First note that
![x=0](https://tex.z-dn.net/?f=x%3D0)
is a regular singular point; in particular
![x=0](https://tex.z-dn.net/?f=x%3D0)
is a pole of order 1 for
![\dfrac2x](https://tex.z-dn.net/?f=%5Cdfrac2x)
.
We seek a solution of the form
![y=\displaystyle\sum_{n\ge0}a_nx^{n+r}](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%7D)
where
![r](https://tex.z-dn.net/?f=r)
is to be determined. Differentiating, we have
![y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}](https://tex.z-dn.net/?f=y%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2Br%29a_nx%5E%7Bn%2Br-1%7D)
![y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}](https://tex.z-dn.net/?f=y%27%27%3D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7D%28n%2Br%29%28n%2Br-1%29a_nx%5E%7Bn%2Br-2%7D)
and substituting into the ODE gives
![\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5Csum_%7Bn%5Cge0%7D%28n%2Br%29%28n%2Br-1%29a_nx%5E%7Bn%2Br-2%7D%2B2%5Csum_%7Bn%5Cge0%7D%28n%2Br%29a_nx%5E%7Bn%2Br-1%7D-x%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%7D%3D0)
![\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bn%5Cge0%7D%28n%2Br%29%28n%2Br-1%29a_nx%5E%7Bn%2Br-1%7D%2B2%5Csum_%7Bn%5Cge0%7D%28n%2Br%29a_nx%5E%7Bn%2Br-1%7D-%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%2B1%7D%3D0)
![\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bn%5Cge0%7D%28n%2Br%29%28n%2Br%2B1%29a_nx%5E%7Bn%2Br-1%7D-%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%2B1%7D%3D0)
![\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%28r%2B1%29a_0x%5E%7Br-1%7D%2B%28r%2B1%29%28r%2B2%29a_1x%5Er%2B%5Csum_%7Bn%5Cge2%7D%28n%2Br%29%28n%2Br%2B1%29a_nx%5E%7Bn%2Br-1%7D-%5Csum_%7Bn%5Cge0%7Da_nx%5E%7Bn%2Br%2B1%7D%3D0)
![\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%28r%2B1%29a_0x%5E%7Br-1%7D%2B%28r%2B1%29%28r%2B2%29a_1x%5Er%2B%5Csum_%7Bn%5Cge2%7D%28n%2Br%29%28n%2Br%2B1%29a_nx%5E%7Bn%2Br-1%7D-%5Csum_%7Bn%5Cge2%7Da_%7Bn-2%7Dx%5E%7Bn%2Br-1%7D%3D0)
![\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%28r%2B1%29a_0x%5E%7Br-1%7D%2B%28r%2B1%29%28r%2B2%29a_1x%5Er%2B%5Csum_%7Bn%5Cge2%7D%5Cbigg%28%28n%2Br%29%28n%2Br%2B1%29a_n-a_%7Bn-2%7D%5Cbigg%29x%5E%7Bn%2Br-1%7D%3D0)
The indicial polynomial,
![r(r+1)](https://tex.z-dn.net/?f=r%28r%2B1%29)
, has roots at
![r=0](https://tex.z-dn.net/?f=r%3D0)
and
![r=-1](https://tex.z-dn.net/?f=r%3D-1)
. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.
When
![r=0](https://tex.z-dn.net/?f=r%3D0)
, we have the recurrence
![a_n=\dfrac{a_{n-2}}{(n+1)(n)}](https://tex.z-dn.net/?f=a_n%3D%5Cdfrac%7Ba_%7Bn-2%7D%7D%7B%28n%2B1%29%28n%29%7D)
valid for
![n\ge2](https://tex.z-dn.net/?f=n%5Cge2)
. When
![n=2k](https://tex.z-dn.net/?f=n%3D2k)
, with
![k\in\{0,1,2,3,\ldots\}](https://tex.z-dn.net/?f=k%5Cin%5C%7B0%2C1%2C2%2C3%2C%5Cldots%5C%7D)
, we find
![a_0=a_0](https://tex.z-dn.net/?f=a_0%3Da_0)
![a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}](https://tex.z-dn.net/?f=a_2%3D%5Cdfrac%7Ba_0%7D%7B3%5Ccdot2%7D%3D%5Cdfrac%7Ba_0%7D%7B3%21%7D)
![a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}](https://tex.z-dn.net/?f=a_4%3D%5Cdfrac%7Ba_2%7D%7B5%5Ccdot4%7D%3D%5Cdfrac%7Ba_0%7D%7B5%21%7D)
![a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}](https://tex.z-dn.net/?f=a_6%3D%5Cdfrac%7Ba_4%7D%7B7%5Ccdot6%7D%3D%5Cdfrac%7Ba_0%7D%7B7%21%7D)
and so on, with a general pattern of
![a_{n=2k}=\dfrac{a_0}{(2k+1)!}](https://tex.z-dn.net/?f=a_%7Bn%3D2k%7D%3D%5Cdfrac%7Ba_0%7D%7B%282k%2B1%29%21%7D)
Similarly, when
![n=2k+1](https://tex.z-dn.net/?f=n%3D2k%2B1)
for
![k\in\{0,1,2,3,\ldots\}](https://tex.z-dn.net/?f=k%5Cin%5C%7B0%2C1%2C2%2C3%2C%5Cldots%5C%7D)
, we find
![a_1=a_1](https://tex.z-dn.net/?f=a_1%3Da_1)
![a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}](https://tex.z-dn.net/?f=a_3%3D%5Cdfrac%7Ba_1%7D%7B4%5Ccdot3%7D%3D%5Cdfrac%7B2a_1%7D%7B4%21%7D)
![a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}](https://tex.z-dn.net/?f=a_5%3D%5Cdfrac%7Ba_3%7D%7B6%5Ccdot5%7D%3D%5Cdfrac%7B2a_1%7D%7B6%21%7D)
![a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}](https://tex.z-dn.net/?f=a_7%3D%5Cdfrac%7Ba_5%7D%7B8%5Ccdot7%7D%3D%5Cdfrac%7B2a_1%7D%7B8%21%7D)
and so on, with the general pattern
![a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}](https://tex.z-dn.net/?f=a_%7Bn%3D2k%2B1%7D%3D%5Cdfrac%7B2a_1%7D%7B%282k%2B2%29%21%7D)
So the first indicial root admits the solution
![y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%20a_0%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%7D%7D%7B%282k%2B1%29%21%7D%2Ba_1%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%2B1%7D%7D%7B%282k%2B2%29%21%7D)
![y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%20%5Cfrac%7Ba_0%7Dx%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%2B1%7D%7D%7B%282k%2B1%29%21%7D%2B%5Cfrac%7Ba_1%7Dx%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%2B2%7D%7D%7B%282k%2B2%29%21%7D)
![y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%20%5Cfrac%7Ba_0%7Dx%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%2B1%7D%7D%7B%282k%2B1%29%21%7D%2B%5Cfrac%7Ba_1%7Dx%5Csum_%7Bk%5Cge0%7D%5Cfrac%7Bx%5E%7B2k%2B2%7D%7D%7B%282k%2B2%29%21%7D)
which you can recognize as the power series for
![\dfrac{\sinh x}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csinh%20x%7Dx)
and
![\dfrac{\cosh x}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ccosh%20x%7Dx)
.
To be more precise, the second series actually converges to
![\dfrac{\cosh x-1}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ccosh%20x-1%7Dx)
, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When
![r=-1](https://tex.z-dn.net/?f=r%3D-1)
, we may seek a second solution of the form
![y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n](https://tex.z-dn.net/?f=y%3Dcy_1%5Cln%20x%2Bx%5E%7B-1%7D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Db_nx%5En)
where
![y_1=\dfrac{\sinh x+\cosh x-1}x](https://tex.z-dn.net/?f=y_1%3D%5Cdfrac%7B%5Csinh%20x%2B%5Ccosh%20x-1%7Dx)
. Substituting this into the ODE, you'll find that
![c=0](https://tex.z-dn.net/?f=c%3D0)
, and so we're left with
![y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n](https://tex.z-dn.net/?f=y%3Dx%5E%7B-1%7D%5Cdisplaystyle%5Csum_%7Bn%5Cge0%7Db_nx%5En)
![y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7Bb_0%7Dx%2Bb_1%2Bb_2x%2Bb_3x%5E2%2B%5Ccdots)
Expanding
![y_1](https://tex.z-dn.net/?f=y_1)
, you'll see that all the terms
![x^n](https://tex.z-dn.net/?f=x%5En)
with
![n\ge0](https://tex.z-dn.net/?f=n%5Cge0)
in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form
![y_2=\dfrac1x](https://tex.z-dn.net/?f=y_2%3D%5Cdfrac1x)
. Adding this to
![y_1](https://tex.z-dn.net/?f=y_1)
, we end up with just
![\dfrac{\sinh x+\cosh x}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csinh%20x%2B%5Ccosh%20x%7Dx)
.
This means the general solution for the ODE is