I assume you mean [tex]y_1=t^{-1}[/tex], and not [tex]y_1=t-1[/tex], since this doesn't satisfy the ODE.
Assume a second solution of the form [tex]y_2=vy_1[/tex], where [tex]v[/tex] is a function of [tex]t[/tex]. Then
[tex]{y_2}'=v'y_1+v{y_1}'[/tex]
[tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]
Substituting into the ODE gives
[tex]t^2\left(\dfrac{v''}t-\dfrac{2v'}{t^2}+\dfrac{2v}{t^3}\right)+3t\left(\dfrac{v'}t-\dfrac v{t^2}\right)+\dfrac vt=0[/tex]
[tex]\implies tv''+v'=0[/tex]
[tex]\implies(tv')'=0[/tex]
[tex]\implies tv'=C[/tex]
[tex]\implies v'=\dfrac Ct[/tex]
[tex]\implies v=C_1\ln|t|+C_2[/tex]
[tex]\implies y_2=\dfrac{\ln t}t[/tex]
where we omit the second term because it's already accounted for by [tex]y_1[/tex].
