Question

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

8. e^xy"-(x²-1)y'+2xy=0

Short answer

There is no singularity point that exists in this differential equation for both ^P(X) and Q(X).

Step-by-step solution

Step 1: Ordinary and Singular Points

A point^X0 is called an ordinary point of 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

e^xy"-(x²-1)y'+2xy=0

Dividing the above equation by e^x we get,

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

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

$Q\left(x\right)=\frac{2x}{e^x}$

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

For ^Px) this occurs at no point. In this case both ^Px) and^Q(x) are analytic at all points.

Therefore, there is no singularity point that exists in this differential equation for both ^Px) and^Q(x) .