Differential equation $\omega'+(1/2)rw=a/r^{n-1}$ I'm reading about the fundamental solution to heat equation and there's a technical step I don't understand. The text says that if we assume $\omega= \omega(r)$ is a solution to $\omega'+(1/2)rw=a/r^{n-1}$ satisfying $\lim_{r\to\infty} \omega=\lim_{r\to\infty} \omega' =0$ then $a$ must be equal to $0$. I tried to explain it but I don't realize why is it true. Does anyone have a suggestion about how to show it? Thank you.
 A: The solution to the ODE is
\begin{align}
\omega(r) &= \text{exp}\left(-\frac{r^2}{4}\right)\int_{c_0}^r \frac{a}{\rho^{n-1}}\text{exp}\left(\frac{\rho^2}{2}\right)\,\text{d}\rho \tag{1}\\
&= \text{exp}\left(-\frac{r^2}{4}\right)\left[\omega_0 - \frac{r}{2} \frac{a}{r^{n-1}} \text{Ei}\left(\frac{n}{2},-\frac{r^2}{r}\right)\right], \tag{2}
\end{align}
where Ei is the exponential integral. Either from studying the integral in $(1)$ or by using the properties of the exponential integral (see e.g. here or here), you can see that $\lim_{r \to \infty} \omega(r) = 0$ for any value of $a$, assuming $n>0$. Only when $n=0$, we have $\lim_{r \to \infty} \omega(r) = 2a$.
Using the same arguments, or using the original ODE to express $\omega'(r)$ in terms of $\omega(r)$ and $r$, you can see derive that $\lim_{r \to \infty} \omega'(r) = 0$ for any value of $a$, now assuming that $n\geq 1$. Only when $0<n<1$, the limit of $\omega'(r)$ diverges as $r \to \infty$, and can only be zero when $a=0$. 
To conclude, there is no pressing mathematical reason to assume that, in general, $a=0$ in order to satisfy the boundary conditions at infinity. However, as @MyGlasses pointed out, there might be a physical reason to take $a=0$ -- or, indeed, a physical reason to assume $0\leq n<1$, which would imply that $a=0$.
