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The boundary conditions are $P(r=r_e)=P_e$ and $P(r=r_w)=P_w$
$$\frac{1}{r} \frac{d}{dr}\left(r \frac{dP}{dr}\right)=0.$$
I know I have to integrate it twice but how?
The answer comes out to $$P=P_w+\frac{P_e-P_w}{\ln(r_e/r_w)}\ln(r/r_w).$$
2 Answers
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$\frac{d}{dr}(r\frac{dP}{dr}) = rP^{\prime\prime}+P^\prime$. Let $Q = P^\prime$, so you have $Q^\prime +\frac{1}{r}Q = 0$. Solve the first order ODE for $Q$ then solve back for $P$. Plug in BC's and you should get your solution.
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Since $$\frac{d}{dr}\left(r\frac{dP}{dr}\right)=0$$
Then $$r\frac{dP}{dr}=c$$ where $c$ is a constant. Now separate the variables and integrate, getting $$P=c\ln r+k$$ where $k$ is also a constant.
Now apply the given conditions to obtain the two constants, and the result follows.
