The value of 1/x explodes at x = 0. Thus, it is singular at x = 0. But what happens at x= infinity?
e.g. in the differential equation, y"+(1/x)y=0, does x = infinity is an ordinary point, regular singular point or irregular singular point??
Please help with this.
Nothing happens at infinity; y=1/x is is regular function there, having zero limit.
At x->inflty you equation asymptotically degenerates to y''=0. However, you can try to solve this rather simple equation directly. Maybe you can find something here: http://www.sosmath.com/diffeq/second/second.html
@Yuri Yegorov Sir, I thought on the similar lines. However, if we perform the transformation as x = 1/t, then the differential equation in y(t) has singularity at t = 0. Doesn't that mean the differential equation in y(x) has singularity at x = infinity?
Pranav, was this equation in math or physics? Because it looks like stationary Schrodinger equation for the potential U=1/x. And there should be a lot of literature about that if it does not have exact solution.
You can try other substitutions, like z=y/x. It is a piece of art to solve nonlinear equations exactly. There was a book of Kamke about all known solutions.
http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf
Sir, I obtained similar equation in the domain of applied mechanics. So, I was looking at this particular problem. Thank you for the book and information.
The answers of Yuri are excellent.
I only see a clarification to make regarding regularity, because of your change of variable. Regularity refers to metrics (I do not want to have an abstract topological view of that), and you do not define a metric at infinity (it may be tricky). The usual y''+y=0 is also not regular at infinity with your change of variable (what can be seen since it oscillates at infinity). So your approach of regularity at infinity probably needs some refinement.
Arnaud
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