Question

Find two power series solutions of the given differential equation about the ordinary point x = 0as$\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

$y_{1=}{c}_0,$$y_2=c_1\sum _{n=1}^{\infty }\frac{1}{n}{x}^n$

Step-by-step solution

Definition of the power series solution of the differential equation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable.

The following expression for the power series equation as;

$y=\sum _{n=0}^{\infty }{c}_n{(x-x_0)}^n$

To compute the series of expansion equation

The given differential equation is:

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

localid="1663943544246" $$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ y\text{'}=\sum _{n=1}^{\infty }n{c}_n{x}^{n-1} \\ {y}^{\text{'}\text{'}}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2} \\ x{y}^{\text{'}\text{'}}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-1} \\ \sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-1}-\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}=0 \\ k=n-1\Rightarrow n=k+1 \\ n=2\Rightarrow k=1 \\ \sum _{k=1}^{\infty }(k+1)k{c}_{k+1}{x}^k \end{gathered}$$

localid="1663943559711" $$\begin{gathered} k=n-2\Rightarrow n=k+2 \\ n=2\Rightarrow k=0 \\ \sum _{k=0}^{\infty }(k+2)(k+1)c_{k+2}{x}^k \\ k=n-1\Rightarrow n=k+1 \\ n=1\Rightarrow k=0 \\ \sum _{k=0}^{\infty }(k+1)c_{k+1}{x}^k \end{gathered}$$

$c_1-2c_2+\sum _{k=1}^{\infty }\left[\right(k+1)k{c}_{k+1}-(k+2\left)\right(k+1)c_{k+2}+(k+1\left)c_{k+1}\right]x^k=0$

To compare the constant and coefficient term

$$\begin{gathered} c_1-2c_2=0 \\ -2c_2=-c_1 \\ {c}_2=\frac{c_1}{2} \end{gathered}$$

$$\begin{gathered} (k+1)k{c}_{k+1}-(k+2)(k+1)c_{k+2}+(k+1)c_{k+1}=0 \\ (k+1)c_{k+1}(k+1)-(k+2)(k+1)c_{k+2}=0 \\ {c}_{k+2}=\frac{(k+1)c_{k+1}(k+1)}{(k+2)(k+1)} \\ {c}_{k+2}=\frac{(k+1)c_{k+1}}{(k+2)} \end{gathered}$$

$$\begin{gathered} k=1,c_{1+2}=\frac{(1+1)c_{1+1}}{(1+2)} \\ {c}_3=\frac{2}{3}\left(\frac{c_1}{2}\right) \\ {c}_3=\frac{c_1}{3} \\ k=2,c_{2+2}=\frac{(2+1)c_{2+1}}{(2+2)} \\ {c}_4=\frac{3c_3}{4} \\ {c}_4=\frac{3}{4}\left(\frac{c_1}{3}\right) \\ {c}_4=\frac{c_1}{4} \\ k=3,c_{3+2}=\frac{(3+1)c_{3+1}}{(3+2)} \\ {c}_5=\frac{4c_4}{5} \\ {c}_5=\frac{4}{5}\frac{c_1}{4} \\ {c}_5=\frac{c_1}{5} \end{gathered}$$

$$\begin{gathered} k=4,c_{4+2}=\frac{(4+1)c_{4+1}}{(4+2)} \\ {c}_6=\frac{5c_5}{6} \\ {c}_6=\frac{5}{6}\frac{c_1}{5} \\ {c}_6=\frac{c_1}{6} \end{gathered}$$

The power series expansion of the differential equation.

$$\begin{gathered} c_1,\frac{c_1}{2},\frac{c_1}{3},\frac{c_1}{4},\frac{c_1}{5},\frac{c_1}{6}\Rightarrow c_1\sum _{n=1}^{\infty }\frac{1}{n}{x}^n \\ {y}_1=c_0 \\ {y}_2=c_1\sum _{n=1}^{\infty }\frac{1}{n}{x}^n \end{gathered}$$