Consider $X$ a solution of
$$dX=f(X)dt+g_1(X)dW_1+g_2(X)dW_2$$
whereas $W_1$, $W_2$ are independent standard brownian motions and $f,g_1,g_2:\mathbb{R}\rightarrow\mathbb{R}$ bounded lipschitz-continuous functions.
I want to show that there is a standard brownian motion $B$, such that $X$ is a solution of
$$dX=f(X)dt+\sqrt{g_1^2(X)+g_2^2(X)}dB$$
$\frac{1}{\sqrt{2}}(W_1+W_2)$ is a brownian motion and I guess I have to use Ito's lemma for this, but I do not succeed.
I'd appreciate some help or a hint. Thanks for your attention.
The main idea is that you want to define $dB_t := \frac{1}{\sqrt{g_1^2(X_t) + g_2^2(X_t)}}\bigg( g_1(X_t) dW_1 + g_2(X_t)dW_2 \bigg)$ and use Levy's characterization to show this is a Brownian motion. There's a little bit of a subtlety here in that you need to say what to do if $g_1(X_t)=g_2(X_t)=0$, but since in that case $dX_t = f(X_t)dt$ and $\sqrt{g_1^2(X_t) + g_2^2(X_t)} = 0$, it doesn't really matter how you define $B_t$ there as long as $B$ is still a Brownian motion. For example you could do something like set $dB_t = dW_1$ when $g_1(X_t)=g_2(X_t)=0$.