Exploring Singular Points of a DE: w''+z*w'+kw=0

[Ice Vox]

Hi,
I'm asked to find and classify the singular points of a function w(z) in the differential equation:

w''+z*w'+kw=0 where k is some unknown constant.

The only singular point I notice is \(\displaystyle z=\infty\). Is that right?

I did a transformation x=1/z and examined the singular point at x=0 and found that the limit as \(\displaystyle x\to0\) gives \(\displaystyle 2-1/x^2\) which blows up, meaning the singular point is irregular. Does that mean that the singular point z=infinity is irregular also?

It also asks me to find the first term of the asymptotic solution as \(\displaystyle z\to\infty\) for each of the two solutions. Does this mean just put it in a power series solution \(\displaystyle w=\sum_{n=0}^\infty a_nz^n\) and see what happens?

Thanks!

 

[chisigma]

Hi,
I'm asked to find and classify the singular points of a function w(z) in the differential equation:

w''+z*w'+kw=0 where k is some unknown constant.

The only singular point I notice is \(\displaystyle z=\infty\). Is that right?

I did a transformation x=1/z and examined the singular point at x=0 and found that the limit as \(\displaystyle x\to0\) gives \(\displaystyle 2-1/x^2\) which blows up, meaning the singular point is irregular. Does that mean that the singular point z=infinity is irregular also?

It also asks me to find the first term of the asymptotic solution as \(\displaystyle z\to\infty\) for each of the two solutions. Does this mean just put it in a power series solution \(\displaystyle w=\sum_{n=0}^\infty a_nz^n\) and see what happens?

Thanks!

Wellcome on MHB CWolf!...

In general a second order linear homogeneous DE can be written as...

$\displaystyle y^{\ ''} + f_{1} (x)\ y^{\ '} + f_{2} (x)\ y =0\ (1)$

We have the following possible cases...

a) if $f_{1} (x)$ and $f_{2} (x)$ are both analytic in x=a, then x=a is an ordinary point...

b) if $f_{1} (x)$ has a pole up to order 1 and $f_{2}(x)$ a pole up to order 2 in x=a, then x=a is a regular singular point...

c) in any other case x=a is a singular point...

In Your case is $f_{1}(x) = x$ and $f_{2} = k$, both analytic in $\mathbb R$, so that there are no singular points...

Kind regards

$\chi$ $\sigma$