I'm supposed to prove $y(t)=ln(1+B_t^2)$ is a solution of the following initial value problem: $$dy = (1 - B_t^2)e^{-2y}dt + 2B_te^{-y}dB_t$$ $$y(0)=0$$
Obviously $y(0)=0$. The problem I'm having is with calculating dy. My understanding is that ito's formula says that $$dy = f_tdt + f_xdx+1/2f_{xx}dxdx$$ if $y=f(t,x)$ and $f_x$ is the partial derivative of $f$ with respect to $x$ and so on. So I define $y=f(t,x)=ln(x)$ and $x=1+B_t^2$ so $y=ln(1+B_t^2)$ as it should be, and $dx=2B_tdB_t$. Then I calculate $dy$:
$$ dy=0 + dx/x - 1/2*dxdx/x^2 = 2B_t(1/x)dB_t - 2B_t^2(1/x^2)dB_tdB_t= 2B_te^{-y}dB_t-2B_t^2e^{-2y}dt$$
because $x = e^y \implies 1/x=e^{-y}$ and I am told that $dB_tdB_t = dt$. But from here it seems $-2B_t^2$ isn't $1-B_t^2$, so the coefficient of $dt$ is incorrect. I'm having a similar issue with a nearly identical problem, so I don't think I'm making a simple calculus error, my issue appears to be conceptual. What am I missing?
Starting from this expression $$Y(t)=ln(1+B_t^2)$$ I just apply Ito's formula, letting $B$ be our "$x$".
$$dY_t= \frac{1}{(1+B^2_t)} 2B_t dB_t+\frac{1}{2}\cdot \left(\frac{2(1+B_t^2)-2B_t\cdot 2B_t }{(1+B^2_t)^2}\right)d\langle B\rangle_t$$
$$dY_t= \frac{1}{(1+B^2_t)} 2B_t dB_t+ \left(\frac{(1+B_t^2)- 2B_t^2 }{(1+B^2_t)^2}\right)d\langle B\rangle_t$$
$$dY_t= \frac{1}{(1+B^2_t)} 2B_t dB_t+\left(\frac{(1-B_t^2) }{(1+B^2_t)^2}\right)d\langle B\rangle_t$$
Now you can simplify the expression by noticing that $d\langle B\rangle _t=dt$.
Hence you are left with:
$$dY_t= \frac{1}{(1+B^2_t)} \cdot 2B_t dB_t+\frac{1 }{(1+B^2_t)^2}\cdot (1-B_t^2)dt$$
$$= e^{-Y_t} \cdot 2B_t dB_t+e^{-2Y_t}\cdot (1-B_t^2)dt$$