Answer
$$y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem }}$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) - 2y = 20 \cr
& \frac{{dy}}{{dt}} - 2y = 20 \cr
& \frac{{dy}}{{dt}} = 2y + 20 \cr
& {\text{separating}} \cr
& \frac{{dy}}{{2y + 20}} = dt \cr
& \frac{{dy}}{{y + 10}} = 2dt \cr
& \int {\frac{{dy}}{{y + 10}}} = \int 2 dt \cr
& \ln \left| {y + 10} \right| = 2t + C \cr
& {\text{solve for }}y \cr
& y + 10 = {e^{2t + C}} \cr
& y = A{e^{2t}} - 10 \cr
& {\text{initial condition }}y\left( 0 \right) = 6 \cr
& 6 = A{e^{2\left( 0 \right)}} - 10 \cr
& A = 16 \cr
& y = A{e^{2t}} - 10 \cr
& {\text{replacing }}A \cr
& y = 16{e^{2t}} - 10 \cr
& {\text{then}} \cr
& y = 16{e^{2t}} - 10{\text{ is a solution of the initial value problem}} \cr} $$