# Solve ODE $y''(x) - iy'(x)-\alpha y(x)/x^{2} = 0$

I want to know how to solve the above equation. x is defined in [0,1] and $\alpha$ is a constant. Wolfram Alpha already gives me a solution. I'm trying to solve using Frobenius method, assuming a solution $y(x) = \sum_{n=0}^{\infty}a_{n} x^{r+n}$. I obtained the solution to the indicial equation: $\alpha_{\pm} = \frac{1}{2}(1+\sqrt{1+4\alpha})$ and the recursion formula for the coefficients (for $\alpha_{+})$: $$a_{n} = \frac{-i i^n\left( \sqrt{4\alpha+1}+3/2 \right)_{n-1}}{\Gamma(n+1)\left( \sqrt{4\alpha+1}+2 \right)_{n-1}}a_{0},$$ where $()_{n-1}$ is the Pochhammer symbol. My problem is that I don't know how to write this recursion in terms of Bessel functions. Maybe I should apply first a smart change of variables and write the equation as a Bessel-like one ? Any help is welcome.

• @xpaul, that question isn't very helpful, the equation discussed there differs in the first derivative. – avemaria Jul 26 '17 at 17:01
• @xpaul, it's not a matter of the imaginary unit, but I have a term $x^2$ and they have a term $x$. – avemaria Jul 26 '17 at 17:10
• @xpaul, applying your change of variable I obtain the equation $t^{2}y''(t)-t^{2}y'(t)-\alpha y(t) = 0$, which is not the same in the link you posted. – avemaria Jul 26 '17 at 17:20
• I made a mistake. – xpaul Jul 26 '17 at 20:35