I am trying to solve the following exercise from Lawler's lecture notes on Stochastic Calculus.
Suppose $W_t$ is a standard Brownian motion on $(\Omega, \mathbb{P})$. For each of the following choices of $X_t$, $0 \le t \le 1$, state whether there is an equivalent probability measure $\mathbb{Q}$ such that the $X_t$ is a standard Brownian motion in the new measure. If the answer is yes, give $d\mathbb{Q}/d\mathbb{P}$ at $t = 1$. In all cases assume that $W_0 = 0, X_0 = 0$. $$ \begin{array} dX_t &= 2 dt + dW_t ,\\ dX_t &= 2 dt + 6 dW_t , \\ dX_t &= 2W_t dt + dW_t . \end{array} $$
I would start this exercise by repeating the usual sketch of proof for Girsanov's theorem. Let $X_t$ be a solution of: $$ dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t. $$ Define the stochastic process $M_t$ on $(\Omega, \mathbb{P})$ as: $$ M_t = \exp\left(-\int_0^t \mu(s,X_s)d W_s - \frac12 \int_0^t \mu(s, X_s)^2 ds\right). $$ It is not hard to check that $M_t$ fulfills: $$ dM_t = -\mu(t,X_t)M_tdW_t. $$ Introducing a new stochastic process $Y_t = M_t X_t$, we have $$ dY_t = \mu_t(1-\sigma_t)M_t dt + (\sigma_t - \mu_t X_t)M_t dW_t, $$ therefore if we introduce a new measure $d\mathbb{Q} = M_t d\mathbb{P}$, the representation of $X_t$ in this new measure is $$ dX_t^{\mathbb{Q}} = \mu_t(1 - \sigma_t)dt + (\sigma_t - \mu_t X_t^{\mathbb{Q}})d W_t^{\mathbb{Q}}. $$ It follows that there exists a change of measure where the stochastic process $S_t$ is a standard Brownian motion iff $\sigma_t$ is identically 1 (this is the only way to make the drift term disappear).
If I carry out explicitly the computation for the third stochastic process in the exercise's text, I get (here I replace the constant 2 with a symbolic constant $\mu$): $$ dY_t = (1 - \mu W_t X_t)M_t dW_t, $$ where the Brownian motion $W_t$ is referred to the probability measure $\mathbb{P}$. I find it hard to rewrite the equation above as an equation for the original process $X_t$ in the new measure $\mathbb{Q}$. My natural guess would be: $$ dX_t^{\mathbb{Q}} = (M_t - \mu W_t^{\mathbb{P}}X_t^{\mathbb{Q}}) dW_t^{\mathbb{Q}}, $$ but I don't like it because there is a Brownian motion with respect to $\mathbb{P}$. My question is: is the equation above correct? Should I just accept the fact that this the equation of a driftless Brownian motion with respect to $\mathbb{Q}$, even if the diffusion is a not very clear stochastic process?
I suspect the problem here is that $M_t$ is not a martingale, as the expectation of $\int W_s dW_s$ is $O(t)$, but I do not see clearly how this ruined the last result: it seems the drift term disappeared anyway.