Using the fact that $y_1$=$t^{-1}$ then $y_2$= z $\cdot$ $t^{-1}$
$y_2'= z'\cdot t^{-1} - z \cdot t^{-2}$
$y_2''= z''\cdot t^{-1} - z'\cdot t^{-2} - z'\cdot t^{-2} + 2 \cdot z \cdot t^{-3}$
Substituting into the original equation:
$t^2 (z''\cdot t^{-1} - z'\cdot t^{-2} - z'\cdot t^{-2} + 2\cdot z \cdot t^{-3})-2t(z \cdot t^{-1})$
Which I then simplified into:
$t\cdot z''-4\cdot z'+2z\cdot t^{-1} -2z = 3 t^2 -1$
I want to use the method of reduction of order to find the second solution, by setting $w=z'$, but there are too many orders of $z$ remaining in the equation. Not sure where I keep going wrong!
If $y_1=t^{-1}$ is to be a solution of the homogeneous equation with minimal changes to the given equation, then the equation has to read $$ t^2y''-2y=3t^2-1. $$ Then the homogeneous equation is Euler-Cauchy with characteristic polynomial $0=m(m-1)-2=(m-2)(m+1)$. Thus the second basis solution is $y_2=t^2$.
Then a particular solution can be found in the form $y_0=A\ln(t)t^2+B$.