"the preceding 2 scalar solutions correspond to the vector solutions $x^1(t)=(t,1)^T$ and $x^2(t)=(t^2,2t)^T$ which have the Wronskian
$$W(t)=\det\left[\begin{array}{lr} \mbox t & t^2\\ \mbox 1 & 2t \\ \end{array}\right]=t^2$$
on the other hand the trace of the matrix of coefficients is $2/t$, so for $t$, $t_0>0$
I am confused about how they got the book got that answer for the trace, I thought the trace was the sum of the diagonal? t+2t=3t? What am I missing?
The ODE $t^2y"-2ty'+2y=0$ can be written as $$\frac{d}{dt}\left(\begin{array}{c} y \\ y' \end{array}\right)=\left(\begin{array}{rr} 0 & 1 \\ -\frac{2}{t^2} & \frac 2t \end{array}\right)\left(\begin{array}{c} y \\ y' \end{array}\right).$$ Here, the matrix of coefficients is $$\left(\begin{array}{rr} 0 & 1 \\ -\frac{2}{t^2} & \frac 2t \end{array}\right)$$ with trace $2/t$ as the book says.