This is the fourth part of a four-part problem.

If the given solutions

$$\boldsymbol{\vec{y}_1}(t) = \left\lbrack \begin{array}{c} t^{2}-2t \\ 2t \end{array} \right\rbrack, \ \ \ \boldsymbol{\vec{y}_2}(t) = \left\lbrack \begin{array}{c} t-1 \\ 1 \end{array} \right\rbrack.$$
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem

$$\boldsymbol{\vec{y}^{\,\prime}} = \left\lbrack \begin{array}{cc} 2t^{-2} & 1-2t^{-1}+2t^{-2} \\ -2t^{-2} & 2t^{-1}-2t^{-2} \end{array} \right\rbrack \boldsymbol{\vec{y}}, \ \ \ \boldsymbol{\vec{y}}(4) = \left\lbrack \begin{array}{c} -31 \\ -21 \end{array} \right\rbrack, \ \ \ t > 0,$$
impose the given initial condition and find the unique solution to the initial value problem for $t > 0$. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks.