WeBWorK Standalone Renderer

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

Let

$\boldsymbol{\vec{y}_1}(t) = \left\lbrack \begin{array}{r} 2e^{3t}+6e^{-t} \\ 3e^{3t}+15e^{-t} \end{array} \right\rbrack, \ \ \ \boldsymbol{\vec{y}_2}(t) = \left\lbrack \begin{array}{r} -4e^{3t}+2e^{-t} \\ -6e^{3t}+5e^{-t} \end{array} \right\rbrack.$
Compute the Wronskian to determine whether the functions $\boldsymbol{\vec{y}_1}(t)$ and $\boldsymbol{\vec{y}_2}(t)$ are linearly independent.

Wronskian = $\mathrm{det}$

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

=

These functions are linearly because the Wronskian is for all $t$ . Therefore, the solutions $\boldsymbol{\vec{y}_1}(t)$ and $\boldsymbol{\vec{y}_2}(t)$ to the system

$\boldsymbol{\vec{y}^{\,\prime}} = \left\lbrack \begin{array}{rr} 9 & -4 \\ 15 & -7 \end{array} \right\rbrack \boldsymbol{\vec{y}}$
form a fundamental set (i.e., linearly independent set) of solutions.