Question

In Problems 9-14 use an appropriate infinite series method about x=0to find two solutions of the given differential equation.

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

Short answer

Therefore, the solution is

$$\begin{gathered} y_1\left(x\right)=c_0\left(1+\frac{3}{2}{x}^2+\frac{1}{2}{x}^3+\frac{5}{8}{x}^4+\cdots \right) \\ {y}_2\left(x\right)=c_1\left(x+\frac{1}{2}{x}^3+\frac{1}{4}{x}^4+\cdots \right) \end{gathered}$$

Step-by-step solution

Given Information

The given value is

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

Use Differentiate

$$\begin{gathered} y\left(x\right)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+... \\ =\sum _{n=0}^{\infty }{c}_n{x}^n \end{gathered}$$

Differentiate (*), we get

$y^{\text{'}}=\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}---\left(1\right)$

$y^{\text{'}\text{'}}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}---\left(2\right)$

Replacing the equation (1) and equation (2) in the initial differential equation, we get

$y^{\text{'}\text{'}}\left(x\right)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}$

Find the series

$0=(x-1)\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+3\sum _{n=0}^{\infty }{c}_n{x}^n$

$=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-1}-\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+3\sum _{n=0}^{\infty }{c}_n{x}^n$

For the first series, we let k = n-1. The summation becomes

$\sum _{k=1}^{\infty }(k+1)k{c}_{k+1}{x}^k$

For the second series, we let k = n-2. Then

$\sum _{k=0}^{\infty }(k+2)(k+1)c_{k+2}{x}^k$

For the third series, k = n we letSo,

$\sum _{k=0}^{\infty }{c}_k{x}^k$

Therefore,

$0=-2c_2+3c_0+\sum _{k=1}^{\infty }(k+1)k{c}_{k+1}{x}^k-\sum _{k=1}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+3\sum _{k=1}^{\infty }{c}_k{x}^k$

$=-2c_2+3c_0+\sum _{k=1}^{\infty }\left[(k+1)k{c}_{k+1}-(k+2)(k+1)c_{k+2}+3c_k\right]x^k$

Using identity property

$-2c_2+3c_0=0$

$c_2=\frac{3c_0}{2}$

$(k+1)k{c}_{k+1}-(k+2)(k+1)c_{k+2}+3c_k=0$

$c_{k+2}=\frac{(k+1)k{c}_{k+1}+3c_k}{(k+2)(k+1)}$

Let c₀= 1 and c₁= 0 Then,

$c_2=\frac{3}{2}$

$c_3=\frac{2c_2+3·0}{6}$

= 1/2

$c_4=\frac{6c_3+3·\frac{1}{2}}{12}$

= 5/8

Hence

$y_1\left(x\right)=c_0\left(1+\frac{3}{2}{x}^2+\frac{1}{2}{x}^3+\frac{5}{8}{x}^4+\cdots \right)$

For c₀ = 0 and c₁ = 1 we have:

$c_2=\frac{0}{2}=0$

$c_3=\frac{2·\frac{1}{2}}{6}=\frac{1}{2}$

$c_4=\frac{1}{4}$

$y_2\left(x\right)=c_1\left(x+\frac{1}{2}{x}^3+\frac{1}{4}{x}^4+\cdots \right)$