(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For what values of K does the DE [itex]xy''-2xy'+(K-3x)y=0[/itex] (1) has a bounded solution in [itex](0, \infty)[/itex]?

2. Relevant equations

Not so sure, Frobenius method maybe.

3. The attempt at a solution

First, I check what happens when x tends to infinity. I see that the DE behaves like [itex]\phi ''-2 \phi '-3 \phi =0[/itex]. Where [itex]\phi (x)=c_1e^{3x}+c_2e^{-x}[/itex]. I checked out and this indeed satisfies the DE (2).

I propose a solution to (1) of the form [itex]y(x)=f(x) \phi (x)[/itex]. So I calculated y' and y'' and replaced all this into the DE (1) and after a lot of algebra I reached that the DE (1) is equivalent to the DE [itex]f''(x)+4f'(x)+f(x) \left ( \frac{Kc_1e^{3x}}{x} \right ) \cdot \left ( \frac{1}{c_1e^{3x}+c_2e^{-x}} \right )=0[/itex] (3) where these [itex]c_1[/itex] and [itex]c_2[/itex] differs from the ones given in the expression for [itex]\phi (x)[/itex] but this doesn't matter, they are constants. Now I must say I'm not really eager to use Frobenius's method on the DE (3). Mainly due to the denominator. I don't really know how to proceed... any help will be appreciated as usual.

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# Mathematical methods of Physics, ODE

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