The question is as follows:
For what value of $k \in R $, does the process $$ Y_t = (W_t + t)e^{-W_t+kt}, t\geq0$$ have zero drift?
I've tried solving this using Ito's Lemma 1 and substituting coefficient of $dt$ (which is the drift) to be equal to zero but that just makes the right side $ln(0)$. Any help or suggestion would be appreciated. Thanks!
Computation:
According to Ito’s Lemma, we know
$$\ dF = \frac{\partial F}{\partial W_t}dW_t + \frac{1}{2}\frac{\partial^2 F}{\partial W_t^2}dt$$
Here, the coefficient of dt is the drift so as stated in the question $\frac{1}{2}\frac{\partial^2 F}{\partial W_t^2} = 0$.
$\frac{\partial^2 F}{\partial W_t^2} = (W_t+t-2)e^{kt-W_t}$
When I equate $ \frac{1}{2}(W_t+t-2)e^{kt-W_t} = 0$, there is no way to derive the equation in terms of k as $ln(0)$ is undefined.
The formula which you stated holds for functions $F$ of the form $F(W_t)$. Here, however, we are dealing with a function of the form $F(t,W_t)$, and then Itô's lemma looks slightly different:
$$dF = \frac{\partial F}{\partial x} dW_t + \left(\frac{1}{2} \frac{\partial^2 F}{\partial x^2} + \frac{\partial}{\partial t} F \right) \, dt. \tag{1}$$
As $$F(t,x) = (t+x) e^{-x+kt}$$ implies that \begin{align*} \partial_x F(t,x) &= (1-t-x) e^{-x+kt} \\ \partial_x^2 F(t,x) &= (-2+t+x) e^{-x+kt} \\ \partial_t F(t,x) &= e^{-x+kt} (1+k(t+x)) \end{align*} we see that the drift term in (1) equals zero if $$\frac{1}{2} e^{-W_t+kt} (-2+t+W_t) + e^{-W_t+kt} (1+kt+kW_t)=0, $$ i.e. if $$ \frac{t}{2} + kt + \frac{W_t}{2} + kW_t=0.$$ Consequently, the proper choice for $k$ is $k=-\frac{1}{2}$.