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fluidistic
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Homework Statement
I must find the constant K such that [itex]y''-\left ( \frac{1}{4}+\frac{K}{x} \right )y=0[/itex] for x>0 has a non trivial solution that is worth 0 when x tends to 0 and when x tends to infinity.
Homework Equations
Frobenius method.
The Attempt at a Solution
I proposed a solution of the form [itex]y(x)=\sum_{n=0}^{\infty}a_nx^{n+s}[/itex].
This gave me [itex]y''(x)=\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2}[/itex].
Plugging y and y'' into the DE I get that [itex]\sum _{n=2}^\infty (n+s)(n+s-1)a_nx^{n+s-2} - \frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=0}^{\infty}a_nx^{n+s-1}=0[/itex].
Now I must equate the powers of x inside each series in order to put all the arguments of the series under a single series.
So let's say I want to have as common power of x, n+s.
This gives me [itex]\sum_{n=0}^{\infty}a_{n+2}x^{n+s} (n+s+2)(n+s+1)-\frac{1}{4}\sum_{n=0}^{\infty}a_nx^{n+s}+K\sum_{n=1}^{\infty}a_{n+1}x^{n+s}=0[/itex].
I separate the first term of the series in order to get the indicial equation.
Hence [itex]Ka_0 x^{s-1}+\sum_{n=0}^{\infty}x^{n+s} [a_{n+2}(n+s+2)(n+s+1)-\frac{1}{4}a_n+Ka_{n+1}]=0[/itex].
So that the indicial equation is [itex]Ka_0=0[/itex]. I think this implies that [itex]a_0=0[/itex].
Setting the argument of the infinite series equal to 0 gives me the recurrence relation [itex]a_{n+2}(n+s+2)(n+s+1)-\frac{a_n}{4}+Ka_{n+1}=0[/itex].
Now I know I should get all terms in function of [itex]a_1[/itex], but I don't know how to mathematically show this.
For example if I calculate [itex]a_2[/itex], I get that it's worth [itex]-\frac{Ka_1}{(s+1)(s+2)}[/itex]. I'd get [itex]a_3[/itex] in terms of [itex]a_2[/itex] and [itex]a_1[/itex] and thus of [itex]a_1[/itex] alone, etc.
So how can I proceed further?
EDIT: Nevermind, I've made lots of errors, I'm going to redo all.
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