Answer
$$y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem }}$$
Work Step by Step
$$\eqalign{
& ty'\left( t \right) - 6y = 18 \cr
& t\frac{{dy}}{{dt}} - 6y = 18 \cr
& t\frac{{dy}}{{dt}} = 6y + 18 \cr
& {\text{separating}} \cr
& \frac{{dy}}{{6y + 18}} = \frac{1}{t}dt \cr
& \frac{{dy}}{{y + 3}} = \frac{6}{t}dt \cr
& \int {\frac{{dy}}{{y + 3}}} = \int {\frac{6}{t}} dt \cr
& \ln \left| {y + 3} \right| = 6\ln \left| t \right| + C \cr
& {\text{solve for }}y \cr
& {e^{\ln \left| {y + 3} \right|}} = {e^{6\ln \left| t \right|}}{e^C} \cr
& y = A{t^6} - 3 \cr
& {\text{initial condition }}y\left( 1 \right) = 5 \cr
& 5 = A{\left( 1 \right)^6} - 3 \cr
& 8 = A \cr
& y = A{t^6} - 3 \cr
& {\text{replacing }}A \cr
& y = 8{t^6} - 3 \cr
& {\text{then}} \cr
& y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem}} \cr} $$