I am trying to reason that the(I am quite sure that my reasoning is fallacious but I can't see why) solution of the SDE
$$dX_t= adt+\sqrt{\vert X_t\vert }dW_t$$ with $X_0=2a$ and $a>0$ is non-negative intuitively.
Assuming that a strong solution to this SDE exists( which follows from a result by V. P Zubchenco ), we know that the solution is continuous and hence for every $\omega \in C[0,\infty)$, $X(\omega)$ starts at $a$ and when a negative push by the brownian increment pushes it closer to zero, the variance which is proportional to $X_t$ decreases and at the moment it hits $0$(the brownian perturbation goes out of play), and the positive drift comes into play and pushes $X(\omega)$ away from $0$. Since $X$ is continuous , it cant jump to a negative value.
I know that the brownian perturbation can take very large negative values with small probabilities but continuity of $X$ shouldnt allow it to jump and become negative. I am not sure if I am clear enough and if my question is comprehensible. I want to understand the problem in my argument.