As an illustration of the difficulties that may arise in using the method of undetermined coefficients, consider $$\displaystyle \ \boldsymbol{\vec{y}^{\, \prime}} = \left\lbrack \begin{array}{rr} -2 & -2 \\ -2 & -2 \end{array} \right\rbrack \boldsymbol{\vec{y}} + \left\lbrack \begin{array}{c} e^{-4t} \\ 0 \end{array} \right\rbrack.$$

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. Show that seeking a particular solution of the form $\boldsymbol{\vec{y}}_P(t) = e^{-4t} \boldsymbol{\vec{a}}$, where $\boldsymbol{\vec{a}} = \left\lbrack \begin{array}{c} a_1 \\ a_2 \end{array} \right\rbrack$ is a constant vector, does not work. In fact, if $\boldsymbol{\vec{y}}_P$ had this form, we would arrive at the following contradiction:
   
   $a_2 =$ $\cdot a_1$
   
   and
   
   $a_2 =$ $\cdot a_1 -$ .
   
   
3. Show that seeking a particular solution of the form $\boldsymbol{\vec{y}}_P(t) = t e^{-4t} \boldsymbol{\vec{a}}$, where $\boldsymbol{\vec{a}} = \left\lbrack \begin{array}{c} a_1 \\ a_2 \end{array} \right\rbrack$ is a constant vector, does not work either. In fact, if $\boldsymbol{\vec{y}}_P$ had this form, we would arrive at the following contradiction:
   
   $a_2 =$ $\cdot a_1$
   
   and
   
   $a_1 =$
   
   and
   
   $a_2 =$ .