Answer
$$y = \ln \left| t \right| + 5$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{1}{t},\,\,\,\,y\left( { - 1} \right) = 5 \cr
& {\text{distributing the differentials}} \cr
& dy = \frac{1}{t}dt \cr
& {\text{integrating both sides of the equation}} \cr
& \int {dy} = \int {\frac{1}{t}} dt \cr
& y = \ln \left| t \right| + C\,\,\,\left( {\bf{1}} \right) \cr
& {\text{using the initial value problem }}y\left( { - 1} \right) = 5 \cr
& 5 = \ln \left| { - 1} \right| + C \cr
& 5 = C \cr
& C = 5 \cr
& {\text{substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \ln \left| t \right| + 5 \cr} $$
