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Find the first three non-zero terms in series solutions $y_1(x)$ and $y_2(x)$ for $$x\frac{d^2y}{dx^2}-\frac{dy}{dx}+4x^3y=0$$...

I've seen the solutions as this problem has been presented here before but what I'm really hoping for is for someone to read through what I've done and tell me what I'm doing wrong.

What I've done:

I identified $x=0$ as a regular singular point and so set $y=\sum^{\infty}_{n=0}a_nx^{n+\sigma}$ and this reduced to the indicial equation $\sigma=\sigma (\sigma -1)$ and so $\sigma =0$ or $\sigma =2$.

For $\sigma =0$ I had that $a_1=a_3=0$ and $a_0$ and $a_2$ are arbitrary and $a_{n+4}=-\frac4{(n+2)(n+4)}a_n$, but this gives two linearly independent solutions. So when I put in $\sigma=2$ I'm going to get even more solutions, i.e. more than 2 linearly independent solutions in total. What am I misunderstanding?

Any help is appreciated, thank you

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    the solution containes the Heun-function – 2017-01-06
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    The solution to $\sigma=2$ corresponds to the choice $a_0=0$ and $a_2$ arbitrary. The special thing is there that the difference between the two solutions for $\sigma$ is an integer. – 2017-01-06
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    @Fabian Sorry if I'm being an idiot here but you're saying that for $\sigma =2$ I'll just get one of my solutions again? (I get for $\sigma =0, a_0(1-\frac12x^4+\frac1{24}x^8$ and $a_2(x^2-\frac16x^6+\frac1{120}x^10$ so for $\sigma =2$). Also, what is the significance of $\sigma _1-\sigma _2$ being an integer? In my notes I have that this means that the second solution will be of the form $(x-x_0)^{\sigma 2}\sum^{\infty}_{n=0}b_n(x-x_0)^n+cy_1\log (x-x_0)$ where we could have $c=0$ but I don't see what it's effect is here. Sorry if I'm being frustratingly slow and thank you for your comment. – 2017-01-06
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    @user307463: here you have $c=0$ and the solution with $\sigma_2=2$ just corresponds to the $a_2$ solution written above. If you plug your second ansatz into the differential equation, you will see that $c=0$. If you want to study an equation where the term with $c$ appears, look at the related $x^2 y'' +x y' + x^2 y=0$. – 2017-01-07
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    The significance of $\sigma_2 - \sigma_1$ being an integer is that you can get the terms of $x^{\sigma_2+n}$ from $x^{\sigma_1 + m}$ with m some larger integer. In the approach, you are using this makes everything a bit more cumbersome (one thing is that suddenly the term $c y_1 \log(x-x_0)$ can appear, another thing is that the second solutions can just give you one of the first solutions again. There is a good description of the method in the Chapter 3.3 of [Bender and Orzag](http://www.springer.com/de/book/9780387989310). – 2017-01-07

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