2+6{\mathrm{a}}_3\mathrm{t}+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3+30{\mathrm{a}}_6{\mathrm{t}}^4+...)+(3{\mathrm{a}}_1\mathrm{t}+6{\mathrm{a}}_2{\mathrm{t}}^2+9{\mathrm{a}}_3{\mathrm{t}}^3+12{\mathrm{a}}_4{\mathrm{t}}^4+...)\\ (-4{\mathrm{a}}_1-8{\mathrm{a}}_4\mathrm{t}-12{\mathrm{a}}_3{\mathrm{t}}^2-16{\mathrm{a}}_4{\mathrm{t}}^3-20{\mathrm{a}}_5{\mathrm{t}}^4+...)+(-{\mathrm{a}}_0-{\mathrm{a}}_1\mathrm{t}-{\mathrm{a}}_2{\mathrm{t}}^2-{\mathrm{a}}_3{\mathrm{t}}^3-{\mathrm{a}}_4{\mathrm{t}}^4-...)=0\\ (2{\mathrm{a}}_2-{\mathrm{a}}_0-4{\mathrm{a}}_1)+(6{\mathrm{a}}_3+3{\mathrm{a}}_1-8{\mathrm{a}}_2-{\mathrm{a}}_1)\mathrm{t}+...=0\end{array}$

Substituting t with x+1

$$\begin{gathered} (2{\mathrm{a}}_2-{\mathrm{a}}_0-4{\mathrm{a}}_1)+(6{\mathrm{a}}_3+3{\mathrm{a}}_1-8{\mathrm{a}}_2-{\mathrm{a}}_1)(\mathrm{x}+1)+...=0 \\ 2{\mathrm{a}}_2-{\mathrm{a}}_0-4{\mathrm{a}}_1=0\Rightarrow {\mathrm{a}}_2=\frac{{\mathrm{a}}_0}{2}+2{\mathrm{a}}_1 \\ 6{\mathrm{a}}_3+3{\mathrm{a}}_1-8{\mathrm{a}}_2-{\mathrm{a}}_1=0\Rightarrow {\mathrm{a}}_3=\frac{2{\mathrm{a}}_0}{3}+\frac{7{\mathrm{a}}_1}{3} \end{gathered}$$

The general solution is,

$\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1(\mathrm{x}+1)+{\mathrm{a}}_2(\mathrm{x}+1{)}^2+{\mathrm{a}}_3(\mathrm{x}+1{)}^3+...$

Substituting,

$\begin{array}{l}\mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1(\mathrm{x}+1)+{\mathrm{a}}_2(\mathrm{x}+1{)}^2+{\mathrm{a}}_3(\mathrm{x}+1{)}^3+...\\ \mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1(\mathrm{x}+1)+\left(\frac{{\mathrm{a}}_0}{2}+2{\mathrm{a}}_1\right)(\mathrm{x}+1{)}^2+\left(\frac{2{\mathrm{a}}_0}{3}+\frac{7{\mathrm{a}}_1}{3}\right)(\mathrm{x}+1{)}^3+...\\ \mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0\left[1+\frac{1}{2}{(\mathrm{x}+1)}^2+\frac{2}{3}{(\mathrm{x}+1)}^3+...\right]+{\mathrm{a}}_1\left[(\mathrm{x}+1)+2{(\mathrm{x}+1)}^2+\frac{7}{3}{(\mathrm{x}+1)}^3+...\right]\end{array}$

Hence the final solution is,$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0\left[1+\frac{1}{2}{(\mathrm{x}+1)}^2+\frac{2}{3}{(\mathrm{x}+1)}^3+...\right]+{\mathrm{a}}_1\left[(\mathrm{x}+1)+2{(\mathrm{x}+1)}^2+\frac{7}{3}{(\mathrm{x}+1)}^3+...\right]$.