Consider a Brownian motion $B$, $x\in\mathbb{R}_+$ and the stochastic process $X:=(\sqrt{x}+B)^2$.
I want to show, that $X$ is a solution of the SDE
$$dX = dt +2\sqrt{X} dW$$
where $W$ is some brownian motion.
Applying ito's rule for $f(x)=x^2$ I find
$$ \begin{align} dX=df(\sqrt{x}+B)&=f'(\sqrt{x}+B) d(\sqrt{x}+B)+\frac{1}{2}f''(\sqrt{x}+B)d[\sqrt{x}+B]\\ &=2(\sqrt{x}+B)dB+\frac{1}{2}2d[B]\\ &=2(\sqrt{x}+B)dB+dt\\ \end{align}$$
Now, I'm skeptical, if this already shows, that $X$ is a solution of the SDE, because of following reason:
Shouldn't $\sqrt{X}$ be positive process, since
$$\sqrt{X}=\sqrt{(\sqrt{x}+B)^2}=|\sqrt{x}+B|\quad ?$$
Is there a way to find a brownian motion $W$, such that
$$(\sqrt{x}+B)dB = |\sqrt{x}+B|dW\quad ?$$
Or am I "overinterpreting" the notion of $\sqrt{X}$?
There is a way to find a Brownian motion $W$ such that $(\sqrt{x}+B)dB = |\sqrt{x}+B|dW$. Specifically, let $W_t := \int_0^t \operatorname{sgn}(\sqrt{x}+B_s)dB_s$. Then $W$ is a Brownian motion by Levy's characterization of Brownian motion, and by associativity of stochastic integrals $$|\sqrt{x}+B|dW = |\sqrt{x}+B|\operatorname{sgn}(\sqrt{x}+B)dB = (\sqrt{x}+B)dB.$$