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

4. (x²+x)y"+3y'-6xy = 0

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

The given differential equation is,

(x²+x)y"+3y-6xy = 0

Dividing the above equation by (x²+x)we get,

y"+$\frac{3}{(x^2+x)}y\text{'}-\frac{6x}{(x^2+x)}y=0$

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

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

Q(x) =$-\frac{6x}{x(x+1)}$

=$-\frac{6}{(x+1)}$..................................(1)

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

For ^P(x) this occurs at x = 0,-1 .

We see that ^Q(x) is actually analytic at x = 0 .

Since we can cancel an x in the numerator and denominator of ^Q(x)as shown in (1).

Therefore, ^P(x)is analytic except at x = 0,-1 as well as ^Q(x)is analytic except at x = -1 .

The singular point exists in this differential equation for both ^P(x)and ^Q(x)is at x = 0,-1 .