# Identifying Types of Singularity in Differential Equations

#### QFT25

1. Homework Statement
Identify the type of singularity at x=0 for these differential equations

x*Sin[1/x]*y''[x]+y[x]==0

x^2*y''[x]+Sin[1/x]*y[x]==0

2. Homework Equations
A Singular point is regular if f(x)(x-x_0)^n is defined as x approaches x_0 and is analytic in a near a neighborhood of that singular point. It is irregular if this doesn't hold.

3. The Attempt at a Solution

x/(Sin[1/x]) x is analytic at x=0 and around x=0 to see if this was a regular or irregular singularity. So I wrote out the Taylor series of Sin[x] and plugged in for x 1/x. After doing the it appeared that as x goes to zero x/Sin[1/x] went to zero so I said it was a regular singularity, I also identified other singular points at 1/(Pi*n) where n is a integer. Using the same process for the other ODE I found that as x goes to zero (x^2/(Sin[1/x)]))x^2 does not converge as x goes to zero thus it is a irregular singularity. Can anyone tell me if my approach was right and if my answers were right?

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#### FactChecker

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2018 Award
it appeared that as x goes to zero x/Sin[1/x] went to zero
What did you do about all the divisions by 0?

#### QFT25

What did you do about all the divisions by 0?
The series looks like x/(1/x -1/x^3+1/x^5...) so as x goes to zero the numerator goes to zero and the denominator increases without bound because of all of the (1/x) terms. So in my head I was like zero over infinity is zero.

#### FactChecker

Gold Member
2018 Award
The series looks like x/(1/x -1/x^3+1/x^5...) so as x goes to zero the numerator goes to zero and the denominator increases without bound because of all of the (1/x) terms. So in my head I was like zero over infinity is zero.
For every x = 1/nπ, x / sin(1/x) = (1/nπ) / ( sin(nπ) ) = (1/nπ) / 0. That is a series of points of x approaching 0 where the function is undefined.

#### QFT25

For every x = 1/nπ, x / sin(1/x) = (1/nπ) / ( sin(nπ) ) = (1/nπ) / 0. That is a series of points of x approaching 0 where the function is undefined.
Yes Sin[1/x] as x goes to zero is undefined. However Sin[1/x] ranges between as -1,0 and 1. Also as x/(Sin[x+Pi]) as x goes to zero approaches negative -1. Hmm now that I think I think about it is un defined right because it will go between 0 and -1 right. So it is irregular point right?

#### FactChecker

Gold Member
2018 Award
Think about "is analytic in a neighborhood of that singular point." Is that true?

#### QFT25

Think about "is analytic in a neighborhood of that singular point." Is that true?
No because Sin[1/z] at z=0 is not a isolated singularity so it isn't analytic in in a neighborhood of z or in this case x.

"Identifying Types of Singularity in Differential Equations"

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