Let $dX_t=\mu X_t dt+\sigma X_t dW_t$.
We know that this is a shorthand for integral equation:
$X_t=X_0+\int_0^t\mu X_s ds + \int_0^t\sigma X_s dW_s$
Now: what if our equation looks like this $dX_t=\mu X_t dt+0\cdot dW_t$
According to our notation it is a shorthand for:
$X_t=X_0+\int_0^t\mu X_s ds$
Now, how can I formally arrive at the solution: $X_t=X_0 e^{\mu t}$, knowing that trajectories of $X_t$ might me nowhere differentiable?