(While this question pertains to fluid dynamics in specific, I am posting it here as I think the basis of my confusion lies in the basic mathematics.)
For solving the Navier-Stokes equation for Newtonian fluid flow in a pipe of length $L$ and radius $R$ under the assumptions of laminar, fully-developed flow we have the classic governing equation: $$\frac{\mu}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v}{\partial r}\right)=\frac{\partial P}{\partial z}.$$ From this equation, I would assume that we need two boundary conditions for $v$ (since the highest order derivative of $v$ is $2$) and one boundary condition for $P$ (likewise): $$v(R,z)=0,$$ $$\frac{\partial v}{\partial r}(0,z)=0,$$ $$P(r,0)=P_0.$$
However, from the textbook solution (which is kind of difficult to follow) there is seemingly another boundary condition needed: $$P(r,L)=P_L.$$
The ultimate solution is supposed to be $$v=\frac{(P_0-P_L)R^2}{4\mu L}\left[1-\left(\frac{r}{R}\right)^2\right].$$ (The textbook solves for it by taking both sides of the differential equation equal to the same constant and then solving by classic separation of variables. This textbook is for students who have not necessarily taken a class in partial differential equations; like me.)
If I skip out on the last boundary condition which I originally didn't think was necessary, I get $$v=\frac{(P_0-P)R^2}{4\mu z}\left[1-\left(\frac{r}{R}\right)^2\right].$$
Why do I need two boundary conditions for $P$? Is my assumption based of the order of each independent variable invalid?