Q 2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: In Problems 1–10, determine all the singular points of the given differential equation.

2. x²y"-3y-xy = 0

Step-by-Step Solution

Answer

The only singularity point exists in this differential equation for both ^P(x) and ^Q(x) is at ^{x = 0}.

Step 1: Ordinary and Singular Points.

A point ^$x_0$ is called an ordinary point of equation u"+p(x)y'+q(x)y = 0 if both p and q are analytic at ^$x_0$ . If ^$x_0$ is not an ordinary point, it is called a singular point of the equation.

Step 2: Find the singular points.

The given differential equation is

 x²y"-3y'-xy = 0

 Dividing the above equation by x² we get,

 

$y"-\frac{3}{x^2}y\text{'}-\frac{x}{x^2}y=0$

 

On comparing the above equation with y"+p(x)y'+q(x)y = 0, we find that,

 P(x)=$-\frac{3}{x^2}$

Q(x) =$-\frac{x}{x^2}$

       =$-\frac{1}{x}$            

 

Hence, ^P(x) and ^Q(x) are analytic except, perhaps, when their denominators are zero.

 

For ^P(x) this occurs at ^{x = 0}.

 

We see that ^P(x) is actually analytic at ^{x = 0} as well as ^Q(x) is analytic except at ^{x = 0}.

 

The only singularity point exists in this differential equation for both ^P(x) and Q(x) is at ^{x = 0}.