I have this equation:

$$(p_0 p_1)'=\dfrac{4}{r^2-1}p_0'$$

where $'$ is derivative $\dfrac{d}{dz}$.

I need to solve it (find $p_1$) for case when $r=1$ (I already know $p_0$), and I don't know what to do because in that case I will have zero in denominator. I can make this shape:

$$(r^2-1)(p_0 p_1)'=4p_0'$$

where I will have zero on left side and just $p_0'$ on the right side, but this equation is part of the system, where I already found solution for $p_0$.

Can you tell me please do you see any solution of this?

$$ $$

Before upper equation it was necessary to solve this one:

$$4\beta\dfrac{\partial^2 u}{\partial r^2} + \dfrac{4\beta}{r}\dfrac{\partial u}{\partial r}=\dfrac{\partial p_1}{\partial z}$$

where $\beta$ is constant, $p=f(z)$, $r$ is radial coordinate, $z$ is longitudinal coordinate and $u=u(r,z)$ and with additional conditions:

$r=0: \dfrac{\partial u}{\partial r}=0$;

$r=1: u=0$.

I got solution: $$

u(r,z)=\frac1{16\beta}\,(r^2-1)\,\frac{dp_1}{dz}(z).

$$ from this equation is my part in upper equation $(r^2-1)$.

Because first equation is probably not correct because zero in denominator, can you tell me please did I make mistake in solution for $u(r,z)$?