How did we get to the constant in r\frac{\partial p}{\partial r}=c_1?

[An1MuS]

Hi,

When we have $$\frac{\partial}{\partial r}(r\frac{\partial p}{\partial r})=0$$

and we get

$$r\frac{\partial p}{\partial r}=c_1$$

To get there, did we do this

$$\int\frac{\partial}{\partial r}(r\frac{\partial p}{\partial r}) dr=\int 0 dr$$

or

$$\partial (r\frac{\partial p}{\partial r})=0\partial r$$
$$\int \partial (r\frac{\partial p}{\partial r})=\int 0\partial r$$

and why?

Thanks in advance

 

[statdad]

Well, yes, but you could also think of it this way: This
$$\frac{\partial}{\partial r} \left(r \frac{\partial p}{\partial r} \right) = 0$$

tells you that the derivative wrt $$r$$ of the expression inside parentheses is zero, so that expression must be a constant.