If I have the following ODE:
$$\frac{dEX(Y)}{dt} = 1 + (\sigma^2-1)E(Y) $$
where $ Y(t) = \frac{1}{X(t)}$, $dX = X(1-X)dt +\sigma Ydw$, $dY = (1-Y+\sigma^2Y)dt - \sigma Ydw $
How do I find $E(X^-1(t))$?
I have solution where I first taking the expectation of the Ito stochastic differential equation involving Y(t) and obtain the following: $$\frac{dE(Y)}{dt} = 1 + (\sigma^2 -1)E(Y)$$ and solve subject to the initial condition leads to $$Y(t) = (\frac{1}{x_0} + \frac{1}{\sigma^2 -1})exp[(\sigma^2 -1)t]-\frac{1}{\sigma^2 -1}$$ I understood how to get the first equation related to E(Y) but not the second where it solves to get Y(t). Moreover, the final step is to obtain $$E[exp(\sigma (W(s) - W(t)))] = exp[\frac{1}{2}\sigma^2 (t-s)$$ As I don't understand the second last step, I am confused about the final step as well. May I know if anyone can help. Thanks!