The indicial equation is


We have two possibility

Deal with the larger value $3$ first,


Taking values for $n\geq2$






The second possibility $r=-1$


we know that $c_4 is an arbitrary constant

We know that $c_4$ will make $c_2=0$ from the above equation


We see that it contradicts our initial assumption that $c_0\neq0$







Obviously they are two sides of the same coin.

How to find another linearly independent solution $y_2$. Is there a closed form for the above series?

The thing seems very messy.

  • $\begingroup$ State clearly $y=\sum_{n=0}^{\infty}c_nx^{n+r}$ ... have you lost the factor $(x^3-x)$ in the middle trem of the second line ? $\endgroup$ Jul 29, 2017 at 10:10
  • $\begingroup$ Nope. I did not show that part. It is right i think. $\endgroup$
    – Crazy
    Jul 29, 2017 at 10:12
  • 2
    $\begingroup$ When the roots of the indicial equation differ by an integer, there are problems. The second solution is $\lnx$ times the first solution plus another series. It's so messy that most DE textbooks skip it. See math.stackexchange.com/questions/963934/… $\endgroup$
    – B. Goddard
    Jul 29, 2017 at 13:07

1 Answer 1


Wolfram Alpha gives the solutions as $$ y_1(x) = x e^{-x^2/4}~I_1(x^2/4) = \frac{x^3}{8} \left( 1 - \frac{x^2}{4} + \frac{5 x^4}{128} - \frac{7 x^6}{1536} + \dots\right) $$ which is the one you found in the expansion. The other solution diverges at $x=0$ and is $$ y_2(x) = x e^{-x^2/4}~K_1(x^2/4) $$ $I_1$ and $K_1$ are modified Besselfunctions.


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