Statement

Let f∈Cc∞((a,b))f\in C_c^\infty((a,b)), where 0<a<b0<a<b, and e∈Re\in\mathbb R. The equation

u′(r)+eru(r)=f(r)u'(r)+\frac er u(r)=f(r)

has a solution u∈Cc∞((a,b))u\in C_c^\infty((a,b)) exactly when the ∫abref(r) dr\int_a^b r^ef(r)\,dr is zero. The solution is unique and equals

u(r)=r−e∫arsef(s) ds.u(r)=r^{-e}\int_a^r s^ef(s)\,ds.
Proof and support

Multiplying the equation by rer^e gives (reu)′=ref(r^eu)'=r^ef. Compact support forces the integral of the right side to vanish. Conversely, under that condition the displayed integral vanishes near both endpoints and solves the equation smoothly. A homogeneous solution is cr−ecr^{-e}, which cannot have compact support unless c=0c=0.

The primitive can fill gaps between separated pieces of the source support; it is contained in their radial interval hull, not necessarily their union. The restriction a>0a>0 avoids a separate regularity problem at the axis.