I solved the DE $\displaystyle x'=\frac{1}{t}$. I got $\displaystyle x=\ln (t)+C$.

I've been asked to determine $\displaystyle C$ if $\displaystyle t=-1$.

So I think that the solution of the DE should be $\displaystyle x=\ ln |t|+C$ so that I can reach $\displaystyle C=2$.

I just want to be sure we can talk about negative values of $\displaystyle t$.