Consider the initial value problem $$\displaystyle \ \boldsymbol{\vec{y}^{\, \prime}} = \left\lbrack \begin{array}{rr} -3 & 3t \\ t^{2} & 3 \end{array} \right\rbrack \boldsymbol{\vec{y}} + \boldsymbol{\vec{g}}(t), \quad \quad \boldsymbol{\vec{y}}(1) = \left\lbrack \begin{array}{c} 4 \\ 0 \end{array} \right\rbrack.$$
Suppose we know that $$\displaystyle \ \boldsymbol{\vec{y}}(t) = \left\lbrack \begin{array}{c} -2t + \alpha \\ 2t^{2} + \beta \end{array} \right\rbrack$$ is the unique solution to this initial value problem. Find $\boldsymbol{\vec{g}}(t)$ and the constants $\alpha$ and $\beta$.

$\alpha =$

$\beta =$

$\boldsymbol{\vec{g}}(t) =$

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