Question

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

Short answer

Therefore, the solution is;

$y=c_0(1-\frac{x^3}{6}+\frac{x^6}{45}-.....)+c_1(x-\frac{x^4}{6}+\frac{5x^7}{\left(252\right)}-....)$

Step-by-step solution

Definition of Power series solution of differential equation.

A differential equations 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 power series equation as:

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

To compute the series of expansion equation

The given differential equation is:

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

Substitute the expression in the problem as,

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

From the first term of the expansion is:

$$\begin{gathered} k=n-2 \\ k+2=n \end{gathered}$$

The values of$n=2,k=2-2\Rightarrow k=0$.

Substitute these values are:

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

From the second term of the expansion is:

$$\begin{gathered} k=n+1 \\ k-1=n \end{gathered}$$

The values of $n=1\Rightarrow k=2$

Substitute these values are:

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

From the third term of the expansion is;

$k=n+1$

If $n=0$, then:

$k=1\Rightarrow k-1=n$

Substitute these values are;

$\sum _{k=1}^{\infty }{c}_{k-1}{x}^k=0$

Rearrange the terms and expansion as

$$\begin{gathered} \sum _{k=0}^{\infty }(k+2)(k+2-1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \\ \sum _{k=0}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \end{gathered}$$

Substitute the values of$k=0,1$ in the expression as;

$$\begin{gathered} (0+2)(0+1)c_{0+2}{x}^0+(1+2)(1+1)c_{1+2}{x}^1+\sum _{k=2}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+c_{1-1}{x}^1+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \\ \left(2\right)\left(1\right)c_2+\left(3\right)\left(2\right)c_3{x}^1+\sum _{k=2}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+c_0{x}^1+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \end{gathered}$$

To simplify the term of the expansion as;

$$\begin{gathered} 2c_2+6c_{3}x+\sum _{k=2}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+c_{0}x+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \\ 2c_2+6c_{3}x+c_{0}x+\sum _{k=2}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \\ 2c_2+6c_{3}x+c_{0}x+\sum _{k=2}^{\infty }(k+2)(k+1)c_{k+2}{x}^k+\sum _{k=2}^{\infty }(k-1)c_{k-1}{x}^k+\sum _{k=2}^{\infty }{c}_{k-1}{x}^k=0 \end{gathered}$$

Take the common value as$x$ and$x^k$.

$$\begin{gathered} 2c_2+(6c_3+c_0)x+\sum _{k=2}^{\infty }\left[\right(k+2\left)\right(k+1)c_{k+2}+(k-1)c_{k-1}+c_{k-1}]x^k=0 \\ 2c_2+(6c_3+c_0)x+\sum _{k=2}^{\infty }\left[\right(k+2\left)\right(k+1)c_{k+2}+k{c}_{k-1}-c_{k-1}+c_{k-1}]x^k=0 \\ 2c_2+(6c_3+c_0)x+\sum _{k=2}^{\infty }\left[\right(k+2\left)\right(k+1)c_{k+2}+k{c}_{k-1}]x^k=0 \end{gathered}$$

To compare the constant and coefficient term.

For constant term as$2c_2=0$

For$x$as;

$$\begin{gathered} 6c_3+c_0=0 \\ 6c_3=-c_0 \\ {c}_3=\frac{-c_0}{6} \end{gathered}$$

For $x^2$as $(k+2)(k+1)c_{k+2}+k{c}_{k-1}=0$

$$\begin{gathered} (k+2)(k+1)c_{k+2}=-k{c}_{k-1} \\ {c}_{k+2}=\frac{-k}{(k+2)(k+1)}{c}_{k-1},k=2,3,4 \end{gathered}$$

For $k=2,c_{2+2}=\frac{-2}{(2+2)(2+1)}{c}_{2-1}$

$$\begin{gathered} c_4=\frac{-2}{\left(12\right)}{c}_1 \\ {c}_4=\frac{-1}{\left(6\right)}{c}_1 \end{gathered}$$

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

$$\begin{gathered} c_5=\frac{-3}{\left(5\right)\left(4\right)}{c}_2 \\ {c}_5=\frac{-3}{20}{c}_2 \end{gathered}$$

Substitute the values as$c_2=0$

$$\begin{gathered} c_5=\frac{-3}{20}\left(0\right) \\ {c}_5=0 \end{gathered}$$

For$k=4,c_{4+2}=\frac{-4}{(4+2)(4+1)}{c}_{4-1}$

$$\begin{gathered} c_6=\frac{-4}{\left(6\right)\left(5\right)}{c}_3 \\ {c}_6=\frac{-4}{30}{c}_3 \end{gathered}$$

Substitute these values are $c_3=-\frac{c_0}{6}$

$$\begin{gathered} c_6=\frac{-4}{30}(-\frac{c_0}{6}) \\ {c}_6=\frac{1}{45}{c}_0 \end{gathered}$$

For$k=5,c_{5+2}=\frac{-5}{(5+2)(5+1)}{c}_{5-1}$

$$\begin{gathered} c_7=\frac{-5}{\left(7\right)\left(6\right)}{c}_4 \\ {c}_7=\frac{-5}{\left(42\right)}{c}_4 \end{gathered}$$

Substitute the values are$c_4=-\frac{1}{6}{c}_1$

$$\begin{gathered} c_7=\frac{-5}{\left(42\right)}(-\frac{1}{6}{c}_1) \\ {c}_7=\frac{5}{\left(252\right)}{c}_1 \end{gathered}$$

The power series expansion of the differential equation.

$$\begin{gathered} y=\sum _{n=0}^{\infty }{c}_n{x}^n \\ y=c_0+c_1{x}^1+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+c_6{x}^6+c_7{x}^7+...... \\ y=c_0+c_1{x}^1+\left(0\right)x^2+(-\frac{c_0}{6})x^3+(-\frac{1}{6}{c}_1)x^4+\left(0\right)x^5+\left(\frac{1}{45}{c}_0\right)x^6+\frac{5}{\left(252\right)}{c}_1{x}^7+...... \\ y=c_0+c_{1}x-\frac{c_0}{6}{x}^3-\frac{1}{6}{c}_1{x}^4+\frac{1}{45}{c}_0{x}^6+\frac{5}{\left(252\right)}{c}_1{x}^7+...... \\ y=c_0-\frac{c_0}{6}{x}^3+\frac{1}{45}{c}_0{x}^6+c_{1}x-\frac{1}{6}{c}_1{x}^4+\frac{5}{\left(252\right)}{c}_1{x}^7+...... \end{gathered}$$

$y=c_0(1-\frac{x^3}{6}+\frac{x^6}{45}-.....)+c_1(x-\frac{x^4}{6}+\frac{5x^7}{\left(252\right)}-....)$