Question

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

$\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore the solution is

$$\begin{gathered} y_1\left(x\right)=C_1{x}^3\left(1+\frac{1}{4}x+\frac{1}{20}{x}^2+\frac{1}{120}{x}^3+\cdots \right) \\ {y}_2\left(x\right)=C_2\left(1+x+\frac{1}{2}{x}^2\right) \end{gathered}$$

Step-by-step solution

Given Information

The given value is

$x{y}^{\text{'}\text{'}}-(x+2)y^{\text{'}}+2y=0$

Use Differentiate

$y\left(x\right)=\sum _{n=0}^{\infty }{c}_n{x}^{n+r}$

Differentiating (*), we get

$y^{\text{'}}=\sum _{n=0}^{\infty }(n+r)c_n{x}^{n+r-1}----\left(1\right)$

$y^{\text{'}\text{'}}=\sum _{n=0}^{\infty }(n+r)(n+r-1)c_n{x}^{n+r-2}----\left(2\right)$

Substitute the equation

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

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

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

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

$x^r\left[r(r-3)c_0{x}^{-1}+\sum _{n=1}^{\infty }(n+r)(n+r-3)c_n{x}^{n-1}-\sum _{n=0}^{\infty }(n+r-2)c_n{x}^n\right]=0$

For the first summation, we let k = n - 1. For the second summation, we let Hence

$x^r\left[r(r-3)c_0{x}^{-1}+\sum _{k=0}^{\infty }(k+1+r)(k+r-2)c_{k+1}{x}^k+\sum _{k=0}^{\infty }(k+r-2)c_k{x}^k\right]=0$

$x^r\left[r(r-3)c_0{x}^{-1}+\sum _{k=0}^{\infty }\left[(k+1+r)(k+r-2)c_{k+1}+(k+r-2)c_k\right]x^k\right]=0$

Identity property

r(r-3)=0

$r_1=3\vee r_2=0$and

${\text{(k + 1 + r)(k + r - 2)c}}_{k+1}{\text{+ (k + r - 2)c}}_k\text{= 0}$

For c₀=0 we have

$k=0\Rightarrow c_1=0$

$k=1\Rightarrow c_2=0$

$k=2\Rightarrow c_3=0$

$k=3\Rightarrow c_4=\frac{1}{4}{c}_3$

$k=4\Rightarrow c_5=\frac{1}{20}{c}_3$

$k=5\Rightarrow c_6=\frac{1}{120}$

For c₃ = 0 we have

c₀= 0

$k=0\Rightarrow c_1=0$

$k=1\Rightarrow c_2=\frac{1}{2}$

$k=2,3,4,5\dots \Rightarrow c_k=0$

$y_2\left(x\right)=C_2\left(1+x+\frac{1}{2}{x}^2\right)$