Deriving SDE from generator of Markov process

Question

Suppose that $U$ is a Markov process. I was able to calculate it's generator which is

$$Gf(u)=-\lambda f'(u)+\frac{1}{2}\lambda f''(u)$$

Is it true, that $U$ satisfies the SDE

$$dU = -\lambda dt+\sqrt{\lambda}dB$$

for some Brownian motion $B$? Does there exist a theorem that applies to this situation?

Any help is appreciated! Thank you in advance!