Question

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

Short answer

Therefore, the solution is:

$y=c_0(1-x^2+\frac{1}{2}{x}^4+...)+c_1(x-\frac{2c_1}{3}{x}^3+\frac{4c_1}{15}{x}^5+....)$

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 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 problem is;

$y+2xy\text{'}+2y=0$

$f(y\text{'}\text{'},y\text{'},y)=y\text{'}\text{'}+2xy\text{'}+2y=0$

The power series expression for the differential equation as $x=x_0$has two linearly independent solutions has the form.

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

At$x_0=0$

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

At $a_0=0$

$y\text{'}=\sum _{n=0}^{\infty }{c}_{n}n{x}^{n-1}$

At $a_0=a_1=0$

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

$y\text{'}=\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}$

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

Substitute the differential equation as:

$$\begin{gathered} f(y^{\text{'}\text{'}},y^{\text{'}},y)=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}+2x\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}+2\sum _{n=0}^{\infty }{c}_n{x}^n \\ f(y^{\text{'}\text{'}},y^{\text{'}},y)=\underset{a_2\to x^0}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}}}+\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }2c_{n}n{x}^n}}+\underset{a_0\to x^0}{\underbrace{\sum _{n=0}^{\infty }2c_n{x}^n}} \\ f(y^{\text{'}\text{'}},y^{\text{'}},y)=\left[2c_2{x}^0+\underset{a_3\to x^1}{\underbrace{\sum _{n=3}^{\infty }{c}_{n}n(n-1)x^{n-2}}}+\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }2c_{n}n{x}^n}}+2c_0{x}^0\underset{a_1\to x^1}{\underbrace{\sum _{n=1}^{\infty }2c_n{x}^n}}\right] \end{gathered}$$

Shift and Replace the summation powers of the expansion.

Replace by summation index from n to n+2,

$$\begin{gathered} f(y^{\text{'}\text{'}},y^{\text{'}},y)=2c_2{x}^0+2c_0{x}^0+\sum _{n+2=3}^{\infty }{c}_{n+2}(n+2)(n+2)x^{n+2-2}+\sum _{n=1}^{\infty }2c_{n}n{x}^n+\sum _{n=1}^{\infty }2c_n{x}^n \\ f(y^{\text{'}\text{'}},y^{\text{'}},y)=2x^0(c_2+c_0)+\sum _{n=1}^{\infty }{c}_{n+2}(n+2)(n+1)x^n+\sum _{n=1}^{\infty }2c_{n}n{x}^n+\sum _{n=1}^{\infty }2c_n{x}^n \\ f(y^{\text{'}\text{'}},y^{\text{'}},y)=2x^0(c_2+c_0)+\sum _{n=1}^{\infty }\left(c_{n+2}\right(n+2\left)\right(n+1)+2c_{n}n+2c_n\left)x^n \\ f\right(y^{\text{'}\text{'}},y^{\text{'}},y)=2x^0(c_2+c_0)+\sum _{n=1}^{\infty }\left(c_{n+2}\right(n+2\left)\right(n+1)+2c_n(n+1\left)x^n \\ f\right(y^{\text{'}\text{'}},y^{\text{'}},y)=2x^0(c_2+c_0)+\sum _{n=1}^{\infty }\left(c_{n+2}\right(n+2)+2c_n)(n+1)x^n \end{gathered}$$

To compare the constant and coefficient term

$$\begin{gathered} 2c_2+2c_0=0 \\ {c}_2=-c_0 \end{gathered}$$

$$\begin{gathered} c_{n+2}(n+2)+2c_n)=0 \\ {c}_{n+2}=-\frac{2c_n}{n+2}, n=1,2,3.... \end{gathered}$$

$n=1,c_{1+2}=-\frac{2c_1}{1+2}\Rightarrow c_3=-\frac{2c_1}{3}$

$n=2,c_{2+2}=-\frac{2c_2}{2+2}\Rightarrow c_4=-\frac{2c_2}{4}$

$c_2=-c_0$

$c_4=-\frac{2(-c_0)}{4}\Rightarrow c_4=\frac{c_0}{2}$

$n=3,c_{3+2}=-\frac{2c_3}{3+2}\Rightarrow c_5=-\frac{2c_3}{5}$

$c_3=-\frac{2c_1}{3}$

$c_5=-\frac{2(-2c_1)}{5\times 3}\Rightarrow c_5=\frac{4c_1}{15}$

The power series expansion of the differential equation 

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ y=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+.... \\ y=c_0+c_{1}x-c_0{x}^2-\frac{2c_1}{3}{x}^3+\frac{c_0}{2}{x}^4+\frac{4c_1}{15}{x}^5.... \\ y=c_0-c_0{x}^2+\frac{c_0}{2}{x}^4+c_{1}x-\frac{2c_1}{3}{x}^3+\frac{4c_1}{15}{x}^5+.... \\ y=c_0(1-x^2+\frac{1}{2}{x}^4+...)+c_1(x-\frac{2c_1}{3}{x}^3+\frac{4c_1}{15}{x}^5+....) \end{gathered}$$