Question

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

1. (x+1)y"-x²y'+3y = 0

Short answer

The only singularity point exists in this differential equation for both P(x) and Q(x) is at x = -1.

Step-by-step solution

Ordinary and Singular Points

A point x₀ is called an ordinary pointof equation y"+p(x)y'+q(x)y = 0 . if both pand qare analytic at x₀. If x₀ is not an ordinary point, it is called a singular point of the equation.

Find the singular points

The given differential equation is

(x+1)y"-x²y'+3y = 0

Dividing the above equation by (x+1)we get,

$y"-\frac{x^2}{(x+1)}y\text{'}+\frac{3}{(x+1)}y=0$

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

P(x)$=\frac{x^2}{(x+1)}$

Q(x)$=\frac{3}{(x+1)}$

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

For P(x) this occurs at x = -1.

We see that P(x) is actually analytic at x = 0 . Therefore, P(x) is analytic except at x = -1 as well as Q(x) is analytic except at ^{x = -1}.

The only singularity point exists in this differential equation for both P(x) and Q(x) is at x = -1.