Consider the initial value problem $$\displaystyle \ \boldsymbol{\vec{y}^{\, \prime}} = \left\lbrack \begin{array}{rr} 5 & -6 \\ 3 & -4 \end{array} \right\rbrack \boldsymbol{\vec{y}} + \left\lbrack \begin{array}{c} e^{-3t} \\ 0 \end{array} \right\rbrack, \quad \quad \boldsymbol{\vec{y}}(0) = \left\lbrack \begin{array}{c} 1 \\ 1 \end{array} \right\rbrack.$$

1. Form the complementary solution to the homogeneous equation.
   
   

   $\boldsymbol{\vec{y}}_C(t) = c_1$

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

   $+ \ c_2$

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

   
   
   
2. Construct a particular solution by assuming the form $\boldsymbol{\vec{y}}_P(t) = e^{-3t} \boldsymbol{\vec{a}}$ and solving for the undetermined constant vector $\boldsymbol{\vec{a}}$.
   
   

   $\boldsymbol{\vec{y}}_P(t) =$

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

   
   
   
3. Form the general solution $\boldsymbol{\vec{y}}(t) = \boldsymbol{\vec{y}}_C(t) + \boldsymbol{\vec{y}}_P(t)$ and impose the initial condition to obtain the solution of the initial value problem.
   
   

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $y_1(t)$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$ $y_2(t)$

   $=$

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