Hest equation in radial coordinates Here is the heat equation in radial coordinates: $\lambda \frac{1}{r} \frac{d}{dr}(r \frac{dT}{dr})+S(T)=0$ with S(T) being const and the boundary contions: T(r1) =T1 T(r2) =T2 ![enter image description here]( Can someone help me to find the analytic solution ?

Let $T'(r)=y(r)$, then $$\frac{d}{dr}(ry(r))+\frac{S}{\lambda}r=0$$ or $$y'(r)+\frac{1}{r}y(r)+\frac{S}{\lambda}=0$$ The integrating factor is $$I=e^{\int\frac{1}{r}dr}=r$$ So, $$y(r)=\frac{c_{1}}{r}-\frac{1}{r}\int{r}\frac{S}{\lambda}dr=\frac{c_{1}}{r}-\frac{Sr}{2\lambda}$$ and $$T(r)=\int{y(r)}dr+c_{2}=c_{2}+c_{1}\ln(r)-\frac{Sr^{2}}{4\lambda}$$ Now the bc's $$T_{1}=c_{2}+c_{1}\ln(r_{1})-\frac{Sr_{1}^{2}}{4\lambda}$$ $$T_{2}=c_{2}+c_{1}\ln(r_{2})-\frac{Sr_{2}^{2}}{4\lambda}$$ $$c_{1}=\frac{1}{\log(r_{1}/r_{2})}\Big(\frac{S(r_{1}^{2}-r_{2}^{2})}{4\lambda}+(T_{1}-T_{2})\Big)$$ $$c_{2}=T_{2}-\frac{\ln(r_{2})}{\log(r_{1}/r_{2})}\Big(\frac{S(r_{1}^{2}-r_{2}^{2})}{4\lambda}+(T_{1}-T_{2})\Big)-\frac{Sr_{2}^{2}}{4\lambda}$$