I can't figure out where : \begin{align} \delta^2\,dt\\ \end{align}
comes from.
Consider the process $$ d\sqrt{v} = = (\alpha - \beta\sqrt{v})\,dt + \delta \,dW $$ Here $\alpha, \beta,$ and $\delta$ are constants. Using Ito's Lemma show that $$ dv = (\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\,dt + 2\delta\sqrt{v}\,dW $$
Notice that $v = f(\sqrt{v})$ for $f(x)=x^2$. We have $f'(x)=2x$ and $f''(x)=2$.
Ito's lemma yields: $$ dv = f'(\sqrt{v})\,d\sqrt{v} + \frac{1}{2}f''(\sqrt{v})\,d\langle\sqrt{v}\rangle. $$
Hence \begin{align} dv &= 2\sqrt{v}\times\left((\alpha-\beta\sqrt{v})\,dt + \delta \,dW\right) + \frac{1}{2}\times 2\times\delta^2\,dt\\ &=(\delta^2 + 2\alpha\sqrt{v} - 2\beta v)\,dt + 2\delta \sqrt{v}\, dW. \end{align}
If $\mathrm dX_t=a_t\mathrm dW_t+b_t\mathrm dt$ then $\mathrm d\langle X\rangle_t=a_t^2\mathrm dt$. Here $a_t=\delta$, hence the $\delta^2\mathrm dt$ term.