Question

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

7. (sinx)y"+(cosx)y =0

Short answer

The singular point exists in this differential equation for^Q(x) is at ^x=$n\pi$ where ⁿ is any integer

Step-by-step solution

Ordinary and Singular Points.

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.

Find the singular points.

The given differential equation is,

(sin x)y"+(cos x )y= 0

Dividing the above equation by we get,

y" +$\left(\frac{\cos x}{\sin x}\right)y=0$

Observing the above equation, we find that,

P(x)= 0 ,

Q(x) =$\frac{\cos x}{\sin x}$

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

For ^Q(x)this occurs at sin x = 0which implies x =$n\pi$ where n is any integer.

Therefore, ^Q(x)is analytic except at x =$n\pi$ .

The singular point exists in this differential equation for ^Q(x)is at x =$n\pi$where ⁿ is any integer.