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

If the given solutions

$$\boldsymbol{\vec{y}_1}(t) = \left\lbrack \begin{array}{r} 2e^{3t}+6e^{-t} \\ 3e^{3t}+15e^{-t} \end{array} \right\rbrack, \ \ \ \boldsymbol{\vec{y}_2}(t) = \left\lbrack \begin{array}{r} -4e^{3t}+2e^{-t} \\ -6e^{3t}+5e^{-t} \end{array} \right\rbrack.$$
form a fundamental set (i.e., linearly independent set) of solutions for the initial value problem

$$\boldsymbol{\vec{y}^{\,\prime}} = \left\lbrack \begin{array}{rr} 9 & -4 \\ 15 & -7 \end{array} \right\rbrack \boldsymbol{\vec{y}}, \ \ \ \boldsymbol{\vec{y}}(0) = \left\lbrack \begin{array}{r} -6 \\ -31 \end{array} \right\rbrack,$$
impose the given initial condition and find the unique solution to the initial value problem. If the given solutions do not form a fundamental set, enter NONE in all of the answer blanks.