I would like to solve for an differential-integro operator $\hat{O}$ (involving the variable $x$ only) which acts on an exponential function $\exp(-a x)$ to bring out a factor of $a^{-1/2}$: $$ \hat{O} e^{- a x} = a^{-1/2} e^{-a x}.$$ I am not sure how to approach such a problem. Obviously if I needed $$ \hat{O} e^{- a x} = a e^{-a x},$$ the answer would be $\hat{O} = -d/dx$, and if I needed $$ \hat{O} e^{- a x} = \frac{1}{a} e^{-a x}, $$ the answer would be $ \hat{O} = - \int dx. $
What approach can I use to get $\hat{O}$ for the problem in the title? I am guessing some convolution would do it. If my original problem doesn't have a clear solution, is there a name for this type of functional calculus which solves for unknown operators on continuous functions?