I am looking to numerically solve the ODE $$\vec{Y^\prime}(z)=\left[\begin{array}{cc}0& \frac{1}{\beta(z)\gamma(z)}\\-\frac{\gamma^{\prime\prime}(z)}{\beta(z)} & 0\end{array}\right]\cdot \vec{Y}(z),$$ where $\beta=\sqrt{1-1/\gamma^2}$ and $\beta(z)$ is the relativistic velocity as a function of position which is known beforehand. This describes the transverse position of a beam of charged particles in an accelerating electric field.
My problem is that I care about solutions when $\beta(z=0) = 0$ when the coefficients of the ODE blow up. For example, an electric field of the form $E_z(z) \propto 1+z$ yields $\beta(z)$ which when inserted gives: $$\vec{Y^\prime}(z)=\left[\begin{array}{cc}0& \frac{1}{\sqrt{z(8+8z+4z^2+z^3)}}\\-\frac{1}{2\sqrt{1-4/(2+2z+z^2)^2}} & 0\end{array}\right]\cdot \vec{Y}(z),$$ The upper right element clearly diverges preventing solution by numerical methods.
However, for similar analytically tractable cases, the equation can be solved and the pole integrated over symbolically.
My question is: what is the name of this type of issue and is there any robust way of working around it?
My current solution is to solve the problem with a different independent variable (physically interpreted as time here) $z^{\prime}(t)\propto\beta(t)$ which gives the numerically solvable equation: $$\vec{Y^\prime}(t)=\left[\begin{array}{cc}0& \frac{1}{\gamma(t)}\\-\gamma^{\prime\prime}(z(t)) & 0\end{array}\right]\cdot \vec{Y}(t),$$ with the additional component $z^{\prime}(t)= c\beta(t)$, of course. The reason I don't like this solution is that I can't choose at what $z$ I receive the solution vector $\vec{Y}$, I only get it at the $t$ that I specify.
My other ideas were to solve for some sort of transformed version of $\vec{Y}$ or to make some prior known independent variable transform $z\to t^2$ which solves the problem as long as the electric field can be modeled as nearly linear at $z=0$, but now I have a known one to one map from $t$ to $z$ (for $t>0$ at least) and can pick the values I care about before numerically solving.