Question

In Problems 23 and 24 use the procedure in Example 8 to find two power series solutions of the given differential equation about the ordinary point

$\begin{array}{rcl}x& =& 0\\ y\text{'}\text{'}+\left(sinx\right)y& =& 0\end{array}$

Short answer

Hence, the two linearly independent solutions are:

$$\begin{gathered} y_1\left(x\right)=1-\frac{x^3}{6}+\frac{x^5}{120}+\dots \\ {y}_2\left(x\right)=x-\frac{x^4}{12}+\dots \end{gathered}$$

Step-by-step solution

Given Information

The given differential equation is:

$y\text{'}\text{'}+\left(\sin x\right)y=0$

Plug into the differential equation

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

Differentiating $(*)$, we obtain

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

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

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

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

We use $\sin x=\left(x-\frac{x^3}{x!}+\frac{x^5}{5!}+\dots \right)$

$\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2}+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)\sum _{n=0}^{\infty }{c}_n{x}^n=0$

After multiplying a few terms above; we obtain

$$\begin{gathered} \left(2c_2+6c_{3}x+12c_4{x}^2+20c_5{x}^3+\dots \right)+\left(c_{0}x+c_1{x}^2+c_2{x}^3-c_0\frac{x^3}{6}+\dots \right)=0 \\ 2c_2+x\left(6c_3+c_0\right)+x^2\left(12c_4+c_1\right)+x^3\left(20c_5+c_2-\frac{1}{6}{c}_0\right)+\dots ..=0 \end{gathered}$$

Identity property

$\begin{array}{rcl}2c_2& =& 0\\ c_2& =& 0\\ & & \end{array}$

$\begin{array}{rcl}6c_3+c_0& =& 0\\ c_3& =& -\frac{c_0}{6}\\ & & \end{array}$

$\begin{array}{rcl}12c_4+c_1& =& 0\\ c_4& =& -\frac{c_1}{12}\\ & & \end{array}$

$\begin{array}{rcl}20c_5+c_2-\frac{1}{6}{c}_0& =& 0\\ c_5& =& \frac{1}{120}{c}_0\\ & & \end{array}$

Continuing the same calculations and replacing them in$\left(*\right),$we obtain

$\begin{array}{rcl}y& =& c_0+c_{1}x+0·x^2-\frac{c_0}{6}{x}^3-\frac{c_1}{12}{x}^4+\frac{c_0}{120}{x}^3+\dots \\ & =& \underset{}{\underbrace{\left(1-\frac{x^3}{6}+\frac{x^5}{120}+\dots \right)}}+\underset{}{\underbrace{c_1\left(x-\frac{x^4}{12}+\dots \right)}}\\ & & \end{array}$

Hence, the two linearly independent solutions are:

$$\begin{gathered} y_1\left(x\right)=1-\frac{x^3}{6}+\frac{x^5}{120}+\dots \\ {y}_2\left(x\right)=x-\frac{x^4}{12}+\dots \end{gathered}$$