<s>[INST] Factoring derivatives Well I have $$r\frac{d^2T}{dr^2} + \frac{dT}{dr} = -\frac{qr}{k}$$
I wonder how to do algebraic manipulation to go from this step to
$$\frac{d}{dr}(r\frac{dT}{dr}) = \frac{-qr}{k}$$  
Is it even possible ?
I wonder how to write a second derivative in terms of a first derivative ?
 [/INST] Use
\begin{align}
\frac{d}{dx}(f(x) g(x)) = f(x) \, \frac{d g(x)}{dx} + \frac{d f(x)}{dx} \, g(x)
\end{align}
to put
\begin{align}
x \frac{d^{2} f}{dx^{2}} + \frac{d f}{dx} = - \frac{q x}{k}
\end{align}
into the form
\begin{align}
\frac{d}{dx} \left( x \frac{df}{dx} \right) = - \frac{q x}{k}.
\end{align}
For the sake of solution integrating both sides of this last equation leads to
\begin{align}
x \frac{d f}{dx} = - \frac{q x^{2}}{2 k} + c_{1}
\end{align}
and can be continued to be of the form
\begin{align}
f(x) &= \int \left[ - \frac{q x}{ 2 k } + \frac{c_{1}}{x} \right] \, dx \\
&= - \frac{q x^{2}}{4 k} + c_{1} \ln(x) + c_{0}
\end{align}
</s>