Question

Without actually solving the differential equation$\left(cosx\right)y^{\text{'}\text{'}}+y^{\text{'}}+5y=0,$find the minimum radius of convergence of power series solutions about the ordinary point$x=0$.About the ordinary point$x-1$

Short answer

The minimum radius of convergence is about $x=0$and $x=1$is$\pi /2$ and$0.5708,$ respectively.

Step-by-step solution

Given Information

The given differential equation is:

$\left(\cos x\right)y\text{'}\text{'}+y\text{'}+5y=0$

Find the minimum radius of convergence of the power series solution of the given differential equation.

First, let's assume we have a differential equation with the following form
$a_1\left(x\right)y\text{'}\text{'}+a_2\left(x\right)y\text{'}+a_3\left(x\right)y=0$

$y\text{'}\text{'}+\frac{a_2\left(x\right)}{a_1\left(x\right)}y\text{'}+\frac{a_3\left(x\right)}{a_2\left(x\right)}y=0$

[Dividing byrole="math" localid="1655138728255" $a_1\left(x\right)$]

$y\text{'}\text{'}\text{'}+P\left(x\right)y\text{'}+Q\left(x\right)y=0 \dots \left(1\right)$

Where we have;

$P\left(x\right)=\frac{a_2\left(x\right)}{a_1\left(x\right)}$and $Q\left(x\right)=\frac{a_3\left(x\right)}{a_1\left(x\right)} \dots \left(2\right)$

Equation (l) is the standard form, upon which we will build our analysis. We recall the following definition.

Definition 1.

A point $x=x_0$is said to be an ordinary point of the differential equation, if $p\left(x\right)$and$Q\left(x\right)$ are analytic at that point, otherwise$x_0$ is said to be a singular point.

To Taylor Series

Investigating (2), we will be able to deduce that $p\left(x\right)$and$Q\left(x\right)$are not analytic at the points, which will make the denominator equal to zero, after common factors reduction. Now, we put the given differential equation in the standard form.

$\frac{\cos x}{\cos x}y\text{'}\text{'}+\frac{1}{\cos x}y\text{'}+\frac{5}{\cos x}y=0$

yields,

$P\left(x\right)=\frac{1}{\cos x}$and $Q\left(x\right)=\frac{5}{\cos x} \dots \left(3\right)$

Now, we equate the denominator by 0 and find the values which satisfy this condition.

Therefore,$p\left(x\right)$and $Q\left(x\right)$are not analytic at$\text{x}=\pi /2$and $\text{x}=\pi /2,$which implies that the differential equation has a singular point at $\text{x}=\pi /2,$and $x=1$Now, we have the singular points. We are left with finding the minimum radius of convergence about$x=0$and $x=1$

About $x=0$

$R=\left|\frac{\pi }{2}-0\right|=\left|-\frac{\pi }{2}-0\right|=\frac{\pi }{2}$

Therefore, the minimum radius of convergence about $x=0$is$\text{x}=\pi /2.$

About$x=0$

Therefore, the minimum radius of convergence about$x=1$ is$0.5708.$

The minimum radius of convergence is about$x=0$ and $x=1$is$\pi /2$ and$0.5708,$ respectively.