Question

Find two power series solutions of the given differential equation about the ordinary point x = 0 as $\mathbf{(}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{)}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

Therefore, the solution is:

$y=c_0(1+\frac{1}{4}{x}^2-\frac{1}{24}{x}^3+\frac{1}{480}{x}^5-\frac{1}{1440}{x}^6+.....)+c_{1}x$

Step-by-step solution

Definition of Power series solution of 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 dependant variable.

The following expression for the power series equation as

$y=\sum _{n=0}^{\infty }{c}_n{x}^n,y\text{'}=\sum _{n=1}^{\infty }n{c}_n{x}^{n-1},y\text{'}\text{'}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}$

Compute radius.

The given problem is;

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

Substitute the expression in the problem as;

$$\begin{gathered} (x+2)\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+x\sum _{n=1}^{\infty }n{c}_n{x}^{n-1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ x\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+\sum _{n=1}^{\infty }n{c}_n{x}^{n-1+1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ \sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2+1}+2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+\infty \sum _{n=1}^{}n{c}_n{x}^{n-1+1}-\sum _{n=0}^{\infty }{c}_n{x}^n=0 \\ \underset{k=n-1}{\underbrace{\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-1}}}+\underset{k=n-2}{\underbrace{2\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}}}+\underset{k=n}{\underbrace{\sum _{n=1}^{\infty }n{c}_n{x}^n}}-\underset{k=n}{\underbrace{\sum _{n=0}^{\infty }{c}_n{x}^n}}=0 \end{gathered}$$

From the first term of the expansion is

k = n-1

If n=2, then:

$k=2-1\Rightarrow k=1$

$n=k+1$

Substitute these values are;

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

From the second term of the expansion is

k = n-2

If n = 2 then;

k = 0

k + 2 = n

Substitute these values are

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

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

From the third term of the expansion is;

k = n

If n = 1, then;

k = 1

Substitute these values as;

$\sum _{n=1}^{\infty }n{c}_n{x}^n=\sum _{k=1}^{\infty }k{c}_k{x}^k$

$\sum _{n=0}^{\infty }{c}_n{x}^n$

From the fourth term of the expansion is

n = 0

k = 0

Substitute these values are:

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

Rearrange the terms and expansion as 

$$\begin{gathered} \sum _{k=1}^{\infty }(k+1)\left(k\right)c_{k+1}{x}^k+2\sum _{k=0}^{\infty }(k+2)\left(\right(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }k{c}_k{x}^k-\sum _{k=0}^{\infty }{c}_k{x}^k=0 \\ \sum _{k=1}^{\infty }(k+1)\left(k\right)c_{k+1}{x}^k+2(0+2)(0+1)c_{0+2}{x}^0+2\sum _{k=1}^{\infty }(k+2)\left(\right(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }k{c}_k{x}^k-c_0{x}^0-\sum _{k=1}^{\infty }{c}_k{x}^k=0 \\ \sum _{k=1}^{\infty }(k+1)\left(k\right)c_{k+1}{x}^k+4c_2+2\sum _{k=1}^{\infty }(k+2)\left(\right(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }k{c}_k{x}^k-c_0-\sum _{k=1}^{\infty }{c}_k{x}^k=0 \\ 4c_2-c_0+\sum _{k=1}^{\infty }(k+1)\left(k\right)c_{k+1}{x}^k+2\sum _{k=1}^{\infty }(k+2)\left(\right(k+1)c_{k+2}{x}^k+\sum _{k=1}^{\infty }k{c}_k{x}^k-\sum _{k=1}^{\infty }{c}_k{x}^k=0 \\ 4c_2-c_0+\sum _{k=1}^{\infty }\left[\right(k+1\left)\right(k)c_{k+1}{x}^k+2(k+2)\left(\right(k+1)c_{k+2}{x}^k+k{c}_k-c_k]x^k=0 \\ 4c_2-c_0+\sum _{k=1}^{\infty }\left[\right(k+1\left)\right(k)c_{k+1}{x}^k+2(k+2)\left(\right(k+1)c_{k+2}{x}^k+(k-1\left)c_k\right]x^k=0 \end{gathered}$$

To compare the constant and coefficient term  

For Constant $4c_2-c_0=0$role="math" localid="1663945471654" $$\begin{gathered} 4c_2=c_0 \\ {c}_2=\frac{c_0}{4} \\ (k+1)\left(k\right)c_{k+1}+2(k+2)\left(\right(k+1)c_{k+2}+(k-1)c_k=0 \\ 2(k+2\left)\right((k+1)c_{k+2}=-(k+1)\left(k\right)c_{k+1}-(k-1)c_k \\ {c}_{k+2}=\frac{-(k+1)\left(k\right)c_{k+1}-(k-1)c_k}{2(k+2)\left(\right(k+1)} \\ {c}_{k+2}=-\frac{(k+1)\left(k\right)c_{k+1}+(k-1)c_k}{2(k+2)\left(\right(k+1)},k=1,2,3,.... \end{gathered}$$

For $k=1,c_{1+2}=-\frac{(1+1)\left(1\right)c_{1+1}+(1-1)c_1}{2(1+2)\left(\right(1+1)}$

$c_3=\frac{-2c_2}{2\left(3\right)\left(2\right)}$

$$\begin{gathered} c_3=\frac{-c_2}{6} \\ {c}_2=\frac{c_0}{4} \\ {c}_3=\frac{-c_0}{24} \\ fork=2,c_{2+2}=-\frac{(2+1)\left(2\right)c_{2+1}+(2-1)c_2}{2(2+2)\left(\right(2+1)} \\ {c}_4=\frac{6c_3+c_2}{2\left(4\right)\left(3\right)} \\ {c}_4=\frac{6c_3+c_2}{24} \\ {c}_4=\frac{6c_3}{24}+\frac{c_2}{24} \\ {c}_4=\frac{1}{4}\left(c_3\right)+\frac{c_2}{24} \\ {c}_4=\frac{1}{4}\left(\frac{-c_0}{24}\right)+\frac{1}{24}{c}_2 \\ {c}_4=\frac{-c_0}{96}+\frac{1}{24}\frac{c_0}{4} \\ {c}_4=\frac{-c_0}{96}+\frac{c_0}{96} \\ {c}_4=0 \\ fork=3,c_{3+2}=-\frac{(3+1)\left(3\right)c_{3+1}+(3-1)c_3}{2(3+2)\left(\right(3+1)} \\ {c}_5=-\frac{\left(12\right)c_4+\left(2\right)c_3}{2\left(5\right)\left(4\right)} \\ {c}_5=-\frac{12c_4}{40}-\frac{2c_3}{40} \\ {c}_5=0-\frac{2}{40}\left(\frac{-c_0}{24}\right) \\ {c}_5=\frac{c_0}{480} \\ fork=4,c_{4+2}=-\frac{(4+1)\left(4\right)c_{4+1}+(4-1)c_4}{2(4+2)\left(\right(4+1)} \\ {c}_6=-\frac{(4+1)\left(4\right)c_{4+1}+(4-1)c_4}{2(4+2)\left(\right(4+1)} \\ {c}_6=-\frac{20c_5+3c_4}{60} \\ {c}_6=\frac{-20c_5}{60}-\frac{3c_4}{60} \\ {c}_6=\frac{-1}{3}\frac{c_0}{480} \\ {c}_6=\frac{-c_0}{1440} \end{gathered}$$

The power series expansion of the differential equation

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ y=c_0+c_1{x}^1+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+..... \\ y=c_0+c_{1}x+\frac{c_0}{4}{x}^2\frac{-c_0}{24}{x}^3+0x^4+\frac{c_0}{480}{x}^5+\frac{-c_0}{1440}{x}^6+..... \\ y=c_0+c_{1}x+\frac{c_0}{4}{x}^2\frac{-c_0}{24}{x}^3+\frac{c_0}{480}{x}^5+\frac{-c_0}{1440}{x}^6+..... \\ y=(c_0+\frac{c_0}{4}{x}^2\frac{-c_0}{24}{x}^3+\frac{c_0}{480}{x}^5+\frac{-c_0}{1440}{x}^6+.....)+c_{1}x \end{gathered}$$