An exercise in my textbook is the following: Let $J_n (z)$ denote the Bessel function of order n. Set $$\theta(z,t)=\sum_{n=-\infty}^{+\infty}J_{n}(z)t^n$$ and assume that the series converges.
Obtain a first order DE relating $\frac{\partial \theta}{\partial z}$ and $\theta(z,t)$.
I hope anybody can help, because I am totally stuck at this moment.
$$\frac{\partial \theta}{\partial z}=\sum_m J'_m(z)t^m$$ Now using the derivative properties for the Bessel function, one has \begin{equation} \frac{\partial \theta}{\partial z}=\frac{1}{2}\sum \left[J_{m-1}(z)-J_{m+1}(z)\right]t^m \end{equation} It remains to shift the index $m$ in the summations to obtain \begin{align} \frac{\partial \theta}{\partial z}=\frac{1}{2}\left[t\sum_m J_m(z)t^m-\frac{1}{t}\sum_m J_m(z)t^m\right]\\ &=\frac{1}{2}\left( t-\frac{1}{t} \right)\theta(z,t) \end{align} Actually, $\theta$ is a generating function for the Bessel functions.