Series solution about $x=0$ of $xy''-y'+4xy=0$.

I want to find at least one solution of the differential equation $$xy''-y'+4xy=0$$ about the point $$x=0$$. I identified that $$x=0$$ is a regular singular point and thus Frobenius Theorem is applicable.

Now, assuming a solution of the form $$y(x)={x^r}\sum_{n=0}^\infty {c_n x^n}=\sum_{n=0}^\infty {c_n x^{n+r}}$$

I get $$y'=\sum_{n=0}^\infty {(n+r)c_n x^{n+r-1}}$$ and $$y''=\sum_{n=0}^\infty {(n+r-1)c_n x^{n+r-2}}$$.

Substituting, and simplifying, I end up with the equation: $$\sum_{n=0}^\infty {(n+r)(n+r-1)c_n x^{n+r-1}} - \sum_{n=0}^\infty {(n+r)c_n x^{n+r-1}} + \sum_{n=2}^\infty {4 c_{n-2} x^{n+r-1}}$$

$$\implies r(r-2)c_0 x^{r-1} + (r^2-1)c_1x^r +\sum_{n=2}^\infty {[(n+r)(n+r-2)c_n + {4}c_{n-2}]x^{n+r-1}}=0$$

In this case, I have two indicial equations, $$r(r-2)=0$$ and $$r^2-1=0$$, giving $$r = 2, 0, 1, -1$$.

At this point, I would have expected to only have two values for $$r$$, but now I have four values.

How should I proceed in terms of using the values of $$r$$ to obtain a series solution? Is it correct to simply set $$c_1=0$$ and just use the indicial equation $$r(r-2)=0$$, and using the larger root $$r=2$$? If yes, then is there a general rule for which indicial equation to use when faced with more than two values of $$r$$ in the case of a second order differential equation?

Note: Maple gave one solution to be $$y_1(x) = x^2\{1 - \frac{1}{2}x^2 + \frac{1}{12}x^4+O(x^6)\}$$

• The case $r^2-1=0$ gives an index-shifted of the true indicial equation $r(r-2)=0$, so it is (always) sufficient to only consider the lowest-degree coefficient. – Lutz Lehmann Apr 21 '19 at 19:54

Note that $$\sum_{n=0}^\infty c_n x^{n+r}=\sum_{n=0}^\infty a_nx^n,$$ where $$a_n=0$$ for $$n and $$a_n=c_{n-r}$$ otherwise. Thus, there's no particular benefit to the multiplication by $$x^r,$$ in the first place.
If we skip that part, instead taking that $$y(x)=\sum_{n=0}^\infty c_n x^n,$$ so that $$y'(x)=\sum_{n=0}^\infty c_n nx^{n-1}=\sum_{n=1}^\infty nc_nx^{n-1}=\sum_{n=0}^\infty(n+1)c_{n+1}x^n$$ and so $$y''(x)=\sum_{n=0}^\infty(n+1)nc_{n+1}x^{n-1}=\sum_{n=1}^\infty(n+1)nc_{n+1}x^{n-1}=\sum_{n=0}^\infty(n+2)(n+1)c_{n+2}x^n,$$ then we obtain
$$\begin{eqnarray}0 &=& xy''-y'+4xy\\ &=& \sum_{n=0}^\infty(n+2)(n+1)c_{n+2}x^{n+1}-\sum_{n=0}^\infty(n+1)c_{n+1}x^n+\sum_{n=0}^\infty 4c_n x^{n+1}\\ &=& -c_1+\sum_{n=0}^\infty(n+2)(n+1)c_{n+2}x^{n+1}-\sum_{n=1}^\infty(n+1)c_{n+1}x^n+\sum_{n=0}^\infty 4c_n x^{n+1}\\ &=& -c_1+\sum_{n=0}^\infty(n+2)(n+1)c_{n+2}x^{n+1}-\sum_{n=0}^\infty(n+2)c_{n+2}x^{n+1}+\sum_{n=0}^\infty 4c_n x^{n+1}\\ &=& -c_1+\sum_{n=0}^\infty\bigl[(n+2)(n+1)c_{n+2}-(n+2)c_{n+2}+4c_n\bigr]x^{n+1}\\ &=& -c_1+\sum_{n=0}^\infty\bigl[(n+2)nc_{n+2}+4c_n\bigr]x^{n+1}.\end{eqnarray}$$ Thus, $$c_1=0,$$ and $$(n+2)nc_{n+2}=-4c_n$$ for all $$n.$$ Readily, then, we see that $$c_n=0$$ for all odd $$n$$ by induction. Letting $$c_0=c,$$ we can then use induction to prove a formula for all even $$n.$$
Edit: The above is based on my assumption that $$r$$ was a nonnegative integer, which is not necessarily the case. However, there is still an important take away.
Note that if $$c_m$$ is the first non-zero coefficient, then $$x^r\sum_{n=0}^\infty c_nx^n=x^{r+m}\sum_{n=0}^\infty a_nx^n,$$ where $$a_n=c_{m+n}$$ for all $$n.$$ Thus, we will always assume that the first coefficient is non-zero. That is, $$c_0\ne 0.$$
Thus, since we need $$r(r-2)c_0=0,$$ then we have $$r(r-2)=0,$$ so $$r=0$$ (in which case the work I did above is correct) or $$r=2$$ (in which case the solution is some scalar multiple of $$y_1$$). From there, we immediately see that in either case, $$c_1=0,$$ and in general find that $$c_n=-\frac{4}{(n+r)(n+r-2)}c_{n-2}$$ for $$n\ge 2.$$