I have the following SDE: $$dY_{t}=A\left(\frac{W_{t}^{1}}{\sqrt{t}},\frac{Y_{t}}{\sqrt{t}}\right)dW_{t}^{1}+B\left(\frac{W_{t}^{1}}{\sqrt{t}},\frac{Y_{t}}{\sqrt{t}}\right)dW_{t}^{2}$$
where $W_{t}^{1}$ and $W_{t}^{1}$ are two independent brownian motions and $$A(x,y)=a\frac{\Phi(y)\Phi(-y)e^{0.5(y^2-x^2)}+\Phi(x)\Phi(-x)e^{0.5(x^2-y^2)}}{1+a(1-2\Phi(x))(1-2\Phi(y))}, \ \ \ \ \ \ \ \ B=\sqrt{1-A^2}$$
It seems unlikely, but I was wondering if there may be an explicit solution to this stochastic differential equation. Perhaps it can be reduced to a linear SDE with a suitable function?
Any help would be greatly appreciated.
As observed in the comments, the quadratic variation of $Y_{t}$ is $t$
$$d\langle Y,Y \rangle_{t}=A^{2}dt+(1-A^{2})dt=dt\Rightarrow \langle Y,Y \rangle_{t}=t.$$
I am assuming $a>0$ is small enough to prevent a zero denominator. If so the coefficients are Lipschitz and so we have a strong solution. Since there is no drift, it is a continuous martingale Why is a Itô integral w.r.t. Brownian motion a martingale?. Therefore, by Levy's theorem it is a Brownian motion
$$Y_{t}=B_{t}.$$
From here if you really want, you can decompose into two independent Brownian motions
$$B_{t}=a W^{1}_{t}+\sqrt{1-a^{2}}W^{2}_{t},$$
for any $a\in [0,1]$. To recover the original $W^{i}$ doesn't seem possible because as you can see by taking expectation with one of them, we get nonlinear integral equations:
$$E[B_{t}|W^{2}]=\int^{t} E[A(W^{1}_{s}/\sqrt{s},B_{s}/\sqrt{s})|W^{2}]dW^{2}_{s}$$
and
$$E[B_{t}|W^{1}]=\int^{t} E[B(W^{1}_{s}/\sqrt{s},B_{s}/\sqrt{s})|W^{1}]dW^{1}_{s}.$$