I'm trying to solve the following problem (ref: Advanced engineering mathematics - Wylie, Barrett 5th Ed. Ch.10, Ex.10.17, p.579)
By assuming a solution of the form $v=\sum_{n=0}^{\infty}a_n t^{-n}$, show that when $\nu=0$ the equation: $$\frac{d^2v}{dt^2}\pm 2i\frac{dv}{dt}+\frac{1-4\nu^2}{4t^2}v=0$$
is satisfied, formally, by the function:
$$v(t)=\theta(t)\mp i\phi(t)$$
where:
$$\theta(t)=1^2-\frac{1^23^2}{2^6 2!}t^{-2} + \frac{1^23^25^27^2}{2^{12} 4!}t^{-4} -\ldots$$
$$\phi(t)=\frac{1^2}{2^3}t^{-1}-\frac{1^23^25^2}{2^9 3!}t^{-3} + \frac{1^23^25^27^29^2}{2^{15} 5!}t^{-5} -\ldots$$
Before showing my attempt. Is there any special meaning to the "formally satisfied" thing?
Also, by inspection I determined that the coefficients of real and imaginary part, $\theta$ and $\phi$, of $v(t)$ are given by the even and odd terms,respectively, of $$\frac{((2n)!)^2}{2^{3n}n!2^{2n}(n!)^2}$$ starting with $n=0$, in an alternating sign fashion on both.
My attempt:
With $v(t)=\sum_{n=0}^{\infty}a_n t^{-n}$ we have
$$\frac{dv}{dt}=\sum_{n=0}^{\infty}-na_n t^{-n-1}$$
$$\frac{d^2v}{dt^2}=\sum_{n=0}^{\infty}n(n+1)a_n t^{-n-2}$$
And with $\nu=0$ the differential equation is: $$\frac{d^2v}{dt^2}\pm 2i\frac{dv}{dt}+\frac{1}{4t^2}v=0$$
Replacing here the series:
$$\sum_{n=0}^{\infty}n(n+1)a_n t^{-n-2} \pm 2i\sum_{n=0}^{\infty}-na_n t^{-n-1} +\frac{1}{4t^2}\sum_{n=0}^{\infty}a_n t^{-n} = 0$$
$$\sum_{n=0}^{\infty}n(n+1)a_n t^{-n-2} \mp i\sum_{n=0}^{\infty}2na_n t^{-n-1} +\sum_{n=0}^{\infty}\frac{a_n}{4}t^{-n-2} = 0$$
And from here all I can see and do is to change indexing in the left and middle series, but since there's no imaginary part on the right side of equation everything seems that the coefficients $a_n$ are all zero. Also, combining left and right series leads to:
$$\sum_{n=0}^{\infty}n(n+1)a_n t^{-n-2}+\frac{a_n}{4}t^{-n-2} =\sum_{n=0}^{\infty}\left[ n(n+1)+\frac{1}{4} \right] a_n t^{-n-2}$$ from where I can't figure out how to obtain a recurrence relation to determine the final coefficients $a_n$. In fact, it also lead to all zeroes values.
I bet I'm forgetting something super essential and basic here, but I'd still appreciate some hint.