Prove that $$X_{t} = X_{0}\cdot \text{exp}((\alpha - \beta^2/2)t + \beta W_{t}) $$ is a solution to the equation $\dot{X_{t}} = \alpha X_{t} + \beta X_{t} \dot{W_{t}}.$ Moreover, calculate $\mathbb{E}[X_{t}]$ if $X_{0} = x$.
I haven't been able to solve this problem. I thought that the best way to go about it would be to just differentiate $X_{t}$ (with Ito differentiation) and plug it into the equation we have. But, I've been having trouble doing so. I've read several sources on Ito differentiation, and I still haven't been able to solve this problem.
I would really appreciate some help.
Take logs of both sides to get $$ \log X_{t}-\log X_{0}=\left(\alpha-\frac{1}{2}\beta^{2}\right)t+\beta W_{t}. $$ Note that the RHS of the above can be written as sum of integrals: $$ \log X_{t}-\log X_{0}=\int_{0}^{t}\left(\alpha-\frac{1}{2}\beta^{2}\right)ds+\int_{0}^{t}\beta dW_{s}. $$ Equivalently, in differential form, $$ dY_{t}=\left(\alpha-\frac{1}{2}\beta^{2}\right)dt+\beta dW_{t} \qquad \text{where} \qquad Y_t \equiv \log X_t. $$ Note, in particular, that $Y$ satisfies an Ito drift-diffusion and hence we are in a position to apply Ito's lemma to $t \mapsto f(Y_{t})\equiv\exp Y_{t}$ to get $$ d(\exp Y_{t})=\exp Y_{t}dY_{t}+\frac{1}{2}\exp Y_{t}d[Y,Y]_{t}. $$ Now, substitute $Y_{t}=\log X_{t}$, the equation for $dY_{t}$, and $d[Y,Y]_{t}=\beta^{2}dt$ into the above and simplify to get the desired result.