I need to solve this differential equation, $\beta$ is constant:
$0=4\beta\dfrac{\partial^2 u}{\partial r^2}+\dfrac{4\beta}{r}\dfrac{\partial u}{\partial r}$
I don`t know what to do with $\dfrac{1}{r}$ in the second term on the right side, because partial differential is with $r$ in the same term?
$$0=4\beta\dfrac{\partial^2 u}{\partial r^2}+\dfrac{4\beta}{r}\dfrac{\partial u}{\partial r}$$ First, why don't you simplify ? (I suppose $\beta\neq 0$. If not, this is trivial). $$0=\dfrac{\partial^2 u}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}$$ Second, there is only one variable $r$. Why don't you write it as a simple ODE ? $$0=\dfrac{d^2 u}{d r^2}+\dfrac{1}{r}\dfrac{d u}{d r}$$ $v=\frac{d u}{d r} \quad\to\quad 0=\dfrac{d v}{d r}+\frac{1}{r}v \quad\to\quad \frac{dv}{v}=-\frac{dr}{r}\quad\to\quad v=\frac{c_1}{r}$ $$u=\int v\,dr=\int \frac{c_1}{r}\,dr$$
$$u(r)=c_1\ln|r|+c_2$$ $c_1$ and $c_2$ are any constants.
Supposing that there is a second hidden variable (for example $t$) : $$u(r,t)=c_1(t)\ln|r|+c_2(t)$$ $c_1(t)$ and $c_2(t)$ are any functions of $t$. $$ $$
IN ADDITION, after the comments :
The PDE to be solved is : $$4\beta\dfrac{\partial^2 u(r,z)}{\partial r^2}+\dfrac{4\beta}{r}\dfrac{\partial u(r,z)}{\partial r} =\dfrac{\partial p(z)}{\partial z} \tag 1$$ Why do you write $\dfrac{\partial p(z)}{\partial z}$ instead of $\frac{dp}{dz}$ ? Since $p(z)$ is a function of one variable only.
Let $\frac{dp}{dz}=g(z)$ $$4\beta\dfrac{\partial^2 u(r,z)}{\partial r^2}+\dfrac{4\beta}{r}\dfrac{\partial u(r,z)}{\partial r} =g(z)$$ There is no derivative with respect to $z$, thus this is an ODE involving the variable $r$ only.
Let $U(r)=u(r,z)$ with $z$ not a variable, but a simple parameter : $$4\beta U''+\dfrac{4\beta}{r}U' =g$$ It is easy to solve this second order linear ODE. $$U(r)=\frac{1}{16\beta}r^2g+c_1\ln(r)+c_2$$ The general solution of thev PDE $(1)$ is : $$u(r,z)=\frac{1}{16\beta}r^2\frac{dp}{dz}+c_1(z)\ln(r)+c_2(z)$$ $c_1(z)$ and $c_2(z)$ are arbitrary functions, to be determined according to the boundary conditions.
If the conditions are : $r=0: \dfrac{\partial u}{\partial r}=0$ and $r=1: u=0$
The first condition implies $c_1(z)=0$.
The second condition $0=\frac{1}{16\beta}1^2\frac{dp}{dz}+c_2(z)$ implies $c_2(z)=-\frac{1}{16\beta}\frac{dp}{dz}$
So, according to the conditions, the solution is : $$u(r,z)=\frac{1}{16\beta}r^2\frac{dp}{dz}-\frac{1}{16\beta}\frac{dp}{dz}$$ $$u(r,z)=\frac{1}{16\beta}(r^2-1)\frac{dp}{dz}$$