In the ODE, solve for [tex]y''[/tex]:
[tex]t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}[/tex]
The Wronskian is then
[tex]W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2[/tex]
Differentiating the Wronskian gives
[tex]W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2[/tex]
Substitute [tex]y_1,y_2[/tex] into the equation for [tex]y''[/tex], then substitute [tex]{y_1}'',{y_2}''[/tex] into [tex]W'[/tex]:
[tex]W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}[/tex]
[tex]\implies W'=\dfrac{2W}{t^2}[/tex]
which is another separable ODE; we have
[tex]\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}[/tex]
Given that [tex]W(y_1,y_2)(2)=3[/tex], we find
[tex]3=Ce^{-2/2}\implies C=3e[/tex]
so that
[tex]W(y_1,y_2)(t)=3e^{1-2/t}[/tex]
and so
[tex]W(y_1,y_2)(6)=\boxed{3e^{2/3}}[/tex]
