Consider the system:
$$ \begin{split} \dot W &= i u(t)W(t)\\ \dot S &= e^{-rt}\sqrt{(1-u(t))W(t)} \end{split} $$
I would like to examine a system: $$ \begin{split} \dot x &= u(t)x(t)\\ \dot y &=\sqrt{(1-u(t))x(t)} \end{split} $$
Can someone tell me the transformations between $x(t)$ and $W(t)$, also between $y(t)$ and $S(t)$?
On
Assume that $x$ is only a function of $W(t)$, so $x(W(t))$. Differentiating this yields
\begin{align} \frac{dx}{dt} &= \frac{\partial x}{\partial W} \frac{dW}{dt} = \frac{\partial x}{\partial W} i\,u(t)\,W(t), \\ &= u(t)\,x(W(t)). \end{align}
This can be simplified to
$$ \frac{\partial x}{\partial W} i\,W(t) = x(W(t)). $$
This can be solved using separation of variables, which yields
$$ i\frac{dx}{x} = \frac{dW}{W} \to \int i\frac{dx}{x} = \int \frac{dW}{W}, $$
$$ i\log(x) + c_1 = \log(W) \to c_2 = \frac{W}{x^i}. $$
Choosing $c_2 = 1$ yields
$$ \boxed{ x(W) = W^{1/i} = \sqrt[i]{W} \Leftrightarrow W(x) = x^i }. $$
However I am not sure if there is a function for $y$ which is just a function of $S$, since $\dot{S}$ is a function of time and $W$, while $\dot{y}$ should be not a function of time, but is of $x$ instead of $W$.
I do not see any particular reason why you would want to transform your differential equation. Why not solve it in the given formulation?
If $u(t)$ is a known input you can solve the first equation
$$\dfrac{dW}{W} = i u(t) dt \implies \ln |W|=\int_{t_0}^{t}iu(\tau)d \tau$$ $$\implies |W(t)| = \exp\left[ \int_{t_0}^{t}iu(\tau)d \tau\right]$$ $$\implies W(t) = \pm\exp\left[ \int_{t_0}^{t}iu(\tau)d \tau\right].$$
Then use this in the second differential equation and integrate the differential equation.
$$S(t)=\pm\int_{t_0}^{t}\exp(-r\tau)\sqrt{(1-u(\tau))W(\tau)}d\tau$$