Write the given third order linear equation as an equivalent system of first order equations with initial values. $$y'\cos\!\left(t\right)-3y'''-\left(t^{4}+2t^{2}\right)y = 3t \\ {\ }\qquad \qquad \mathrm{with}\qquad y(-2)=0,\ \ y'(-2)=2,\ \ y''(-2)=3$$ Use $x_1 = y$, $x_2 = y'$, and $x_3 = y''$.

$\displaystyle \boldsymbol{\vec{x}\,'} =$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

$\displaystyle \boldsymbol{\vec{x}}$ +

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

with initial values

$\qquad \boldsymbol{\vec{x}} \biggl($

$\biggr) =$

$\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

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Hint: