Question

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

(Cos X) y" + y = 0

Short answer

Therefore, the solution is

$\begin{array}{l}{\mathrm{y}}_1\left(\mathrm{x}\right)=\mathrm{x}-\frac{1}{6}{\mathrm{x}}^3-\frac{1}{60}{\mathrm{x}}^5+..\\ {\mathrm{y}}_2\left(\mathrm{x}\right)=1-\frac{1}{2}{\mathrm{x}}^2+\frac{1}{720}{\mathrm{x}}^6+..\end{array}$

Step-by-step solution

Given Information

The given value is

$(\mathrm{Cos}\mathrm{X})\mathrm{y}"+\mathrm{y}=0$

Use Differentiate

Differentiate$\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ to get${\mathrm{y}}^{\text{'}}\left(\mathrm{x}\right)$ and${\mathrm{y}}^{\text{'}\text{'}}\left(\mathrm{x}\right).$

We have

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

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

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

We will also use:

$\mathrm{cosx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}}$

$=1-\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}-\frac{{\mathrm{x}}^6}{6!}+\dots$

Substitute the equation

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

$\cos \left(\mathrm{x}\right){\mathrm{y}}^{\text{'}\text{'}}+\mathrm{y}=0$

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

$\left(1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{24}-\frac{{\mathrm{x}}^6}{720}+\dots \right)+(2{\mathrm{c}}_2+6{\mathrm{c}}_3\mathrm{x}+12{\mathrm{c}}_4{\mathrm{x}}^2+20{\mathrm{c}}_5{\mathrm{x}}^3+30{\mathrm{c}}_6{\mathrm{x}}^4+\dots )+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0$

$\left(2{\mathrm{c}}_2+6{\mathrm{c}}_3\mathrm{x}+(12{\mathrm{c}}_4-{\mathrm{c}}_2){\mathrm{x}}^2+(20{\mathrm{c}}_5-3{\mathrm{c}}_3){\mathrm{x}}^3+\left(30{\mathrm{c}}_6-6{\mathrm{c}}_4+\frac{1}{12}{\mathrm{c}}_2\right)\right){\mathrm{x}}^4+\dots )+({\mathrm{c}}_0+{\mathrm{c}}_1\mathrm{x}+{\mathrm{c}}_2{\mathrm{x}}^2+{\mathrm{c}}_3{\mathrm{x}}^3+{\mathrm{c}}_4{\mathrm{x}}^4+{\mathrm{c}}_5{\mathrm{x}}^5+..)=0$

role="math" localid="1668573995497" $\left(2{\mathrm{c}}_2+{\mathrm{c}}_0+({\mathrm{c}}_1+6{\mathrm{c}}_3)\mathrm{x}+(12{\mathrm{c}}_4-{\mathrm{c}}_2+{\mathrm{c}}_2){\mathrm{x}}^2+(20{\mathrm{c}}_5-3{\mathrm{c}}_3+{\mathrm{c}}_3){\mathrm{x}}^3+\left(30{\mathrm{c}}_6-6{\mathrm{c}}_4+\frac{1}{12}{\mathrm{c}}_2+{\mathrm{c}}_4\right){\mathrm{x}}^4+\dots \right)=0$

$\left(2{\mathrm{c}}_2+{\mathrm{c}}_0+({\mathrm{c}}_1+6{\mathrm{c}}_3)\mathrm{x}+\left(12{\mathrm{c}}_4\right){\mathrm{x}}^2+(20{\mathrm{c}}_5-2{\mathrm{c}}_3){\mathrm{x}}^3+\left(30{\mathrm{c}}_6-5{\mathrm{c}}_4+\frac{1}{12}{\mathrm{c}}_2\right){\mathrm{x}}^4+\dots \right)=0$

Identity property

$c_0+2c_2=0\to c_2=-\frac{c_0}{2}$

$c_1+6c_3=0\to c_3=-\frac{c_1}{6}$

$c_4=0$

$c_5=\frac{1}{60}$

$c_6=0$

${\mathrm{y}}_1\left(\mathrm{x}\right)=\mathbf{x}-\frac{1}{6}{\mathbf{x}}^3-\frac{1}{60}{\mathbf{x}}^5+\dots$

For${\mathbf{c}}_0=\mathbf{1}$ and${\mathbf{c}}_1=\mathbf{0}$ we have:

${\mathrm{c}}_2=-\frac{1}{2}$

${\text{c}}_3=0$

$c_4=0$

$c_5=0$

${\mathrm{c}}_6=\frac{1}{720}$

${\mathrm{y}}_2\left(\mathrm{x}\right)=1-\frac{1}{2}{\mathrm{x}}^2+\frac{1}{720}{\mathrm{x}}^6+..$