Consider the following SDEs :
(E) : $dX_t = (\alpha b(X_t) + {1\over2}b(X_t)b'(X_t))dt + b(X_t)dB_t$
(E') : $dY_t = \alpha dt + dB_t $
prove that E can be transformed to E' using :
$ Y_t = h(X_t) $ , where h is the premitive function of : $x \to {1\over b(x)}$
a hint ?
For SDEs of the form
$$dZ_t = \alpha(X_t) \, dt+ \beta(X_t) \, dB_t,$$
i.e. SDEs which do not explicitely depend on the time $t$, there are several sufficient conditions/approaches how to transform them into a linear SDE (which, in turn, can be solved explicitely); see e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes.
Here, we can simply follow the hint: We set $$h(y) := \int_0^y \frac{1}{b(x)} \, dx$$ and apply Itô's formula:
$$\begin{align*} h(X_t)-h(X_0) &= \int_0^t \frac{1}{b(X_s)} \, dX_s- \frac{1}{2} \int_0^t \frac{b'(X_s)}{b(X_s)^2} \, d\langle X \rangle_s \\ &= \int_0^t \left(\alpha + \frac{1}{2} b'(X_s) \right) \, ds + \int_0^t 1 \, dB_s - \frac{1}{2} \int_0^t b'(X_s) \, ds \\ &= \int_0^t \alpha \, ds + \int_0^t dB_s \end{align*}$$
using that the quadratic variation $\langle X \rangle_t$ is given by
$$\langle X \rangle_t = b^2(X_t) \, dt;$$
this follows directly from the stochastic differential $dX_t$.