MHB What Are the Singular Points of the Differential Equation w'' + z*w' + kw = 0?

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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 $$z=\infty$$. Is that right?

I did a transformation x=1/z and examined the singular point at x=0 and found that the limit as $$x\to0$$ gives $$2-1/x^2$$ 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 $$z\to\infty$$ for each of the two solutions. Does this mean just put it in a power series solution $$w=\sum_{n=0}^\infty a_nz^n$$ and see what happens?

Thanks!
 
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CWolf said:
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 $$z=\infty$$. Is that right?

I did a transformation x=1/z and examined the singular point at x=0 and found that the limit as $$x\to0$$ gives $$2-1/x^2$$ 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 $$z\to\infty$$ for each of the two solutions. Does this mean just put it in a power series solution $$w=\sum_{n=0}^\infty a_nz^n$$ 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

$\chi$ $\sigma$
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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