Question

I don't know how to start,
Solve near point 0 ;
x^2y'' -(x-2/9) y =0
Please, help

Answer 1

\[x^2y''-\left(x-\frac{2}{9}\right)y=0\]
or
\[x^2y''-\frac{x-2}{9}y=0~~?\]

Answer 2

the first one

Answer 3

You WILL need a plan of attack. Series? Systems? Laplace Transform? What?

Answer 4

we don't have initial condition , how to use Laplace?

Answer 5

@tkhunny, "near point x=0", so I would assume series. That, and the fact that OP has posted similar questions.

Answer 6

It doesn't look like there are any non-zero solutions. At least, according to what I've worked through.

Answer 7

@SithsAndGiggles the problem doesn't have Q(x), or y' =0, how can I test whether 0 is singular regular point by taking limit?

Answer 8

@SithsAndGiggles I agree. Just trying to get @Loser66 to have an opinion and offer some reassurance that there is a chance of solving the problem.

Answer 9

@Loser66, \(x=0\) could be an irregular singular point: https://en.wikipedia.org/wiki/Regular_singular_point#Examples_for_second_order_differential_equations

Answer 10

anyone, please

Answer 11

so far, I got, but don't know how to do next