I know that the Geometric Brownian Motion, with the expression $dX_t = v X_t dt + \sigma X_t dW_t$ 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 $$X_t = X_{t_1} e^{\sigma W_t-W_{t_1}+ (v-\frac{\sigma ^2}{2})(t-t_1)}$$
Using Ito's Lemma we have
$$d\log X_t=\left(v-\frac12\sigma^2\right)dt+\sigma dW_t \tag 1$$
Integrating $(1)$ between $t_1$ and $t_2$ yields
$$\begin{align} \log(X_{t_2}/X_{t_1})&=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\int_{t_1}^{t_2}dW_t\\\\ &=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right) \end{align}$$
from which we have
$$X_{t_2}=X_{t_1}e^{\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right)}$$