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$$x^2\frac{d^2y}{dx^2}+(x^3-x)\frac{dy}{dx}-3y=0$$

$$\sum_{n=0}^{\infty}(n+r)(n+r-2)c_nx^{n+r}+\sum_{n=2}^{\infty}(n+r-2)c_{n-2}x^{n+r}-3\sum_{n=0}^{\infty}c_nx^{n+r}=0$$

$$r(r-2)c_0x^r+(r^2-1)c_1x^{1+r}-3c_0x^r-3c_1x^{r+1}+\sum_{n=2}^{\infty}[(n+r)(n+r-2)c_n+(n+r-2)c_{n-2}-3c_n]x^{n+r}=0$$

The indicial equation is

$$(r-3)(r+1)=0$$

We have two possibility

Deal with the larger value $3$ first,

$$c_n=-\frac{(n+1)c_{n-2}}{n^2+4n}$$

Taking values for $n\geq2$

$$c_2=-\frac{c_0}{4}$$

$$c_3=0$$

$$c_4=-\frac{5c_2}{32}$$

$$c_5=0$$

$$y_1(x)=C_1x^3(1-\frac{x^2}{4}+\frac{5x^4}{128}+...)$$

The second possibility $r=-1$

$$c_n=-\frac{(n-3)c_{n-2}}{n^2-4n},n\neq4$$

we know that $c_4 is an arbitrary constant

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

$$c_2=-\frac{c_0}{4}$$

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

$$c_3=0$$

$$c_5=-\frac{2c_3}{5}=0$$

$$c_6=-\frac{c_4}{4}$$

$$c_7=0$$

$$c_8=-\frac{5c_6}{32}$$

$$y_2=(x^{-1})c_4(x^4-\frac{x^6}{4}+\frac{5x^8}{128}+...)$$

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.

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  • $\begingroup$ 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 ? $\endgroup$ – Donald Splutterwit Jul 29 '17 at 10:10
  • $\begingroup$ Nope. I did not show that part. It is right i think. $\endgroup$ – Crazy Jul 29 '17 at 10:12
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    $\begingroup$ 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 '17 at 13:07
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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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