Q 7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: 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 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,

 (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.