I have two noncentral chi-square distributions (and also equivalent expressions in terms of the modified Bessel function of the first kind $I_N(.)$) with $\mu$ and $\nu$ being different parameters:
$$ \begin{align*} f(y|\mu) &= \frac{2N}{\sigma^2} \sum \limits_{i=0}^{\infty} \dfrac{\exp(-\frac{\mu N \bar{E}}{\sigma^2})(\frac{\mu N \bar{E}}{\sigma^2})^{i}}{i!} f_{\chi^2}(\frac{2N}{\sigma^2} y; 2N+2i) \\ &= \frac{N}{\sigma^2} \exp\Bigg(-\frac{N}{\sigma^2}y-\frac{N \mu \bar{E}}{\sigma^2}\Bigg) \Bigg(\frac{4y}{\mu \bar{E}}\Bigg)^{\frac{N-1}{2}} I_{N-1}\Bigg(\frac{2N}{\sigma^2} \sqrt{\mu \bar{E} y}\Bigg) \end{align*} $$ and $$ \begin{align*} f(y|\nu) &= \frac{2N}{\sigma^2} \sum \limits_{i=0}^{\infty} \dfrac{\exp(-\frac{\nu N \bar{E} }{\sigma^2})(\frac{\nu N \bar{E}}{\sigma^2})^{i}}{i!} f_{\chi^2}(\frac{2N}{\sigma^2} y; 2N+2i)\\ &= \frac{N}{\sigma^2} \exp\Bigg(-\frac{N}{\sigma^2}y-\frac{N \nu \bar{E}}{\sigma^2}\Bigg) \Bigg(\frac{4y}{\nu \bar{E}}\Bigg)^{\frac{N-1}{2}} I_{N-1}\Bigg(\frac{2N}{\sigma^2} \sqrt{\nu \bar{E} y}\Bigg) \end{align*} $$
How can I evaluate the intersection points of these two distributions in a closed form expression ? I know zero is a trivial solution but I need the other point.