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Find two linearly independent series solutions to $$2xy''+y'+xy=0$$ ($x_0$ is a regular singular point),
Edit
So, $$y=\sum^\infty_{n=0}a_nx^{n+r}$$$$y'=\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}$$ and $$y''=\sum^\infty_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}$$
So sum works out to; $$\sum^\infty_{n=0}2(n+r)(n+r-1)a_nx^{n+r-1}+\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}+\sum^\infty_{n=0}a_nx^{n+r+1}$$ $$=$$ $$\sum^\infty_{n=0}2(n+r)(n+r-1)a_nx^{n+r-1}+\sum^\infty_{n=0}(n+r)a_nx^{n+r-1}+\sum^\infty_{n=2}a_{n-2}x^{n+r-1}$$ Equating coeffiecents, when $n=0$, $$2r(r-1)+r=0$$ $$r=0, r=\frac{1}{2}$$ When $n=1$ $a_1=0$ for $r=\frac{1}{2}$ and $a_1$ is $0$ for $r=0$ Since $ a_1$ is equal to zero, then there are no odd term in the series. I then got $ a_n = \frac {a_02^{\frac {n}{2}}(-1)^n}{(2n+1)! (2n)!}$ But I can't think of a what I'd do next...

Edit

i've answered the qs

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Codefailure
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  • I think it is "natural" to do $;n=0;$ in the derivatives of the series: it is not a mere power series but rather the variable's exponent is $;n+r;$ ...! If it were $;n;$ then yes: it is pointless to begin at $;n=0;$ in the derivative as that'd equal zero. – DonAntonio Oct 31 '16 at 10:28
  • @DonAntonio. May be pointless but, from my point of view, very useful since making my life easier when playing with the same powers of $x$. – Claude Leibovici Oct 31 '16 at 10:38
  • @Am I think you may have some mistake in your differential equation (2nd. line), as you seem to have done something different when writing it with series. – DonAntonio Oct 31 '16 at 10:40
  • @ClaudeLeibovici I really don't think it makes life easier in general, but if that helps you then why not? – DonAntonio Oct 31 '16 at 10:40
  • @DonAntonio. This is exactly what I was meaning ! But, remembering my old times, this is also what we learnt (at that time) in France. – Claude Leibovici Oct 31 '16 at 10:49
  • @Am Shouldn't the differential equation be equal to zero, for example? – DonAntonio Oct 31 '16 at 10:52
  • @DonAntonio. Oh, i see, thanks for clearing that up. When i made $n=0$ i got a somewhat different answer.... In the 2nd line? Do you mean that its $n+r-1$ and not $n+r-2$?? – Codefailure Oct 31 '16 at 10:56
  • @DonAntonio yh, its equal to 0 i just forgot to put it in – Codefailure Oct 31 '16 at 10:57
  • @AmM Not only that: the diff. eq. seems to be $;2xy''+y'+xy=0;$ , and you wrote $;...+ x\color{red}{y'}=0;$ there – DonAntonio Oct 31 '16 at 10:59
  • @DonAntonio oooh, that was a mistake as well haha, i'll edit that now, thanks! – Codefailure Oct 31 '16 at 11:06

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