Answer
$$y = {e^{\sqrt {2{t^2} - 7} }}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = \frac{{2ty}}{{\ln y}},\,\,\,\,\,\,y\left( 2 \right) = e \cr
& {\text{Separate the variables}} \cr
& \frac{{\ln y}}{y}dy = 2tdt \cr
& {\text{Integrating}} \cr
& \int {\frac{{\ln y}}{y}} dy = \int {2t} dt \cr
& \frac{{{{\left( {\ln y} \right)}^2}}}{2} = {t^2} + C \cr
& {\text{Use the initial condition }}\,y\left( 2 \right) = e \cr
& \frac{{{{\left( {\ln e} \right)}^2}}}{2} = {2^2} + C \cr
& \frac{1}{2} = 4 + C \cr
& C = - \frac{7}{2} \cr
& ,{\text{ then}} \cr
& \frac{{{{\left( {\ln y} \right)}^2}}}{2} = {t^2} - \frac{7}{2} \cr
& {\text{Solve for }}y \cr
& {\left( {\ln y} \right)^2} = 2{t^2} - 7 \cr
& \ln y = \sqrt {2{t^2} - 7} \cr
& y = {e^{\sqrt {2{t^2} - 7} }} \cr} $$
