Problem Statement: Determine the equlibrium temperature distribution inside a circular annulus $r_1\leq r \leq r_2$. If the outer radius is at temperature $T_2$ and inner radius at temp $T_1$.
So my teacher told us to use the formula $\Delta u=\frac{1}{r}\frac{\partial}{\partial r}(r \frac{du}{dr})$ and solve the equation $$\frac{1}{r}\frac{\partial}{\partial r}(r \frac{du}{dr})=0$$ So I have $$\int \frac{1}{r}\frac{\partial}{\partial r}(r \frac{du}{dr})dr=\int 0\ dr$$ which i tried applying integration by parts on with $$a=\frac{1}{r},\ da = \ln r\ dr,\ dv =\frac{\partial}{\partial r}(r \frac{du}{dr})dr,\ v=r \frac{du}{dr}$$
So then I get $$\frac{du}{dr}-\int r\ln r \frac{du}{dr}dr=\int 0\ dr$$
Here is where I think I messed up (if I hadn't already made a mistake). I thought it would transform to: $$\frac{du}{dr}-\int r\ln r\ du=\int 0\ dr$$
resulting in:
$$\frac{du}{dr}-r\ln r\ u=C$$
I thought this was a linear equation with $$\mu (r)=e^{-\int r\ln r dr}=e^{-r^2(\frac{1}{2}\ln r-\frac{1}{4})}=r^{-\frac{1}{2}r^2}e^{\frac{1}{4}r^2}$$ This left me with: $$e^{-r^2(\frac{1}{2}\ln r-\frac{1}{4})}u=C\int e^{-r^2(\frac{1}{2}\ln r-\frac{1}{4})}dr$$
but this is way to complicated to solve with elementary functions.
Where did I go wrong?