I am trying to solve pde:
$4\beta\dfrac{\partial^2 u}{\partial r^2} + \dfrac{4\beta}{r}\dfrac{\partial u}{\partial r}=\dfrac{\partial p}{\partial z}$,
where $\beta$ is constant, $p=f(z)$, $r$ is radial coordinate, $z$ is longitudinal coordinate and $u=u(r,z)$.
Additional conditions:
$r=0: \dfrac{\partial u}{\partial r}=0$;
$r=1: u=0$
I have no idea how special is this equation because of $\dfrac{1}{r}$ in the second term on the left side, so may I substitute that $u=e^{ar}$, where $a$ is constant and solve it as homogenuous equation?
This is no pde, as there is no $z$ on the left. For each $z $, you can see the equation as a first order linear ode on $\partial u/\partial r$.
Fix $z$. Write $v=\frac{\partial u}{\partial r}(r,z)$, $q=p'/4\beta$. Then the equation is $$ v'+\frac1r\,v=q. $$ The integrating factor is $e^{\int 1/r}=r$, which allows us to rewrite the equation as $(rv)'=rq$. Then $$ v=\frac{r}2\,q+\frac cr $$ for some constant $c$. We have the initial condition $v(0)=0$, which is satisfied if $c=0$. As $v$ is the derivative of $u$, we get $$ u(r,z)=\frac{r^2}4\,q+d. $$ The other initial condition was $$ 0=u(1,z)=\frac{q}4+d. $$ So $d=-q/4$. This gives $$ u(r,z)=\frac1{16\beta}\,(r^2-1)\,\frac{dp}{dz}(z). $$