I know that the Geomtric Brownian Motion, with the expresion $dX_t = v X_t dt + \sigma X_t dW_t$ has the next solution $$X_t = X_0 e^{\sigma W_t+ (v-\frac{\sigma ^2}{2})t}$$ on the interval [0,t]. But, what would be the solution on a general interval $[t_1,t_2]$?
Would it be, $X_{t_2} = X_{t_1} e^{\sigma (W_{t_2}-W_{t_1})+ (v-\frac{1}{2}\sigma^2)(t_2-t_1)}$?
let $Y_t=\ln X_t$ by application of Ito lemma we have \begin{align} dY_t=\frac{1}{X_t}dX_t-\frac{1}{2X_t^2}d[X,X](t)=(v-\frac{1}{2}\sigma^2)dt+\sigma\,dW_t \end{align} by integration on $[t_1,t_2]$, we have \begin{align} Y_{t_2}=Y_{t_1}+(v-\frac{1}{2}\sigma^2)(t_2-t_1)+\sigma\,(W_{t_2}-W_{t_1}) \end{align} then \begin{align} X_{t_2}=X_{t_1}exp\left((v-\frac{1}{2}\sigma^2)(t_2-t_1)+\sigma\,(W_{t_2}-W_{t_1})\right) \end{align}
