Properties of $r^{n-1}w'+\frac{1}{2}r^nw=a$ 
Suppose $r^{n-1}w'+\frac{1}{2}r^nw=a$ for some constant $a\in{\Bbb R}$ and $n$ is a positive integer. Assume that 
  $$
\lim_{r\to\infty}w(r)=0,\quad \lim_{r\to\infty}w'(r)=0.
$$
  Then we must have $a=0$. 

This a step of the derivation of the fundamental solution for the heat kernel in Evan's Partial Differential Equations. I don't see at all why this is true even in the case $n=1$. Could anyone come up with some ideas?
 A: Dividing, you have 
$$w' + {rw\over 2} = {a\over r^{n-1}}.$$
Now introduce the integrating factor $e^{r^2/4}$ to get
$$w'e^{r^2} + {rwe^{r^2/4}\over 2} = {ae^{r^2/4}\over r^{n-1}}.$$
The form on the left is exact and we have
$$\left({we^{r^2/4} }\right)' =  {ae^{r^2/4}\over r^{n-1}}$$
Integrating,
$$we^{r^2/4} - w(0) = a\int_0^r {e^{s^2/4}\, ds\over s^{n-1}}   $$
If $n\ge 2$, the integral on the right will not integrate finitely, because of its bad behavior at $0$.  
If $n = 1$, you have
$$w = w(0)e^{-r^2/4} + a\int_0^r e^{(r^2 - s^2)/4} ds $$
Now what do the conditions at $\infty$ say?
A: i will look at the case $n = 1.$ we have $$\frac{dw}{dr} + \frac12 rw = a$$ 
taking the limit we get $$\lim_{r \to \infty} rw = 2a.\tag 1$$
using the integral representation gives you 
$$w(r) = e^{(1-r^2)/4}w(1) + a\int_1^r e^{-(r^2-s^2)/4}\, ds \tag 2$$
we know that 
$$ \int_1^r e^{-(r^2-s^2)/4}\, ds < \int_0^\infty e^{-s^2/4} ds \tag 3$$ 
letting $r \to \infty$ in $(2)$ and using $(3)$ shows that $$a = 0.$$
